Fisica Classica

Bekir Karaoglu
Classical Physics
A Two-Semester Coursebook
Classical Physics
Bekir Karaoglu
Classical Physics
A Two-Semester Coursebook
Bekir Karaoglu
Istanbul, Turkey
Translated from Turkish by Mehmet Karaoglu
ISBN 978-3-030-38455-5 ISBN 978-3-030-38456-2 (eBook )

https://doi.org/10.1007/978-3-030-38456-2
© Springer Nature Switzerland AG 2020

This work is subject to copyright. All rights are reserved by the Publisher, whether the whole or part of the material is
concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction
on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation,
computer software, or by similar or dissimilar methodology now known or hereafter developed.
The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not
imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and
regulations and therefore free for general use.
The publisher, the authors, and the editors are safe to assume that the advice and information in this book are believed
to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty,
expressed or implied, with respect to the material contained herein or for any errors or omissions that may have been
made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.
Cover picture: The Three Country Bridge at night, between France, Germany and Switzerland, by Wladyslaw Taxiarchos228
This Springer imprint is published by the registered company Springer Nature Switzerland AG.
The registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland
V
PREFACE
This book has been prepared as a textbook for use in the Introductory Physics courses
given in the first year of the Science, Engineering, Education and Medicine schools of
universities. It is calculus-based and can be taught entirely in two semesters.
Over the last decade, many universities around the world have been reducing un-
dergraduate degree credits in favor of Master’s degree programs. Introductory Physics
courses that once spanned 4 semesters were reduced to 2 semesters. Subsequently, a
serious problem arose in the use of standard textbooks. Of course, one cannot find fault
with the physics covered in these books, which I, too, have used in the past. However, it
is also a fact that they are not fully useful in Physics courses that have been shortened to
2 semesters with reduced hours. The main reason for this is the excessive volume of these
books. I have personally observed the panic and despair of students when they first see
these books, which run at least 1600 pages. Because of this volume problem, instructors
also have difficulty in deciding which parts they will teach, while still informing the
students about the omitted parts. Thus, the coherence and flow of the text are lost.
Sometimes, less is more. You may consider this book as a trimmed-down version of
standard textbooks for a two-semester course. Whichever topics you choose to teach,
you will find the same calculus-based approach and the same essential concepts covered,
and all this without worrying about continuity and time limits.
THE BOOK’S APPROACH TO PHYSICS
No single book can be expected to meet the needs of each and every course, with their
varied aims and different student bodies. I would like briefly to explain the philosophy of
the book and answer other possibly relevant questions.
An important problem that holds back the teaching of physics in many universities
is that the calculus courses given in parallel with physics cannot keep abreast of it. Such
topics as integrals, vectors, differential equations, complex analysis, etc., come too late to
be useful. Hence, it is necessary to teach physics topics either by keeping mathematics at
a reasonable level or by developing some mathematics yourself.
In accordance with this viewpoint, the book develops some vital mathematics (inte-
grals and vectors) along the way and omits minor topics that require advanced mathe-
matics. However, in the treatment of each subject, care has been taken to ensure that the
conclusions and concepts reached are based on the fundamental laws of physics. Thus,
the student will be able to grasp the unifying concepts of physics and retain the results
without the need for memorization.
ABOUT USING THE BOOK
In accordance with the above understanding, the book’s distinctive features are:
(1) Each theoretical development in the book should be considered together with the
worked examples that follow. The examples extend the theory a little further, so that the
volume of the book and the duration of the lecture can be reasonable. For each topic, the
number of examples is more than you may need.
(2) Students’ ability to solve problems is an integral part of learning in physics. One can
avoid rote learning only by solving problems. For this purpose, problems and multiple-
choice questions are given at the end of each chapter, at a level that can be solved by
the student, and hence can be assigned as homework. The numerical answer to each
problem is given below it so that the student can easily check his/her own solution. In
this approach, the steps of solving are more important than the numerical result.
Lastly, I want to address the students: You live in a wonderful World, rich in detail,
sometimes enigmatic, but sometimes dangerous as well. The least you can do is to be
aware of its basic laws, knowledge that can take you to new heights of understanding
and wisdom.
VI PREFACE
ACKNOWLEDGMENTS
In the long writing phase of this book, many people from all parts of the world
have helped. Many colleagues have contributed by using the book in their courses or by
reading and correcting it. Students were in touch, informing us of the mistakes they had
spotted. Moreover, it was a pleasant surprise to discover that all of the people of whom
we asked permission to use their visual materials wanted to contribute to such a book
project without expecting any financial compensation.
Among these people of goodwill whose help I gratefully acknowledge are Hayrettin
Sönmez, Muzaffer Adak, Allen Crockett, Christophe Ehlen, Cesur Ekiz, Erik Forsberg,
Murat Gökbayrak, Dave Jackson, Taner Kalayci, Orhan Kamer, Birtan Kavanoz, Ahmet
Köroglu, Sedat Ozsoy, Suat Ozkorucuklu, Hakan Omer Oztek, Eren Ozturk, Wolfgang
Rueckner, Hakan Koray Tutkun, Yaşar Yılmaz and Boris Veytsman.
I thank Mehmet Karaoglu for capably translating the book, Ismail Çam for single-
handedly drawing all the figures, and Marc Beschler for editing of the manuscript. Their
time and effort are greatly appreciated.
I thank Seçkin Publishing House for the successful publication of the book in Turkish,
its director Koray Seçkin, and its staff, Mesut Yildirim and Musa Gündogan, for their help.
I also thank Springer for believing in the value of the book, in particular, Marina
Forlizzi, Barbara Amorese and Suraj Kumar for their help and support.
Finally, I would like to thank in advance all lecturers and students who will con-
tact us (karabekirus@gmail.com or customerservice@springernature.com) with their
suggestions for a better book.
Bekir Karaoglu
6.2 LAW OF CONSERVATION OF MOMEN-
TUM . . . . . . . . . . . . . . . . . . . . 97
6.3 COLLISIONS IN ONE DIMENSION . . . 99
6.4 COLLISIONS IN TWO DIMENSIONS . . 102
6.5 CENTER OF MASS . . . . . . . . . . . . 104
Contents 6.6 ROCKET MOTION . . . . . . . . . . . .
MULTIPLE-CHOICE QUESTIONS . . . . . . .
107
109
PROBLEMS . . . . . . . . . . . . . . . . . . . . 110
7 ROTATIONAL MOTION 113

PREFACE . . . . . . . . . . . . . . . . . . . . . V 7.1 ANGULAR KINEMATICS . . . . . . . . 114
7.2 TORQUE (MOMENT OF A FORCE) . . . 118
1 UNITS AND VECTORS 1 7.3 ROTATIONAL DYNAMICS . . . . . . . 120
1.1 DIMENSIONS AND UNITS . . . . . . . 1 7.4 MOMENT OF INERTIA CALCULATIONS 123
1.2 PRECISION AND SIGNIFICANT FIGURES 5 7.5 ROLLING MOTION . . . . . . . . . . . 126
1.3 VECTORS . . . . . . . . . . . . . . . . . 8 7.6 ANGULAR MOMENTUM AND ITS
MULTIPLE-CHOICE QUESTIONS . . . . . . . 18 CONSERVATION . . . . . . . . . . . . . 128
PROBLEMS . . . . . . . . . . . . . . . . . . . . 20 MULTIPLE-CHOICE QUESTIONS . . . . . . . 130
PROBLEMS . . . . . . . . . . . . . . . . . . . . 132
2 MOTION IN A STRAIGHT LINE 23
2.1 POSITION, DISPLACEMENT, VELOC-
8 STATIC EQUILIBRIUM 135
ITY AND ACCELERATION . . . . . . . 24
8.1 TWO CONDITIONS OF STATIC EQUI-
2.2 MOTION WITH CONSTANT ACCELER-
LIBRIUM . . . . . . . . . . . . . . . . . 136
ATION . . . . . . . . . . . . . . . . . . . 28
8.2 APPLICATIONS . . . . . . . . . . . . . . 137
2.3 FREE FALL . . . . . . . . . . . . . . . . 31
PROBLEMS . . . . . . . . . . . . . . . . . . . . 140
MULTIPLE-CHOICE QUESTIONS . . . . . . . 34
PROBLEMS . . . . . . . . . . . . . . . . . . . . 35
9 HARMONIC MOTION 143
9.1 SIMPLE HARMONIC MOTION . . . . . 143
3 TWO-DIMENSIONAL MOTION 39
9.2 PENDULUM MOTION . . . . . . . . . . 151
3.1 POSITION AND DISPLACEMENT VEC-
TORS . . . . . . . . . . . . . . . . . . . . 40 9.3 DAMPED HARMONIC MOTION . . . . 153
3.2 PROJECTILE MOTION . . . . . . . . . . 42 9.4 DRIVEN HARMONIC MOTION – RESO-
3.3 UNIFORM CIRCULAR MOTION . . . . 45 NANCE . . . . . . . . . . . . . . . . . . 155
3.4 RELATIVE MOTION . . . . . . . . . . . 48 MULTIPLE-CHOICE QUESTIONS . . . . . . . 156
MULTIPLE-CHOICE QUESTIONS . . . . . . . 51 PROBLEMS . . . . . . . . . . . . . . . . . . . . 158
PROBLEMS . . . . . . . . . . . . . . . . . . . . 52
10 WAVES 161
4 NEWTON’S LAWS OF MOTION 55 10.1 GENERAL PROPERTIES OF WAVES . . 162
4.1 NEWTON’S LAWS . . . . . . . . . . . . 56 10.2 INTERFERENCE, REFLECTION AND
4.2 TYPES OF FORCES IN MECHANICS . . 59 TRANSMISSION OF WAVES . . . . . . . 166
4.3 APPLICATIONS OF NEWTON’S LAWS 65 10.3 DOPPLER EFFECT - SHOCK WAVES . . 172
4.4 CIRCULAR MOTION . . . . . . . . . . . 69 MULTIPLE-CHOICE QUESTIONS . . . . . . . 176
PROBLEMS . . . . . . . . . . . . . . . . . . . . 73
11 FLUIDS 179
5 WORK AND ENERGY 77 11.1 GENERAL PROPERTIES OF FLUIDS . . 180
5.1 WORK . . . . . . . . . . . . . . . . . . . 78 11.2 ARCHIMEDES’ PRINCIPLE . . . . . . . 185
5.2 POWER . . . . . . . . . . . . . . . . . . 81 11.3 SURFACE TENSION AND CAPILLARITY 187
5.3 KINETIC ENERGY . . . . . . . . . . . . 83 11.4 VISCOSITY . . . . . . . . . . . . . . . . 189
5.4 POTENTIAL ENERGY . . . . . . . . . . 84 11.5 BERNOULLI’S EQUATION . . . . . . . . 190
5.5 LAW OF CONSERVATION OF ENERGY 87 MULTIPLE-CHOICE QUESTIONS . . . . . . . 195
PROBLEMS . . . . . . . . . . . . . . . . . . . . 92
12 TEMPERATURE AND HEAT 199
6 IMPULSE AND MOMENTUM 95 12.1 THERMODYNAMIC EQUILIBRIUM
6.1 IMPULSE AND MOMENTUM . . . . . . 96 AND TEMPERATURE . . . . . . . . . . 200
VII
VIII CONTENTS
12.2 HEAT . . . . . . . . . . . . . . . . . . . 202 18.4 COMBINATION OF RESISTORS . . . . 310

12.3 THERMAL EXPANSION . . . . . . . . . 205 18.5 ELECTRICAL MEASURING INSTRU-
12.4 HEAT TRANSFER . . . . . . . . . . . . 207 MENTS . . . . . . . . . . . . . . . . . . 312
12.5 IDEAL GAS . . . . . . . . . . . . . . . . 210 MULTIPLE-CHOICE QUESTIONS . . . . . . . 314
PROBLEMS . . . . . . . . . . . . . . . . . . . . 214
19 MAGNETIC FIELD 319
13 THE LAWS OF THERMODYNAMICS 217 19.1 MAGNETIC FORCE . . . . . . . . . . . 320
13.1 WORK DONE BY A GAS . . . . . . . . . 218 19.2 MAGNETIC FORCE ON A CURRENT-
13.2 INTERNAL ENERGY – FIRST LAW OF CARRYING WIRE . . . . . . . . . . . . 323
THERMODYNAMICS . . . . . . . . . . 220 19.3 MAGNETIC TORQUE ON A CURRENT
13.3 APPLICATIONS OF THE FIRST LAW . . 221 LOOP – THE ELECTRIC MOTOR . . . . 324
13.4 KINETIC CALCULATION OF PRESSURE 225 MULTIPLE-CHOICE QUESTIONS . . . . . . . 327
13.5 HEAT ENGINES . . . . . . . . . . . . . 229 PROBLEMS . . . . . . . . . . . . . . . . . . . . 328
13.6 SECOND LAW OF THERMODYNAMICS
– THE CARNOT CYCLE . . . . . . . . . 232 20 SOURCES OF MAGNETIC FIELD 331
MULTIPLE-CHOICE QUESTIONS . . . . . . . 234 20.1 MAGNETIC FIELD OF A CURRENT . . 332
PROBLEMS . . . . . . . . . . . . . . . . . . . . 235 20.2 MAGNETIC FIELD CALCULATIONS . . 333
20.3 FORCE BETWEEN PARALLEL CUR-
14 THE ELECTRIC FIELD 239 RENTS – AMPERE UNIT . . . . . . . . 337
14.1 ELECTRIC CHARGE . . . . . . . . . . . 240 20.4 AMPERE’S LAW . . . . . . . . . . . . . 339
14.2 COULOMB’S LAW . . . . . . . . . . . . 242 20.5 MAGNETIC PROPERTIES OF MATTER 342
14.3 ELECTRIC FIELD . . . . . . . . . . . . . 244 MULTIPLE-CHOICE QUESTIONS . . . . . . . 346
PROBLEMS . . . . . . . . . . . . . . . . . . . . 254
21 FARADAY’S LAW – INDUCTION 351
21.1 FARADAY’S LAW . . . . . . . . . . . . . 352
15 GAUSS’S LAW 257
21.2 GENERATORS AND TRANSFORMERS . 355
15.1 ELECTRIC FLUX (Φ) . . . . . . . . . . 257
21.3 INDUCTANCE – MAGNETIC ENERGY 358
15.2 GAUSS’S LAW . . . . . . . . . . . . . . 258
21.4 RLC CIRCUITS . . . . . . . . . . . . . . 361
15.3 APPLICATIONS OF THE GAUSS’S LAW 260
15.4 ELECTRIC FIELD IN CONDUCTORS . . 264
PROBLEMS . . . . . . . . . . . . . . . . . . . . 363
PROBLEMS . . . . . . . . . . . . . . . . . . . . 266
22 GEOMETRIC OPTICS 367
16 ELECTRIC POTENTIAL 269
22.1 THE WAVE NATURE OF LIGHT . . . . 368
16.1 ELECTRIC POTENTIAL . . . . . . . . . 270
22.2 REFLECTION AND REFRACTION . . . 371
16.2 SYSTEM OF POINT CHARGES . . . . . 272 22.3 IMAGE BY REFLECTION – MIRRORS . 375
16.3 POTENTIAL OF CONTINUOUS 22.4 IMAGE BY REFRACTION– LENSES . . 379
CHARGE DISTRIBUTIONS . . . . . . . 276 22.5 OPTICAL INSTRUMENTS . . . . . . . . 384
16.4 CONDUCTORS AND EQUIPOTENTIAL MULTIPLE-CHOICE QUESTIONS . . . . . . . 387
SURFACES . . . . . . . . . . . . . . . . 277 PROBLEMS . . . . . . . . . . . . . . . . . . . . 389
PROBLEMS . . . . . . . . . . . . . . . . . . . . 281 23 WAVE OPTICS 391
23.1 YOUNG’S DOUBLE-SLIT EXPERIMENT 392
17 CAPACITORS AND DIELECTRICS 283 23.2 INTERFERENCE IN THIN FILMS . . . . 394
17.1 CAPACITANCE . . . . . . . . . . . . . . 284 23.3 DIFFRACTION FROM A SINGLE SLIT . 396
17.2 COMBINATIONS OF CAPACITORS . . 287 PROBLEMS . . . . . . . . . . . . . . . . . . . . 398
17.3 DIELECTRICS . . . . . . . . . . . . . . . 291
MULTIPLE-CHOICE QUESTIONS . . . . . . . 296 APPENDICES 401
PROBLEMS . . . . . . . . . . . . . . . . . . . . 298 A. PHYSICAL CONSTANTS . . . . . . . . . . 401
B. USEFUL MATHEMATICAL RELATIONS . . 401
18 CURRENT, RESISTANCE AND CIR- C. TRIGONOMETRIC TABLE . . . . . . . . . 402
CUITS 301 D. PERIODIC TABLE . . . . . . . . . . . . . . 403
18.1 ELECTRIC CURRENT . . . . . . . . . . 301 PICTURE CREDITS . . . . . . . . . . . . . . . 404
18.2 OHM’S LAW AND RESISTANCE . . . . 304
18.3 DIRECT CURRENT CIRCUITS . . . . . 307 INDEX 405
1
UNITS AND VECTORS
The Russian Soyuz spaceship

and the Earth in the background
as viewed from the International
Space Station (ISS). These two
spaceships docked at 400 km al-
titude from the ground in Jan-
uary 2011.
The taking of this picture may
have lasted one second, but it
was only made possible with
centuries of development in
Physics.
1.1 DIMENSIONS AND UNITS

The natural sciences started with measurement. According to historians,
measurement may have started in around 3000 BC in Ancient Egypt. The Nile
river used to flood and fields would become submerged. There then arose the
problem of finding the former boundaries of the fields after the flood withdrew.
You may easily guess how they solved this problem: They first identified a fixed
rock or a tree, and then they agreed on a unit of length. They measured and
recorded the distance of the field boundaries to the origin in terms of this unit.
You can see the overall structure of measurement from this example:
First, we observe physical quantities that describe a physical property or
condition, such as distance, area, speed, force, pressure . . . .
These physical quantities have dimensions, i.e., common characteristics
with respect to measurements: For example, although width, length, diameter,
© Springer Nature Switzerland AG 2020 1

B. Karaoglu, Classical Physics, https://doi.org/10.1007/978-3-030-38456-2_1
2 1. UNITS AND VECTORS
perimeter, etc., are different, their common aspect is that they are a kind of length,
i.e., they have the dimension of length. Let’s take a look at some examples:
Physical quantity Dimension

distance, width,
)
length
depth, length . . .
day, month, year,

)
time
season, period,. . .
The dimension of certain quantities can be expressed in terms of more basic

dimensions. For example:
Surface area = width×length = (length)2

Volume = width×length×height = (length)3
Then, a dimension standard or unit is established for each dimension. For

example, various units such as yards, leagues, meters, feet, etc., have been used
throughout history for the dimension of length.
The act of measurement is thus the act of determining the amount that a
physical quantity corresponds to in terms of its own dimensional standard, i.e.,
unit. For example, to measure the width or length of a table, we take a meter ruler
and count how many units the length of the table takes: 3.45 meters, 0.86 meters,
etc.
We must emphasize one important point here: The result of each measurement
should be expressed in terms of units. Although, in daily life, one may say, “My
height is 1.67", this is actually incorrect, the correct expression being “my height
is 1.67 meters".
The International System of Units (SI)
Which units among many should be accepted as the basic units? This se-
lection has differed over various countries throughout history. However, to-
day, the internationally accepted International System of Units (SI = Systeme
Internationale), formerly known as the MKS System of Units, is used.
The SI System of Units consists of 7 basic units. All physical quantities can
thus be completely expressed and measured in terms of these basic units.
The basic units of the SI system

Dimension Unit Abbreviation
Time second s
Length meter m
Mass kilogram kg
Electrical current ampere A
Temperature kelvin K
Light magnitude candela cd
Amount of matter mole mol
Three of these units, the Meter, the Kilogram and the Second are sufficient
for mechanical topics. The other units shall be defined in due course.
Now let us define these basic units:
1.1. DIMENSIONS AND UNITS 3
• Meter : The distance traveled by light in vacuum in 1/299 792 458 seconds.
• Second : 9 192 631 770 times the vibration period of the Cs133 atom.
Note that these two units are defined with atomic methods. The units of meter
and second were defined differently in the past: The meter was taken as a
fraction of the length of the meridian or the length of a special platinum stick
stored in France. Likewise, fractions of the Earth’s movement around the Sun
(year) or around its own axis (day) were used for the second. However, these
became insufficient as the precision of measurements in science gradually
increased. Atomic definitions are both very precise and are repeatable at
anywhere around the world without the necessity of traveling to France.
Figure 1.1: The laser assembly

that determines the meter unit
(left). The cesium atomic clock
that provides the second unit
(center). The platinum-iridium
cylinder stored in France as the
kilogram standard (right).
• Kilogram: The mass of a cylindrical prototype of the platinum-iridium alloy

manufactured in 1889 and stored at the Bureau International des Poids et
Mesures (BIPM) institution in Paris.
Now, this was true until 2019. However, variations up to 50 micrograms
were observed between this prototype and the copies sent to other countries
over the many years that passed. After long discussion, it was decided to
define it in terms of the Planck constant h arising in phenomena at the
atomic scale. Accordingly, once the Planck constant is defined as
h = 6.62607015 × 10−34 kg.m2 /s
the kilogram is then defined in terms of meters or seconds. The new atomic
definition became valid as of May 2019. But it will take a long time to
implement it in laboratories, hence the prototypes will still be in use.
Prefixes for Multiples and Fractions
Exponents are used in the scientific representation of numbers. For example,
1 350 000 is expressed as 1.35 × 106 or 0.000047 as 4.7 × 10−5 . However, it is
difficult to express this in daily verbal language. Instead, fractions and multiples
may easily be expressed by placing prefixes before each unit. The most commonly
used prefixes are shown in the following table:
Multiples Fractions
name symbol quantity name symbol quantity
kilo k 103 milli m 10−3
mega M 106 micro µ 10−6
giga G 109 nano n 10−9
tera T 1012 pico p 10−12
peta P 1015 femto f 10−15
These prefixes have entered daily life: Examples include kilometer (km),
gigahertz (GHz) and terabyte (TB).
Derived Units
The 3 basic units (meter, kilogram and second) mentioned above are sufficient
to derive the units of all other quantities in mechanics. Here are some examples:
Some derived units

quantity definition unit abbreviation
Area width×length (meter)2 m2
Volume width×length×height (meter)3 m3
Speed distance/time meters/second m/s
Acceleration speed/time meters/(second)2 m/s2
Force mass×acceleration kilogram×meter/(second)2 kg·m/s2
Work force×distance kilogram×meter2 /(second)2 kg·m2 /s2
Consistency of Units
It is not sufficient merely to have the same numbers on both sides of the
equation in a physical formula. The dimensions and units must also be consistent.
In other words, “you cannot compare apples and oranges."
Only the quantities with the same dimension can appear on both sides
of physical formulas.
For example, let us take a look at the accelerated motion formula you know
from high school:
x = v0 t + 12 a t2
The left-hand side of this equation has the dimension of length. Thus, each of the
terms on the right-hand side must also have the length dimension. Let us plug in
the dimensions of each quantity explicitly:
meter meter
meter = second
×
+ second
× 2
= meter
second second 2

This gives us a clue from the start about whether or not the formula is consistent.
We shall reconsider examples on units in the next section after learning how
to carry out numerical calculations.
Other Systems of Units
It will be useful to know two systems of units other than SI.
• In the CGS system, the basic units are cm, gram and second. It is easy to
convert into the SI system.
• The Imperial System of Units is used in the Anglo-Saxon world; and, with
slight differences, in the United States. The unit of length is the foot, the unit
of mass is the pound and the unit of time is the second. The fraction of the
unit of length is the inch and the multiples are the yard and the mile.
These units are converted into the SI system using the following formulas:
1 foot (ft) = 12 inch (in) = 0.3048 meter

1 pound (lb) = 16 ounce (oz) = 0.453 kg
1.2. PRECISION AND SIGNIFICANT FIGURES 5
1.2 PRECISION AND SIGNIFICANT FIGURES

The degree of precision is the degree of closeness of measurement of a
quantity to that quantity’s true value. The precision is limited by the capability
of the measuring instruments. The difference between the real value and the
measured value of a quantity is called the margin of error. Each measurement
has a margin of error.
In science and engineering, the results of measurements and calculations
should be expressed properly by taking into consideration the margin of error.
In other words, they should be given with the correct number of significant
figures. For example, it would be incorrect to express the result of weighing
an item on a grocery store scale as 352.8461 grams or the result of a distance
measurement with a ruler as 25.87346 cm .
Absolute Error – Relative Error
The minimum interval measurable by a measuring apparatus is called the
absolute error. Consider that you are measuring the length of your book using
a ruler. The smallest division on the ruler is 1 millimeter. It is not possible to
measure lengths smaller than a millimeter using this ruler (Figure 1.2). Therefore,
the absolute error of the measurements with this ruler is ∆L = 1 mm . (Actually,
the absolute error of a ruler is 0.5 mm , as it is possible to see further as to which
side the reading is closer. We will take it as 1 mm here for the sake of simplicity.)
For example, let the measured length of the book be L = 294 mm . Then, the Figure 1.2: The red rectangle
measurement result can be expressed as follows, taking into consideration the measures somewhere between
margin of error: 1.7 and 1.8 cm long. Here, the
L ± ∆L = 294 ± 1 mm absolute error is 1 mm since that
is the smallest division we can
Accordingly, the length will be within the range (293 mm < L < 295 mm) . see.
The ratio ∆x/x is called the relative error, and it is expressed in terms of
percentage (%). For example, if the mass of an object is given as m = (35 ± 1) g ,
its relative error is calculated as follows:
∆m 1
= = 0.029 ≈ 3 %
m 35
Both absolute and relative errors propagate when calculations are made with
these measurements. For example, after measuring the width, length and height
of the book using the aforementioned ruler, it will be necessary to calculate the
volume V of the book and again provide a V ± ∆V margin of error.
How do errors propagate in calculations? There are two simple rules for
calculating the margin of error in the results:
• In additions and subtractions, absolute errors are added:
z=a±b =⇒ ∆z = ∆a + ∆b
• In multiplications and divisions, relative errors are added:
∆y ∆a ∆b
(
ab
y= =⇒ = +
a/b y a b
These two rules are sufficient to find the margin of error in complex calculations.
For example, the margin of error in the expression z = a3 (b + c)2 is found as
follows:
z = a3 (b + c)2
∆z ∆(a3 ) ∆(b + c)2
= +
z a3 (b + c)2
∆a ∆a ∆a ∆(b + c) ∆(b + c)
= + + + +
a a a b+c b+c
∆z ∆a ∆b + ∆c
= 3 +2
z a b+c
It is also possible to guess the margin of error from the expression of data.
For example, if two separate mass measurements are given as 45 g and 45.0 g ,
this means that, in the first measurement, nothing is known after the last digit,
in other words, the absolute error is ∆m = 1 g . In the second measurement, the
decimal of gram was also measured and found to be 0, in other words, absolute
error is ∆m = 0.1 g .
Significant Figures
The precision of physical data is not always given by its absolute error. Some-
times it is understood from the number of significant figures expressing the
data.
For example, if the mass of an object is given as 76.4 g , the number of signifi-
cant figures is 3. Nothing changes when we express this as 0.0764 kg ; the number
of significant figures is still 3. The number of significant figures of a number is
found without taking into consideration the leading zeros. For example,
1.2398 Number of significant figures: 5
(Notice, on the last row, that the zero on the right-hand side was written explicitly.
This means it was measured and found to be zero. Hence, the number of significant
figures is 3.) The higher the number of significant figures, the more precisely that
quantity is known.
How many significant figures should be kept in the result after arithmetic
operations are carried out with two numbers? For example, if a moving object
travels 8.0 meters in 3.0 seconds , the speed value goes on as v = 8.0/3.0 =
2.6666 . . . . Where should we cut this value?
There are, again, two simple rules for keeping the correct number of significant
figures in calculation results:
• In additions and subtractions, the lowest of the number of decimal
places is kept:
3.2339 + 5.4 = 8.6339 = 8.6
9.12 − 5.4917 = 3.6283 = 3.63
(In the last row above, the round off rule was applied when discarding two
digits. According to this rule, if the first number of the discarded part is 5
or greater, the last kept number is rounded up. Here, the first number of the
discarded part is 8, hence the last kept number was rounded up from 2 to 3.)
1.2. PRECISION AND SIGNIFICANT FIGURES 7
• In multiplications and divisions, the lowest number of significant

figures is kept:
3.4567 × 2.7 = 9.33309 = 9.3

15.67 × 0.00012 = 0.0018804 = 0.0019
Of course, the natural numbers ( 1, 2, 3, 4 . . . ) that are used for counting do

not affect the number of significant figures. For example, if the mass of a book is
285.6 g , then the mass of 3 books is 3 × 285.6 = 856.8 g , in other words, it does
not lose any decimal places.
According to these rules, the speed of the object in the example above should
be taken as v = 8.0/3.0 = 2.666666 . . . = 2.7 m/s . In this course, we shall
adopt the following rule: The number of significant figures shall be taken as 3 in
intermediate calculations and 2 in results unless specified otherwise.
Example 1.1 ∆A 0.1 0.2

= + = 0.043
43.2 5.2 8.3
The dimensions (a, b) of a carpet are given in two different
ways: ∆A = 43.2 × 0.043 = 1.87 ≈ 2 m2
(a) a = (5.2 ± 0.1) m and b = (8.3 ± 0.2) m . Calculate the
In the last row, we reduced the surface area error down to
surface area of the carpet and give the result with a margin
one digit, because the width and length margins of error had
of error.
one digit. Accordingly, the surface area of the carpet, with
(b) Dimensions are given only as a = 5.2 m and b = 8.3 m .
its margin of error, is written as:
Once again, calculate the surface area and give the result
with the correct number of significant figures.
A ± ∆A = (43 ± 2) m2
Answer
(a) First, calculate the surface area of the carpet: (b) The number of significant figures of the width and length
is two. Accordingly, only 2 significant figures are kept in the
A = ab = 5.2 × 8.3 = 43.16 m2 product of the two:
The relative error formula for multiplication is used: A = ab = 5.2 × 8.3 = 43.16 ≈ 43 m2
∆A ∆a ∆b
= +
A a b
Example 1.2
shall be kept with 2 significant figures as well. If 1 mil =
The speed of an automobile is given as 34 mph (miles/hours). 1.609 km ,
Calculate the speed in terms of km/hour and m/s . ( 1 mile = 34 mph = 34 × 1.609 = 54.706 = 55 km/hour
1 609 m ). Likewise, if 1 hour = 60 minutes and 1 minute = 60 seconds ,
Answer 55 × 1000
55 km/hour = = 15.27 = 15 m/s
Speed is given with 2 significant figures. Thus, the results 3600 s
Example 1.3
Answer
The conversion of mass into energy is calculated with Einstein’s The number of significant figures of the material m = 1.0 g
famous formula: is 2, taking into account the zero on the right. Therefore, the
result should be given with 2 significant figures. Knowing
E = m c2 this, it will be sufficient to round the speed of light down to 3
digits:
Here, c = 2.997925 × 108 m/s is the speed of light, E is energy
and its unit is joule (J) in the SI system. ( 1 J = 1 kg·m2 /s2 ). E = mc2 = (1 × 10−3 kg) × (3.00 × 108 )2
Calculate the energy of 1.0 gram of material. E = 9.0 × 1013 J
1.3 VECTORS
Physical quantities can be divided into two distinct groups with respect to
measurement: Some quantities can be completely specified just by giving a nu-
merical value. We call these scalars. For example, saying “the temperature of this
room is 18◦ degrees” or “the mass of that table is 25 kg ” is sufficient. Quantities
of the scalar type include mass, energy, volume, temperature, electrical resistance,
refractive index, etc. Scalars can be manipulated by algebraic calculations.
For certain quantities, merely giving a numerical value is not sufficient; one
should also specify its direction. For example, when we say, “the speed of this
ship is 10 km/hour", they will ask us: “But 10 km/hour in which direction?” If
we reply, “10 km/hour in the north-east direction,” we will have fully specified
the velocity of the ship.
Quantities that have both a magnitude and a direction are called vectors. Vec-
tors are manipulated according to certain rules of addition. Velocity, acceleration,
~
Figure 1.3: Notation of the A force, momentum, electric field, etc., are among vector quantities. Vectors are
vector. ~ ~F . . . .
shown with an arrow ( → ) over letters, such as A,
The magnitude (or, the norm) of a vector is a positive scalar number shown
~
as |A| or, briefly, as A .
In diagrams, a vector is represented by an arrow drawn in the direction of
the vector. The length of the arrow is a measure of the magnitude of the vector.
Multiplication of a Vector with a Scalar Number
The product of the number c and the vector A ~ is the vector cA
~ which has a
magnitude of cA .
~ and cA
A ~ are in the same direction if c is positive,
~ is in the opposite direction if c is negative.
cA
For example, the vector 3A ~ is shown with an arrow that is 3 times as long in the
~
direction of the vector A whereas the vector −2A ~ is shown with an arrow twice
~ 3A
Figure 1.4: Vectors A, ~ and as long in the opposite direction (Figure 1.4).
~
−2A . Addition of Two Vectors
Vectors are not added by rules of arithmetic, because these are quantities with
directions. For example, when a ship is displaced 3 km to the east and then 4 km
to the north, its total displacement is not 3 + 4 = 7 km , but only 5 km. Therefore,
vector algebra is different from scalar numbers and all algebraic operations must
be redefined.
The vector A ~ +B ~ and B
~ , which is the sum of vectors A ~ is defined with the
triangle rule or the parallelogram rule. These rules are shown in Figures 1.5 and
1.6.
Figure 1.5: The parallelogram

rule
In the parallelogram rule, the two vectors are first shifted by preserving
their directions such that both of their tail points are in the same location. Then,
a parallelogram is formed by drawing lines from the head points of each vector
parallel to the other. The diagonal between the vectors of this parallelogram is
the vector A ~ +B~.
1.3. VECTORS 9
Figure 1.6: The triangle rule.
In the more useful triangle rule, one of the vectors ( A ~ or B

~ ) is shifted in
parallel to itself to the head point of the other vector (Figure 1.6). The vector
~ ) to the head of the second vector ( B
drawn from the tail of the first vector ( A ~ ) is
~ +B
A ~.
Figure 1.7: (a) Adding many vec-

tors is easier using the triangle
rule. (b) What do you think the
sum of these five vectors would
be?
Both rules give the same result. However, the parallelogram rule is not
suitable for adding more than two vectors, as things immediately get complicated.
However, the sum of more than two vectors can be drawn immediately using
the triangle rule (Figure 1.7a). After the vectors are lined up head-to-tail, it is
sufficient to draw a vector from the tail of the first vector to the head of the last
vector.
Vector Subtraction
~ −B
Vector subtraction is performed using the same rule. The difference A ~ is
~
nothing but the sum of the vectors A and −B ~:
Figure 1.8: Difference of two
~ −B
A ~ + (−B
~ =A ~) (1.1) vectors.
~ −B
Observing Figure 1.8, you will notice that the head of the vector A ~ ends at
~
the positive-signed vector ( +A ).
Example 1.4
Use the triangle rule to calculate the following for the vectors ~ B
(b) The difference of two vectors is written as A− ~
~ = A+(− ~)
B
~ B
A, ~ shown above:
~ , and C and turned into a sum. Here, the vector −B ~ is in opposite
(a) the sums A ~ +B~ and B ~ +C ~, direction to B~ . The results shown in the following figure are
~ ~
(b) the differences A − B and C ~ −B ~. obtained when the triangle rule is applied again:
Answer
According to the triangle rule, one of the two vectors is shifted
in parallel to itself until it comes into contact with the tail
point of the other vector. The vector drawn from the tail of
the fixed vector to the head of the shifted vector is the sum.
These sum vectors are shown below:
Components of a Vector
Vector components are defined in order to perform addition and other vector
operations algebraically and not graphically. The components are simply algebraic
numbers.
Consider the familiar rectangular coordinate system on the plane. We call
this the cartesian coordinate system. This system consists of an origin O and
two coordinate axes denoted as x - and y -. Positive coordinates are located in
the half of the axes between the origin and the direction of the arrow, and the
negative coordinates are located in the other half.
Let us draw parallel lines from the head points of the vector A~ to the x - and
y - axes (Figure 1.9).
Figure 1.9: Components of a The lengths intersected by these parallels are the x - and y -components of
~.
vector A the vector A ~ and are indicated with A x and Ay respectively. This is shown as,
~ : (A x , Ay )
A
Now, the components themselves can be used to specify the vector, instead of
magnitude and direction.
More generally, in three-dimensional space with three coordinate axes x -, y -
and z - the components are defined as follows: A perpendicular is dropped from
the head of A ~ onto the xy -plane. At the intersection point, two lines are drawn
parallel to the x - and y -axes. The lengths that these parallels intersect with the
axes constitute the A x and Ay components of the A ~ vector. The perpendicular
drawn down to the plane or its projection on the z -axis is the Az component:
Figure 1.10: Vector compo-
nents in 3 dimensions. ~ : (A x , Ay , Az )
A
Now let us find the formula that gives us the components of a vector A ~ in
a plane. Let the magnitude of this vector be A and its angle with the +x axis
be θ . The positive direction of the angle θ is accepted as the anticlockwise rotation
direction.
Let us remember the trigonometric formulas of a right triangle:
b a b
sin θ = , cos θ = , tan θ = (1.2)
c c a
~ (refer to Figure 1.9), we get
Applying these formulas to the components of A
Figure 1.11: Right triangle.
Ax
cos θ = −→ A x = A cos θ
A
Ay
sin θ = −→ Ay = A sin θ
A
In the opposite case, if the components are given, the magnitude and direction
of a vector are found with the following formulas (remember the Pythagorean
formula):
q
A = A2x + A2y
Ay
tan θ =
Ax
1.3. VECTORS 11
We can thus use these formulas to specify a vector either in terms of its compo-
nents or in terms of its magnitude and direction.
Let us summarize these important formulas together:
q
A x = A cos θ A= A2x + A2y
Ay (1.3)
Ay = A sin θ tan θ =
Ax
Notice on calculating the angle θ : The expression for tan θ in formula (1.3)
may not always give the correct result when θ is calculated as an inverse trigono-
metric function on a calculator. It may sometimes give the complement of that
angle, because (−3)/4 and 3/(−4) have the same value of −0.75 for the calcula-
tor.
However, the result of the calculator can be corrected if the signs of the
components A x and Ay are taken into account separately. For example, if A x =
−3, Ay = 4 then the vector A ~ is in the 2nd quadrant, in other words, [90◦ < θ <
180◦ ] . Likewise, if Bx = 2, By = −5 then the vector B ~ is in the fourth quadrant,
in other words, [−90 < θ < 0 ] . (It is incorrect to say that [270◦ < θ < 360◦ ] here;
◦ ◦
angles greater than 180◦ are measured from the negative side.)
Example 1.5
Find out the components of the vectors ~a, ~b, ~c and ~d shown in Answer
the figure. (Each division has a unit of 1 .) You can read from the graph by taking note of the signs of
the components:
ax = 5 ay = 2
bx = 7 by = −2
c x = −4 cy = 0
d x = −3 dy = −2
Example 1.6
Answer
~ ~ ~ ~
Calculate the components of the vectors A, B, C and D whose We use formulas (1.3), taking care to measure the angles anti-
magnitudes and angles are shown below. (You may use Ap- clockwise from the +x axis:
pendix C at the end of the book for unknown sinus and cosine A x = A cos(90 − 30)◦ = 10 × cos 60◦ = 10 × 0.5 = 5
values.) Ay = A sin 60◦ = 10 × 0.87 = 8.7
Bx = B cos(90 + 53)◦ = 6 × (− cos 37◦ ) = −6 × 0.8 = 4.8
By = B sin(90 + 53)◦ = 6 × sin 37◦ = 6 × 0.6 = 3.6
C x = −C cos 25◦ = −8 × 0.91 = −7.3
Cy = −C sin 25◦ = −8 × 0.42 = −3.4
If a vector is on one of the axes, this means its other compo-
nent is zero:
D x = 0 and Dy = −3
Example 1.7
Find the magnitudes and directions of these vectors.
The components of the vectors ~F and G
~ are given as follows:
Answer
F x = 3, G x = −5
Fy = −4, Gy = −12 We use Eqs. (1.3) that give magnitude and direction,
q
θ = −53◦ .
p
F= F 2x + Fy2 = 32 + (−4)2 = 5
We use the same ~
Fy −4 p method in calculating G :
tan θ = = = −1.33 G = (−5) + (−12) = 13
2 2
Fx 3 −12
Two angles with tangents equal to -1.33, are +127◦ and −53◦ . tan θ = = 2.4
−5
We decide which is correct by taking into account the signs Two angles with tangents equal to 2.4 are +64◦ and −113◦ .
of the components. Since ~F has a negative y-component, this ~
This angle is in the 3rd quadrant, as both components of G
angle must be in the 4th quadrant. Therefore, the answer is are negative. Therefore, the answer is θ = −113◦ .
Unit Vectors
Vectors of unit length (1) are defined along the coordinate axes in order to be
able to manipulate vectors easily. First, the ı̂ vector of unit length is taken along
the x -axes in the positive direction. Let us write its components:
ı̂ : (1, 0, 0)
Here, the hat ( ˆ ) sign is used to denote that a vector has a unit length. A unit
Figure 1.12: Unit vectors. vector can be defined for every vector. For example, when we write Â , we
understand the unit vector in the direction of A ~.
Likewise, the unit vector in the direction of the y -axis is defined as ̂ and
the unit vector along the z -axis is defined as k̂ . And their components are as
follows:
̂ : (0, 1, 0) , k̂ : (0, 0, 1)
Now let us construct a vector A ~ on a plane using what we have learned so far.
First, let us try to see what the A x ı̂ product is (Figure 1.13). This is a vector of
unit length multiplied with the number A x . Its magnitude is A x and its direction
is ı̂, in other words, in the direction of the +x -axis. Likewise, the product Ay ̂ is
a vector in the +y direction with length Ay .
Now let us look at the sum of these two vectors: Placing these end-to-end
starting from the origin, we get the A ~ vector according to the triangle rule:
Figure 1.13:
~ = A x ı̂ + Ay ̂
A
This expression can easily be generalized to three-dimensional space:
~ = A x ı̂ + Ay ̂ + Az k̂
A (1.4)
This expression will be used quite often. When vectors are written in this form,
all addition and multiplication operations performed on vectors can be carried
out as arithmetic operations on the components. Likewise, the components of
the vectors given in such form can be identified immediately. For example, the
expression
~ = 3ı̂ − 5 ̂ + 6 k̂
D
↓ ↓ ↓
D x Dy Dz
shows us that D x = 3, Dy = −5, and Dz = 6 .

1.3. VECTORS 13
Addition Using Vector Components

~ and B
Let us write two vectors such as A ~ in terms of their components and
unit vectors:
~ = A x ı̂ + Ay ̂ + Az k̂
A
~ = Bx ı̂ + By ̂ + Bz k̂
B
~ =A
Using them, let us form the vector C ~ +B
~:
~ = A
C ~ +B
~
= (A x ı̂ + Ay ̂ + Az k̂) + (Bx ı̂ + By ̂ + Bz k̂)
= (A x + Bx )ı̂ + (Ay + By ) ̂ + (Az + Bz ) k̂
~ = C x ı̂ + Cy ̂ + Cz k̂
C
Comparing the last two rows, we see that the sum of two vectors is a vector whose
components are the arithmetic sum of the corresponding components of the two
vectors. In other words,



 C x = A x + Bx
~ =A~ +B
~

Cy = Ay + By (1.5)

C ⇐⇒




 Cz = Az + Bz


Working on components through the use of conventional addition and subtraction

is much easier than using graphical methods. This method also applies to vector
expressions with more than two vectors or those multiplied with scalars.
Example 1.8
Bx = 2 By = 0 Bz = −3
~ = 3ı̂ − 4 ̂ + 7 k̂ , B
The vectors A ~ = 2ı̂ − 3 k̂ and C~ = 8 ̂ Cx = 0 Cy = 8 Cz = 0
are given. (b) When making additions and subtractions using compo-
(a) Find the components of each vector. nents, the coefficients of unit vectors are algebraically added
(b) Calculate the sum A ~ +B ~ and the difference B ~.
~ −C or subtracted:
~ ~
(c) Calculate the expression 3A − 8B + 9C . ~ ~ +B
A ~ = (3ı̂ − 4 ̂ + 7 k̂ ) + (2ı̂ − 3 k̂ )
= 5ı̂ − 4 ̂ + 4 k̂
~ −C
B ~ = (2ı̂ − 3 k̂ ) − (8 ̂ ) = 2ı̂ − 8 ̂ − 3 k̂
Answer
(a) The coefficients of the unit vectors ı̂, ̂ , k̂ are the x -, y - (c) Likewise, the coefficients are calculated using algebraic
and z -components respectively. The components are identi- rules:
fied from the given expressions: ~ − 8B
3A ~ = 3(3ı̂ − 4 ̂ + 7 k̂ ) − 8(2ı̂ − 3 k̂ ) + 9(8 ̂ )
~ + 9C
Ax = 3 Ay = −4 Az = 7 = −7ı̂ + 60 ̂ + 45 k̂
Example 1.9
terms of unit vectors,
(b) Calculate the vector R~ which is R
~ = 3A
~ − 2B~.
~.
(c) Find the magnitude and direction of the vector R
Answer
(a) The components:
A x = 5 cos 53◦ = 3 Ay = 5 sin 53◦ = 4
The magnitudes and directions of the vectors A ~ and B~ are Bx = −8 cos 37 = −6.4
◦
By = −8 sin 37◦ = −4.8
given in the figure. Accordingly, in terms of the unit vectors:
(a) Calculate the components and express these two vectors in ~ = 3ı̂ + 4 ̂
A
√
~ = −6.4ı̂ − 4.8 ̂
q
B R = R2x + R2y = 21.82 + 21.62 = 31
~ is calculated as follows:
(b) The vector R Its direction is calculated as its angle with respect to the +x
~ ~
R = 3A − 2B ~ = 3(4ı̂ + 3 ̂ ) − 2(−6.4ı̂ − 4.8 ̂ ) axis:
~ = 21.8 ı̂ + 21.6 ̂ Ry 21.6
R tan θ = = ≈1
R x 21.8
~ is calculated as follows:
(c) The magnitude of the vector R The angle is θ = 45◦ as both components are positive.
Scalar Product
Many formulas in physics can be expressed as products of vectors. Two types
of product are defined for this purpose:
(1) Scalar product,
(2) Vector product.
As can be understood from their names, the first results in a scalar number and
the second in a vector. First, let us consider the scalar product.
The scalar product of two vectors A ~ and B~ is a scalar number defined as
Figure 1.14: Scalar product.
~ ·B
A ~ = A B cos θ (Scalar product) (1.6)
where A, B are magnitudes, and θ is the angle between the two vectors.
Let us underline the important properties of a scalar product:
• Commutation: ~ ·B
A ~ =B ~
~ ·A
• Distribution: ~ · (B
A ~ =A
~ + C) ~ ·B ~ ·C
~ +A ~
• If two vectors form an angle θ = 90◦ , in other words, if the two vectors are
perpendicular to each other, the scalar product is zero. ( cos 90◦ = 0 ). This
feature is often used in calculations as the condition for perpendicularity.
~ ·A
• A ~ = A A cos 0◦ = A2 or the scalar product of a vector by itself gives the
square of its magnitude.
• The sign of the scalar product comes from the cos θ term. The product is
positive if the angle between the two vectors is less than 90◦ and negative if
greater.
Expression of Scalar Product in Terms of Its Components
First, let us find the scalar products of the unit vectors (ı̂, ̂ , k̂ ) with each
other:
ı̂ · ı̂ = 1.1. cos 0 = 1
ı̂ · ̂ = 1.1. cos 90◦ = 0
Likewise, the scalar products of the vectors ̂ and k̂ with themselves is 1 and
with other combinations is zero. Therefore,
ı̂ · ı̂ = ̂ · ̂ = k̂ · k̂ = 1
(1.7)
ı̂ · ̂ = ̂ · k̂ = k̂ · ı̂ = 0
Now let us use these results in the scalar product of two vectors given with
components:
~ ·B
A ~ = (A x ı̂ + Ay ̂ + Az k̂ ) · (Bx ı̂ + By ̂ + Bz k̂ )
1.3. VECTORS 15
As the scalar products in the brackets are expanded, the components, which are
numbers, become coefficients, and only the scalar products of the unit vectors
remain:
~ ·B
A ~= A x Bx (ı̂ · ı̂) + A x By (ı̂ · ̂) + A x Bz (ı̂ · k̂) +
+Ay Bx ( ̂ · ı̂) + Ay By ( ̂ · ̂) + Ay Bz ( ̂ · k̂) +
+Az Bx ( k̂ · ı̂) + Az By ( k̂ · ̂) + Az Bz ( k̂ · k̂)
As the scalar products of the vectors in this expression will either be 1 or 0, the
result simplifies as follows:
~ ·B
A ~ = A x Bx + Ay By + Az Bz
In particular, the scalar product of a vector with itself gives the square of its
magnitude:
~ ·A
A2 = A ~ = A2x + A2y + A2z (1.8)
Consequently, also recalling the definition (1.6), we can calculate the scalar
product in two different ways:
A B cos θ
(
Scalar Product : ~ ·B
A ~= (1.9)
A x Bx + Ay By + Az Bz
A good application of a scalar product is in finding the angle between two

vectors. Combining the two aforementioned expressions for cos θ ,
A x Bx + Ay By + Az Bz A x Bx + Ay By + Az Bz
cos θ = = q q (1.10)
AB
A2x + A2y + A2z B2x + B2y + B2z
Scalar product will be used in many definitions in this course, such as work,
electric potential, magnetic flux, etc.
Example 1.10
~ B
Find the scalar products of the vectors A, ~ with each Answer
~ and C
other. Calculate the angles between the vectors from the figure
when using the formula A.~ B
~ = AB cos θ :
~
A·B ~ = AB cos(180◦ − 37◦ ) = 3 × 5 × (− cos 37◦ )
~ ·B
A ~ = 15 × (−0.8) = −12
~
A·C ~ = AC cos(90◦ − 30◦ ) = 3 × 4 × cos 60◦
~ ·C
A ~ = 12 × (0.5) = 6
~
~ · C = BC cos(90◦ + 67◦ ) = 5 × 4 × cos 157◦
B
~ ·C
B ~ = 20 × (− cos 23◦ ) = 20 × (−0.92) = −18
Example 1.11
Answer
(a) Use the component expression of scalar product:
~p = 3ı̂ − 8 ̂ , ~q = 8ı̂ + 7 ̂
~p · ~q = p x q x + py qy = 3 × 8 + (−8) × 7 = −32
(a) Find the scalar product ~p · ~q ,
(b) Find the angle between these two vectors. (b) First, find the magnitudes of the vectors:
q p √ ~p · ~q −32
p= p2x + p2y = 32 + (−8)2 = 73 ≈ 8.5 cos θ = = = −0.34
pq 8.5 × 11
√ √ θ = 110◦
q= 82 + 72 = 113 ≈ 11
The result is an obtuse angle, as the scalar product is negative.
This data is used in formula (1.10):
Example 1.12
Answer
The scalar product must be zero if the two vectors are per-
A~ = A x ı̂ + 12 ̂ , B
~ = 4ı̂ + 5 ̂
pendicular. Write this condition in terms of the components,
What should the value of the unknown component A x be such ~ ·B
A ~ = A x Bx + Ay By = 0
that these two vectors are perpendicular? A x × 4 + 12 × 5 = 0 → A x = −60/4 = −15
Vector Product
There are many cases in physics in which operations on two vectors result in
a new vector. Hence, it is convenient to define a vector product.
Definition: The vector product of the vectors A ~ and B~ with an angle θ
between them, is a new vector denoted as
~ =A
C ~ ×B
~
and its magnitude and direction are:
• Magnitude: C = A B sin θ ,
~ and B
• Direction: Perpendicular to the plane formed by A ~ and given by
the right-hand rule.
Right-Hand Rule: The right-hand rule needs to be well understood, as it

will be used in many topics throughout this course. As in Figure (1.15), four
~ ) and the palm is
fingers of the right hand are pointed towards the first vector ( A
turned towards the second vector ( B~ ), thus the thumb gives the direction of the
~
vector C .
Figure 1.15: The direction of This rule may be described differently in various textbooks. Although the
vector product is given with the expression we provide here is quite common, you may continue to use any other
right-hand rule. technique that suits you.
Let us underline the important properties of a vector product:
• Anticommutativity: B ~ ×A ~ = −A ~ ×B~.
Hence, the order matters in a vector product.
• Distribution: ~ × (B
A ~ =A
~ + C) ~ ×B ~ ×C
~ +A ~
• If the two vectors are parallel ( θ = 0 ) or anti-parallel ( θ = 180◦ ), then the

vector product will be zero, as the sines will have zero value.
In particular, the vector product of a vector with itself is zero: A~ ×A ~ =0
1.3. VECTORS 17
Expression of Vector Product in Terms of Components

First, let us find the vector products of the unit vectors (ı̂, ̂ , k̂ ) with each
other. By definition, the product of each vector with itself will be zero,
ı̂ × ı̂ = ̂ × ̂ = k̂ × k̂ = 0
Now let us look at the product ı̂ × ̂ : As the magnitudes are 1 and the angles are
90◦ , the result of the vector product will be a vector of magnitude 1.1. sin 90◦ =1 .
Its direction will be in the +z direction according to the right-hand rule. Hence,
this is just our unit vector k̂ .
Also, as the vector product is anticommutative, ̂ × ı̂ = − k̂ .
The vector products of unit vectors are thus as follows:
ı̂ × ı̂ = ̂ × ̂ = k̂ × k̂ = 0
(1.11)
ı̂ × ̂ = k̂, ̂ × k̂ = ı̂, k̂ × ı̂ = ̂
Now let us use these results in the vector product of two vectors given with
components:
~ =A
C ~ ×B
~ = (A x ı̂ + Ay ̂ + Az k̂ ) × (Bx ı̂ + By ̂ + Bz k̂ )
If the brackets are expanded and the component are taken outside of the vector
products,
~ = A x Bx (ı̂ × ı̂) + A x By (ı̂ × ̂) + A x Bz (ı̂ × k̂) +
C
+Ay Bx ( ̂ × ı̂) + Ay By ( ̂ × ̂) + Ay Bz ( ̂ × k̂) +
+Az Bx (|{z}
k̂ × ı̂ ) + Az By ( k̂ × ̂ ) + Az Bz (|{z}
k̂ × k̂)
|{z}
̂ −ı̂ 0
The result simplifies as follows when the vector product expressions of unit
vectors are used:
~ =A
C ~ ×B
~ = (Ay Bz − Az By ) ı̂ + (Az Bx − A x Bz ) ̂ + (A x By − Ay Bx ) ı̂ (1.12)
| {z } | {z } | {z }
Cx Cy Cz
The circular permutation technique is used to memorize this formula. Consider

that the indexes rotate after each other as follows:
x → y → z, y → z → x, z→x→y
Accordingly, when writing the C x component, follow it with the y -component of

~ multiplied by the z -component of B
A ~ and then switch the indexes and subtract. Figure 1.16: Circular permuta-
The circular permutation of the indexes is performed for the other components: tion.
C x = Ay Bz −Az By , Cy = Az Bx −A x Bz , Cz = A x By −Ay Bx
| {z } | {z } | {z }
x→y→z y→z→x z→x→y
Another way of expressing vector product is to write it as a determinant:

ı̂ ̂ k̂
~ ×B
A ~ = det A x Ay Az (1.13)

Bx By Bz
If you expand this determinant according to the minors of the first row, the
expression (1.12) of the vector product follows.
Quantities expressed as vector products in physics include torque, angular
momentum, magnetic force, etc.
The limited vector information we provided here is vital to understanding
subsequent topics. It would be wrong to continue without understanding these
matters.
Example 1.13
Answer
According to the right-hand rule, if four fingers are pointed
towards the first vector and the palm is turned towards the
second vector, then the thumb gives the direction of the prod-
uct. Accordingly, the results are as follows:
The figure shows vectors drawn at the corners of a cube with

unit length (1) edges.
Find only the directions of the vector products ~a × ~b, ~c × ~d and
~e × ~f and mark them on a figure.
Example 1.14
Answer
Let us first find the magnitude of the vector ~c . It can be
seen from the figure that the angle between the vectors is
90 − (23 + 30) = 37◦ . Accordingly,
c = a b sin 37 = 15 × 12 × 0.6
c = 108
The magnitudes of vectors ~a and ~b shown on the xy -plane are The direction of ~c should be perpendicular to the xy -plane,
a = 15 and b = 12 units respectively. Find the magnitude and in other words, along the z -axis. The −z direction is found
direction of the vector ~c which is the product ~c = ~a × ~b . according to the right-hand rule.
Example 1.15
remember the components.
~E = 7ı̂ − 3 k̂ are given.
~ = 3ı̂ − 5 ̂ ,
The vectors D F x = Dy Ez − Dz Ey = −5 × (−3) − 0 × 0 = 15
Calculate the components of the vector ~F = D~ × ~E .
Fy = Dz E x − D x Ez = 0 × 7 − 3 × (−3) = 9
Answer
Fz = D x Ey − Dy E x = 3 × 0 − 5 × 7 = −35
Use the formula (1.12) that gives the components. The circu-
lar permutation of the component indexes allows us to easily Accordingly, ~F = 15ı̂ + 9 ̂ − 35 k̂ .
Multiple-choice Questions
1. How many significant figures does the number 0.003804

2. How should the result of 1.2 + 0.222 =? be expressed?
have?
(a) 1.2 (b) 1.22 (c) 1.4 (d) 1.422
(a) 2 (b) 3 (c) 4 (d) 5
MULTIPLE-CHOICE QUESTIONS 19
3. The mass of a metal coin is (8.2 ± 0.1) grams. What is ~ ·A

12. The scalar product of a vector ( A ~ ) with itself is:
the mass of two metal coins? (a) Zero.
(a) 16.4 (b) 16.4±0.2 (c) 16.6 (d) 16±0.1 (b) Equal to its magnitude.
(c) Equal to the square of its magnitude.
4. The relative error in the measurement of the edge of a (d) Equal to the square root of its magnitude.
cube is 1 % . What is the relative error of the volume of
the cube? 13. Which of the following equalities is true?
(a) 1 % (b) 2 % (c) 3 % (d) 4 % ~ ·B
(a) A ~ =B ~ ·A~
~ ×B
(b) A ~ = −B ~
~ ×A
5. An object has a mass of 8.888 g and volume of 2.0 cm3 . ~ ~
(c) A × A = 0
What is the density of this object in g/cm3 ?
(d) All of the above
(a) 4 (b) 4.4 (c) 4.44 (d) 4.444
14. The magnitude of vector ~a is 5 units and the magni-
6. Which of the following quantities is not a vector?
tude of vector ~b is 3 units. Accordingly, which of the
(a) Velocity
following cannot be the magnitude of ~a + ~b ?
(b) Force
(c) Electric field (a) 1 (b) 3 (c) 5 (d) 7
(d) Volume
15. Which of the following propositions is true?
7. Which of the following is true if two vectors are perpen- (a) The sum of two vectors with different magnitudes
dicular? may be zero.
(a) The vectors have the same magnitude. (b) The magnitude of a vector is equal or greater than
(b) Their scalar product is zero. its perpendicular components.
(c) Their vector product is zero. (c) The sum of the vectors (ı̂ + ̂) is also a unit vector.
(d) Their components are the same. (d) If ~a = ~b + ~c then a > b and a > c are always true.
16. Which of the following vector algebra operations does

not comply with the rules?
~ · (B
(a) A ~
~ × C)
~ × (B
(b) A ~
~ + C)
~ × (B
(c) A ~
~ · C)
8. Which sum is correct for the three vectors in Diagram ~ × (B
(d) A ~
~ × C)
1?
17. Which of the following vector equalities is false?
(a) ~a = ~b + ~c (b) ~b = ~a + ~c (c) ~c = ~a + ~b (d) None.
~ · (B
(a) A ~ =A
~ + C) ~ ·B ~ +A~ ·C~
9. Which difference is correct for the three vectors in Dia- ~
(b) A × (B ~ ~
~ + C) = A × B ~
~ +A×C ~
gram 2? ~ · (A
(c) A ~ ×B~) = 0
(a) ~a = ~b − ~c (b) ~b = ~a − ~c (c) ~c = ~a − ~b (d) None. ~ ~ ~
(d) A · (B × B) = B2 A~
10. Which equality is true for the three vectors in Diagram 18. Which of the following is true if the sum of three vectors
3? is zero?
(a) ~c = ~a + ~b (a) All three vectors are on the same plane.
(b) ~a − ~b = ~c (b) The vectors are perpendicular.
(c) ~a + ~b + ~c = 0 (c) The vectors have the same magnitude.
(d) The sum of three vectors cannot be zero.
(d) ~b = ~a + ~c
11. Which of the following is true if the vector product of 19. Which of the following is true if the scalar product of
two vectors is zero? two vectors is negative?
(a) The vectors are parallel. (a) The vectors have equal magnitude.
(b) The vectors are perpendicular. (b) The vectors are in opposite directions.
(c) The magnitudes of the vectors are equal. (c) Their angle is obtuse.
(d) None of the above. (d) The vectors are perpendicular.
20. If ~a ·~b = ~a ·~c , then which of the following is always true? (b) ~b and ~c are in the same direction.
(c) ~b and ~c are perpendicular.
(a) ~b = ~c (d) The projections of ~b and ~c along ~a are equal.
Problems
1.1 Dimensions and Units 1.3 Vectors

1.1 The fuel consumption of an automobile is given as
60.0 mil/gallon . Convert this into the unit km/liter . ( 1 mil =
1.609 km and 1 gallon = 3.788 liter .)
[Answer: 25.5 km/L.]
1.2 The unit light-year, a unit of length used in astronomy,

is the distance traveled by light in one year with the speed
c = 2.998 × 108 m/s . Another unit of length, the Astronomic Problem 1.9
Unit (AU), is the average distance between the Earth and the 1.9 For the vectors shown on a millimeter graph paper in
Sun and 1 AU = 1.50 × 108 km . (a) How many meters is one the figure, find the sums ~a + ~b , ~b + ~c and ~a + ~c by drawing
light-year? (b) How many AUs is one light-year? them according to the triangle rule.
[A: (a) 9.45 × 1015 m , (b) 6.3 × 104 AU .]
1.10 The magnitudes and angles to the +x axis of three vec-
1.3 A wall of 6.0 m2 can be completely covered with 1.0 liters tors are given as follows: (a) 33 m/s, 60◦ , (b) 128 m/s, 150◦ ,
of paint. What is the thickness of the paint? (c) 22 m/s, −145◦ . Calculate the components of these vectors.
[A: 0.17 mm .] [A: (a) 17, 29 , (b) −111, 64 , (c) −18, −13 .]
1.4 Express the following data using prefixes of basic units:

3 × 10−9 m , 8 × 1013 bytes , 5 × 10−6 seconds .
[A: 3 nm , 80 terabytes , 5 µs .]
1.2 Precision and Significant Figures

1.5 The radius of a sphere is measured as 6.5 ± 0.2 cm . Cal-
culate the surface area and volume of this sphere, and express Problem 1.11
the results with a margin of error. (The surface area of a 1.11 Calculate the components of the vectors A ~,B~ and C ~,
sphere is A = 4πr2 and its volume is V = (4/3)πr3 .) whose magnitudes and directions are given in the figure.
[A: A = (531 ± 30) cm2 , V = (1150 ± 100) cm3 .] [A: A x =52, Ay =30 , Bx = − 24, By =32 , C x =0, Cy = − 30 .]
1.6 A watch brand claims that its watches produce 8 seconds
1.12 Calculate the magnitudes and directions (angles to the
of error in one year. (a) What is the relative error of this
x axis) of the following three vectors specified with their
watch? (b) How much error is there at the end of a 90-minute
components: a x =5, ay =12 , b x = − 4, by = − 3 , c x = − 3, cy =1 .
football match? [A: (a) 3 × 10 , (b) 0.001 s .]
−7 √
[A: a = 13, 67◦ , b = 5, −143◦ , c = 10, 162◦ .]
1.7 The length of a rectangular plate is measured with a
millimetric ruler and found to be 18 mm . Then, its width is
measured more precisely with a micrometer and found to be
3.5 mm . (a) Find the relative errors of the width and length.
(b) Calculate the surface area of the plate with a margin of
error. [A: (a) 0.06, 0.03. (b) (63 ± 6) mm2 .] Problem 1.13
1.8 According FIFA rules, the width of a football field should 1.13 (a) Write the vectors A ~ and B~ shown in the figure
be within the range 65 − 75 m and length within the range in terms of the unit vectors ( ı̂, ̂ ). (b) Find the vector
100 − 110 m . (a) What are the relative errors for the width ~ = 2A
C ~ − 3B ~ . (c) Calculate the magnitude and direction
and the length? (b) If we wish to cover this field with grass, of the vector C ~.
what would the surface area of the field with a margin of ~ ~ = 99ı̂+94 ̂ .
~ = −17ı̂−10 ̂ . (b) C
[A: (a) A = 24ı̂+32 ̂ , B
error be? [A: 7 % 5 % , (b) 7350 ± 882 m2 .] (c) C = 137, θ = 44 .]◦
PROBLEMS 21
1.14 The vectors ~F = −2ı̂ + 3 ̂ + 6 k̂ and G~ = 4ı̂ − 7 ̂ − 4 k̂

are given. (a) Calculate the magnitudes of the vectors.
(b) Calculate the vector ~F − G~.
[A: (a) 7 and 9, (b) ~F − G
~ = −6ı̂ + 10 ̂ + 10 k̂ .]
1.15 (a) Calculate the magnitudes of the vectors A ~ = 2ı̂ −

~
2 ̂ + k̂ and B = 6ı̂ + 2 ̂ − 3 k̂ . (b) Calculate their scalar
product. (c) Find the angle between the two vectors.
[A: (a) 3 and 7, (b) 5, (c) 76◦ .]
Problem 1.19
1.16 The vectors ~a = 2ı̂ − 5 k̂ , ~b = 3 ̂ − 4 k̂ and ~c = 5ı̂ + 2 ̂ 1.19 For the vectors indicated in the figure, show only the
are given. Calculate the scalar product ~a · (~b − ~c) . [A: 10.] directions of the products ~a × ~b , ~c × ~d , ~e × ~f and ~g × h ~ on
the figure.
1.17 Find a vector ~b perpendicular to the vector ~a = 3ı̂ − 5 ̂
such that the component b x is 4 units. [A: ~b = 4ı̂ + 2.4 ̂ .] 1.20 (a) Calculate the magnitudes of the vectors A ~ = 2ı̂+2 ̂ −
~
k̂ and B = 4ı̂ + 4 ̂ + 7 k̂ . (b) Find the angle between them.
1.18 The magnitudes of the vectors A ~ and B ~ are known to (c) Find the components of the vector product C ~ =A ~ ×B ~.
be A = 1.2 and B = 5 . The vector products of these two (d) Calculate the magnitude of C ~.
~ ×B
vectors is A ~ = 3ı̂ − 4 ̂ . Find the angle between these two [A: (a) 3 and 9, (b) 71◦ , (c) C x = 18 , Cy = −18 , Cz = 0 ,
√
vectors. [A: 53◦ .] (d) C = 18 2 .]
2
MOTION IN A STRAIGHT
LINE
The masterpiece of Japanese

public transportation system,
Shinkansen (the bullet train)
passing by Mount Fuji Yama en
route from Tokyo to Osaka.
The average speed of
Shinkansen can reach up
to 320 km/h. How can we
estimate the position of this
train at a future time?
Motion is one of the main topics of interest in physics. The branch of physics
called Mechanics studies all moving objects, from tiny pebbles to automobiles,
from cannon shells to planets and massive stars.
Mechanics comprises two parts: Kinematics is the branch of Mechanics
that studies the relations between the positions and speeds of objects with time,
regardless of the reasons for the motion. Dynamics sets the laws of motion and
examines the motion under physical forces. In this chapter, we will examine
one-dimensional kinematics. Vector concepts shall not be required here, but the
basic concepts that we will develop will later form the basis of vector expressions
in two and three-dimensional motions.

24 2. MOTION IN A STRAIGHT LINE
2.1 POSITION, DISPLACEMENT, VELOCITY AND ACCELERATION

Position and time are the two most fundamental concepts in physics. Time is
a uniform quantity that flows at the same rate for everyone. Because of these prop-
erties, time is the independent variable in kinematics, as it is not possible
to alter it. It is indicated in equations with the symbol t .
Position and Displacement
The position of an object is its location on a coordinate system. In three-
dimensional space, position is specified with the Cartesian coordinates ( x, y, z ).
These coordinates vary as a function of time during a general motion:
x = x(t), y = y(t), z = z(t)
Only one coordinate axis is required in one-dimensional space. For this purpose,
an infinite line is taken and the positive direction is marked by an arrow at the
end. Then, an origin (O), in other words, a yardstick, is chosen to mark the place
from which the measurements are to be made. This can be a tree or a pole that is
known to everyone. The origin separates the positive region from the negative
one on the x -axis.
Figure 2.1: The position of an
object in a one-dimensional coor-
dinate system.
Therefore, the position of an object at point P is its x coordinate at time t :
x = x(t) (Position) (2.1)
The property that determines motion is the change in position. Displacement

is defined for this purpose.
Figure 2.2: Displacement ∆x .
Definition: If the position of an object is x1 at time t1 and its position is x2

at a later time t2 , the difference
∆x = x2 − x1 (Displacement) (2.2)
is called the displacement (Figure 2.2). Its unit is the meter (m). (In physics,
the ∆ sign before a symbol means the change in that quantity. For example:
∆a = alast − afirst .)
The sign of displacement determines the direction of motion. If ∆x > 0 then
x2 > x1 , in other words, the object moves in the positive direction. Likewise, if
∆x < 0 then x2 < x1 , and the object moves in the negative direction.
Average and Instantaneous Velocity
Anyone can run 100 meters, but some of us run it in a shorter time. It is
important to know the amount of displacement in a given time interval. Velocity
is defined for this purpose.
2.1. POSITION, DISPLACEMENT, VELOCITY AND ACCELERATION 25
Definition: If the position of an object at time t1 is x1 and its position at a

later time t2 is x2 , then the ratio
displacement x2 − x1 ∆x
vav = = = (average velocity) (2.3)
elapsed time t2 − t1 ∆t
is called the average velocity. Its unit is the meter/second (m/s).

The direction of motion is always determined by the sign of velocity. As ∆t
is always positive, only the sign of the displacement ∆x matters. Accordingly,
if displacement ∆x > 0 , then the velocity is also positive and the object moves
in the +x direction. In contrast, if ∆x < 0 , then the velocity is negative and the
object moves in the −x direction.
Average velocity is not useful for calculating in physics, because it requires
measurement at two points and it is not possible to know the velocity until the
object reaches the final point x2 . Instead, we would like to know the velocity at a
given instant. The concept of instantaneous velocity is defined for this purpose.
Let us remember the concept of the derivative in mathematics. Consider
placing two sticks at the positions x1 and x2 along a path so as to measure
average speed. Let us measure the times t1 and t2 when the moving object passes
through these sticks. This measurement will give us the average speed in the
interval [x1 , x2 ] . Now, let us bring the stick at x2 closer to x1 , in other words,
let ∆x get smaller. In this case, the value ∆t will also be smaller. However, the
ratio ∆x/∆t may not be small.
Therefore, let us gradually bring x2 closer to x1 such that ∆x → 0 . In
such a case, the time interval will also approach zero. Yes, ∆x → 0, ∆t → 0
but surprisingly, the ratio ∆x/∆t remains at a finite value. The operation of
“approaching zero without setting equal to zero” is known in mathematics as the
limit. And the ratio ∆x/∆t is called “the derivative of x with respect to t ". The
first derivative is shown as dx/dt or x0 .
Definition: The limit of the average velocity is called the instantaneous
velocity (or simply the velocity):
x2 − x1 ∆x dx
v = lim = lim = (velocity) (2.4)
t2 →t1 t2 − t1 ∆t→0 ∆t dt
We thus know the velocity at a given time t . The direction of motion again
depends on the sign of the velocity v .
Speed
Speed is the absolute value (or, the magnitude) of the velocity, shown as | v | .
In one-dimensional motion, the velocity seems merely to be an algebraic
quantity, it can be either positive or negative. But, in reality, it is a vector quantity,
as will be clear in two- and three-dimensional motion. The speed of a vector
velocity has the same definition as the magnitude.
Brief information on derivatives
The concept of derivative and derivation techniques are examined extensively
in calculus courses. Here, let us briefly review the derivatives of the most fre-
quently encountered types of functions without proof. The derivatives of most
useful functions y(x) with respect to the independent variable x are as follows:
Derivatives of certain functions

dy dy
function y derivative y0 = function y derivative y0 =
dx dx
x 1 sin x cos x
x3 3x2 cos x − sin x
x−5 −5x−6 tan x 1/ cos2 x
√ √
x 1/ 2 x ex ex
xn n xn−1 ln x 1/x
In mechanics, the independent variable should be taken as t when using these

formulas.
Example 2.1
the figure:
The positions of a moving object at various times t are shown During the interval [t1 , t2 ] : ∆x = x2 − x1 = 6 − 2 = 4 m
on the x axis: (a) Determine the displacements ∆x = x2 − x1 The object was displaced by 4 m in the positive direction.
and ∆x = x3 − x2 . (b) Determine the average velocities at the During the interval [t2 , t3 ] : ∆x = x3 − x2 = −4 − 6 =
time intervals ∆t = t2 − t1 and ∆t = t3 − t2 . −10 m
The object was displaced by 10 m in the negative direction.
(b) The average velocity is the ratio of these displacements to
the elapsed time:
∆x 4
During the interval [t1 , t2 ] : vav = = = 4 m/s
∆t 2−1
Answer ∆x −10
During the interval [t2 , t3 ] : vav = = = −3.3 m/s
(a) We read the positions of the object at various times from ∆t 5−2
Example 2.2 ∆x 8 − 4
(b) vav = = = 1 m/s
∆t 5−1
The position of an object is given as a function of time as:
(c) The general expression of velocity is the derivative dx/dt :
x = t3 − 5t2 + 8 (meters) dx
(a) Find the positions of the object at times t = 1 s and t = 5 s , v= = 3t2 − 10t
dt
(b) Calculate the average velocity of the object during this The values of this expression at the requested times t are the
time interval, instantaneous velocities:
(c) Find the velocities of the object at times t = 1 s and t = 5 s , The velocity at t = 1 : v(1) = 3 × 12 − 10 × 1 = −7 m/s
(d) Find the positions at which the velocity of the object is The velocity at t = 5 : v(5) = 3 × 52 − 10 × 5 = 25 m/s
zero.
(d) When the velocity is zero, we have v(t) = 0 :
Answer v = 3t2 − 10t = 0
(a) The x values of the given function at times t = 1 s and The velocity is zero at the roots of this expression, which
t = 5 s give the positions: are t = 0 and t = 10/3 . We calculate the x positions that
For t = 1 s : x(1) = 13 − 5 × 12 + 8 = 4 m correspond to these t values:
For t = 5 s : x(5) = 53 − 5 × 52 + 8 = 8 m x(0) = 8 m and x (10/3) = −11 m
Average and Instantaneous Acceleration

Velocity gives us very important information about the motion of an object.
However, we sometimes also need to know the rate of change of velocity with
respect to time. We might ask, in regard to two cars, “Which one speeds up more
quickly?” The rate of change of velocity is called the acceleration.
2.1. POSITION, DISPLACEMENT, VELOCITY AND ACCELERATION 27
Definition: If the velocity of an object is v1 at time t1 , and v2 at a later time

t2 , then the ratio
Figure 2.3: Average accelera-

tion.
v2 − v1 ∆v
aav = = (average velocity) (2.5)
t2 − t1 ∆t
is called the average acceleration. Its unit is the meter/second 2 ( m/s2 ).

It is important to understand the sign of the acceleration. We sometimes hear
definitive statements such as “acceleration is positive if the object speeds up and
negative if it slows down.” This is incorrect. The expression for acceleration above
shows that the sign depends on the chosen coordinate system. For example, if the
object moves in the positive direction ( v1 , v2 > 0 ) and its velocity is increasing
( v2 > v1 ), then surely ∆v > 0 and therefore aav will be positive. However,
consider now the case in which the object is moving in the negative direction
( v1 , v2 both negative). If its speed is increasing, for negative numbers, this means
that v2 < v1 (e.g., −5 < −3 ). In this case, ∆v = v2 − v1 < 0 and therefore aav is
negative!
The correct interpretation of the sign of acceleration is as follows: Objects that
speed up in the positive direction or slow down in the negative direction have positive
acceleration. And, vice versa, objects that slow down in the positive direction or
speed up in the negative direction have negative acceleration.
Average acceleration is, again, not a very useful quantity, because it requires
measurement at two points and it is not possible to know until the object reaches
the final velocity v2 . Instead, we would like to know the acceleration at a given
instant.
Definition: The limit of the average acceleration is called the instantaneous
acceleration (or simply the acceleration):
∆v dv d2 x
a = lim = = 2 (acceleration) (2.6)
∆t→0 ∆t dt dt
In other words, acceleration is the derivative of velocity with respect to time.
Also, as velocity is the derivative of position, we may regard acceleration as the
second derivative of position. In mathematics, the second derivative is shown
with d2 x/dt2 or x00 .
We can thus know the acceleration at a given time t . Acceleration is not often
used in daily life. However, we sometimes hear the performance of sports cars
expressed as “reaching 100 km/h speed in 6 seconds”, etc. This is, in fact, another
expression of acceleration.
Example 2.3
The velocities of a moving object at various times are shown on

the x axis.
(a) Determine the velocity differences ∆v = v2 − v1 and
∆v = v3 − v2 ,
(b) Determine the average accelerations at the time intervals
∆t = t2 − t1 and ∆t = t3 − t2 . (b) The average acceleration is the ratio of these differences

Answer in velocity to the elapsed time:
(a) Reading the velocities of the object at various times from ∆v 3
During the interval [t1 , t2 ] : aav = = = 3 m/s2
the figure: ∆t 2 − 1
During the interval [t1 , t2 ] : ∆v = v2 − v1 = 6 − 3 = 3 m/s −8
During the interval [t2 , t3 ] : aav = = −2.7 m/s2
During the interval [t2 , t3 ] : ∆v = v3 −v2 = −2−6 = −8 m 5 −2
Example 2.4
v = x0 = 3t2 − 10t
The position of an object as a function of time is given by a = v0 = 6t − 10
(b) For t = 1 , v = 3 × 12 − 10 × 1 = −7 m/s ,
x = t3 − 5t2 + 4 (meters) .
(a) Determine the velocity and acceleration functions using for t = 5 , v = 3 × 52 − 10 × 5 = 25 m/s .
derivative, (c) These velocities are used in the definition of average ac-
celeration:
(b) Find the velocities of the object at times t = 1 s and t = 5 s ,
(c) Calculate the average acceleration of the object during this ∆v 25 − (−7)
aav = = = 8 m/s2
time interval, ∆t 5−1
(d) We use the values t = 1 and 5 in the acceleration expres-
(d) Find the accelerations of the object at times t = 1 s and
sion:
t = 5 s,
a(1) = 6 × 1 − 10 = −4 m/s2
(e) Find the time when the acceleration is zero.
a(5) = 6 × 5 − 10 = 20 m/s2
Answer (e) We calculate the time when the acceleration expression is
(a) The first derivative of x is the velocity v and its second zero:
derivative is the acceleration a : 0 = 6t − 10 → t = 1.7 s .
2.2 MOTION WITH CONSTANT ACCELERATION

There is accelerated motion whenever the velocity of an object traveling on a
straight line changes. Examining the most general accelerated motion is difficult
and requires advanced mathematics. However, if the rate of change of velocity is
constant, in other words, if the change in velocity is always the same at equal time
intervals, then it is a motion with constant acceleration. This is the simplest
accelerated motion.
Let us remember the formulas for average acceleration and average velocity:
v2 − v1 x2 − x1
aav = , vav =
t2 − t1 t2 − t1
Let us simplify the notation here: Let the object start moving at t1 = 0 from an
initial position x0 with an initial velocity of v0 and let its final velocity be v at
the final time t2 = t at the final position x2 = x . Since the average of a constant
quantity is itself, making these changes in the formulas and rearranging, we get
v − v0 x − x0
aav = a = , vav =
t−0 t−0
v = v0 + a t , x = x0 + vav t
We can eliminate the average velocity vav in the last expression as follows: The
average of the velocity that changes uniformly from the value v0 to the value v
will be,
v + v0
vav =
2
Also, after using this expression, the position formula can finally be written:
x = x0 + v0 t + 21 a t2
2.2. MOTION WITH CONSTANT ACCELERATION 29
Notice that the position x changes with the square of time ( t2 ) in motion with
constant acceleration.
If the time t in the velocity expression is extracted as t=(v−v0 )/a and used in
the position expression, we find a very useful velocity formula without time:
v2 − v20 = 2 a (x − x0 )
These formulas allow us to calculate the position and velocity at time t of the an
object with given initial conditions, in other words, initial position x0 and initial
velocity v0 , moving with acceleration a .
Let us summarize all the formulas for motion with constant acceleration:
v = v0 + a t
x = x0 + v0 t + 12 a t2 (motion with constant acceleration) (2.7)
v2 − v20 = 2 a (x − x0 )
Motion with Constant Velocity

A special case of accelerated motion is the motion at constant velocity along
a line. There is no need to derive separate formulas for this. Acceleration is zero
if the velocity is constant. We find the formulas for uniform linear motion if we
take a = 0 in the aforementioned formulas:
x = x0 + v0 t (uniform linear motion) (2.8)
If we wish to know the distance traveled by the object rather than its final position,
we can write this formula in terms of the distance traveled s using s = x − x0 , as
follows:
s = x − x0 = v t (distance traveled) (2.9)
Example 2.5
v = v0 + at = 0 + 2t = 2t
A car starts from rest and accelerates at a constant rate of x = x0 + v0 t + 12 t2 = 0 + 0 + 12 2t2 = t2
2 m/s2 . We can solve each item using these equations:
(a) The velocity and position at time t=3 s are found by plug-
(a) What will its velocity and distance traveled be after 3 s ?
ging t = 3 in the expressions that we found for v and x :
(b) In how many seconds will it reach the velocity 10 m/s ?
v = 2 × 3 = 6 m/s and x = (3)2 = 9 m .
(c) In how many seconds will it cover a distance of 64 m ?
(b) t is found by taking v = 10 m/s in our velocity expression:
Answer 10 = 2t −→ t = 5 s .
If we choose the origin at the initial position of the object, (c) The t value giving x = 64 m in the position expression is
then x0 = 0 . Since it starts from rest, v0 = 0 . As the acceler- calculated:
ation is given as a = 2 m/s2 , using Eqs. (2.7), 64 = t2 −→ t = 8 s .
Example 2.6
velocities given in terms of km/h into m/s units:
A car traveling at 72 km/h is slowed down through the pressing v0 = 72 km/h = 72000/3600 m/s = 20 m/s
of its brakes and its velocity is reduced to 36 km/h in 5 s . v = 36 km/h = 10 m/s
(a) What is the constant acceleration of the car? (a) We use the time t = 5 s that passes between these two
(b) How much distance does it travel during that time? velocities to find the acceleration:
(c) How much time passes from the start until the car fully v = v0 + at → 10 = 20 + a × 5 → a = −2 m/s2
stops? As the car traveling in positive direction slows down, its
Answer First, it is necessary to convert the initial and final acceleration is negative.
(b) If we take the initial position of the car as the origin, then (c) The car coming to a full stop means that its final velocity
x0 = 0 . Accordingly, is v = 0 . Using this value,
x = x0 + v0 t + 2 at = 0 + 20 × 5 + 2 (−2) × 5 = 75 m .
1 2 1 2 v = v0 + at → 0 = 20 − 2t → t = 10 s
Example 2.7
started to accelerate, then x0 = 0 and v0 = 4 m/s . It will
A runner cruising at a velocity of 4 m/s suddenly starts to be suitable to use the velocity formula without time, as the
accelerate and reaches a velocity of 7 m/s over a distance of first and final velocities are given. Substituting the position
11 m . x = 11 m and the velocities, we get,
(a) What is the constant acceleration of the runner? v2 − v20 = 2a(x − x0 ) → 72 − 42 = 2a × 11 → a = 1.5 m/s2
(b) How much time elapsed during this acceleration? (b) Both the v and the x formula can be used to find the time
Answer t . The velocity formula gives a quicker result:
(a) If we choose the origin as the point at which the runner v = v0 + at → 7 = 4 + 1.5t → t = 2 s
Example 2.8
final velocity with v M and position with x M , the accelerated
motion formulas are as follows:
v M = v M0 + at = 0 + 3t → v M = 3t
x M = x M0 + v M0 t + 12 at2 = 0 + 0 + 1.5t2 → x M = 1.5t2
An automobile is traveling at a constant speed of 90 km/h when The automobile performs uniform linear motion at a constant
it passes a police motorcycle parked under a tree. Spotting a velocity of 25 m/s , in other words, its acceleration is zero.
traffic violation, the policeman starts the chase on his motor- The initial position of the automobile is 100 m at t = 0 . If
cycle when the automobile is 100 m away and accelerates at a we indicate its final position with xA ,
rate of 3 m/s2 . vA = vA0 = 25 m/s = constant
(a) Choose a coordinate system and starting time to write xA = xA0 + v0A t → xA = 100 + 25t
the equations giving the positions and velocities of the In this systematic approach, the solution follows easily once
automobile and motorcycle. the equations of motion are written down:
(b) How long will it take the policeman to catch the automo-
(b) When the motorcycle catches the automobile, they will
bile?
both have the same position. The mathematical expression
(c) How far will the motorcycle be from its original position?
of this is as follows:
(d) What is the velocity of the motorcycle when it catches up
x M = xA
to the automobile?
Let us substitute the expressions we found for x M and xA :
Answer 1.5t2 = 100 + 25t
In problems with two moving objects, a good approach is From this, we find a quadratic equation:
to choose a single coordinate system and write both of their 3t2 − 50t − 200 = 0
equations in that same system. This systematic approach will
The roots of these equation are t = −3.3 and t = 20 . As the
be much more efficient in the long run.
negative root is not physically valid, the result is t = 20 s .
First, let us convert the velocity of the automobile into m/s :
(c) Substituting t = 20 s in the expression for x M , the distance
90 km/h = 90 000/3 600 = 25 m/s .
traveled by the motorcycle is found as follows:
(a) Let us take the position where the motorcycle sets off
as the origin and also start the clock when the motorcycle x M = 1.5t2 = 1.5 × 202 = 600 m
begins moving. Accordingly, the initial velocity and position (d) The value t = 20 s is used in the expression v M = 3t :
of the motorcycle are zero: x M0 = v M0 = 0 . If we indicate its v M = 3 × 20 = 60 m/s
Example 2.9
(a) Choose a coordinate system for writing the equations of
motion for each car.
(b) When do the cars meet?
(c) What is the distance of the meeting point to the origin?
(d) What is the velocity of each car when they meet?
Two cars 100 m apart, start moving toward each other at the
same instant. The first car starts with a speed of 4 m/s and Answer
increases its speed at a rate of 1 m/s2 . The second car starts in This problem could be solved by choosing a separate coordi-
the opposite direction with a speed of 6 m/s and increases its nate system for each moving object. In that case, two origins
speed at a rate of 3 m/s2 . and two positive directions would be necessary. Yet, in the
2.3. FREE FALL 31
method we used in the previous problem, it is more consis- x1 = x2

tent to choose only a single coordinate system and write both This equality does not mean that they traveled the same dis-
motions in that same system. tance. It is the condition of both of them being at the same
(a) Let us choose the origin at the position where the first car coordinate. Therefore, if we substitute the expressions we
sets off and choose the positive direction towards the other found for x1 and x2 ,
car. Accordingly, we can immediately write the equations of 4t + 12 t2 = 100 − 6t − 23 t2
motion for the first car from the given data: From here, we find a quadratic equation for t :
v1 = v10 + a1 t = 4 + 1 × t t2 + 5t − 50 = 0
x1 = x10 + v10 t + 12 a1 t2 = 0 + 4t + 12 t2 The roots of this equation are −10 and 5 . As negative time
The second car starts off at a distance of 100 m away and its is meaningless, the solution is t = 5 s .
velocity is in the negative direction. Its acceleration is also (c) The position of the meeting place is found by taking t = 5
negative as it gets faster in the negative direction. Therefore, in either of the expressions for x1 or x2 . Using x1 ,
the equations of motion of the second car are as follows: x1 = 4 × 5 + 12 (5)2 = 32.5 m = x2 .
v2 = v20 + a2 t = −6 − 3t (d) We substitute the value t = 5 in the expressions for v1
x2 = x20 + v20 t + 12 a2 t2 = 100 − 6t + 21 (−3)t2 and v2 that we found above:
It is easier to solve the problem after writing these equations. v1 = 4 + 5 = 9 m/s
(b) The cars are at the same position when they meet: v2 = −6 − 3 × 5 = −21 m/s
2.3 FREE FALL

A most familiar example of motion with constant acceleration is falling objects.
The fall of a stone dropped from a height, a ball thrown upwards decelerating and
falling back down, etc. Galileo was the first person to discover that all of these
motions occurred under the same constant gravitational acceleration. Until
the 16th century, everyone believed that “heavier objects fall faster,” as stated
in Aristotle’s book. Galileo showed that this was not true by dropping various
objects from different heights and showing that all objects fall with the same
constant acceleration if air friction is neglected.
Today, Galileo’s experiment can be easily repeated in a vacuum tube in a
laboratory (Figure 2.4). A feather and a coin are observed to fall with the same
acceleration in a vacuum tube. The motion that occurs under the effect of only Figure 2.4: All objects fall with
gravity is called free fall. the same acceleration in vac-
The Earth has a constant acceleration near its surface that accelerates all uum.
objects towards its center. This is called gravitational acceleration, and its
absolute value is indicated by g . Although it varies slightly according to geo-
graphical location, it is approximately
g = 9.81 m/s2 .
The approximate value of g ≈ 10 m/s2 shall be used in solving problems in this

book. The 2 % error arising from this approximation is negligible.
When using this gravitational acceleration g in the formulas of motion with
constant acceleration (2.7), what should be used for a ? a = +g or a = −g ? The
answer to this question is: It depends on how the y -axis is chosen. If the y -axis is
chosen upwards, a = −g regardless of which direction the object is thrown. This is
because the gravitational acceleration is towards the center of the earth, in other
words, in opposite direction to our y -axis. But, if the y axis is chosen downwards,
a = +g regardless of the direction of throwing, because the y -axis is in the same
direction as the acceleration.
This may not be important in simple problems. However, if the object moves
in both directions or if the motion of two separate objects is examined simultane-
ously, then working with a single y axis and not changing it provides the correct
solutions.
Therefore, the choice of the y -axis must also be specified when applying the
constant acceleration formulas (2.7). We summarize them for free fall motion:
y-axis is upwards y-axis is downwards

a = −g a = +g
v = v0 − g t v = v0 + g t (free fall) (2.10)
y = y0 + v0 t − 21 g t2 y = y0 + v0 t − 21 g t2
v2 − v20 = −2g(y − y0 ) v2 − v20 = 2g(y − y0 )
These formulas apply both when the object is moving upwards or downwards,
as long as we do not change the coordinate system. For example, we can use the
same equation to examine the motion of a stone thrown upwards until it falls to
the ground. We will discuss this in worked examples below.
Maximum height
Equations (2.10) are sufficient to solve all free fall problems; it is not necessary
to memorize formulas for special cases. As an example, let us calculate the
maximum height reached by an object thrown vertically upwards with an initial
speed v0 . When the object reaches the peak point, it stops for an instant and then
Figure 2.5: The distances trav- starts moving downwards. In other words, the characteristic of the maximum
eled by a falling object in equal height is that the velocity there is zero:
time intervals.
v=0 (at maximum height)
In the last one of Eqs. (2.10, upwards) we take v = 0 and then solve for y − y0 =
hmax :
v2
hmax = 0
2g
Example 2.10
(b) The velocity and position at the value t = 1 s can be
A ball is thrown from the ground upwards with a speed of directly calculated:
15 m/s . v = 15 − 10 × 1 = 5 m/s
(a) Choose a coordinate system to write the equations giving y = 15 × 1 − 5 × 12 = 10 m
the velocity and position of the ball. (c) The characteristic of maximum height is that the velocity
(b) What will be its position and velocity 1 s after being is zero there: v = 0 . We use this condition to find the time:
thrown? 0 = 15 − 10t → t = 1.5 s
(c) In how many seconds does it reach maximum height? (d) We find the maximum height by using the time t = 1.5 s
(d) What is the maximum height? in the y equations:
(e) At what times does it pass through 10 m above the ground? y = 15 × 1.5 − 5 × 1.52 = 11.3 m
(f) After how many seconds from being thrown will it fall (e) The y equation should satisfy the value y=10 m :
back to the ground? 10 = 15t − 5t2 → t2 − 3t + 2 = 0
Answer (a) If we choose the +y direction upwards, the ac- The two solutions to this equation are t=1 s (going upward)
celeration will be −g . Choosing the place where the ball and t = 2 s (coming downward).
is thrown as the origin (y0 = 0) , the equations of motion (f) As the time taken to reach the maximum height will be
for a ball thrown in the positive direction with a velocity of equal to the time it takes to fall back to the ground, we take
v0 = 15 m/s are as follows: twice the time found in (c). However, if we had not known
v = v0 − gt = 15 − 10t this property, we could have directly found the answer by
y = y0 + v0 t − 21 gt2 = 15t − 5t2 using the y equation above. The equation y = 15t − 5t2 is
2.3. FREE FALL 33
valid regardless of whether the ball is moving upwards or Both solutions are valid. The t = 0 solution is the starting
downwards. In particular, y = 0 should be true when the time and the t = 3 s solution is the falling time. In other
ball is at the ground level. Using this condition, words, the ball is at y = 0 height at both times.
0 = 15t − 5t2 = t(15 − 5t) → t = 0 and t = 3 s .
Example 2.11
This is the maximum height from the origin, in other words,
A stone is thrown upwards with a velocity of 30 m/s from the from the ground. If we wish to find the height from where it
roof of a building at a height of 35 m from the ground. was thrown, it is y − y0 = 80 − 35 = 45 m .
(a) Choose a coordinate system to write the velocity and posi- (c) The stone falling to the ground means it reaches the posi-
tion equations of the stone. tion y = 0 :
(b) Calculate the maximum height and the time to reach it. 0 = 35 + 30t − 5t2 = 0 → t = −1 and 7 s
(c) How many seconds will it take to fall to the ground? The negative solution is not considered and the answer is
(d) With what velocity will it hit the ground? t = 7 s . Notice here that we used the same equation through-
Answer (a) Taking the ground level as the origin and up- out. Since we remain in the same coordinate system, we
wards as the positive direction, we get y0 = 35 m and a = −g . are able to solve all of the problems with a single equation
Accordingly, the equations of motion are as follows: without separating the motion into two parts of upwards and
v = v0 + at = 30 − 10t downwards.
y = y0 + v0 t + 12 at2 = 35 + 30t − 5t2 (d) The velocity of the stone when it falls to the ground is
We can answer all of the questions using these two equations. found by taking t = 7 in the v equation:
(b) At maximum height v = 0 : v = 30 − 10t = 30 − 10 × 7 = −40 m/s .
0 = 30 − 10t → t = 3 s The negative sign here indicates that the velocity is in the
Plugging this value of t into the y equation, opposite direction to the direction that we chose as positive,
y = 35 + 30 × 3 − 5 × 32 = 80 m in other words, it is downwards.
Example 2.12
chronometer will show 1 second less than the first one during
The different times technique. Water is dripping in each stage. In other words, the relation between the two
1 second intervals from the roof of a building at a height of times will be as follows:
45 m from the ground. What is the height of the second drop t2 = t1 − 1
when the first drop hits the ground? Therefore, the expression of the y2 coordinate will be
y2 = 12 g(t1 − 1)2 .
Answer It would be too complicated to try to solve this prob-
Now, when the first drop reaches the ground, it will have
lem using our familiar methods. However, the solution will
traveled a distance of y1 = 45 m . We use this value to find
be simpler if we write each drop in terms of a separate time t
the time t1 :
with its own chronometer.
45 = 5t12 → t1 = 3 s
Let us take the origin as being at the roof and the y direction
downwards. Let us start the chronometer when the first drop The distance traveled by the second drop at the end of this
sets off and show the measured time with t1 . Accordingly, time t1 :
the equation of motion will be: y2 = 12 g(t1 − 1)2 = 5 × 22 = 20 m
y1 = 12 gt12 The difference between the two distances shows how far
When the second drop sets off 1 second later, we start time behind the second drop is:
t2 in a separate chronometer. Its motion will be y2 = 12 gt22 . y1 − y2 = 45 − 20 = 25 m .
As the drops set off in one-second intervals, the second The second drop is 25 m above the ground.
Example 2.13
the ground, a second ball is thrown downwards with a velocity
of 4 m/s from the roof of a building at a height of 32 m .
(a) Using the same coordinate system, write the equations of
motion for both balls.
(b) When will the two balls meet?
(c) At what height will they meet?
(d) What are their velocities when they meet?
Answer
When a ball is thrown upwards with a velocity of 12 m/s from In this problem, we will again examine the motion of two
objects in the same coordinate system. y1 = y2

(a) If we choose the origin to be at the ground and the pos- 12t − 5t2 = 32 − 4t − 5t2 → 16t = 32
itive direction upwards, the initial position of the first ball The solution to this equation tells us that the balls meet at
thrown from the ground is y10 = 0 and its initial velocity time t = 2 s .
is v10 = 12 m/s . For the second ball thrown downwards (c) Using this value of t in either y1 or y2 , we find the height:
from the top of the building, these values are y20 = 32 m y1 = 12 × 2 − 5 × 22 = 4 m = y2
and v20 = −4 m/s . Accordingly, let us write the equations of (d) We find the velocities by again using this time t in the
motion for both balls side by side: velocity equations:
v1 = 12 − 10 × 2 = −8 m/s
1st ball 2nd ball v2 = −4 − 10 × 2 = −24 m/s
v1 = 12 − 10t v2 = −4 − 10t Notice that the velocity v1 is negative at the time of the
y1 = 0 + 12t − 5t2 y2 = 32 − 4t − 5t2 collision. In other words, the 1st ball went up to its maximum
height and returned, and the meeting took place on its way
(b) Their y positions will be the same when the balls meet: down.
1. In a coordinate system, which of the following is incor-
rect for an object with an acceleration of a = +2 m/s2 : 7. A stone is thrown upwards from the ground. Which of
the following is correct if the +y axis is chosen upwards?
(a) It is getting faster if it is traveling in the positive
direction. (a) Its acceleration is +g when traveling upwards.
(b) It is getting slower if it is traveling in the negative (b) Its acceleration is zero at maximum height.
direction. (c) Its acceleration is +g when moving downwards.
(c) Its velocity is always different from zero. (d) Its acceleration is −g throughout the motion.
(d) Its velocity changes by equal amounts in equal time
intervals. 8. A stone dropped from a height falls to the ground in 5
seconds. What is its average acceleration in m/s2 ?
2. Which of the following is correct if an object traveling
(a) 0 (b) 2 (c) 5 (d) 10
in the positive direction has negative acceleration?
(a) The velocity of the object increases.
9. A stone is thrown upwards. Which of the following is
(b) The object stops at some point and turns back.
correct at maximum height?
(c) The velocity of the object remains constant.
(d) None of the above. (a) Velocity and acceleration are zero.
(b) Velocity is zero, acceleration is maximum.
3. Which is true for accelerated motion? (c) Velocity is maximum, acceleration is zero.
(a) Acceleration is zero if the velocity is zero. (d) Velocity is zero, acceleration is the gravitational
(b) Acceleration is always positive if the velocity is acceleration.
always positive.
(c) An object with constant acceleration may stop. 10. A car traveling at speed v can stop at a distance d when
(d) An object with constant acceleration cannot stop. the brakes are applied. At what distance will it stop
when it travels at a velocity of 2v ?
4. How many seconds does it take for an object dropped
from a height of 20 m to fall to the ground? (a) d/2 (b) d (c) 2d (d) 4d
(a) 1 (b) 2 (c) 3 (d) 4
11. The velocity of an object traveling in the positive direc-
5. What distance does an object dropped from a height tion increases by 8 m/s in 5 s . What is its acceleration
travels in 5 s ? in m/s2 ?
(a) 50 (b) 75 (c) 100 (d) 125 (a) 1.0 (b) 1.2 (c) 1.4 (d) 1.6
6. The velocity of an object dropped from a height is 40 m/s

12. The position of an object varies as x = t3 (meters). What
after 4 s . What is its average velocity in this time inter-
is its acceleration at time t = 1 s in m/s2 ?
val?
(a) 0 (b) 10 m/s (c) 20 m/s (d) 40 m/s (a) 2 (b) 4 (c) 6 (d) 8
PROBLEMS 35
13. The gravitational acceleration of the Moon is less than 17. What are the ratios of the distances traveled by an ob-
that of the Earth. Which of the following is true for a ject getting faster with constant acceleration in 1, 2, 3
stone dropped from a height on the Moon? seconds?
(a) It falls to the ground in less time than on the Earth. (a) 1:2:3 (b) 1:3:5 (c) 1:4:8 (d) 1:4:9
(b) It falls to the ground in more time than on the Earth.
(c) It falls to the ground in the same time.
(d) None of the above. 18. A stationary object speeds up with constant acceleration.
What are the ratios of the distances it travels at the 3rd
14. A bicycle rider traveling at 10 m/s presses the brakes and 4th seconds?
and stops in 5 s . How many meters does the bike travel (a) 3/4 (b) 5/7 (c) 7/9 (d) 9/16
before stopping?
(a) 10 (b) 15 (c) 20 (d) 25
19. Two stones are thrown upwards from the ground with
15. The positions of an object at times t = 1, 2, 3, 4 s are velocities v0 and 2v0 . What is the ratio of the maximum
x = 2, 7, 9, 15 m , respectively. What is the average ve- height of the faster one with respect to the other?
locity of this object in the [2, 4] s interval in m/s ? √ √
(a) 2/1 (b) 4/1 (c) 2/1 (d) 1/ 3
(a) 1 (b) 2 (c) 3 (d) 4
16. Which of the following propositions is true? 20. Two objects start to move from the same position at
(a) The acceleration of an object traveling at constant the same time and in the same direction. The object
velocity is zero. A travels at constant velocity of 10 m/s . The object B
(b) The acceleration of an object slowing in the negative accelerates at a rate of 4 m/s2 . After how many seconds
direction is positive. will the two objects be in the same position again?
(c) The acceleration of an object getting faster in the (a) 2 (b) 3 (c) 4 (d) 5
negative direction is negative.
(d) All of the above.
Problems
2.1 Position, Displacement, Velocity and Ac-

celeration
Problem 2.3
2.3 The velocities of an object at various times are shown
in the figure above. (a) Find the average accelerations in the
[t1 , t2 ] and [t2 , t3 ] time intervals.
Problem 2.1 [A: aav = 2 m/s2 and aav = −3.3 m/s2 .]
2.1 The positions of an object at various times are shown in 2.4 The velocity of an object varies as v(t) = 20 − 3t2 .
the figure above. (a) Find the displacements in the [t1 , t2 ] and (a) Calculate its average acceleration in the t = [1, 2] s time
[t2 , t3 ] time intervals. (b) Calculate the average velocities in interval. (b) Find its instantaneous accelerations at times
the same intervals. t = 1 and t = 2 s .
[A: (a) ∆x = 4 m ve ∆x = −2 m , (b) vav = 4 m/s [A: (a) aav = −9 m/s2 , (b) a = −6, −12 m/s2 .]
and vav = −0.67 m/s ]
2.5 A ball hits a wall perpendicularly with a speed of 24 m/s
2.2 The position of a moving object varies in time as x = 5t2 and is rebounded back at the same speed. The ball has con-
(meters). (a) What are its positions at times t = 1 s and tacted the wall for 0.03 s . (a) What is the change in velocity?
t = 3 s ? (b) What is its average velocity in the t : [1, 3] time (b) What is the average acceleration during the collision?
interval? (c) What are its instantaneous velocities at times [A: (a) 48 m/s , (b) 1600 m/s2 .]
t = 1 s and t = 3 s ?
[A: (a) 5 and 45m , (b) 20 m/s , (c) 10, 30 m/s .]
2.2 Motion With Constant Acceleration 2.13 An automobile is trying to pass a trailer truck that has a
length of 20 m . At the start, the automobile and the truck are
2.6 In order to make a safe landing, an airplane has to touch
in the same lane and traveling at the same constant velocity
the ground at a maximum speed of 120 m/s and slow down
of 15 m/s . Once the opposite lane is empty, the automobile
with a maximum acceleration of 5 m/s2 . (a) In how many
passes to the left lane and starts to accelerate at 5 m/s2 , man-
seconds can the plane stop? (b) What should the minimum
aging to pass the truck once it reaches its front end. (a) How
length of the runway be? [A: (a) 24 s , (b) 1440 m .]
long does it take for the automobile to pass the truck? (b)
2.7 An object moving with constant acceleration passes What length would the empty lane need to be for this pass-
through two points 25 m apart at a velocity of 10 m/s from ing? (Note: The length of the automobile is added to that of
the first and 15 m/s from the second. (a) What is the accel- the truck and can be neglected.) [A: (a) 2.8 s , (b) 62 m .]
eration of the object? (b) In how many seconds will it travel
between the two points? [A: (a) a = 2.5 m/s2 , (b) t = 2 s .]
2.8 An object moving with a constant acceleration passes

between two points 10 m apart in 2 s . It has a velocity of
8 m/s when passing through the second point. Find the ac- Problem 2.14
celeration of the objects and its velocity at the first point. 2.14 A motorcycle waiting at a red lights starts to acceler-
[A: a = 3 m/s2 , v0 = 2 m/s .] ate at 5 m/s2 once the light changes to green. Just at that
2.9 A train departing from a station accelerates at a rate of moment, a truck traveling at a constant velocity of 8 m/s con-
2 m/s2 for 12 s . It then travels for 5 s at constant velocity. tinues on its way through the green light without changing
Finally, it decelerates at 4 m/s2 and stops at the second sta- its velocity. (a) When will the motorcycle catch up with the
tion. What is the distance between the two station in meters? truck? (b) How much distance has the motorcycle traveled
[A: 336 m .] once it catches up? [A: (a) t = 3.2 s , (b) x = 25.6 m .]
2.10 A jet airplane can accelerate at a maximum rate of 2.3 Free Fall
4 m/s2 on the runway and can takeoff once it reaches a ve-
locity of 80 m/s . However, if the pilot changes his/her mind 2.15 A stone is thrown downwards with a velocity of 20 m/s
about the takeoff, the plane can be slowed down at a max- from the roof of a building at a height of 105 m . (a) How long
imum rate of 5 m/s2 . In an emergency, the pilot changes will it take to reach the ground? (b) With what velocity will
his/her mind about takeoff when the plane is at the takeoff it hit the ground? [A: (a) 3 s , (b) 50 m/s .]
velocity. What should the minimum total length of the run-
way be so that the airplane can stop safely? [A: 1440 m .] 2.16 A stone is dropped into a water well and the sound of
the stone hitting the water’s surface is heard 4.5 s after it
2.11 The maximum acceleration of a train is 8 m/s2 . This was dropped. Since the propagation speed of sound in air is
train should travel between two stations 1800 m apart in 340 m/s , how deep down is the surface of the water at the
the shortest time possible. For this purpose, it accelerates in bottom of the well? [A: 90 m .]
the first half of the path and decelerates in the second half,
stopping at the station. Calculate the maximum velocity that 2.17 After throwing a ball upwards, a juggler runs to and
the train can reach and how many seconds it takes to travel then back from a door at a distance of 9 m with a velocity of
the path. [A: v = 120 m/s and t = 30 s .] 5 m/s , catching the ball before it falls to the ground. At what
minimum velocity should he/she throw the ball upwards to
be able to do this? [A: 18 m/s .]
Problem 2.12
2.12 The driver of a sports car traveling at velocity 34 m/s
suddenly sees a truck 30 m ahead, traveling in the same direc- Problem 2.18
tion with a constant velocity of 4 m/s , and applies the brakes. 2.18 A ball is dropped from the roof of a building. A man
If the car can decelerate at a maximum rate of 6 m/s2 , will it standing behind a window on one of the lower floors sees the
crash into the truck from behind? (Hint: Solve the equivalent ball pass his window in 0.2 seconds. The height of the win-
problem of a stationary truck and a car with different veloc- dow is 1.80 m . How high is the roof from the upper edge of
ity.) the window? (Hint: The given information is sufficient to find
[A: It crashes. The car needs at least 95 m to slow down the velocity of the stone at the upper edge of the window.)
before crashing.] [A: 3.2 m .]
PROBLEMS 37
a velocity of 20 m/s . At what distance d between the truck

and the tree should the monkey start to drop in order to land
on the truck? [A: 22 m .]
2.20 A juggler throws a ball such that it barely touches the

ceiling of the hall, which is 5 m high. (a) What is the initial
velocity of the ball? (b) If he throws a second ball upwards
with the same velocity when the first ball is at the level of
Problem 2.19 the ceiling, when and at what height will the two balls meet?
2.19 A monkey sitting in a tree by the road on a branch 6 m [A: (a) 10 m/s , (b) 0.5 s and 3.75 m .]
above the ground wants to drop on a truck approaching with
3
TWO-DIMENSIONAL
MOTION
An acrobatic stunt at a mo-

tocross race.
The racer is moving in both
the horizontal and vertical di-
rections. Can we examine both
of these movements within the
same framework?
A bird flying through the air, a fish swimming in water, a free throw in
basketball, a spacecraft flying past a planet, etc. How can the motion of all these
objects be examined? We cannot apply the method that we discussed in the
previous chapter directly because these objects do not move in a straight line.
In a three-dimensional Cartesian coordinate system, the coordinates (x, y, z) of
these objects simultaneously vary depending on time t :
x = x(t) , y = y(t) , z = z(t)
Notice that, when an object is moving in space, each of the coordinates travels
along its own axis, in other words, it performs linear motion! Therefore, we can
apply the concepts that we developed for linear motion to each of the (x, y, z)
coordinates. Then, we can unify them using vectors.
In this course we will only examine two-dimensional motion. The methods
that we will develop can easily be extended to three-dimensional motion.

40 3. TWO-DIMENSIONAL MOTION
3.1 POSITION AND DISPLACEMENT VECTORS

Let (x, y) be the coordinates of a point P in the xy coordinate system. The
vector drawn from the origin to this point is called the position vector and is
indicated by ~r . As seen in the figure, the components of this vector ~r are the
coordinates of the point P :
~r = x ı̂ + y ̂ (3.1)
Figure 3.1: Position vector. Displacement Vector

If an object located at point P1 with position ~r1 at time t1 , is located at point
P2 with position ~r2 at a later time t2 (Figure 3.2), the vector
∆~r = ~r2 − ~r1 = (x2 − x1 ) ı̂ + (y2 − y1 ) ̂ (3.2)
−−−−→
is called the displacement vector. As seen in the figure, the vector ∆~r = P1 P2
is the secant line that intersects the trajectory at P1 and P2 .
The average velocity vector ~vav of the object during this time interval ∆t
is the ratio of the displacement vector to the time difference:
Figure 3.2: Displacement vec- ∆~r
tor. ~vav = (3.3)
∆t
This is the application of the definition in one-dimensional motion to both com-
ponents. If we write more explicitly,
~r2 − ~r1 y2 − y1 ∆x ∆y
! !
x2 − x1
~vav = = ı̂ + ̂ = ı̂ + ̂
t2 − t1 t2 − t1 t2 − t1 ∆t ∆t
∆x ∆y
vav,x = , vav,y =
∆t ∆t
Likewise, the instantaneous velocity vector (or, simply, the velocity vector), is
obtained by taking the derivative as in one dimension:
∆~r d~r
~v = lim = (3.4)
∆t→0 ∆t dt
We understand the following from the derivative of a vector: The derivatives of
the x and y components of that vector are taken separately, and these become
the components of the derivative vector:
d~r dx dy
~v = = ı̂ + ̂ = v x ı̂ + vy ̂
dt |{z}
dt dt
vx vy
|{z}
The magnitude (absolute value, norm) of the velocity vector and its angle with
the +x axis are found using familiar formulas:
q vy
v = | ~v | = v2x + v2y , tan θ = (3.5)
vx
What is the direction of the velocity vector? In order to determine this, let us
−−−−→ −−−−→
consider the secant ∆~r = P1 P2 . The P1 P2 vector is in the direction of motion
(Figure 3.3). Also, as ∆t → 0 , P2 will gradually approach the point P1 , and in the
end, the secant P1 P2 will come to the tangent line. Therefore, in two-dimensional
motion, the velocity vector is always tangent to the trajectory.
3.1. POSITION AND DISPLACEMENT VECTORS 41
Figure 3.3: (a) Limit direction of

the displacement vector, (b) di-
rection of the ~v velocity vector.
Acceleration Vector
Acceleration is the rate of change of velocity with respect to time. In order
to extend this to two-dimensional motion, we use the same method for the
components.
Definition: If the velocity vector of an object at time t1 is ~v1 and its velocity
vector at a later time t2 is ~v2 (Figure 3.4), then the average acceleration vector
is
∆~v ~v2 − ~v1
~aav = = (3.6)
∆t t2 − t1
Figure 3.4: ∆~v = ~v2 − ~v1 vector.
or, in terms of the components,

∆v x v2x − v1x ∆vy v2y − v1y
aav,x = = , aav,y = =
∆t t2 − t1 ∆t t2 − t1
Accordingly, the instantaneous acceleration vector (or, simply, the accelera-
tion vector) is found by taking the limit of average acceleration:
∆~v d~v
~a = lim = (3.7)
∆t→0 ∆t dt
or, more explicitly in terms of the components,
d~v dv x dvy
~a = = ı̂ + ̂ = a x ı̂ + ay ̂
dt |{z}
dt dt
|{z}
ax ay
As the velocity vector is the derivative of position, we can also write the accelera-
tion as the second derivative of position:
d~v d2~r
~a == 2 (3.8)
dt dt
The magnitude and the angle of the acceleration vector are,
q ay
a = | ~a | = a2x + a2y , tan θ = (3.9)
ax
In two-dimensional motion, the direction of the acceleration vector does not
necessarily have to be along the path and may be in any direction depending on
the type of motion.
Example 3.1 dv x
ax = = 6t − 4 (m/s2 )
dt
The coordinates of an object moving on a plane are given as, dvy
ay = =6 (m/s2 )
x = t3 − 2t2 + 5 (m) dt
y = 3t2 − 4t + 4 (m) (b) We find the components by taking t = 1 in these expres-
sions for x, v, a :
(a) Express velocity and acceleration as functions of time.
(b) Calculate the components of position, velocity and accel- x=4 y=3
eration at time t = 1 s . v x = −1 vy = 2
(c) Find the magnitudes of velocity and acceleration at time ax = 2 ay = 6
t = 1 s. (c) We calculate the magnitudes of the vectors with the given
(d) Find the angle between velocity and acceleration at time components:
t = 1 s. q p √
v = v2x + v2y = (−1)2 + 22 = 5 = 2.2 m/s
Answer q √ √
(a) Derivatives of the position components with respect to a = a2x + a2y = 22 + 62 = 40 = 6.3 m/s2
time give the velocity components:
vx =
dx
= 3t2 − 4t (m/s) (d) The angle between the two vectors is found using the
dt scalar product method (Eq. 1.10):
dy √
vy = = 6t − 4 (m/s) ~v · ~a (−1) × 2 + 2 × 6 2
dt cos θ = = √ √ =
Derivatives of the velocity components give the acceleration v a 5 40 2
components: From here, the angle is found to be θ = 45◦ .
3.2 PROJECTILE MOTION

The first example of two-dimensional motion that we examine will be the
motion of objects thrown into the air near the Earth’s surface under gravitational
acceleration. Neglecting air resistance and other small effects, the acceleration is
always downwards and towards the center of the Earth. Choosing the coordinate
system shown in Figure 3.5, its components can be written as,
ax = 0
)
−→ ~a = −g ̂ (3.10)
ay = −g
The components of motion can easily be deduced from this: We observe uniform
linear motion in the horizontal x direction and motion with constant acceleration
in the vertical y direction We examined both types of motions in Chapter 2. Now,
we can write those formulas for each component.
Figure 3.5: Coordinates in pro-

jectile motion. As the horizontal
x component does uniform lin-
ear motion, the vertical y com-
ponent does free fall motion.
Let us consider an object thrown from the surface of the earth with an initial
velocity of v0 with an angle of θ to the horizontal (Figure 3.5). The components
of the initial velocity can be written as follows:
v0x = v0 cos θ v0y = v0 sin θ (3.11)

3.2. PROJECTILE MOTION 43
Choosing the place where the object was thrown as the origin and the y direction
as upwards, we can write the projectile motion formulas:
v x = v0 cos θ x = (v0 cos θ) t

(projectile motion) (3.12)
vy = v0 sin θ − gt y = (v0 sin θ) t − 21 g t2
Trajectory equation: If we wish to find the trajectory, or the curve drawn

by the object in the air regardless of its position at any given time, we eliminate t
from these equations. Taking t = x/v0 cos θ from the x position formula and
substituting it into the y equation and simplifying, we get the trajectory equation:
g
y = (tan θ) x − x2 (Trajectory equation) (3.13)
2v20 cos2 θ
This is the equation of a parabola. (Figure 3.5).

If the object is thrown not from the origin, but from any initial position
(x0 , y0 ) , it is sufficient to add x0 and y0 to the equations above:
x = x0 + (v0 cos θ) t
(3.14)
y = y0 + (v0 sin θ) t − 12 g t2
These formulas are valid for all types of projectile motion in which the air
resistance can be neglected. It is not necessary to develop separate formulas
for the maximum height or the horizontal range. For example, since θ = 0 for
a horizontal projectile, it is sufficient to have sin θ = 0 and cos θ = 1 in the
formulas:
v x = v0 vy = −gt
(horizontal projectile) (3.15)
x = x0 + v 0 t y = y0 − 12 g t2
Likewise, it is sufficient to know that the y component of velocity is zero (vy = 0)

at the maximum height.
Example 3.2
of motion (3.12) follow:
A ball kicked by a football player rises with an initial velocity v x = v0 cos 37 = 24 vy = v0 sin 37◦ − gt = 18 − 10t
◦
of 30 m/s at an angle of 37 to the horizontal. x = v0 cos 37 t = 24 t y = v0 sin 37◦ t − 12 g t2 = 18t − 5t2
(a) Write the equations of motion of the ball.
(b) Plugging t = 1 into these expressions,
(b) What are the position and velocity components of the ball
v x = 24 m/s vy = 18 − 10 × 1 = 8 m/s
at time t = 1 s ?
x = 24 m y = 18 − 5 = 13 m
(c) In how many seconds does it reach maximum height?
(c) At maximum height the motion becomes horizontal, in
(d) What is the maximum height of the ball?
other words, y component of velocity vanishes:
(e) How far away will the ball land on the ground?
vy = 18 − 10t = 0 → t = 1.8 s
Answer (d) The y value at this t = 1.8 s gives the maximum height:
(a) First, calculate the components of the initial velocity: y max = 18 × 1.8 − 5 × 1.8 = 16.2 m
2
(e) The range of the ball, in other words, the horizontal dis-
v0x = v0 cos 37◦ = 30 × 0.8 = 24 m/s tance of the falling point is the value of the x coordinate at
v0y = v0 sin 37◦ = 30 × 0.6 = 18 m/s t = 2 × 1.8 = 3.6 s . The range is shown as R :
We choose the origin at the kicking point, and the equations R = xmax = 24 × 3.6 = 86 m
Example 3.3
A ball is thrown at an angle of 53◦ towards a wall 9 m away. Answer

The ball collides with the wall at height of 7 m . What is the Since no time is involved, the trajectory equation will give
initial velocity of the ball? the quickest solution. Using the values x = 9 m , y = 7 m ,
cos 53◦ = 0.6 and tan 53◦ = 4/3 in the trajectory formula
(3.13), we get
g
y = (tan θ) x − 2 x2
2v0 cos2 θ
4 10
7= ×9− 2 × 92
3 2v0 × 0.62
v20 = 225 → v0 = 15 m/s .
Example 3.4
jumping point and the y axis upwards. Since the motorcycle
jumps horizontally, we take θ = 0◦ in the projectile motion
formulas (3.12):
v x = v0 x = v0 t
vy = −gt = −10t y = y0 − 21 gt2 = 20 − 5t2
(a) The y coordinate should be zero when the motorcycle
falls to the ground:
0 = 20 − 5t2 → t = 2 s
(b) In order to pass the river, its x coordinate should be greater
than or equal to 9 m at the time t = 2 s when it touches the
A motorcycle rider wants to jump off of a cliff at a height of ground:
20 m with a horizontal velocity of v0 and get to the other side x = v0 t → 9 = v0 × 2 → v0 = 4.5 m/s
of a river that is 9 m wide. (c) We calculate the velocity components at time t = 2 s :
(a) What is the time of flight of the motorcycle?
v x = v0 = 4.5 m/s vy = −10t = −20 m/s
(b) What should the minimum initial velocity of the motorcy-
The negative sign of the vy component shows that it is down-
cle be in order to cross the river?
wards. We calculate the magnitude of the velocity using these
(c) With what velocity will the motorcycle touch the ground?
components:
Answer
q p
v = v2x + v2y = (4.5)2 + (−20)2 = 21 m/s
Let us chose the origin at the ground level directly below the
Example 3.5
(b) What will the time of flight of the ball be until it falls to
the ground?
(c) How far away from the building will the ball land?
(d) What should the acceleration of the kid be so that he/she
can catch the ball?
Answer
We first calculate the initial velocity components of the ball:
v0x = v0 cos 53◦ = 5 × 0.6 = 3 m/s
v0y = v0 sin 53◦ = 5 × 0.8 = 4 m/s
(a) Taking the origin at the place where the kid starts, the x
A ball is thrown with at an angle of 53◦ and initial velocity and y coordinates of the ball are as follows:
5 m/s from the roof of a building that is 33 m high. Waiting x1 = v0x t = 3 t
directly below on the ground, a kid starts to run with constant y1 = y10 + v0y t − 21 g t2 = 33 + 4t − 5t2
acceleration a at the same time and in the same direction as The kid accelerates in the x direction and does not move in
the ball. the y direction:
(a) Choose a coordinate system and write the equations of x2 = 12 at2 y2 = 0
motion for the ball and for the kid. (b) When the ball reaches the ground level, its y coordinate
3.3. UNIFORM CIRCULAR MOTION 45
should be zero: x1 = 3 t = 9 m
y1 = 33 + 4t − 5t2 = 0 → t = −11/5 and + 3 s (d) The kid should arrive to the same position x = 9 m at the
The solution is the positive value t = 3 s . same time t = 3 s as the ball:
(c) We find the x position of the ball at this time t : x2 = 21 at2 → 9 = 12 a × 32 → a = 2 m/s2
Example 3.6
ball will collide with the elevator?
The outdoor elevator of a building starts to descend with a con- Answer
stant speed u from a height of 8 m from the ground. At the We first calculate the initial velocity components of the ball:
same time, a kid waiting on the ground at a horizontal distance v0x = v0 cos 53◦ = 20 × 0.6 = 12 m/s
of 6 m from the building throws a ball towards the elevator v0y = v0 sin 53◦ = 20 × 0.8 = 16 m/s
with a velocity of 20 m/s and at an angle of 53 . ◦
(a) Taking the origin as the place where the ball is thrown,
the equations of the ball are as follows:
x1 = v0x t = 12 t y1 = v0y t − 12 g t2 = 16t − 5t2
The horizontal distance of the elevator is constant and it is
moving in the y direction with a constant downward speed:
x2 = 6 m y2 = y20 − ut = 8 − ut
(b) The ball will have traveled 6 m in the x direction when
it hits the wall:
6 = 12 t → t = 0.5 s
(a) Choose a coordinate system and write the equations of
(c) We calculate the y position of the ball at this time t :
motion of the elevator and of the ball.
y1 = 16 × 0.5 − 5 × (0.5)2 = 6.75 m
(b) After how many seconds will the ball hit the wall?
(c) At what height will the ball hit the wall? (d) The elevator should be at the same y position at that time:
(d) What should the velocity u of the elevator be so that the 6.75 = 8 − u × 0.5 → u = 2.5 m/s
Example 3.7
The solution x = 0 is where the object was thrown. The
Maximum range. What is the optimal angle for an object that other solution is the range and is indicated by R . Us-
is always thrown at the same speed v0 so that its horizontal ing the trigonometric identities tan θ = sin θ/ cos θ and
range is maximum? 2 sin θ cos θ = sin 2θ , we find the range formula as follows:
v2 sin 2θ
Answer R= 0
Let us use the trajectory equation to find the range formula. g
For the same v0 initial velocities, the sine function should be
y = 0 should be true when the ball is at the ground level.
maximum for R to be maximum. The sine function takes its
This condition is used in the (3.13) trajectory formula:
g maximum value 1 at 90◦ . Therefore,
0 = (tan θ) x − 2 x2 2θ = 90◦ → θ = 45◦
2v0 cos2 θ
2v2 cos2 θ tan θ For maximum range, objects should be thrown at an angle of
Solution: x = 0 and x = 0 45◦ .
g
3.3 UNIFORM CIRCULAR MOTION

Centripetal Acceleration
The second important example of two-dimensional motion is the motion of an
object rotating at constant speed around a circle. Although the speed is constant,
the object has an acceleration, because the velocity changes direction. Let us try
to determine how this happens.
Let an object rotating with a constant speed v on a circle with radius r have
a velocity vector ~v1 at position P1 at time t1 and then a velocity vector ~v2 at
position P2 at a later time t2 (Figure 3.6a).
The magnitudes of these two velocities are the same: |~v1 | = |~v2 | = v . However,
they should be considered as two different vectors, as their directions are different.
Figure 3.6: Velocity vectors and

∆~v difference vector in circular
motion.
Now let us remember the definitions of the average acceleration,
~v2 − ~v1 ∆~v

~aav = =
t2 − t1 ∆t
and the acceleration vector ~a as the limit of this expression when ∆t → 0 :
∆~v
~a = lim (3.16)
∆t→0 ∆t
Since this acceleration is a vector, we must determine both its magnitude and its
direction. We start with the magnitude.
Now, what is this vector ∆~v ? To get an idea about it, let us shift both vectors
to the midpoint of the arc on the circle, keeping their directions unchanged
(Figure 3.6b). The equal sides of the isosceles triangle formed are ~v1 and ~v2 while
the opposite side is ∆~v .
Let us compare this triangle with the triangle OP1 P2 formed in the original
circle. These are similar triangles, because both are isosceles and both of their
apex angles are equal to ∆θ . (When the object turns by ∆θ , its velocity vector
also turns by the same amount.) Therefore, the ratios of similar sides are equal:
| ∆~r | | ∆~v |
=
r v
Here, we shall introduce a small error by using the length of the arc ∆s instead
of the length of the secant | ∆~r | , but the difference will approach zero later when
we take the limit ∆t → 0 . Therefore, substituting | ∆~r | ≈ ∆s ,
v ∆s
| ∆~v | =
r
and, taking the limit ∆t → 0 we get,
| ∆~v | v ∆s v ∆s
a = lim = lim = lim
∆t→0 ∆t ∆t→0 r ∆t r ∆t→0 ∆t
where we took the constants outside of the limit. The limit in this expression is
just the definition of speed v with which we are familiar, because it is the ratio
of the distance taken ∆s to the elapsed time: lim∆t→0 ∆s/∆t = v . Therefore, the
magnitude of the acceleration is as follows:
v2
a=
r
This magnitude is constant, because the radius r and the speed v are constant.
Next, we find the direction of the acceleration ~a . In Eq. (3.16), since ∆t is just
a scalar, the direction of ~a will be the same as the limit direction of ∆~v . Look
at Figure 3.6b again. As ∆θ → 0 , or velocity ~v2 gets closer to ~v1 , they align
3.3. UNIFORM CIRCULAR MOTION 47
as almost parallel, with ∆~v becoming almost perpendicular to both. In terms

of angles, as ∆θ → 0 (Figure 3.7), the base angles of the isosceles triangle will
approach 90◦ . The vector ∆~v , hence ~a , being perpendicular to the velocity ~v ,
will point towards the center of the circle.
Therefore, in uniform circular motion, a constant acceleration will be formed
that is always directed towards the center (Figure 3.8). This is called centripetal
Figure 3.7: As ∆θ → 0 , the ∆~v
acceleration and is denoted by ar :
vector is directed towards the
center of the circle.
v2
ar = (centripetal acceleration) (3.17)
r
This formula can also be written in terms of period T , the time for one complete
cycle. As the distance traveled in one complete cycle is the perimeter length 2πr ,
the speed and the acceleration can also be expressed in terms of the period:
2πr
v =
T
4π2 r
ar = (3.18)
T2
Figure 3.8: In circular motion,
the centripetal acceleration ~ar
Tangential Acceleration is always towards the center at
If, along with the direction, the magnitude of the velocity also changes in every point.
circular motion, a tangential acceleration (Figure 3.9) will arise in addition
to the centripetal acceleration. The expression of tangential acceleration is the
change in the magnitude of velocity:
dv
at = (3.19)
dt
We will address tangential acceleration later in the discussion of rotational motion
in Chapter 7.
The acceleration ~a is the resultant of these centripetal and tangential acceler-
ations:
Figure 3.9: If the magnitude
~a = ~ar + ~at (3.20) of velocity also changes, a tan-
Therefore, in the most general case, the acceleration ~a does not necessarily point gential acceleration ~at is also
to the center, but may be in any direction. formed.
Example 3.8
circle is 2πr , the distance traveled in 150 cycles is divided by
An object revolving in a circular orbit with a radius 80 cm is 1 minute:
observed to make 150 cycles per minute. distance 150 × 2πr
v= = = 5πr = 12.5 m/s
(a) What is the velocity of the object? time 60 s
(b) What is the centripetal acceleration of the object? (b) The velocity found above is used in the centripetal accel-
eration formula (π2 ≈ 10) :
Answer
(a) The constant velocity v is directly calculated as the ratio v2 (5πr)2
ar = = = 25π2 r = 200 m/s2
of the traveled distance to time. As the circumference of the r r
Example 3.9
(b) Calculate the centripetal acceleration of a GPS satellite.
The Global Positioning System (GPS) consists of around 20 satel- Answer
lites revolving at an altitude of 20 000 km from the surface of
the Earth. Each satellite circles the Earth twice a day. The (a) The period of a satellite, in other words, the time it takes
Earth’s radius is approximately 6 400 km . to complete one full cycle is half a day. We calculate this in
terms of seconds:
T = 12 × day = 21 × 24 × 60 × 60 = 43 200 s
(b) We use the formula that expresses centripetal accelera-
tion in terms of the period T . Here, r is the distance mea-
sured from the center of the circle. If the Earth’s radius is
R and the height of the satellite from the surface is h , then
r = R + h = 6400 + 20000 = 26400 km = 26.4 × 106 m . We
substitute these values ( π2 ≈ 10 ):
4π2 r 40 × 26.4 × 106
(a) Calculate the period of a GPS satellite. ar = 2 = = 0.57 m/s2 .
T (43.2 × 103 )2
3.4 RELATIVE MOTION

Consider a rower pulling oars to move a boat along a river. From the shore, we
observe that the same rower moves faster when traveling in the same direction
as the current and slower when rowing against the current. In fact, the boat
may even be observed to go backwards when trying to go against the current.
Likewise, on an escalator, even if we stay still, people on the ground see us as
moving forward.
These observations remind us that concepts such as velocity and position are
relative, in other words, “they depend on the observer making the measurements.”
If the observer is moving, the velocity measured by him/her will be different from
the velocity measured by an observer at rest. Therefore, it is important to know
the relation between such various measurements.
In one dimension, it is easy to see the relation between velocities. When a
swimmer with speed v in still water starts to swim in a river with a current speed
u , he/she will have a speed of v + u when swimming in the same direction as the
current, and v − u when swimming against the current.
To calculate relative velocities in two dimensions, let us consider two objects
positioned at points A, B at time t , and an observer standing at rest at the origin
Figure 3.10: The positions of O. From Figure 3.10, we can see the following relations between these vectors:
objects A and B with respect to
−−→ −−→ −−→
the observer O . OB = OA + AB
But these are just position vectors, namely:

−−→
OB = ~rBO : The position of object B with respect to origin O;
−−→
OA = ~rAO : The position of object A with respect to origin O;
−−→
AB = ~rBA : The position of object B with respect to object A.
Hence, while these positions change with time, we will always have:
~rBO = ~rAO + ~rBA = ~rBA + ~rAO (3.21)

3.4. RELATIVE MOTION 49
where we switched the order of the last two terms. By taking the derivative of
this expression with respect to time, we get velocities:
d~rBO d~rBA d~rAO
= +
dt dt dt
~vBO = ~vBA + ~vAO
where each term can easily be identified:
~vBO = The velocity of B with respect to origin O;

~vAO = The velocity of A with respect to origin O;
~vBA = The velocity of B with respect to A.
Accordingly, the relation between the velocities is as follows:
~vBO = ~vBA + ~vAO (relative velocity addition rule ) (3.22)
Here, we stress two points that are important for calculations:

• Notice that, when we switch the subscripts, we get a negative sign. For
example,
d~rAB d(−~rBA )
~vAB = = = −~vBA
dt dt
which is true for all velocities, ~vOB = −~vBO , etc.
• It is very easy to remember the velocity addition rule, Eq. (3.22). Notice
the ordering of the subscripts. We see that the inner subscripts ( A ) on the
right-hand side are the same. And the outer subscripts ( B, O ) are just those
on the left-hand side.
Hence, any velocity that you wish to obtain can be separated into two terms
into which you insert the missing subcript. For example, if we want ~vAB (the
velocity of A with respect to B), we write:
~vAB = ~vAO + ~vOB = ~vAO − ~vBO
where, in the last term, we reversed subscripts to get the velocity of B with
respect to origin, hence we picked up a negative sign.
The relative velocity addition rule given in Eq. (3.22) is valid only in classic
physics, in which speeds are small compared to the speed of light. When the
speed of light is approached, we must use The Theory of Relativity developed by
Einstein.
Relative Acceleration
Let us reconsider Eq. (3.22) which gives us the velocity addition rule:
~vBO = ~vBA + ~vAO
The objects shall have acceleration if their velocities are changing in time. The
second derivative of this equation with respect to time will give the relation
between the accelerations:
~aBO = ~aBA + ~aAO (3.23)

We will not use this general expression in this course. However, in order to draw
an important conclusion, let us assume that only the velocity of the object B
changes and the object A moves at constant velocity with respect to the origin.
In this case, ~aAO = 0 and we find that
~aBO = ~aBA (3.24)
This is an important results: Observers moving at constant velocities with respect

to each other observe the same acceleration in other objects. This result will make it
easier to understand the laws of dynamics later.
Example 3.10
The speed of the boat with respect to the water:
A boat has a speed of 20 km/h in still water. The captain is vBW = 20 km/h
trying to cross a river that is 2 km wide. The current speed in The velocity of the boat with respect to the ground: vBO =?
the river is 5 km/h . Velocity addition formula: ~vBO = ~vBW + ~vWO
This vector equality is valid for both questions.
(a) We get the situation observed in Figure (a) when the ve-
locity of the boat with respect to the water ~vBW is kept per-
pendicular to the water. Writing the hypotenuse and angle
formulas in this right triangle,
q √
vBO = v2BW + v2WO = 202 + 52 = 21 km/h
tan θ = vWO /vBW = 5/20 = 0.25 → θ = 14◦ .
(a) What will the speed and direction of the boat be with re- (b) We get the situation observed in Figure (b) when the veloc-
spect to the ground if the captain heads straight across the ity of the boat with respect to the river ~vBW is kept at some
river? angle, such that the resultant velocity ~vBO is ensured to be
(b) The captain wants to cross the river perpendicular to the perpendicular to the shore. The triangle formulas give:
shore. In which direction should he keep the bow of the q √
boat so that its velocity observed from the ground is per- vBO = v2BW − v2WO = 202 − 52 = 19 km/h
pendicular? In such a case, what will its velocity be with sin θ = vWO /vBW = 5/20 = 0.25 → θ = 15◦ .
respect to the ground?
(c) If we denote the width of the river as L , the diagonal path
(c) Calculate the time to cross the river in both cases.
taken in part (a) will be L/ cos θ . Accordingly,
Answer L 2
ta = = = 0.10 hour = 6 minutes .
In order to apply the velocity addition formula, let us desig- vBO cos 14◦ 21 × 0.97
nate the river water with W , the boat with B , and the ground The path taken in part (b) will be L . The crossing time is
with O . Accordingly, the given data are: calculated accordingly:
The speed of the water with respect to the ground: L 2
tb = = = 0.11 hour = 6.5 minutes .
vWO = 5 km/h vBO 19
Example 3.11
ship B as observed by the captain of ship A ?
Answer
If we denote the velocity of ship B with respect to A as
~vBA , we may write the relative velocity addition formula (by
following the subscripts correctly) as follows:
~vBA = ~vBO + ~vOA
Here, ~vOA is the velocity of the ground with respect to the
The velocities of two ships in the Mediterranean Sea are mea- ship A and if we reverse it as ~vOA = −~vAO the formula we
sured from the ground. Ship A is observed to be sailing at seek is:
25 km/h in the South-North direction and ship B at 40 km/h ~vBA = ~vBO − ~vAO
in the East-West direction. The difference between these two vectors is shown in the
What are the magnitude and direction of the velocity of figure. If we denote the angle at which the ship A observes
the vector ~vBA as θ , we can calculate the velocity vBA and vBO 40
tan θ = = = 1.6 → θ = 58◦
the angle using vAO 25
q the right triangle
√
formulas:
vBA = v2BO + v2AO = 402 + 252 = 47 km/h
Example 3.12
~vAW , the velocity of the wind with respect to the ground ~vWO ,
The pilot of an airplane with a speed of 60 m/s in still air is we find the velocity of the airplane with respect to the ground
flying by keeping the nose of the airplane directed towards the in windy air ~vAO :
North in a region where the wind is 20 m/s in the South-East ~vAO = ~vAW + ~vWO
direction. What is the velocity of the airplane with respect to These velocity vectors are shown in the figure.
the ground and its angle with the North? We use the Sine and Cosine theorems (Appendix B) to
solve this triangle:
√
c = a2 + b2 − 2ab cos C (The cosine theorem)
sin A sin B sin C
= = (The sine theorem)
a b c
By comparing these two triangles, we calculate the requested
velocity vAO and its direction:
q
vAO = v2AW + v2WO − 2vAW vWO cos 45
q √
Answer = 602 + 202 − 2 × 60 × 20 × 2/2 = 48 m/s
Let the airplane be A , the wind W and the ground O . If we √
sin θ 2/2
add to the velocity of the airplane with respect to the wind = → θ = 36◦
20 48
1. Which of the following is true for an object moving in a (a) Velocity is zero.
plane with a constant magnitude of velocity? (b) The horizontal component of velocity is zero.
(a) The acceleration of the object may not be zero. (c) The vertical component of velocity is zero.
(b) The velocity of the object is parallel to the trajectory. (d) Velocity is maximum.
(c) The object may have centripetal acceleration. 6. Object A is revolving around a circle with radius r at a
(d) All of the above. constant velocity v . Object B is revolving around a circle
2. Which is true for a motion on any curved trajectory? with radius 2r at velocity 2v . Which one has higher
centripetal acceleration?
(a) Acceleration and velocity are in the same direction.
(b) The velocity vector is tangent to the trajectory. (a) A (b) B (c) Equal (d) None.
(c) The velocity is perpendicular to the trajectory. 7. Object A is revolving around a circle with radius r at
(d) Acceleration and velocity are perpendicular. a constant velocity v . Object B is revolving around a
circle with radius 2r at velocity v/2 . Which one has
3. Two balls are thrown from the same height at the same higher centripetal acceleration?
instant. Ball A is dropped in free fall, while ball B is (a) A (b) B (c) Equal (d) None.
thrown in a horizontal direction. Which is true?
8. Which of the following is incorrect for an object per-
(a) A reaches the ground first.
forming accelerated motion on a plane?
(b) B reaches the ground first.
(c) They arrive at the same time. (a) The velocity vector may be constant.
(d) It is impossible to tell. (b) The velocity magnitude may be constant.
(c) The velocity direction may be constant.
4. Which of the following remains constant in projectile (d) Velocity may change direction.
motion? 9. In how many seconds will an object dropped from a
(a) The magnitude of velocity. height of 80 meters reach the ground?
(b) The horizontal component of velocity. (a) 1 (b) 2 (c) 3 (d) 4
(c) The vertical component of velocity.
10. Two objects are traveling in the same direction along a
(d) The angle of velocity.
straight line. Object A has a speed of 10 m/s and object
5. Which is correct at the maximum height in projectile B has a speed of 4 m/s . Which of the following will be
motion? observed by object A?
√ √
(a) Object B moving away at 6 m/s. (a) 3t (b) 3t (c) t/ 3 (d) 9t
(b) Object B moving away at 14 m/s.
(c) Object B approaching at 6 m/s. 16. Which of the following is incorrect if the direction and
(d) Object B approaching at 14 m/s. magnitude of velocity are changing in circular motion?
11. Two objects are traveling towards each other in the oppo- (a) There is centripetal acceleration.
site directions along a straight line. Object A has a speed (b) There is tangential acceleration.
of 10 m/s and object B has a speed of 4 m/s . Which of (c) Acceleration is perpendicular to velocity.
the following will be observed by object A? (d) Acceleration can be in any direction.
(a) B moving away at 6 m/s. 17. You want to swim to the point right across you in a river
(b) B moving away at 14 m/s. where the current runs from the left to the right. In
(c) B approaching at 6 m/s. which direction should you swim?
(d) B approaching at 14 m/s. (a) Right ahead.
12. What is the cause of the centripetal acceleration in uni- (b) Towards the left.
form circular motion? (c) Towards the right.
(a) A change in the magnitude of velocity. (d) Halfway towards left and then towards the right.
(b) A change in the direction of velocity. 18. A boy who can swim at 4 m/s in still water tries to swim
(c) A change in position. perpendicular to the shore in a river where the current
(d) Velocity being tangent to the trajectory. is 3 m/s . What will be its velocity and direction be as
13. A stone is dropped from the window of a moving train. observed from the ground?
Which of the following is incorrect? (a) 4 m/s perpendicular to the shore.
(a) Observer on the train sees the stone in free fall. (b) 5 m/s perpendicular to the shore.
(b) Observer on the ground sees the stone in free fall. (c) 5 m/s and 37◦ wide.
(c) Observer on the ground sees a horizontal projectile (d) 5 m/s and 53◦ wide.
motion. 19. A man wants to get upstairs with an escalator. The man
(d) None of the above. walks up the stairs in 6 s when the escalator is still. The
14. Which of the following is incorrect for projectile mo- man gets upstairs in 3 s when the escalator is moving
tion? and he is at rest on the steps. In how many seconds will
(a) The time it takes to rise is equal to the time it takes the man get upstairs by walking when the escalator is
to fall. moving?
(b) Speeds are the same in two directions at the same (a) 1 (b) 1.5 (c) 2 (d) 2.5
height.
(c) The speed on the ground is maximum. 20. Two objects are thrown at the same angle from the same
(d) The speed at the maximum height is equal to the point. Object B reaches a maximum height that is 3 times
vertical component of the velocity. that of object A. What is the ratio of the initial velocity
of B with respect to A?
15. In a horizontal projectile motion, the object falls to the √ √
ground in t seconds. What will the time of flight be if (a) 3 (b) 3 (c) 1/ 3 (d) 9
the height is tripled?
Problems
3.2 Projectile Motion building will it hit the ground? (c) What are its velocity com-
ponents when the ball hits the ground?
[A: (a) 4 s , (b) 69 m , (c) v x = 17 vy = −30 m/s .]
Problem 3.1
3.1 A ball is thrown from the roof of a 40 m high building,
Problem 3.2
with a speed of 20 m/s at an angle of 30◦ to the horizontal.
(a) What is the time of flight? (b) How far away from the 3.2 A stone is thrown horizontally towards the sea from a
PROBLEMS 53
45 m high cliff. There is a 20 m wide beach extending to- standing next to the platform wall on the ground starts to
wards the sea under the cliff. (a) What is the time of flight? run with acceleration a . What should the acceleration a be
(b) What should the minimum speed of the stone be in order so that the kid can catch the ball? What is the time of flight
to land in the sea? [A: (a) 3 s , (b) 6.7 m/s .] of the ball? [A: a = 2 m/s2 , t = 4 s .]
3.3 An athlete jumps up off the ground with a speed of 8 m/s 3.8 A basketball player shoots the ball at the basket, which
and lands 6.2 m away horizontally. (a) At what angle has he is 4 m away and 2 m high from the level of his hand. Since
jumped? (b) How long does the jump take in seconds? (Hint: the throwing angle is 53◦ , what should the initial velocity be
Use the trajectory equation.) [A: (a) 37◦ , (b) 1 s .] in order to make a basket? [A: 8.2 m/s .]
Problem 3.9
3.9 A gunner located on a 140 m high hill can fire shells
with an initial speed of 100 m/s and at an angle 37◦ with the
horizontal. Observing that a tank is approaching with a con-
Problem 3.4
stant speed of 10 m/s on a horizontal path on the ground, the
3.4 A rescue airplane flying horizontally at a speed of 80 m/s
gunner fires when the tank is at a distance d . What should
at an altitude of 50 m drops a relief package when a boat on
the distance d be for the shell to hit the tank? What is its
the sea is right underneath it. (a) In how many seconds will
time of flight? [A: d = 1260 m and 14 s .]
the package fall to the sea? (b) How far away from the boat
will it fall? [A: (a) 3.2 s , (b) 256 m .]
Problem 3.5
Problem 3.10
3.5 A ball thrown horizontally from a height hits a wall that 3.10 An outdoor elevator of a building is descending at a con-
is 20 m away at a point 15 m below. (a) What is the time of stant speed of 7 m/s . As the elevator passes through height
flight? (b) What is the initial
√ speed of the ball?
√ h , a boy located on the ground 36 m away from the building
[A: (a) 3 = 1.7 s , (b) 20/ 3 = 11.5 m/s .] throws a ball towards the elevator with an initial speed of
20 m/s and at an angle of 53◦ .
What should the height h be in order for the ball to hit the
elevator? What is the time of flight of the ball?
[A: h = 24 m and 3 s .]
3.3 Uniform Circular Motion
Problem 3.6
3.6 A ball rolls off of the edge of a 30◦ inclined roof whose 3.11 What is the centripetal acceleration of a point on the
edge is 35 m above the ground. The ball falls to the ground Earth’s surface due to its rotation about its own axis? The
in 2 s . (a) What is its initial speed? (b) How far away from Earth’s radius is 6400 km . [A: 0.03 m/s2 .]
the edge of the roof will it fall? [A: (a) 15 m/s , (b) 26 m .] 3.12 A racing car must have a maximum centripetal accelera-
tion of 9 m/s2 in order to go around a curve without slipping.
At what maximum speed can it take a circular curve with a
radius of 150 m ? [A: 37 m/s .]
3.13 The human body can withstand a maximum of 9g of
acceleration. What is the minimum radius of the loop that a
pilot can make in the air with a fighter jet having a speed of
Problem 3.7
900 km/h ? [A: 694 m .]
3.7 A ball is thrown at an angle of 37◦ to the horizontal and
with a speed of 20 m/s from one end of a horizontal platform, 3.14 Protons are accelerated at the European Nuclear Re-
which is 48 m long and 32 m high. At the same time, a kid search Center (CERN) and caused to collide at very high
energies. According to the most recent data, protons make

11 000 revolutions in 1 second in a circular trajectory whose
perimeter is 27 km long. What is the centripetal acceleration
of the protons? [A: 21 × 1012 m/s2 .]
3.15 The Earth rotates around the Sun in 365 days in an orbit
with an approximate radius of 15 × 107 km . The planet Mer- Problem 3.18
cury rotates in 88 days around an orbit with an approximate
radius of 6 × 10 km . Calculate the ratio of the centripetal 3.18 The pilot of an airplane with a speed of 200 km/h in
7
acceleration of Mercury with respect to Earth. still air wants to fly in the South-North direction on a day
[A: ar (Mercury)/ar (Earth) = 7 .] when the wind is 60 km/h in the East-West direction. In
which direction should he keep the nose of the airplane and
3.4 Relative Motion what will be the velocity of the airplane with respect to the
3.16 The velocity of a boat in a river is 10 m/s when it sails ground? ◦
in the same direction as the current, and 4 m/s when it sails [A: 17 with the North towards the East and 191 km/h .]
against the current. Calculate the speeds of the boat and the
current. [A: 7 m/s and 3 m/s .] 3.19 The velocities of two airplanes in the air are measured
from the ground. Airplane A is observed to be flying at
250 km/hour in the South-North direction and airplane B at
200 km/hour in the East-West direction. What are the mag-
nitude and direction of the velocity of plane A as observed
by the pilot of B?
[A: 320 m/s and 39◦ with the North towards the East.]
Problem 3.17
3.17 The pilot of an airplane with a speed of 60 m/s in still 3.20 The captain of a ship A sailing at a velocity of
air is flying by keeping the nose of the airplane directed to- 60 km/hour in the South-North direction observes another
wards the North in a region where the wind is 20 m/s in the ship B as sailing with a velocity of 50 km/hour in the West-
North-East direction. What is the velocity of the airplane East direction. What is the velocity and direction of ship B
with respect to the ground and its angle with the North? with respect to the ground?
[A: 75 m/s and θ = 11◦ .] [A: 78 km/hour and 40◦ with the North towards the East.]
4
NEWTON’S LAWS OF
MOTION
The Ariane 5 rocket starts its

space journey on April 22, 2011.
Developed by the European
Space Agency (ESA), this rocket
broke a record in carrying 9.6
tons of payload into space by
freeing it from the gravitational
pull of the Earth.
The fundamental cause that gen-
erates motion is the force. But
what exactly is the connection
between force and motion?
In the previous chapters, we have examined the motion in terms of position,

velocity and acceleration, but have not discussed the question of what causes
objects to move? We know from our daily lives that a force is necessary to set
objects into motion. Every object acts in a different manner, depending on its
own mass under such forces.
Throughout history, many scientists have examined the effect of forces on
the motion of objects. Galileo was the first to establish many of the features
of motion. Ultimately, English scientist Isaac Newton (1642–1727), building on
the Galileo’s results, was able to lay the foundations for the modern science of
Mechanics, in his book Principia Mathematica.
Newtonian mechanics was believed to be the absolute truth for about 200

56 4. NEWTON’S LAWS OF MOTION
years. Around the 1900s, these laws were found to give incorrect results both for
particles at the atomic scale and for particles traveling close to the speed of light.
Two modern theories were later developed, one for microscopic scales (Quantum
Mechanics) and the other for large speeds (Relativistic Mechanics). However,
Newtonian mechanics continues to remain valid for macroscopic objects.
4.1 NEWTON’S LAWS

Newtonian Mechanics is based on three simple and reasonable laws. By
accepting these laws without proof, you can explain the motion of all objects,
from stones and balls to cars, from rockets and bullets to stars and planets.
Let us state and discuss Newton’s laws in order.
Newton’s First Law
Discovered by Galileo, the first law regards objects upon which no net force
is acting:
Newton’s First Law

If the net force on an object is equal to zero, the object will either
remain at rest, or continue in a straight line with the same constant
velocity:
~Fnet = 0 ⇐⇒ ~a = 0 (4.1)
The force ~Fnet in this law is the vector sum of many forces and is called the
net force or resultant force:
~Fnet = ~F1 + ~F2 + ~F3 + · · · = ~Fi

X
Until Galileo, people believed in the following law by Aristotle: “Objects move
when a force is applied, and they stop if the force is removed.” Even today, many
people unaware of physics think in this manner. Indeed, doesn’t a book on a
tabletop that we push with our hand stop as soon as we withdraw our hand?
What is wrong with that?
Galileo answered this question by showing that an invisible friction force
was causing the object to stop. He devised very fine experiments by perfectly
polishing wood and marble surfaces in order to reduce friction. Then, although
Figure 4.1: Why are these they did not immediately stop when the driving force was removed, objects went
sportswomen sweeping in front further, depending on how polished the surface was. He thus discovered the first
of the moving curling stone? law.
We emphasize the following points about the first law:
• It is important that the net (resultant) force be zero in this law: Many forces
may be acting together on the object, but the law shall apply if their resultant
is zero.
• It is also easy to understand the “either, or” expression of the law: The
acceleration of the object is zero if it is either at rest or in uniform linear
motion.
4.1. NEWTON’S LAWS 57
• The first law is actually the definition of the force. Notice that the arrow is
bidirectional in Eq.(4.1) expression of the law. It means that the reverse
statement is also true: If an object is accelerating, there must be a net force
acting on it. In other words, acceleration is a sign of the presence of force.
Newton’s Second Law
If there is a net force acting on an object, it will perform accelerated motion.
The amount of this acceleration gives us the second law.
Newton’s Second Law

The acceleration of an object is directly proportional to the net
force acting on it, and inversely proportional to the mass of the
object. The acceleration is in the same direction as the net force:
~Fnet = m ~a (4.2)
Galileo again found this law as a result of the fine experiments that he con-
ducted using inclined planes. Galileo was able to show that the acceleration of
an object placed on an inclined plane, which he had polished to reduce friction,
increased as its slope was increased.
Let us stress some important points about the second law:
• The second law fixes the unit of force. This is a derived unit named the
Newton and abbreviated with (N) in the SI system. Writing the units of both
sides in the law’s expression, we get
1 N = 1kg·m/s2
More clearly, a force that gives 1 m/s2 acceleration to a 1-kg object will have
a magnitude of 1 Newton.
• The second law is a vector equation. In other words, this equation must be
true for each component in the xy -coordinate system:
F x,net = m a x
(
~Fnet = m ~a ⇐⇒ (4.3)
Fy,net = m ay
• At first glance you may think that the first law is a special case of the second
law. Indeed, if we set ~a = 0 in the second law (4.2), we get ~Fnet = 0 which
gives us the first law, right? This thinking is incorrect, because the second
law is actually the definition of mass. If the first law did not specify what the
force was, the second law could not relate this to mass and acceleration.
• The second law defines mass as a measure of the object’s resistance to accel-
eration. In other words, if the same force is applied to two objects, it will be
more difficult to accelerate the one with the larger mass (Figure 4.2). This
property of objects “to resist acceleration” is called inertia. In other words, Figure 4.2: Which bucket is
mass is the measure of the inertial characteristic of objects. harder to push?
• For which observers are Newton’s laws valid? For example, an accelerating
observer will see a still object on the ground as accelerating in the reverse
direction and find an incorrect solution: He/she will observe acceleration
although there is net zero force. In order to reply to this question, let us
remember the topic of relative speed from Chapter 2. We had seen that
(Equation 3.24), observers making uniform linear motion with respect to
each other will measure the same acceleration. Therefore, Newton’s laws are
valid for observers in uniform linear motion with respect to each other. We call
these inertial reference frames.
Newton’s Third Law (Principle of Action and Reaction)

Did you ever think about how a person jumps up? The answer is simple: We
push the ground down with our feet and the ground pushes us up in the opposite
direction. So, we know instinctively that the ground will produce a reaction force.
Isaac Newton was the first to consider that every force always generates a
reaction. The third law determines the relation between this action and reaction.
Newton’s Third Law

If an object applies a force ~F12 on a second object, the second
object will always apply a force ~F21 equal in magnitude and op-
posite in direction on the first object:
~F21 = −~F12 (4.4)
Before rockets were built, some scientists claimed that it would be impossible
to fly in the vacuum of space, where there is no air. However, rockets are able to
fly in space by utilizing the reaction force generated when they discharge their
fuel in the reverse direction with a large thrust. This is possible with the third
law.
The points to pay attention to in applying the third law are as follows:
• Action and reaction are applied on different objects. There can be a contradic-
tion if this distinction is not made.
As an example, consider a horse-drawn carriage (Figure 4.3). How would
you reply to this statement: “If the horse pulls the carriage forward, the
carriage will pull the horse back with the same force in the reverse direction.
Therefore, both forces will cancel each other out and the carriage will not
move.” This reasoning is incorrect, because it ignores the fact that the action
and reaction are applied on different objects. When examining one of these
objects, only the forces acting on that object should be taken into consideration.
The horse manages to not be pulled back by balancing the reaction force of
Figure 4.3: Forces on a horse the carriage with the friction force that it generates on the ground. However,
and carriage. the friction on the wheels of the carriage does not prevent it from going
forward, because it is very low.
• Only the external forces acting on an object are taken into consideration when
examining its motion. This is because internal forces mutually cancel each
other out according to the third law. Is it possible for a person inside of a
boat to make it move by pushing it? No. Let us examine the boat and the
4.2. TYPES OF FORCES IN MECHANICS 59
person inside as a whole. In this case, the action-reaction pair applied by

the person on the boat and by the boat on the person becomes the internal
forces of the system (boat+person) that we are examining, and their sum is
zero according to the third law.
• It is irrelevant to ask which of these forces between two objects is the action
and which is the reaction. Both arise simultaneously, in other words, there
is no causality between them. Only an action-reaction pair is mentioned in
order to state this fact.
4.2 TYPES OF FORCES IN MECHANICS

There are various types of forces that can lead to motion of objects. These
forces include the gravitational force and weight force as its special case, the
friction generated on surfaces and the normal reaction force and the tension force
on ropes. . . . One or more of these forces can be included in each problem. It is
not possible to find the correct solution unless we know how to take them into
consideration.
Now let us review the most important forces.
Weight (Gravitational force)
As we discussed in the free fall and projectile motion problems, any object
released near the Earth’s surface will accelerate with a constant acceleration g
towards the center of the Earth.
Interpreting this with the second law, we conclude that there must be a force
that causes such an acceleration (Figure 4.4). This special force is called weight
and its magnitude is indicated by W . As weight will fulfill the F = ma law with
the acceleration g ,
W = mg (weight force) (4.5)
and it is a vertically downward force. For example, the weight of an object with a
mass of 1 kg is W = 1 × 9.8 ≈ 10 N . This force is present regardless of whether
the object is moving or not.
Some students may confuse the concepts of mass and weight. We can prevent Figure 4.4: Weight is the force
such confusion as follows: Weight is the attraction force that Earth applies to that causes gravitational accel-
objects. Objects will have different weights on other planets and the Moon because eration.
the gravitational attraction forces are different there. However, the object has no
weight in empty space, where there is no other mass nearby.
Mass, on the other hand, is an intrinsic quality of the object and a measure of
its capability for acceleration. The object will have a mass regardless of where it
is in space. Here is a striking example that explains this: You can hold a rock with
a mass of 1 ton in space on your fingertips, because it has no weight. But if you
kick that rock, you may break your foot, because it is very difficult to accelerate,
having a very large mass.
Law of Universal Gravitation
Weight is a special case of a much more general attraction force. It is the
oldest of the four fundamental forces (gravitation, electromagnetism, nuclear
force and weak force). Gravitational force is what makes planets rotate around
the sun, keeps our satellite Moon in orbit around Earth and makes an apple fall
from a tree. Discovered by Isaac Newton, this law of universal gravitation (or
gravitational force) is expressed as follows:
Newton’s Law of Universal Gravitation

Every object in the universe attracts every other object with a force
that is directly proportional to the product of their masses and in-
versely proportional to the square of the distance between them:
m1 m2
Fg = G (4.6)
r2
Figure 4.5: Law of gravitation.

The proportionality coefficient here is G , the universal gravitational con-
stant, and it is one of the fundamental constants of physics:
G = 6.67 × 10−11 N·m2 /kg2 (Universal gravitational constant) (4.7)
Now let us apply this law to an object with mass m near the Earth’s surface
(Figure 4.6). Let one object have mass m and the other object be the Earth, with
mass ME and radius RE . Accordingly,
 
mME  GME 
Fg = G 2 = m  2 
RE RE
The expression inside of the brackets depends only on the mass and radius of the
Earth. The Earth’s mass is ME = 5.97×1024 kg and its radius is RE = 6.38×106 m .
Figure 4.6: Weight is caused by Using these values, we find the gravitational force acting on mass m :
the law of gravitation.
Fg = m × (9.81 m/s2 )
The constant factor has dimensions of acceleration and a value of 9.81 m/s2 . It
is just the gravitational acceleration g that we are familiar with. Hence, this
gravitational force is the weight W itself:
W = Fg = m g (Weight) (4.8)
GME (The gravitational acceleration
g = (4.9)
R2E on the Earth’s surface)
This formula shows us how the gravitational acceleration g decreases with

increasing height. To find the value of g at a height h from the surface of the
Earth, it is sufficient to substitute (RE + h) for RE in this formula. Likewise,
the gravitational acceleration at the surface of the Moon or other planets can be
calculated using this method.
In calculating the gravitational force between objects with volume, the dis-
tance r in the formula is the distance between the centers of mass of the two
objects.
Example 4.1
A stone with mass m = 2 kg is descending in free fall. A vertical

force F is applied on it.
(a) What is the acceleration of the stone if F = 0 ?
(b) What is the acceleration of the stone if F = 5 downwards?
(c) What is the acceleration of the stone if F = 5 upwards?
in the same direction. Using the second law,

Answer Fnet = W + F = ma
(a) Although there is no force F on the stone, there is still
mg + F = ma → a = g + F/m = 10 + 5/2 = 12.5 m/s2
the weight W = mg . According to the second law Fnet = ma ,
if we take W = mg for Fnet , we get a = g . In other words, (c) Writing the second law for the weight W and the force F
the object will continue its free fall with acceleration g . in the opposite direction,
(b) This time, in addition to the weight W , there is a force F mg − F = ma → a = g − F/m = 10 − 5/2 = 7.5 m/s2 .
~)
Normal Force On Surfaces ( N
Consider a book on a desk (Figure 4.7). We know that the book has a weight
and is pulled towards the center of the Earth with a force W=mg . If this force were
acting alone, then the book would have to go through the table and accelerate
downwards according to the F=ma rule.
However, the book is motionless, in other words, has zero acceleration. There-
fore, there should be another force acting upwards, opposite to the weight, so
that the net force can be zero. This force acting perpendicular to the surface and
always outwards from the surface is called the normal force and is indicated by
~ . It only appears when the object is in contact with the surface and disappears
N
when the objects leaves the surface. The source of the normal force is the complex
interactions between the molecules that constitute the table and the book.
Normal force is always perpendicular to the surface and large enough to prevent
the object from entering the surface. For example, in the case of the book on
the table, we had N=W (Figure 4.7). Now, let us consider an additional force
F pressing the book down towards the table. This time, the normal force will
increase to a value that counters both the W and F forces, in other words, we
have N = W + F .
The following question could come to mind here: “Doesn’t the normal force Figure 4.7: The normal reac-
have a reaction force?” Yes, it does. Indeed, there is an opposite force (−N) tion force N generated on the
applied by the book on the table. However, as we are only concerned with the surface of the table is always per-
forces acting on the book, we did not deal with this (−N) force applied on the pendicular to the surface.
table.
This (−N) force is what is measured when you step on a bathroom scale. If
someone presses down on your shoulder at that moment, the scale will add this
to your weight.
Example 4.2
(c) What is the normal reaction force of the table?
A force of F = 12 N is applied at an angle of 37◦ with the Answer
horizontal, on an object with mass m = 3 kg located on a (a) In addition to the force ~F applied on the object, there is the
frictionless plane. downward weight W = mg and the vertical reaction force N
applied by the surface. These forces are shown in the figure.
(b) If we choose the axes as shown in the figure, the accelera-
tion will be in the x -direction. Since there is no acceleration
perpendicular to the table, the net force component is zero in
the y -direction. Therefore, the second law can be written as
follows for the components:
F x,net = ma
(
~Fnet = m~a →
Fy,net = 0
(a) Show all of the forces acting on the object. The first equation gives the acceleration. In the x -direction
(b) What is the acceleration of the object? there is only F x =F cos θ :
F cos 37◦ = ma → 12 × 0.8 = 3a → a = 3.2 m/s2 F should also be included:

N + F sin 37◦ − mg = 0 → N + 12 × 0.6 − 3 × 10 = 0
(c) The vertical component gives the normal reaction force N . N = 22.8 newton
Here, in addition to the weight W , the vertical component of As you can see, the reaction N may not always equal mg .
Friction Force
Let us again consider a book on a tabletop. Let us push this book with a force
F that is parallel to the table. Will the book move along the table? The answer
depends on the magnitude of the force. We know from experience that there
will be a friction force preventing motion and that it will be necessary to apply a
minimum force that is large enough to overcome the friction force.
We experience friction force in every aspect of our daily lives. In fact, we can
say that life would be impossible without friction. We would be unable to walk on
the ground or hold anything in our hands. On the other hand, many technologies
became possible through the reduction of friction. For example, without oil in
the engine of an automobile, it would be necessary to replace its cylinders and
pistons before it had traveled barely one kilometer.
Similar to the normal force, friction force arises only upon contact with a
surface. Both are a type of force called contact force. If we look closely at
the cross-section of the interface between the book and the table (Figure 4.8),
Figure 4.8: Friction force is we will observe rough edges at the microscopic scale, even if they are perfectly
caused by the fact that surfaces polished. The atoms and chemical broken bonds at the tips of these rough surfaces
are rough. immediately interact and make weak bonds. When the object moves, these bonds
are broken and then immediately reformed along the path. They thus oppose the
motion.
Forces that oppose motion are also formed in liquids and gases. These have
different characteristics. We shall only deal with the friction force between solid
surfaces.
Experiments have demonstrated the following features of friction force:
• Friction force is approximately the same regardless of which surface of the
object is in contact if all surfaces have the same roughness.
• The friction force is approximately constant regardless of the velocity of the
object.
Figure 4.9 shows the behavior of friction force as the applied force F increases.
When F is small, the friction force f adjusts itself in the opposite direction so
as to fully counter it, and the object does not move. This balancing continues
until it reaches a maximum value fmax and it can no longer counter the force
F , whereupon motion starts. Experiments show that this maximum value of the
friction force is proportional to the normal force on the surface:
Figure 4.9: Friction force acts fmax = µN (4.10)
differently in static and kinetic
cases. where the dimensionless coefficient µ is the coefficient of friction and depends
on the type and roughness of the surfaces.
The maximum value of the friction force slightly decreases after the object
starts moving (Figure 4.9). In order to describe this behavior, a coefficient of
kinetic friction µk is defined for moving objects and a coefficient of static
friction µ s for objects at rest:
0 ≤ fs ≤ µs N (if the object is at rest)
(4.11)
fk = µk N (if the object is moving)
Static and kinetic coefficients of friction measured between certain surfaces

are given in the following table.
Friction coefficients of certain surfaces

Surface Static friction, µS Kinetic friction, µK
Wood–wood 0.35 0.30
Steel–steel 0.80 0.50
Steel–ice 0.1 0.05
Rubber–dry asphalt 1.0 0.8
Rubber–wet asphalt 0.7 0.5
Figure 4.10: The inclination an-
gle of coal piles depends on fric-
These values are approximate and may vary depending on whether the sur- tion force. If they get wet in a
faces are rough or wet. For this reason, in the problems in this book, no static- rainfall, they may become dan-
kinetic distinction shall be made and only the coefficient µ shall be used. gerous.
Example 4.3
we only have the weight W and the reaction force N in the
vertical direction, we get N = W = mg . We first calculate
the maximum value of the friction force:
fmax = µN = µmg = 0.5 × 3 × 10 = 15 N
(a) The force F1 = 12 N is less than the maximum value of
the friction force. Therefore, the friction force fully counters
A block with mass m=3 kg is on a horizontal plane. The static
F and the block does not move:
and kinetic coefficients of friction between the block and the
f = 12 N
surface are equal and have the value µ = 0.5 .
(b) The block will move because the force F2 = 18 N is
(a) A horizontal force F1 = 12 N is applied on this block.
greater than the maximum value of the friction force. The
What is the friction force?
friction force is at maximum value during motion:
(b) This time, a horizontal force F2 = 18 N is applied. What
f = fmax
is the acceleration of the block?
We can find the acceleration using the second law:
Answer Fnet = ma → F2 − f = ma
The forces acting on the block are shown in the figure. As 18 − 15 = 3a → a = 1 m/s2
Example 4.4
F x,net = F cos 37◦ − f = ma
A block with mass m = 3 kg is on an inclined plane. The static Fy,net = N − F sin 37◦ − mg = 0
and kinetic coefficients of friction between the block and the sur- The reaction force N is calculated from the second equa-
face are equal and have the value µ = 0.3 . A force F = 20 N tion:
with an angle 37◦ below the horizontal is applied on this block. N = mg + F sin 37◦ = 30 + 20 × 0.6 = 42 N
Calculate the normal reaction force and the acceleration of the Now, we must determine if there will be motion. For this
block. purpose, it is necessary to compare the F cos θ component
of the applied force along the direction of motion with the
maximum value of the friction force:
F cos 37◦ = 20 × 0.8 = 16 N
f = µN = 0.3 × 42 = 12.6 N
The block will start moving, because F cos θ > f . We can find
the acceleration by plugging these values into the expression
Answer above for the x -component of the second law:
The forces acting on the block are shown in the figure. If we F cos 37◦ − f = ma → 16 − 12.6 = 3a
write the second law in terms of the components, a = 1.1 m/s2
Example 4.5
must be opposite, hence upward along the inclined plane.
A block of mass m = 5 kg is placed on a plane inclined at 37 . ◦ For the coordinate axes, we take the x -axis along the
The static and kinetic coefficients of friction between the inclined inclined plane and towards the direction of motion, in other
plane and the object are equal and have the value µ = 0.2 . words, downwards. The y -axis is taken as perpendicular to
(a) Show the forces acting on the object on a figure. the inclined plane.
(b) What is the friction force? Accordingly, will the object (b) First, the y -component of the second law is written so as
move? If so, what is its acceleration? to find the friction force from the normal reaction force:
Fy,net = N − mg cos 37◦ = 0 → N = mg cos 37◦
fmax = µN = µmg cos 37◦ = 0.2 × 5 × 10 × 0.8 = 8 N
The force moving the object is the component of weight along
the inclined plane:
mg sin 37◦ = 50 × 0.6 = 30 N > fmax , hence the object
will move.
Answer The x -component of the second law is used to find the accel-
(a) The three forces acting on the block are shown in the eration:
figure. Normal force N is perpendicular to the inclined plane. F x,net = mg sin 37◦ − f = ma → 30 − 8 = 5a
As the object will try to slip downwards, the friction force f a = 4.4 m/s2
Tension Forces on Ropes

The type of force applied by flexible objects such as ropes, cordes or cables is
called the tension force (or, simply, the tension). The most important feature of
the tension force is that it can only pull the object. You cannot push objects with
a rope, because it will bend quickly.
Let us hang a block with mass m from the ceiling by means of a flexible rope
(Figure 4.11). As this block is at rest, this means that the weight mg is balanced
by a tension T that is equal and opposite: T =mg . When a second mass is added,
the tension increases to balance the sum of their weights. Therefore, the tension
is an attractive force that always establishes balance, as long as the rope is not
broken.
Figure 4.11: Tension on a rope. Now, let us proceed towards the end of the rope hung from the ceiling. Ac-
cording to the third law, the lower and upper parts in any cross-section of the
rope will pull each other with equal tension forces in opposite directions. The
tension force at this height is usually not equal to the one at the lower end. This
is because the rope also has a mass, and this tension force is always equal to the
weight of the sum (object + rope below).
However, if the mass of the rope is small enough to be negligible, each cross-
section will keep the same T =mg value, and thus, when we reach the ceiling, the
tension on this end will also be T . Therefore, the tensions on both ends of a rope
with zero mass are equal. The tensions on both ends remain equal even if the rope
is moving.
Free-Body Diagrams
In dynamics, it is important to clearly state the examined system. If there
are multiple forces acting on multiple masses (Figure 4.12), these forces may be
separated into two groups:
1. Internal forces: Forces applied by the masses inside of the examined system
on each other ( T 1 , T 2 , T 3 in the figure). According to Newton’s third law,
Figure 4.12: Internal and exter- these forces will always be in pairs. These are not included in the equations,
nal forces acting on 3 masses. because they are both positive and negative and cancel each other out.
4.3. APPLICATIONS OF NEWTON’S LAWS 65
2. External forces: Forces originating from outside of the system ( Fa , Fb in

the figure).
In order to better understand the issue, let us consider the forces acting on
masses m1 , m2 , m3 separately:
• If we are only considering the mass m1 as our system, each of the forces acting
on it ( T 1 , T 2 , Fa ) are external forces (Figure 4.13) and should be included in
equations.
• If we are considering the masses together (m2 +m3 ) as our system, this time, Figure 4.13: The free-body dia-
the external forces are (T 1 , T 2 , Fb ) . In this system, T 3 is an internal forces gram of mass m1 .
and is not taken into consideration, because it will appear twice in the
equations in opposite directions, hence canceling itself out.
• If the system is the total mass (m1 +m2 +m3 ) , then only Fa and Fb are ex-
ternal forces and the force pairs (T 1 , T 2 , T 3 ) applied by the masses to each
other become internal forces and are not included in the equations.
After taking into consideration all of the external forces acting on a system,
we can show it as isolated, ignoring all other objects in its environment. This is
called a free-body diagram (Figure 4.13). For example, after showing the forces
T 1 , T 2 , Fa acting on mass m1 in the same system with three masses, we no longer
need to show the other masses, because we have already taken their effect into
consideration.
4.3 APPLICATIONS OF NEWTON’S LAWS

In this section, we will use Newton’s laws to solve examples of motion that
objects may undergo under the action of various forces.
Example 4.6
Answer
A painting frame with mass m = 5 kg is held in equilibrium,
Each rope applies an attractive tension along itself. These are
hanging on the wall by two ropes with the angles given in the
shown as T 1 and T 2 in the figure. We apply the first law, as
figure. Calculate the tensions in the ropes.
the object is at rest:
~Fnet = 0
This equality also applies to the components:
F x,net = 0 → T 1 cos 53◦ − T 2 cos 30◦ = 0
Fy,net = 0 → T 1 sin 53◦ + T 2 sin 30◦ − mg = 0
We find T 1 and T 2 from these two equations:
T 1 = 44 N and T 2 = 30 N .
Example 4.7
Atwood’s machine. Two equal blocks with masses

M=10 kg are hung on two ends of a rope passing through a
pulley. The pulley is frictionless and the mass of the rope is
negligible. An additional small mass m=1 kg is placed on one
of the blocks.
(a) Show the forces acting on both blocks in a diagram. (You
do not have to show the forces acting on the pulley.)
(b) For each block, write the equations that give the accelera-
tion and the tension in the rope.
(c) Solve this system of equations to calculate acceleration a For mass (M + m) : (M + m)g − T = (M + m)a
and tension T . 110 − T = 11a
Answer For mass M : T − Mg = Ma
(a) The tensions on both ends of the rope are equal, because T − 100 = 10a
it is assumed that the mass of the rope is negligible and the (c) By solving these two equations, we find acceleration a
pulley is frictionless. For the same reasons, both masses will and tension T :
accelerate with the same acceleration a . a = 0.5 m/s2 and T = 105 N
Hence, the forces are as shown in the figure. Note: This setup is used in operating elevators and funiculars.
(b) The problem is one-dimensional. We apply the second A small mass m can be used to easily accelerate a larger mass
law in the vertical direction to each mass, in their direction M . If we had attempted to pull only the mass M upwards
of motion: with the same acceleration a , we would have had to use a
Fnet = ma force F much larger than the force mg here.
Example 4.8
Two blocks with masses m1 =1 kg and m2 =2 kg on a frictionless Answer (a) The two ends of the rope have equal tension T
horizontal plane are tied to each other with a rope. A horizontal in opposite directions. The normal reaction forces cancel out
force F=5 N is applied on mass m1 . the weights in the vertical direction. The forces are as shown
in the figure above.
(b) The second law is written in the direction of motion for
each block (it is not necessary to write the expressions in the
vertical direction, because there is no friction):
m1 mass : F − T = m1 a 5−T =a
(a) Draw a free-body diagram for each block. m2 mass : T = m2 a T = 2a
(b) Find the acceleration and the tension in the rope by sepa- From here, we find the acceleration and the tension:
rately considering the blocks m1 and m2 . a = 1.7 m/s2 and T = 3.3 N
(c) Consider the blocks together and calculate the acceleration (c) This problem can also be solved by considering the two
of the (m1 + m2 ) system. masses together. This is possible because the accelerations
are equal. Now, the tensions T and −T in the rope in the
two-mass system become internal forces, and, according to
third law, they cancel each other out. Thus, the acceleration
of the whole system is only due to the force F :
F = (m1 + m2 )a → a = 5/3 = 1.7 m/s2
It is not possible to find the tension T in this system, because
T is an internal force.
Example 4.9
Answer
(a) The forces acting on each block are shown above.
A block of mass m1 = 7 kg on a frictionless inclined plane with
Consider first the system as a whole. On one hand, the
a slope of 30◦ is tied to the end of a rope that passes through a
mass m2 g will try to pull the mass m2 down and, on the
pulley. A mass m2 = 5 kg is hanging on the other end of the
other hand, the component of mass m1 parallel to the in-
rope.
clined plane, m1 g sin θ will try to pull it down.
(a) Draw a free-body diagram for each block and determine Let us calculate which of these forces will become dominant:
the direction of motion. m1 g sin 30 = 7 × 10 × 0.5 = 35 N
(b) Separately, write the equations of motion for the masses m2 g = 5 × 10 = 50 N
m1 and m2 . Calculate the acceleration and the tension Since m2 g > m1 g sin θ , the motion will be downwards for m2
force in the rope. and upwards along the inclined plane for m1 .
4.3. APPLICATIONS OF NEWTON’S LAWS 67
(b) Since there is no friction, it is sufficient to use the second follows:

law only for the component along the motion: 50 − 35 = 12a → a = 1.25 m/s2
for m1 : T − m1 g sin θ = m1 a T − 35 = 7a Plugging this value into either one of the equations, we get
for m2 : m2 g − T = m2 a 50 − T = 5a T:
Adding both sides of this equation, we find acceleration a as T = 44 N .
Example 4.10
Answer
The rope in the figure has zero mass and the surfaces and the
(a) In this problem, the accelerations of the blocks are different.
pulley are frictionless. (a) Draw a free-body diagram for each
The forces acting on each one are shown in the figure.
block. (b) What is the relation between the accelerations?
(b) The acceleration of the mass m2 tied to the pulley will be
(c) Calculate the accelerations and the tension in the rope.
less, because, as it descends, some of the rope will be trans-
ferred to the other side. For example, how much does mass
m2 descend when mass m1 goes to the right by 1 m ? The
answer is 1/2 m . Therefore, the relationship is as follows:
a2 = a1 /2
(c) We use the second law for each mass along its direction
of motion:
for m1 : T = m1 a1 → T = a1
for m2 : m2 g − 2T = m2 a2 → 20 − 2T = 2a2
There are three unknowns in these two equations: a1 , a2 , T .
The third equation that we need for the solution is the relation
a1 = 2a2 between the accelerations that we found in item (b).
Solving these three equations, the results are as follows:
a1 = 6.7 m/s2 , a2 = 3.3 m/s2 and T = 6.7 N .
Example 4.11
opposite direction. The forces acting on the object are shown
in the figure.
Let us write the equations of motion:
F x,net = ma → − f = ma
Fy,net = 0 → N − mg = 0
With the friction force f = µN and N = µmg , the accelera-
A block with mass m is thrown with a speed of 4 m/s along a tion a is found to be negative:
horizontal plane on which the coefficient of friction is µ=0.25 . a = −µg = −0.25 × 10 = −2.5 m/s2
What distance will the block travel before it stops? Using the velocity formula without time from kinematics, the
Answer distance traveled until the object stops (v = 0) is calculated
In this problem, there is no force acting in the direction of as follows:
motion, rather the block moves due to its initial speed, and v2 − v20 = 2ax → 0 − 42 = 2(−2.5)x
then it slows down due to the friction force f acting in the x = 3.2 m .
Example 4.12
forces acting on the block. (b) Calculate the acceleration of the
block.
Answer
(a) The friction force will be in the opposite direction to that
of the motion, hence we must first determine in which direc-
A block of mass m = 1 kg is placed on an inclined plane with tion the block will move. Two forces come into play along the
a coefficient of friction of µ = 0.3 and incline angle of 37◦ . A inclined plane: The component mg sin θ of its weight pulling
horizontal force F = 20 N is applied on the block. (a) In what the block downwards, and the component F cos θ pushing
direction will the block move? Using this information, show the it upwards. Let us compare these two:
mg sin 37◦ = 10 × 0.6 = 6 N and the y -axis is taken as perpendicular to the plane, the
F cos 37◦ = 20 × 0.8 = 16 N equations of motion will be as follows:
As F cos θ is greater, the block will try to go upward. It F x,net = ma → F cos 37◦ − mg sin 37◦ − f = ma
may not be able to move, due to the friction force, but, in any Fy,net = 0 → N − mg cos 37◦ − F sin 37◦ = 0
case it will not go downwards. The forces acting on the blockand we have the relation between friction and the normal
are thus as follows: force: f = µN .
We calculate acceleration a from these three equations:
(F cos 37◦ −mg sin 37◦ )−µ(mg cos 37◦ +F sin 37◦ )
a=
m
a = 4 m/s2
If the resulting acceleration were not positive here, we would
(b) If the x -axis is taken upwards along the inclined plane conclude that the block was unable to move.
Example 4.13
not always equal to mg . By definition, the normal reaction
force is one that counters all of the forces perpendicular to
the surface. In this problem, mg has no component perpen-
dicular to the surface; there is only the force F . Therefore,
we have
N−F =0 → N =F
The block can only move downwards. The equations of mo-
A block with a mass of 2 kg can slide downwards when pressed
tion in this direction are as follows:
by a horizontal force F = 30 N against a wall. The coefficient
mg − f = ma
of friction between the block and the wall is µ = 0.5 . Show the
Again, as the friction force is defined with the formula
forces acting on the block and find its acceleration.
f = µN , we get f = µF . We find the acceleration a from
Answer these two equations:
This is a good example that shows that the normal force is a = (mg − µF)/m = (20 − 0.5 × 30)/2 = 2.5 m/s2 .
Example 4.14
this will act on the lower block. This opposite force f is the
force that makes m2 move. Therefore, the forces acting on
the objects are as follows:
The friction force between the blocks with masses m1 = 1 kg

and m2 = 2 kg shown in the figure is µ = 0.3 . There is no
friction between the lower block m2 and the horizontal plane
underneath. A horizontal force F is applied on the block m1 .
(b) Let us write the equations of motion for the lower block:
(a) Show the forces acting on the blocks in separate diagrams.
f = m2 a
(b) What is the maximum acceleration that the two blocks can
travel together without causing the upper block m1 to slip? (c) This friction force f has a maximum value:
What is the force F that provides this maximum acceleration? max f =µN 1 =µm 1 g . This is the maximum force that will act in
the direction of motion on the lower mass. From here, we
Answer find acceleration a :
(a) This example shows that friction may not always be in the µm1 g = m2 a → a = (m1 /m2 )µg = 0.5 × 0.3 × 10 = 1.5 m/s
2
opposite direction to the motion, but may sometimes assist Therefore, the maximum acceleration that both masses
motion. Friction is only between two blocks. As the mass m1 can travel together will be a = 1.5 m/s2 .
will try to move to the right compared to the lower mass, the (c) We also use this maximum acceleration for the upper block
force (− f ) preventing it will be towards the left. However, m1 and find the maximum force F :
according to the third law, a force + f equal and opposite to F − f = m1 a → F = µm1 g + m1 a = 4.5 N .
4.4. CIRCULAR MOTION 69
4.4 CIRCULAR MOTION

In Chapter 3, we had examined circular motion and established that an object
rotating with a constant speed v around a circle with radius r would get a
centripetal acceleration ar towards the center:
v2
ar = (centripetal acceleration) (4.12)
r
According to the second law, there must be a force Fr acting in the same direction
as this acceleration, in other words, towards the center (Figure 4.14): Figure 4.14: There must be
a centripetal force causing the
v2 centripetal acceleration.
Fr = mar = m (centripetal force) (4.13)
r
This centripetal force Fr is the net force in the radial direction causing the
acceleration ar . It could be achieved in various ways. For example, if we are
swirling a stone tied to a rope, it will be the tension T in the rope. If a cart rotates
on rails, it will be the normal reaction force of the rails.
You may still wonder whether there is any need for such a force while the
object is rotating. Let us consider that the rope suddenly breaks while the object
is rotating (Figure 4.15). After that moment, according to the first law, the object Figure 4.15: What will happen
will travel along a linear path, since the tension T causing the rotation is no if the force causing centripetal
longer present. acceleration is removed?
As a car is turning around a bend, the centripetal force is caused by the friction
force on the tires. Likewise, for an object rotating inside of a railed circle, it will
be the normal reaction force N of the rails: N = mv2 /r .
Another example is the gravitational force that allows planets to rotate around
the Sun. For example, if we use the gravitational force for a satellite with mass m
rotating around the Earth on an orbit with radius r , we get
mME v2
Fr = G = m
r2 r
Using this equation, we can find the velocity of the satellite at any height.
Figure 4.16: Examples of cen-

tripetal force.
Let us correct a common misunderstanding here: In many books, a centrifugal

force is mentioned as the force acting outwards from the center. Students may
readily accept this instinctively. Indeed, you may think that, “When the bus enters
a curve, there is a force pushing me outwards from the center,” right?
Such a thinking is incorrect. The correct interpretation is as follows: You
are not pushed outwards when the bus enters a circular trajectory. If you do not
hold onto anything, you would continue to travel in a straight line. Actually, it
is the side of the bus turning inwards that is approaching you. Consider also the
following: When you hold on to a handlebar as the bus turns, in which direction
is the handlebar applying a force on you? Towards the center, of course.
The concept of centrifugal force is a technique developed to solve certain

problems as statics. When writing all of the forces acting on a rotating object,
an outward force of magnitude mv2 /r is added and then the sum of all forces is
set as equal to zero, as if the object is in equilibrium. But this has nothing to do
with reality. This is because, an observer rotating together with the object does not
constitute an inertial reference frame, and thus Newton’s laws do not apply there.
Therefore, we should refrain from using the term centrifugal force.
Example 4.15
N of the surface have no components towards the center;
A block with mass m = 200 g rotating with constant speed at there is only the force T . Therefore,
the end of a rope of length 80 cm on a frictionless horizontal v2
Fr = mar → T = m .
plane makes 500 revolutions in 3 minutes. r
We must first calculate the velocity v . As one rotation
Calculate the centripetal acceleration of the block and the
around a circle with radius r will have a length of 2πr , we
tension in the rope.
divide the total distance by time:
distance traveled 500 × 2π × 0.80 m

v= = = 14 m/s
time duration 3 × 60 s
From here, we calculate the centripetal acceleration and
the tension in the rope:
Answer v2 142
The force towards the center causes the circular motion. As ar = = = 245 m/s2 ,
r 0.8
seen in the figure, the object’s weight and the reaction force T = mar = 0.2 × 245 = 49 N .
Example 4.16
Answer
(a) As seen in the figure, the friction force f causes the cen-
tripetal acceleration:
v2 (72 × 1000/3600 m/s)2
f = m = 1200 × = 2400 N
r 200
(b) The maximum value of the friction force is fmax =µN=µmg ,
hence it is the maximum force directed towards the center.
We can write the centripetal acceleration caused by this force
(a) An automobile with mass 1200 kg is traveling on a horizon- as follows:
tal circular track of radius r = 200 m at a speed of 72 km/h . v2
µmg = m → v2 = µgr .
What is the friction force between the wheels and the road? r
This maximum√velocity is calculated as follows:
(b) If the coefficient of friction on this track is µ = 0.8 , what √
will the maximum speed be at which the automobile can turn v= µgr= 0.8 × 10 × 200 = 40 m/s = 144 km/hour .
without sliding?
Example 4.17
Answer
A car rounds a frictionless curve banked at an angle of 30 This time, there is no friction. So, which is the centripetal
◦
and with a radius of r = 100 m . Calculate the necessary speed force that keeps it on a circular path? As seen in the figure,
v so that the car will not slip up or down the incline. the normal reaction force N of the road will have a compo-
nent towards the center. This is the centripetal force that
causes rotation.
We write the equations of motion along the x - and y -
directions shown in the figure:
x -direction: N sin θ = mv2 /r
y -direction: N cos θ = mg
Dividing the two equations on both sides, we get
tan θ = v2 /gr
From this,
p we calculate the
√ necessary speed: v = 24 m/s = 87 km/hour .
v = gr tan 30◦ = 10 × 100 × 0.58
Example 4.18
In this problem, the centripetal force is the combination of
the weight mg of the pilot and the normal force N of the
The pilot of an airplane executes a loop-the-loop stunt at a
seat. (The centripetal force rotating the airplane is the force
constant speed of 300 m/s in a vertical circle of radius 2 km .
generated by the wings, but we only consider the pilot here.)
(a) Calculate the centripetal acceleration of the airplane.
(a) First, let us calculate the centripetal acceleration in terms
(b) Calculate the normal reaction force applied by the seat to
of multiples of g :
the pilot with mass m at the top and bottom of the circle
ar = v2 /r = 3002 /2000 = 45 m/s2 = 4.5g
in terms of multiples of the weight mg of the pilot.
(b) As seen in the figure, the reaction force N of the seat
is always towards the center at the top and bottom of the
circle. However, the weight mg is towards the center at the
top, but outwards from the center at the bottom. Therefore,
the resultant force towards the center will be written in both
cases:
At the top:
N + mg = mar → N = mar − mg = 4.5mg − mg
N = 3.5mg
At the bottom:
N − mg = mar → N = mar + mg = 4.5mg + mg
N = 5.5mg
Clearly, the most difficult part of the pilot’s maneuver is
Answer at the bottom of the circle.
Example 4.19
motion of the stone.
Therefore, the centripetal force should consist of the sum of
these two:
v2
mg + T = m
r
What will happen if we reduce the stone’s speed v ? The right
side of this expression will decrease. Since weight is constant
A stone tied to the end of a string with radius r is being rotated on the left side, tension T will also decrease. When a certain
around in a vertical circle. What should the speed of the stone speed is reached, we get T = 0 , in other words, the string
be at the top so that it will complete the circle without the string will get loose.
getting loose there?
Therefore, the minimum velocity v will be the velocity at
Answer T = 0:
Both the weight mg of the stone and the tension T of the v2 √
string will be towards the center at the top of the circular mg + 0 = m → v = gr
r
1. A bicycle and truck collide head-on. Which of the fol-

2. Which is the force that makes a falling ball bounce back
lowing is correct?
up from the ground?
(a) The force on the bicycle is larger. (a) Gravitational force
(b) The force on the truck is larger. (b) Weight
(c) The forces are equal and opposite. (c) Friction force
(d) The forces are equal and perpendicular. (d) Normal reaction force
3. Which of the following is incorrect? (a) It will accelerate with g.

(b) It will fall with a constant velocity.
(a) The net force on a car at rest is zero.
(c) It will remain suspended in the air.
(b) The net force on a car traveling at constant speed is
(d) It will accelerate upward.
zero.
(c) The acceleration of a car that gets faster is zero.
11. Which of the following are incorrect?
(d) The acceleration of a car traveling in a straight line
I. The mass of an object in outer space is zero.
at constant velocity is zero.
II. The weight of an object in outer space is zero.
III. The mass of an object is less on the Moon.
4. A car applies brakes and stops. Which force makes the IV. The weight of an object is less on the Moon.
car stop?
(a) I & II (b) III & IV (c) I & III (d) II & IV
(a) Weight
(b) Friction force 12. What is the force F applied on an object with mass
(c) Normal force on the surface 1 kg moving vertically upwards with an acceleration of
(d) Gravitational force 4 m/s2 ?
(a) 6 N (b) 10 N (c) 14 N (d) 18 N
5. Which is correct at the maximum height of a stone in
projectile motion?
13. A man in an elevator releases the suitcase in his hand
(a) The net force on the stone is zero. and the suitcase does not drop to the ground. Which of
(b) There is a horizontal force acting on the stone. the following could be correct?
(c) There is a gravitational acceleration on the stone. I. The elevator is in free fall.
(d) The acceleration of the stone is zero. II. The elevator is descending with constant speed.
III. The elevator is moving downwards with accelera-
6. On a tabletop with a friction force of 6 N , what will tion g .
the acceleration be of an object with mass 2 kg pulled IV. The elevator is moving upwards with acceler-
horizontally with a force of 8 N in units of m/s2 ? ation g .
(a) 1 (b) 2 (c) 3 (d) 4 (a) I & II (b) I & III (c) II & III (d) I & IV
7. Passengers on a train with no windows and traveling at 14. Which of the following is incorrect for friction force?
constant speed observe that a ball on the floor starts to (a) It is always opposite to the direction of motion.
roll forward. Which of the following could be true? (b) It is always proportional to mg.
I. The train could be slowing down. (c) It is always proportional to the normal reaction force
II. The train could be getting faster. N.
III. The train could be going uphill. (d) It is always perpendicular to the surface.
IV. The train could be going downhill.
(a) I & III (b) I & IV (c) II & III (d) II & IV 15. How would a ball tied to the end of a string and rotating
on a frictionless plane move when the string breaks?
8. Two people are playing tug of war. Which of the follow- (a) It will stop.
ing is correct? (b) It will continue rotating.
(a) The one who pulls stronger wins. (c) It will move along a straight line.
(b) The one with the higher mass wins. (d) It will move towards the center.
(c) The one with higher friction force on the ground
wins. 16. An object with mass 10 kg is rotating with a speed of
(d) The one who grabs the rope more tightly wins. 3 m/s on a circular path with radius 2 m . What is the
centripetal force acting on the object in newton units?
9. For which observers do Newton’s laws not apply? (a) 30 (b) 35 (c) 40 (d) 45
(a) Observers at rest.
17. What could cause an automobile to round a curve?
(b) Observers in uniform linear motion.
I. Friction force.
(c) Accelerating observers.
II. The mass of the automobile.
III. The speed of the automobile.
IV. The normal force of the road.
10. An elevator goes into a free fall when its cable breaks.
How would a screw that comes off in the ceiling of the (a) I & II (b) II & III (c) III & IV (d) I & IV
elevator move with respect to an observer in the elevator?
PROBLEMS 73
√
18. If a car is able to turn around a curved road with radius (a) 1 (b) 2 (c) 2 (d) 4
r at a speed v without sliding, at what speed can it turn
without sliding around a curve with radius 2r ? 20. When the speed of a rotating stone at the end of a string
√
(a) v (b) 2v (c) 2v (d) 4v with length r reaches 10 m/s , the string breaks. If a
string with length 4r made of the same material is used,
19. What is the ratio of the centripetal forces F B /F A of an at what speed will the rope break (in units of m/s )?
automobile A with mass m turning around a curve with √ √
(a) 10 (b) 20 (c) 10 2 (d) 10/ 2
radius r at speed v and an automobile B with mass 2m
turning around a curve with radius 2r at speed 2v ?
Problems
4.3 Applications of Newton’s Laws is µ=0.3 . Calculate the accelerations of the blocks and the
tension in the rope. [A: a = 1.3 m/s2 , T = 8.7 N .]
4.1 A bathroom scale measures the normal reaction force
N that it applies on the person standing on it. A boy with
a mass of 50 kg is standing on the scale in an elevator. (a)
How many newtons (N) will the scale show when the eleva-
tor is accelerating upwards at 2 m/s2 ? (b) What will it show Problem 4.5
when the elevator is accelerating downwards with the same 4.5 The blocks with masses m1 =1 kg and m2 =2 kg in the
acceleration? [A: (a) 600 N , (b) 400 N .] figure are tied to each other with a rope passing through a
frictionless pulley. The coefficient of friction between the
blocks and on the ground is µ=0.35 . The lower block is be-
ing pulled with a force F=25 N parallel to the plane. Draw
free-body diagrams for each block and find the acceleration
of the blocks and the tension in the rope.
[A: a = 2.5 m/s2 , T = 6 N .]
Problem 4.2
4.2 In the figure, the forces F1 =18 N and F2 =20 N are ap-
plied on a block with a mass of m=1 kg on a horizontal plane.
The coefficient of friction of the plane is µ = 0.2 . Calculate
the acceleration of the block. [A: 2.5 m/s2 .] Problem 4.6
4.6 The blocks with masses m1 =1 kg and m2 =2 kg in the
figure are placed on two planes inclined at angles 53◦ and
37◦ respectively, and tied to each other with a massless rope
Problem 4.3 passing through a frictionless pulley. The coefficient of fric-
tion of the surfaces is µ=0.1 . (a) Determine the direction of
4.3 A 2 kg block is pressed against a vertical wall with a
motion. (b) Calculate the acceleration of the blocks and the
force F=30 N at an angle of 53 with the horizontal. The co-
◦
efficient of friction between the block and the wall is µ=0.2 .
[A: (a) m2 goes downward, (b) a = 0.6 m/s2 , T = 9.2 N .]
In which direction will the block move and what will be its
acceleration? [A: Upwards a = 0.2 m/s2 .]
Problem 4.7
4.7 The masses of three blocks tied to each other on a friction-
Problem 4.4 less horizontal plane are m1 =1, m2 =2, m3 =3 kg . The mass
4.4 The blocks in the figure with masses m1 =1 kg and m1 is being pulled with the horizontal force F=12 N . Draw
m2 =2 kg have been tied to each other with a rope with negligi- free-body diagrams for each block and calculate the accelera-
ble mass passing through a frictionless pulley. The coefficient tion of the blocks and the tensions in the ropes.
of friction between the horizontal plane and the block m2 [A: a = 2 m/s2 , T 1 = 10, T 2 = 6 N .]
Problem 4.12
4.12 A block with mass m is thrown with an initial speed
Problem 4.8 of v0 =8 m/s upwards along an inclined plane with a slope
4.8 The three masses in the figure are equal and angle of 37◦ from the bottom of the plane. The coefficient of
m1 =m2 =m3 = 1 kg . All the surfaces and pulleys can be as- friction is µ=0.3 . Find the acceleration of the block and the
sumed to be frictionless and the ropes to be massless. Draw distance it travels along the inclined plane until it stops.
free-body diagrams for each block and calculate the accelera- [A: a = 8.4 m/s2 , 3.8 m .]
tion of the blocks and the tensions in the ropes.
[A: a = 1.7 m/s2 , T 1 = 8.3 N and T 2 = 1.7 N .] 4.4 Circular Motion
4.13 A car is driven with constant speed v on a horizontal
circular track with a radius of r=200 m . The coefficient of
friction between the tires and the road is µ=0.8 . What is
the maximum speed with which the car can turn without
slipping? [A: v = 40 m/s .]
Problem 4.9
4.14 Car A with mass m goes around a circular path with ra-
4.9 The block with mass m1 =1 kg in the figure is tied to dius r at speed v . Car B with mass 5m goes around another
the wall with a horizontal rope and placed on another block circular path with radius 3r at speed 2v . What will be the
with mass m2 =3 kg . The coefficient of friction of all of the ratio of the centripetal force F on car B to the centripetal
B
surfaces is µ=0.4 . The block m2 is being pulled away from force F on car A? [A: F B /F A = 6.7 .]
A
the wall with the horizontal force F=30 N . Draw free-body
diagrams for each block and calculate the acceleration of m2
and the tension in the rope. [A: a = 3.3 m/s2 , T = 4 N .]
Problem 4.10 Problem 4.15

4.15 A motorcycle showman with mass m drives through a
4.10 The blocks with masses m1 =1 kg and m2 =4 kg have
vertical circular track with radius r=40 m . What should his
been tied to each other with a pulley and ropes on a horizon-
minimum speed v be at the top point so that the motorcycle
tal plane, as seen in the figure. The mass m2 is being pulled
will not loose contact with the track? [A: 20 m/s .]
away with the horizontal force F=20 N . Draw free-body
diagrams for each block and calculate the accelerations of the
blocks and the tension in the rope tied to m1 .
[A: a1 = 5, a2 = 2.5 m/s2 , T 1 = 5 N .]
Problem 4.16
4.16 The mass m = 2 kg tied to the end of a pendulum with
length 3 m has a speed of 10 m/s at the position where the
rope makes an angle of 60◦ with the vertical. Calculate the
tension in the string. [A: T = 77 N .]
Problem 4.11
4.11 Masses m1 =1 kg and m2 =2 kg are tied to each other
with a rope on an inclined plane with slope angle 53◦ . The
coefficients of friction of the masses are different because
they are made of different materials. The coefficient of fric-
tion between m1 and the inclined plane is µ1 =0.1 and, it is Problem 4.17
µ2 =0.2 for m2 . Calculate the acceleration of the blocks and 4.17 Conical pendulum. A ball with mass m = 1 kg is
the tension in the rope. [A: a = 7 m/s2 , T = 0.4 N .] tied to a string of length 8 m hung from the ceiling. The ball
PROBLEMS 75
is rotating on a circular trajectory on the horizontal plane this chairoplane at an angle of 37◦ with the vertical. Calcu-
where the rope makes an angle of 53◦ with the vertical. late the speed v of the boy and the tension in the chain.
(a) What is the tension in the rope? (b) What is the speed of [A: v = 6.7 m/s , T = 625 N .]
the ball? [A: (a) T = 16.7 N , (b) v = 9.2 m/s .]
Problem 4.19
4.20 One end of a string is tied to a puck of mass m=2 kg on

a frictionless table and the other end passes through a hole
at the center of the table, with another mass M = 3 kg tied
Problem 4.18 to that end (See the figure below). The suspended mass M is
4.18 Two strings with equal lengths L=5 m are tied to a mass able to stay at rest when the mass m on the table is rotating
of m=4 kg and their other ends are fixed to a vertical axis at on a circular trajectory with a radius of 60 cm . What is the
a distance of 8 m from each other. The mass m is rotating speed of the mass m ? [A: 3 m/s .]
with a speed of v=6 m/s on a horizontal plane around the
axis with the strings staying tight. Find the tensions in the
strings. [A: T 1 = 65 N and T 2 = 15 N .]
4.19 In an amusement park, a chairoplane is able to rotate

people at the end of a chain with length 5 m that can freely
swing at the end of a horizontal bar of length 3 m (See the
figure below). A boy with a mass of m = 50 kg is rotating on
Problem 4.20
5
WORK AND ENERGY
In an amusement park the speed

of a roller coaster increases or
decreases as it goes up and
down.
What is it that is stored when
its speed decreases and regained
when its height decreases?
As a principle, all mechanical problems can be solved by using Newton’s

three laws alone. However, this method may often be complicated or difficult to
interpret. On the other hand, using concepts such as work, energy and momentum,
which are closer to our daily intuitions, we can both work in an easier manner
and also make interpretations that are more easily adoptable for technology.
From this chapter onwards, we shall not set forth new laws but merely define
new quantities and seek solutions in terms thereof. The most important feature
of these new quantities is that they can be expressed within general conservation
laws.

78 5. WORK AND ENERGY
5.1 WORK
The concept of ‘work’ borrowed from daily life is the ability of a force to
displace an object. It takes on a precise meaning in physics.
Work Done by a Constant Force
In order to understand the concept of work in an easier manner, let us first
define the work done by a constant force.
Definition: The work done by a constant force F exerted on an object
during displacement d is the scalar quantity
W = Fd cos θ (Work) (5.1)
In this equation, θ is the angle between the force and the displacement. As seen
in the figure, F cos θ is the projection of the force along the path d . Thus, the
work is the projection of the force along the path times the displacement.
Figure 5.1: Work by a force. The unit of work is newton × meters. This derived unit was named as the
‘Joule’ after English scientist James P. Joule, and is abbreviated as J.
Let us emphasize the most important features of the work:
• The work is zero if the object is not displaced ( d = 0 ) despite a force being
applied on it.
This may seem contrary to our daily intuitions. For example, a weightlifter
standing motionless holding 200-kg dumbbells, does zero work according
to our definition. Yet, it would be obvious that the sweating sportsman is
getting exhausted, just like someone who performs work. This should not be
seen as a contradiction, because the concepts of physics are defined within a
precise context.
• The work is zero if the force is perpendicular to the displacement ( cos 90◦ = 0 ).
• If the force forms a wide-angle with the direction of motion, in other words,
if the projection of the force along the path is in the opposite direction to the
displacement, then the work done is negative.
Expression of Work As a Scalar Product
Our definition of work actually fits the form of the scalar product of two
~ and B
vectors that we defined in Chapter 1. Instead of the vectors A ~ used there,
if we take ~F as the force vector and ~d as the displacement vector, then we can
write the expression (5.1) as follows:
W = Fd cos θ = ~F · ~d (5.2)
Example 5.1
Calculate the works done by each of the forces on the object in
the figure for a displacement d=2m in the +x direction.
Answer
We apply the definition of work (Eq.5.1) to each force:
W1 = F1 d cos 30◦ = 15 × 2 × 0.87 = 26 J
W2 = F2 d cos 90◦ = 8 × 2 × 0 = 0
W3 = F3 d cos 180◦ = 3 × 2 × (−1) = −6 J
W4 = F4 d cos(180◦ − 37◦ ) = 6 × 2 × (−0.8) = −9.6 J
5.1. WORK 79
Work Done by a Variable Force

As the concept of work will be used extensively, we should also be able to
calculate the work done by a force whose magnitude varies along the path. The
concept of the integral used in calculus will be useful here. Now, we will show
how to build the integral for the work done by a variable force that is parallel to
the displacement.
Let the object be moving along the x axis from position x = a to position
x = b . Thus, the displacement is d = b − a . Let us assume that the variable force
exerted on the object in this interval varies as a function of position: F = F(x) .
Such a F(x) function is shown in the figure below.
Figure 5.2: The total work done

in the interval [a, b] is the sum
of the small works ∆Wi done
in the small intervals ∆x . If
the distance ∆x decreases grad-
ually ( ∆x → 0 ), the difference
between the rectangles and the
curve vanishes.
Let us divide the path [a, b] into N intervals, each with widths as small as
∆x . Assuming that the force F remains approximately constant in each of these
intervals, such as between xi and xi + ∆x , then the small work performed in this
ith interval is
∆Wi ≈ F(xi ) ∆x i = 1, 2, 3, . . . N
As seen in the figure, ∆Wi is just the area of the thin rectangle drawn in the ith
interval. Here, we commit a small error by assuming that the force has a constant
value of F(xi ) in this interval. However, this error will vanish when the ∆x → 0
limit is taken later.
We may thus write the total work as the sum of the works in these small
intervals:
N
X XN
W= Wi ≈ F(xi ) ∆x
i=1 i=1
If we now use the concept of limit in calculus, in other words, if we gradually

approach the value of ∆x to zero without setting it equal to zero, this sum
expression becomes the definition of the definite integral of the function F(x)
in the interval [a, b] :
N
X Z b
lim F(xi ) ∆x = F(x) dx
∆x→0 a
i=1
As a result, the integral expression of the work done by a variable force F(x) in
the interval [a, b] becomes as follows:
Z b
W= F(x) dx (Work done by a variable force) (5.3)
a
Brief Information on Integrals

The concept of the integral and integrating techniques are examined compre-
hensively in calculus courses. Here, let us briefly review the integrals of the most
frequently encountered functions without proof.
The integral expression that we found for work above has the following
meaning: Let us find such a function Φ(x) such that its derivative is the integrated
function F(x) :
Z
dΦ
Φ(x) = F(x) dx or = F(x)
dx
The expression here without the limits is called the indefinite integral. If this
function Φ is known, the integral can be calculated as follows:
Z b x=b
F(x) dx = Φ(x) = Φ(b) − Φ(a)
a x=a
The result is the value of Φ at the higher limit less its value at the lower limit.
Indefinite integrals of certain functions (c a constant).
function (y) y(x) dx function (y) y(x) dx

R R
1 x+c cos x sin x + c

2x + c − cos x + c
1 2
x sin x
3x + c ex + c
1 3
x2 ex
√
x = x1/2 2 3/2
3x +c 1
x ln x + c
n+1
x
xn n+1 +c ln x x ln x − x + c
You do not need to remember this table if you know derivatives. The integral that
you are looking for should accept the given function F(x) as its derivative.
Example 5.2
Answer We use the integral definition of work (Eq. 5.3):
Z b Z 5 5
Using integrals, calculate the work done by a force varying x4 54 14
W= F(x) dx = x dx =
3
= −
as F(x) = x along the x axis from the position a = 1 to
3
a 1 4 1 4 4
position b = 5 . W = 156 J
Work Done by a Spring Force

We all are familiar with the spiral springs widely used in technology. Steel
springs are used in ball-point pens and clothespins, inside mattresses, the shock
absorbers of cars, etc. When we try to stretch out or compress a spring, it resists
with an opposing force.
Let us fix one end of a spring with a normal length L0 to a wall and then let
us stretch out or compress the other end along the x axis (Figure 5.3). Let the
new length of the spring be L . Let us show the amount of extension of the spring
as
x = L − L0 (5.4)
5.2. POWER 81
The value x is positive for extension and negative for compression.

The expression of the spring force is known as Hooke’s law, in memory of
the English scientist Robert Hooke:
The force exerted by a spring is proportional to the extension and is in the opposite
direction to the extension.
F = −kx (Hooke’s law) (5.5)
The coefficient k in this formula is referred to as the spring constant or the

force constant. Its unit is newton/meters and depends on the type of material
used in the spring.
Let us emphasize the important aspects of the Hooke’s law:
• The spring force is proportional not to the length of the spring, but to
x=L−L0 , the amount of extension or compression. Figure 5.3: Spring force x is in
• The negative sign in the formula indicates that the force is in the opposite opposite direction to its exten-
direction to the extension. If the spring is extended ( x>0 ), then the force sion.
is in the opposite direction, in other words, F < 0 . In the reverse case, if
the spring is compressed ( x < 0 ), then the force is in the positive direction:
F > 0.
• The flexibility of the spring has a certain limit. When extended for more than
a certain x value, the spring gets deformed and loses its flexibility.
We wish to calculate the work done by the spring force when we extend a
spring from its normal length ( x = 0 ) until it reaches the value x = d . The
spring would, of course, not extend by itself; we will have to apply a force of at
least ( −F ). But we are only interested in the work performed by the spring force.
Therefore, if we take the boundaries of the integral as a = 0 and b = d , we may
write the work performed by the spring force as follows:
Z d Z d Z d d
Ws = F dx = (−kx) dx = −k x dx = −k 12 x2
0 0 0 0
Ws = − 12 kd2
As expected, the work performed by the spring force is negative, because it is

in the opposite direction to the extension. We need to perform a positive work
against this in order to extend the spring.
We will use this result later in the topic of potential energy.
Example 5.3
Hooke’s law gives us (the negative sign is not important):
A spring is observed to extend by 10 cm when a force of 30 N F = kx → 30 = k × 0.10
is applied. How much work should be performed to extend this k = 300 N/m
spring by 40 cm ? Using this value of constant k , we can calculate the work done
against the spring force to extend the spring by x=0.40 m :
Answer
We first calculate the spring constant k using the given data. W = −Ws = 12 kx2 = 12 × 300 × 0.42 = 24 J
5.2 POWER
Anyone can carry a load for a distance of 100 meters, but some of us can do it
faster. In daily life, and also in technology, it is important to know the amount of
work performed per unit time. We describe this with the concept of power.
Definition: If the amount of work performed in a time interval ∆t is ∆W ,

then the average power is defined as,
∆W
Pave = (average power) (5.6)
∆t
The instantaneous power at a given instant t is defined as,
∆W dW
P = lim = (instantaneous power) (5.7)
∆t→0 ∆t dt
which is just the derivative of work with respect to time t .
Units of Power
In the SI system, the unit of power is joule/seconds = watt and is abbreviated
as (W). The use of kilowatt ( 1kW = 1000W ) is more common in the industry.
The watt is one of the rare units used both in science and in technology.
If we rewrite the definition above as dW = P dt , we see that the unit of work
(joule) can also be expressed as watt × hours or kilowatt-hours (kWh). When we
talk about our electricity consumption as being in “kilowatts,” we actually mean
kilowatt-hours, as it refers to the energy that we consume.
Figure 5.4: Horsepower was Horsepower is another unit of power used in the automotive industry and
used in the past as a measure is abbreviated as (HP):
of the amount of coal extracted 1 HP=746 watts =0.746 kW
from a coal mine. The horsepower unit’s first historical use was in extracting coal from the mines
of Britain. Back in those days, when Scottish scientist James Watt (1736–1819)
invented the steam engine, he proposed this unit in order to compare the power
of the steam engine with the power of horses.
There is another useful formula for expressing power in terms of force and
velocity in mechanics. In the definition above, let us write the work ∆W as the
work performed by a force ~F over a small displacement of ∆~r . Then,
~F · ∆~r ∆~r
P = lim = ~F · lim
∆t→0 ∆t ∆t→0 ∆t
The limit in the brackets is just the velocity ~v . Then, we can write the power
generated by the force ~F at time t as follows:
P = ~F · ~v (Power) (5.8)
Example 5.4
speed, it is exerting a force F = mg equal and opposite to
the weight mg . We use Eq. (5.8) that we found for power:
A crane is lifting a mass of 2 tons of coal from under the ground
P = F v = mgv = 2000 × 10 × (12000/3600)
to the surface at a speed of 12 km/hour . Calculate the power
= 67000 watt = 67 kW
of the crane engine in units of watts and horsepower.
and use the conversion formula 1 HP = 0.746 kW :
Answer Since the crane is pulling the mass m at constant P = 67/0.746 = 90 HP
Example 5.5
P ∆t :
An air-conditioner is operating at a power of 3 kW . How much ∆W = 3 × (30 days × 24 hours) = 2160 kWh.
energy does it consume in one month? What is the cost for one The monthly cost is found by multiplying this by the energy
month if the price is 20 cents per kilowatt-hours? unit price:
2160 × 0.2 = 432 $ .
Answer The consumed energy can be calculated as ∆W =
5.3. KINETIC ENERGY 83
5.3 KINETIC ENERGY

What really happens when a force performs some work? For example, when
we apply a force to an object, it may gain speed or, be lifted to some height or,
generate electricity, etc. These are different aspects of the energy concept. As the
most important concept defined in physics, energy may roughly be regarded as
“the ability to do work.” Working with the concept of energy both provides conve-
nience and may also be used outside of mechanics, in fields such as electricity,
chemistry and biology.
There are various types of energy in physics, such as kinetic energy, potential Figure 5.5: Where does the
energy, electrostatic energy, magnetic energy, chemical energy, etc. Only kinetic work performed by these racers
and potential energy are used in mechanics. in pushing the sled go?
Definition: The kinetic energy of an object with mass m traveling at speed
v is,
K = 12 mv2 (kinetic energy) (5.9)
Kinetic energy is always positive. An object at rest has zero kinetic energy. Notice
that kinetic energy is also expressed in terms of the work unit:
kg × m
kg × (m/s)2 = × m = newton × m = joule
s2
Work-Energy Theorem
To show the relation between work and kinetic energy, let us calculate the
work performed by the net force exerted on an object with mass m . However,
rather than the most general cases of three-dimensional and variable force, we
will consider the work performed by a one-dimensional constant force.
An object with mass m has a speed of v0 at the location x0 , and, under the
influence of a constant force Fnet , reaches the speed v when it arrives at the
position x (Figure 5.6). Let us write the net work performed by the constant force:
Wnet = Fnet d = Fnet (x − x0 ) Figure 5.6: Work-energy theo-

rem.
We write the force as the second law Fnet = ma and use the formula 2.7 relating
the position to speed [ v2 − v20 = 2a(x − x0 ) ], finding that,
Wnet = ma (x − x0 ) = m a(x − x0 )
| {z }
(v2 − v20 )/2
Wnet = 1
2 mv
2
− 21 mv20 = K − K0 (5.10)
This result is known as the Work-Energy Theorem:

The net work done on an object is equal to the change in the kinetic
energy of the object.
This is a very general result and is always true, even in the cases of three-
dimensional motion or with variable forces:
Z 2
Wnet = ~Fnet ·d~r = 1 mv2 − 1 mv2 (Work-energy theorem) (5.11)
2 2 2 1
1
Here, d~r refers to a small displacement vector with components (dx, dy, dz) .
The work-energy theorem is actually another way of expressing Newton’s

second law, in terms of work and kinetic energy. The theorem shows clearly
that, if the work done by the net force is positive, then the object’s kinetic
energy increases. If the net work is negative, then the kinetic energy decreases.
Another advantage of this theorem is the convenience of working with only scalar
quantities, instead of the vector law ~F = m~a . Some types of problems can be
solved very quickly using this theorem.
Example 5.6
by first calculating acceleration. But it is easier to use the
When a net force F is applied on an object with mass 3 kg work-energy theorem: the work performed by the unknown
traveling at a speed of 5 m/s , its speed is increased to 10 m/s force F produces the change in kinetic energy:
at the end of a distance of 9 m . Fd = 12 mv2 − 12 mv20
Calculate the magnitude of the force. m(v2 − v20 ) 3 × (102 − 52 )
F= = = 12.5 N .
Answer We could have solved this problem the long way, 2d 2×9
Example 5.7
other. Hence, only the friction force remains, and it is oppo-
An object with mass m is thrown along a horizontal surface at site to the motion. The work it performs is negative, and it
a speed of 6 m/s . The coefficient of friction is µ = 0.4 How far makes the final kinetic energy zero ( v = 0 ):
will it travel before coming to rest?
Wnet = 12 mv2 − 12 mv20
Answer − f d = −(µmg)d = 21 mv2 − 21 mv20
Among the forces exerted on the object, the weight and the
normal force are in the vertical direction and balance each d = v20 /(2µg) = 62 /(2 × 0.4 × 10) = 4.5 m .
5.4 POTENTIAL ENERGY

Conservative and Nonconservative Forces
Forces in mechanics are divided into two groups in terms of one specific
feature. The distinction of conservative and nonconservative forces is crucial in
defining potential energy.
To understand this, let us first consider throwing a stone upwards from the
ground with a speed of v (Figure 5.7). Neglecting air friction, the stone is under
the influence of gravity alone when in the air. The kinetic energy of the stone
on the ground is K1 = 12 mv2 . When the stone rises to reach its maximum height,
its speed is zero for a short instant: K2 = 0 . Now, can we say that the kinetic
energy of the object has disappeared? No, because, immediately afterwards, the
Figure 5.7: Kinetic energy may stone resumes its downward motion and recuperates its former kinetic energy
be reduced with the effect of when it reaches the ground. In this example, we say that the gravitation force is
gravity but it does not disappear. conservative, because it stores the kinetic energy somehow and later returns it.
It can be regained. On the other hand, consider a mass m thrown with a speed of v along a
horizontal surface with friction. The net force exerted on this object along the
path is the frictional force. The kinetic energy of the object at the start is again
K = 12 mv2 . The object gradually slows down and its kinetic energy vanishes when
it comes to a stop. Now, we can wait for ages, but we will never see this object
re-accelerate. In this example, we say that the frictional force is nonconservative,
because it irreversibly takes the kinetic energy of the object and never returns it.
(Actually, it is converted into another type of energy: heat.)
5.4. POTENTIAL ENERGY 85
Potential energy is that type of energy that can store and return the work.
According to the above example, potential energy can be defined only for conser-
vative forces.
The following criterion determines whether or not a force is conservative:
A force is conservative if the work done is independent of the path taken between
any two points. Figure 5.8: The work per-
Let us test this criterion in the example of gravitational force. We want to formed by a conservative force
is independent of the traveled
calculate the work performed by gravity on different paths taken by the mass m in
path.
Figure 5.9 from A to B. Since the work is zero when the force is is perpendicular
to the path, there follows:
path ACB: WACB = WAC + WCB = −mg × AC + 0 = −mgh

path ADB: WADB = WAD + WDB = 0 − mg × DB = −mgh
diagonal AB: WAB = (−mg sin θ) × AB = −mg × (AB sin θ) = −mgh
The work along all three paths is the same. This shows that gravity is a conserva-
tive force. If we had made the same calculation for friction force, we would have Figure 5.9: Work done by grav-
found that the work varied according to the traveled path . ity along three different paths.
General Definition of Potential Energy
The work done against a conservative force is equal to the change in
potential energy.
If the conservative force is ~Fc then the force doing work against it will be
~
(−Fc ) . If we denote potential energy as U , then we may write this definition as
follows:
Z 2
− ~Fc · d~r = U2 − U1 (potential energy definition) (5.12)
1
It seems reasonable here to define the increase in potential energy as negative

work. Since the negative work decreases kinetic energy, the potential energy
should be increasing.
Now let us define gravitational and elastic potential energies in accordance
with this general definition.
Gravitational Potential Energy
Consider an object near the Earth’s surface, with mass m traveling from point
A at position (x1 , y1 ) to a point B with position (x2 , y2 ) along some curvilinear
path (Figure 5.10). Let us denote the weight of the object with the vector m~g
and a small displacement along the path with d~r . The angle between d~r and the
y -axis is θ . We now write the work done against the conservative force, in other
words, the work done by the vector (−m~g) , as a scalar product:
Z B Z B
−W = (−m~g) · d~r = (−mg) dr cos(180◦ − θ)
A A
As can be observed in the figure, dr cos(180◦ − θ) = −dr cos θ = −dy . Accord-

ingly, Figure 5.10: Coordinates in a
Z B Z y2
gravitational potential energy.
−W = mg dy = mg dy = mg(y2 − y1 )
A y1
The result shows that the work done against the gravitational force is independent
of the path taken and depends only on the heights ( y -coordinates) of the two
points. Now, we can define the gravitational potential energy using the general
definition of Eq. (5.12):
U2 − U1 = (Work done against gravity) = −W = mg(y2 − y1 )
Comparing both sides of this equation, the potential energy at any place with
coordinate y can be written as,
U(y) = mgy + C
The constant C in this equation is determined after some zero-reference point of

the potential energy is chosen. For example, if we want U = 0 at the level y = 0 ,
then C = 0 . The potential energy takes negative values below this level. As a
result, we may write gravitational potential energy as follows:
U(y) = mgy (gravitational potential energy) (5.13)
The choice of the zero level of the gravitational potential energy is arbitrary and
can be taken at any height. This has no importance, because, as we shall see
later, only the potential energy difference between two heights will appear in
equations. This difference does not change, regardless of where the zero level is
chosen.
Elastic Potential Energy
Next, we find the elastic potential energy due to spring force. As seen in
Figure 5.11, the spring force corresponding to an extension of x along the x -axis
is F = −kx and in the opposite direction. And the force (−~F) performing work
against this force becomes +kx , hence it is in the same direction as x . Thus, the
work done during the extension from the value x1 to the final value x2 is:
Z x2 Z x2 Z x2 x2
−W = (−F) dx = (+kx) dx = k x dx = k 12 x2 = 12 kx22 − 12 kx12
x1 x1 x1 x1
Using the general definition of potential energy (Eq. 5.12), we find that
Figure 5.11: Coordinates in
elastic potential energy. U2 − U1 = 1 2
2 kx2 − 12 kx12
U(x) = 1 2
2 kx +C
The constant C is again determined by arbitrarily choosing the place where the
potential energy is zero. If we take it to be zero at the normal length of the spring,
in other words, at x = 0 , then C = 0 . As a result, we can write the elastic
potential energy as follows:
U(x) = 12 kx2 (elastic potential energy) (5.14)

5.5. LAW OF CONSERVATION OF ENERGY 87
Gravitational Potential Energy (General Case)

If the height continuously increases on the surface of the Earth, then the
gravitational acceleration g ceases to be a constant and the expression U = mgy
becomes invalid. Instead, we should return to Newton’s gravitational law, which
we had introduced in Chapter 4, and calculate the gravitational potential energy
therefrom.
Let us rewrite the gravitational force between the Earth (with mass ME ) and
an object of mass m at a distance of r from the center:
mME
F =G (5.15)
r2
We again calculate the potential energy as the work performed against the gravi-
tational force. For this purpose, we need to apply the force −~F against this force
~F that is exerted when the mass m travels from a distance of r1 to a distance of
r2 (Figure 5.12). Accordingly,
Z r2 Z r2 Z r2 Figure 5.12: Coordinates in a
−W = ~
(−F) · d~r = − ~
F · d~r = − F dr cos 180◦ gravitational potential energy.
r1 r1 r1
It can be seen from the figure that the angle between the attractive force ~F and
the displacement vector d~r is 180◦ . Since cos 180◦ = −1 , and if the expression
for F is substituted, then
Z r2 1 r2
dr
−W = GmME 2
= GmM E −
r1 r r r1
1 1
= −GmME −
r2 r1
Again, if we write that the change in the potential energy is equal to this opposing
work, then 1 1
−W = U2 − U1 = −GmME −
r2 r1
From here, the potential energy at any distance r can be written as:
GmME
U(r) = − +C
r
How do we choose the constant C ? It is impossible to choose the origin ( r = 0 ),
as the expression diverges ( U → ∞ ). Instead, most naturally, U = 0 is chosen at
the point where objects are too far apart to interact, in other words, when r → ∞ .
Then, C = 0 and the gravitational potential energy is obtained:
GmME
U(r) = − (gravitational potential energy) (5.16)
r
It is seen from this formula that potential energy is zero at infinity.
5.5 LAW OF CONSERVATION OF ENERGY

Combining the concepts of work, kinetic and potential energies, we can now
formulate the most important conservation law in physics.
According to the Work-Energy theorem, the work performed by a net force

acting on an object is equal to the increase in kinetic energy (Equation 5.11):
Z 2
~Fnet · d~r = 1 mv2 − 1 mv2 = K2 − K1
2 2 2 1
1
We have also learned that we could separate the forces exerted on an object into
two groups as conservative and nonconservative. Then, we can divide the work
they perform into the following two terms:
~Fnet = ~Fc + ~Fnc
Z 2
(~Fc + ~Fnc ) · d~r = K2 − K1
1
Now let us recall the definition of potential energy: The work performed against
conservative forces is equal to the increase in potential energy:
Z 2 Z 2
~Fc · d~r + ~Fnc · d~r = K2 − K1
1
| {z } | {z } 1
−(U2 − U1 ) Wnc
In this expression, the work performed by nonconservative forces is denoted as
Wnc . Rearranging the terms on both sides, we arrive at the Law of Conservation
of Energy:
(K1 + U1 ) + Wnc = K2 + U2 (Law of Conservation of Energy) (5.17)
Here, K= 21 mv2 is the kinetic energy and U can be any of the gravitational, elastic
or other potential energies.
The law of conservation of energy states that, after we subtract the work of
nonconservative forces from the total initial energy (kinetic+potential), whatever
remains will be equal to the total final energy (kinetic+potential).
Here, we only took into consideration the conservation of mechanical energy.
Actually, the conservation of energy is still valid when all other types of energy
(electrical, magnetic, nuclear, etc.) are taken into consideration; it is a universal
law of physics.
Special case: If there are no nonconservative forces exerted on an object or
if their work is somehow zero, then,
Wnc = 0 −→ K1 + U1 = K2 + U2 (Special Energy Conservation) (5.18)
In such a case, no mechanical energy is lost and the sum of the energies (ki-
netic+potential) remains constant throughout the motion.
The sum of kinetic and potential energies is called total mechanical energy
and is denoted with E :
E = K + U = 21 mv2 + U (total mechanical energy) (5.19)

Figure 5.13: When a roller
coaster rushes downward, its po- The law of conservation of energy shows us that total mechanical energy is
tential energy decreases to be conserved even if the kinetic or potential energies vary. In the equation (5.17), if
converted into kinetic energy.
we state that K1 + U1 = E1 and K2 + U2 = E2 , then
5.5. LAW OF CONSERVATION OF ENERGY 89
General : E1 + Wnc = E2
If Wnc = 0 : E1 = E2
The terms in the law of conservation of energy are only scalar quantities.
Considering the difficulty of working with vector equations such as Newton’s
second law ~F = m~a , energy conservation formulas provide great convenience.
Example 5.8
A block with a mass of 4 kg travels at a speed of 3 m/s on Answer

a frictionless horizontal plane and hits a spring with one end Since no friction occurs, the initial and final total energies
fixed to a wall. The spring constant is k = 100 N/m . How are the same:
much does the block compress the spring? E1 = E2
1
2 mv 2
+ 0 = 0 + 21 kx2
r r
m 4
x= v= 3 = 0.60 m
k 100
Example 5.9
The ABC section of the path in the figure is frictionless, while Answer
the coefficient of friction is µ = 0.5 beyond C. A ball with a (a) Total mechanical energy is conserved at points A, B and
mass of m = 2 kg is thrown along the path with a speed of C, as there is no friction:
2 mvA + mgyA = 2 mv B + mgy B = 2 mvC + mgyC
1 2 1 2 1 2
3 m/s from a point A at a height of 2 m above the ground.
(a) Find the speeds of the ball at points B and C. From these equalities, we find the speeds vB and vC using
(b) How far will it travel after point C before coming to a stop? the valuesqyA = 2, yB = 0, yC = 1.2 m :
√
vB = v2A + 2gyA = 32 + 2 × 10 × 2 = 7 m/s
q √
vC = v2B − 2gyC = 72 − 2 × 10 × 1.2 = 5 m/s
(b) The kinetic energy of the object decreases from the point
C onward, due to the work done by friction:
2 mvC − f d = 2 mvC − (µmg) d = 0
1 2 1 2
d = vC /(2µg) = 5 /(2 × 0.5 × 10) = 2.5 m .

2 2
Example 5.10
Here, W f = − f d is the work performed by the friction force
and is f = µmg cos 37◦ . Thus
2 mvA + mgyA − (µmg cos 37 ) d = 2 mv B + mgy B
1 2 ◦ 1 2
If we choose the zero level of potential energy at point A,

then yA = 0 and the height of point B at distance d becomes
yB = d sin 37◦ . The speed vB is found when the other values
are also substituted:
102 − 16 = v2B + 24 → vB = 7.7 m/s
(b) The block stops when it comes to some point C where
A spring with constant k = 100 N/m is fixed at the top of a the compression of the spring is at a maximum: vC = 0 .
plane inclined at an angle of 37◦ . The coefficient of friction of We write the conservation of energy between A and C. This
the plane is µ = 0.5 . A block with a mass of 1 kg is thrown time, we have the elastic potential energy in addition to the
with an initial speed of 10 m/s along the plane from the point gravitational potential energy. If we denote the maximum
A located at the lower end and at a distance of 2 m from the compression of the spring as x , then
free end of the spring.
2 mvA − µmg(d + x) = 0 + mgyC + 2 kx
1 2 1 2
(a) What is the speed of the block when it reaches point B? The height of point C is yC = (d + x) sin 37◦ and the numer-
(b) How much does the block compress the spring? ical values are substituted:
Answer (a) We write the general energy conservation be- 100 − 20(2 + x) = 100x2 → 5x2 + x − 3 = 0
tween points A and B: The positive root of this 2nd degree equation is the result:
EA + W f = EB x = 0.68 m
Example 5.11 s !
1 1
v2 = v21 + GME−
The speed of a meteor approaching the Earth is measured as r2 r1
100 m/s at a distance of 900 km from the sea level. At what (The product GME will appear very often in problems with
speed does this meteor crash on the Earth’s surface? gravitational potential energy.) Here, the distances r1 and
(For the Earth, GME =4 × 10 m /kg·s and RE =6400 km .) r2 must be measured from the center of the Earth, while the
14 3 2
data is given as measured from the surface of the Earth. We
Answer calculate them first:
The energy conservation is written as: r1 = 6400 + 900 = 7300 km = 7.3 × 106 m
GME m 1 2 GME m r2 = 6400km = 6.4 × 106 m
1
2 mv 2
1 − = 2 mv2 − From these values, we find the crash speed of the meteor:
R1 r2
We simplify as follows: v2 = 2780 m/s
Example 5.12
fall back to Earth? When the object is slowing down, if its
speed becomes zero at some finite distance, it will fall back.
Therefore, it should reach infinity when its speed becomes
zero: v∞ = 0 . Accordingly, we write the energy conservation
for an object with mass m between the Earth’s surface and
infinity:
GME m 1 GME m
Escape speed from the Earth. A rocket or stone thrown from 1 2
2 mvesc − = 2 m × 02 −
RE ∞
the surface of the Earth must have a minimum speed to move As 1/∞ = 0r, the escape speed formula is,
infinitely far away and to not fall back to the Earth. This
2GME
is called the escape speed. Calculate the value of this speed. vesc =
(GME = 4 × 1014 m3 /kg·s2 and RE = 6400 km .) RE
Substituting the numerical values, we get:
Answer What should the condition be for the object to not vesc = 11.2 × 103 m/s = 11.2 km/s
Example 5.13
Answer In example 4.19, we saw that the tension T at point
B should be at a minimum ( T =0 ) so that the rope would not
get loose. Accordingly, the centripetal acceleration at point
B is only due to the weight mg . We thus find the minimum
speed vB :
v2
Fr = mar → mg = m B
L
v2B = gL
Now, we write the energy conservation between points A
and B. As the height of point B is 2L ,
An object with mass m is tied to the end of a rope with length
2 mvA + 0 = 2 mv B + mg(2L)
1 2 1 2
L = 50 cm and is thrown with a horizontal speed vA from the
point A at the bottom. With what minimum speed vA should Using the expression v2B that we found above, the speed vA
it be thrown to ensure that the rope does not get loose when is found as follows:
v2A = gL + 4gL = 5mgL vA = 5gL = 5 m/s
p
passing through the point B at the top?
Example 5.14
Blocks m1 =1 kg and m2 =2 kg are tied to each other with a

weightless rope passing through a frictionless pulley. The mass
m1 is tied to a spring with k=100 N/m on a surface with a co-
efficient of friction µ=0.4 and the mass m2 hangs freely. Both
blocks are released when the spring is at normal length. Calcu-
late the maximum distance h that the mass m2 will descend.
Answer We could have solved this problem the long way

again by formally writing the conservation of energy formula. m2 gh = 12 kh2 + µm1 gh

However, a quick thinking simplifies the problem: Since the (When the mass m2 descends a distance of h , the spring also
blocks are at rest both at the beginning and at the end, the extends for h and the mass m1 travels a distance of h .)
elastic potential energy gained by the spring and the nega- The height h is calculated from there:
tive work done by the friction are solely due to the potential 2(m2 − µm1 )g 2(2 − 0.4 × 1) × 10
energy loss m2 gh of the mass m2 . Hence, we write: h= = = 0.32 m
k 100
Example 5.15
tential energy is 12 (k1 + k2 )x2 . Hence, we may consider this
system as a single spring with a spring constant of (k1 + k2 ) .
If the block is released from rest with springs extended by x0
at the start, then the speed v at a later extension x is found
from the conservation of energy:
A block with mass m = 1 kg is connected to two springs with
2 (k1 + k2 )x0 = 2 mv + 2 (k1 + k2 ) x
1 2 1 2 1 2
spring constants k1 = 30 N/m and k2 = 70 N/m on a fric- s
tionless plane. The springs are initially at normal length when (k1 + k2 )(x02 − x2 )
the block is initially at rest. The block is pulled a distance v=
m
of x0 = 50 cm to the right from its equilibrium position and Calculating for x = 0.4 m , we find that v = 3 m/s .
released. (b) For the speed to be maximum, according to the expression
(a) What is the speed of the block when it is at a distance of above, the factor (x02 − x2 ) in the square root must be maxi-
40 cm from the equilibrium position? mum. As x0 was given at the beginning, x must be minimum,
(b) What is the maximum speed of the block and at what in other words, x = 0 . Thus, the block reaches maximum
point is it reached? speed when passing through the equilibrium position. The
Answer maximum speed is found when we set x = 0 in this formula:
(a) When one of the springs is extended by x , the other is
r
k1 + k2
compressed by the same amount, and the total elastic po- vmax = x0 = 5 m/s
m
1. What is the kinetic energy of an object with mass 2 kg (a) Its kinetic energy remains constant.
and speed 2 m/s ? (b) Its potential energy remains constant.
(a) 2 J (b) 4 J (c) 6 J (d) 8 J (c) Its kinetic energy increases, while its potential en-
ergy decreases.
(d) Its kinetic energy decreases, while its potential en-
ergy increases.
2. Which is incorrect?
(a) Force perpendicular to the displacement performs 5. Which of the following is correct for a simple pendulum?
no work.
(b) Force opposite to the displacement performs nega- (a) Potential is minimum at the highest point.
tive work. (b) Kinetic energy is maximum at the highest point.
(c) Work is the product of the parallel component of (c) Potential energy is minimum at the lowest point.
the force to the displacement with the displacement. (d) Total energy is minimum at the lowest point.
(d) Work is the product of the perpendicular component
of the force with the displacement. 6. What is the power of a crane if it is able to pull a mass
of 100 kg upwards at a speed of 2 m/s ?
(a) 1 kW (b) 2 kW (c) 3 kW (d) 4 kW
3. What is the speed of an object with mass 1 kg and ki-
netic energy 8 J ? 7. Which of the following performs the most work?
(a) 3 m/s (b) 4 m/s (c) 5 m/s (d) 6 m/s (a) A man carrying a stone with a force of 10 N for 2 m.
(b) A man carrying a stone with a force of 5 N for 3 m.
(c) A weight-lifter keeping dumbbells of 200 kg fixed.
(d) A kid who raises a mass of 20 kg to a height of 1 m.
4. If the work performed by the net force on an object is
zero, then which of the following is correct?
8. Which of the following is incorrect? 15. Which of the following is correct for the gravitational
(a) The work performed by a conservative force is in- potential energy?
dependent of the path. (a) It is inversely proportional to the square of the dis-
(b) The work performed by a nonconservative force is tance.
dependent on the path. (b) It is proportional to the total mass of the two objects.
(c) Potential energy can be defined for a conservative (c) It is inversely proportional to the distance.
force. (d) It is inversely proportional to the mass.
(d) Potential energy can be defined for a nonconserva-
tive force. 16. The rotation radius is doubled for an artificial satellite
rotating on an orbit around the Earth. How does its
9. An elevator is being pulled upwards at a constant speed.
potential energy change?
The work performed on the elevator is:
(a) Zero. (a) It halves.
(b) Positive. (b) It doubles.
(c) Negative. (c) It decreases to one fourth.
(d) Constant. (d) It increases by 4 times.
10. If the work done by a net force on an object at rest is 17. An car applies the breaks to stop on a horizontal road.
doubled, its final speed increases by how many times? Where is the kinetic energy that it lost spent?
√
(a) 1 (b) 2 (c) 2 (d) 4 (a) In the work performed by the friction force.
(b) To increase its potential energy.
11. How much should the net force acting on an object at (c) To the heating of the engine.
rest be increased such that its final speed increases 4 (d) All of the above.
times?
(a) 2 (b) 4 (c) 8 (d) 16 18. Which of the following is correct for an object sliding
along a path?
12. If an amount of work 5 J is needed to extend the length
of a spring by 1 cm , then how much work should be (a) The friction force does not perform any work.
done to extend it by 2 cm ? (b) The normal force performs positive work.
(a) 6 J (b) 10 J (c) 15 J (d) 20 J (c) The normal force performs negative work.
(d) The friction force performs negative work.
13. Which of the following is correct for the gravitational
potential energy of the Earth? 19. Which of the following is incorrect?
(a) It is zero at the center of the earth. (a) Friction force is conservative.
(b) It is zero at infinity. (b) Gravitational force is conservative.
(c) It is zero on the surface of the earth. (c) Spring force is conservative.
(d) It is zero at the center of the Sun. (d) Friction force is nonconservative.
14. Which of the following is correct for kinetic energy?

20. Two satellites are rotating in orbits around the Earth at
(a) It depends on the direction of motion of the object. radii r1 < r2 . Which of the following is correct for the
(b) It may be negative. gravitational potential energy?
(c) It is always positive.
(d) It is the same for all observers. (a) U1 < U2 (b) U1 = U2 (c) U1 > U2 (d) None.
Problems
5.1–3 Work, Kinetic Energy, Power
5.1 Calculate the work performed by the forces in the figure

on the right, along the 5 m horizontal displacement.
[A: 43 J, −18 J, 0 .] Problem 5.1
PROBLEMS 93
5.2 A block with weight W=10 N is pulled upward on a

37◦ inclined plane with a force F=9 N that is parallel to the
inclined plane at a constant speed over a distance of 2 m . (a)
What is the work performed by the force F ? (b) The work
performed by the gravitational force? (c) The work performed
by the friction force? [A: (a) 18 J , (b) −12 J , (c) −6 J .] Problem 5.11
5.11 Calculate the gravitational potential energy of the mass
m = 3 kg at the points 1, 2, 3 and 4 shown in the figure. (In-
dicate where you choose your zero level.)
[A: zero level on the ground: 480, 0, 360, 210 J .]
5.12 A telecommunication satellite with a mass of 3 tons was

placed into orbit at a distance of 36 000 km from the surface
Problem 5.3 of the Earth. How much energy is required to lift this satellite
5.3 Find the work performed by the variable force F(x) in to such a height? (Use the values GME =4 × 10 m /kg·s
14 3 2
[A: 4.5 J .] and RE =6400 km .) [A: 1.6 × 10 J .]

11
the figure.
5.5 Law of Conservation of Energy
5.4 A force exerted on an object is given as:
F(x) = 6x2 − 2x (newton)
Calculate the work performed by this force within the range
x : [0.5 m] . [A: 225 J .)
5.5 5 J work is performed to extend a spring by 10 cm . How

much work should be done to further extend it by 30 cm ?
[A: 80 J .]
Problem 5.13
5.6 An object at rest with mass 2 kg starts moving and 5.13 A mass of m1 =1 kg is at rest on the bottom of a fric-
reaches a speed of 8 m/s over a distance of 5 m . What is tionless plane inclined at 37◦ . It is tied to another mass of
the constant net force acting on this object and how much m2 =2 kg with a rope that passes through a frictionless pulley.
work has it performed? [A: 12.8 N and 64 J .] At the start, the mass m2 is released from a height of 3 m
from the ground. What is the speed of mass m2 when it
5.7 A cannon has a barrel-length of 6 m . When a shell of reaches the ground? [A: 5.3 m/s .]
mass 20 kg is fired, it leaves the barrel at a speed of 200 m/s .
(a) What is the kinetic energy of the shell? (b) What is the
the work done on the shell by the average force inside of
the barrel? (c) What is the average force exerted on the shell
inside of the barrel? [A: (a) 400 kJ , (b) 400 kJ , (c) 67 kN .]
5.8 An athlete with a mass of 60 kg climbs up a hanging rope Problem 5.14

of length 10 m in 20 s . Calculate the power of the athlete in 5.14 A block with mass m is thrown at a speed of 5 m/s up
units of watts and horsepower. [A: 300 W, 0.4 HP .] from the lower end of a plane inclined at an angle of 30◦ and
with a coefficient of friction µ=0.5 . How much distance does
5.9 The engine of a car is able to drive it at a constant speed it travel along the inclined plane until it stops? (Note: The
of 72 km/hour while delivering a power of 100 HP . As the mass m will be canceled at the end of the calculation.)
traction force of the car is 1400 N , what percent of the power [A: 1.34 m .]
of the engine is lost to friction and air resistance?
[A: 62 % .]
5.4 Potential Energy

5.10 The normal length of a spring attached to the ceiling is
1 m . The length of this spring becomes 105 cm when a mass
of 200 g is hung to its free end. (a) Calculate the spring con- Problem 5.15
stant. (b) How much will the spring extend when its potential 5.15 A spring with a constant of k=200 N/m is fixed after
energy is 5 J ? [A: (a) 40 N/m , (b) 50 cm .] point C along the frictionless path ABC in the figure. A ball
with mass 2 kg is released from rest at the point A. (a) What

will be the speed of the ball at points B and C? (b) How much
will the ball compress the spring?
[A: (a) vB = 6, vC = 5 m/s , (b) x = 0.5 m .]
Problem 5.18
5.18 A ball is compressed against a spring fixed on a fric-
tionless table and released. We would like the ball to fly off
of the edge of the table and fall inside a bowl on the floor
at a distance of 2 m . In a first trial in which the spring is
compressed by 20 cm , the ball falls 40 cm short of the bowl.
How much should the spring be compressed so that the ball
Problem 5.16 falls into the bowl in the second trial? (Hint: There is no need
to completely solve the horizontal projectile problem. The
ball’s time of flight will be the same in both trials.)
5.16 A spring with a constant of k=200 N/m is fixed to the [A: 25 cm .]
lower end of a plane inclined at an angle of 37◦ and a coef-
ficient of friction of µ=0.4 . A block with mass m = 1 kg is
thrown from point A downwards at a speed of 5 m/s . The
free end B of the spring is at a distance of AB=2 m . (a) At
what speed will the block hit the free end of the spring? (b)
What is the maximum compression of the spring, in other Problem 5.19
words, what is BC = x ? [A: (a) 6 m/s , (b) 0.44 m .]
5.19 A spring gun with a spring constant k=500 N/m is able
to shoot bullets of 50 g . It is observed that a bullet fired by
this gun at an angle of 37◦ to the horizontal reaches a height
of 180 cm . (a) Find the initial speed of the bullet, (b) Find the
amount of compression of the spring.
[A: (a) 10 m/s , (b) 0.1 m ]
Problem 5.17
5.17 A portion of a frictionless rail track is bent as a circular Problem 5.20

loop with radius R . A block with mass m is released from 5.20 A ball released at the peak point of a frictionless spheri-
rest at a height h from one end of the rail. From what height cal surface with a radius of R starts to slide down the surface.
h should the block be released such that it can complete the At what angle θ will it leave the surface? (Hint: The normal
loop without leaving the rails? (Hint: As long as the block reaction force is zero where it leaves the surface. The cen-
touches the rails, there is some normal force acting on it. tripetal force needed during the circular motion is simply the
What should be its minimum value at the peak point?) radial component of the weight.)
[A: h = 5R/2 .] [A: cos θ = 2/3, θ = 48◦ .]
6
IMPULSE AND
MOMENTUM
The forces generated during the

collision of billiard balls have a
very complex structure and are
difficult to examine.
However, there is one quantity
that always remains the same
before and after the collision.
We will find out what this is in
this chapter. (Photo: Dave Jack-
son)
Collision problems are one of the cases in which direct application of Newton’s
laws is difficult. When two billiard balls collide, action-reaction forces arise
between them during that very short contact. The details of these forces are
complicated and very difficult to examine. However, these two balls have some
well-defined velocities before and after the collision. Is it possible to find a relation
between the initial and final states without looking into the internal mechanisms
of the collision?
In this chapter, we will define the concepts of impulse and momentum, which
are easy to interpret for the interaction of two or more particles. Furthermore,
the momentum concept leads to another law of conservation that will make it
easier to solve many problems.

96 6. IMPULSE AND MOMENTUM
6.1 IMPULSE AND MOMENTUM

Let us consider Newton’s second law. We know that acceleration in this law
is the derivative of velocity:
~F = m ~a = m d~v = d(m~v)
dt dt
Mass m can be included in the derivative because it is constant. This expression
shows that force itself is also the derivative of another quantity.
Definition: The momentum vector of an object with mass m and velocity
vector ~v is
~p = m ~v (Momentum) (6.1)
The unit of momentum is kg.m/s and does not have a particular name. Newton’s
second law can thus also be written as follows (Newton had actually expressed
the second law in this form):
~
Figure 6.1: Momentum vector. ~F = dp (6.2)
dt
Let the same object with mass m initially have a momentum ~p = m~v and
then, let a constant force ~F act on it during a short time interval ∆t . After the
interaction, the object may have a different momentum ~p 0 = m~v 0 . If we rewrite
Eq. (6.2) with ∆t ,
~ ~0 ~
~F = ∆p = p − p
∆t ∆t
or
~p + ~F ∆t = ~p 0
(6.3)
m~v + ~F ∆t = m~v 0
This expression resembles the work-energy theorem that we discussed in Chapter
5. There, the increase in kinetic energy was equal to the work performed by the
force ~F . And, in this formula, the increase in momentum is equal to the product
of force with time. It is defined as a new quantity:
Definition: The integral of a variable force ~F over a finite interval [t1 , t2 ] ,
Z t2
~J = ~F dt (Impulse) (6.4)
t1
is called the impulse of the force ~F . It has the same unit as momentum ( kg.m/s ),
and is mostly used for forces that act for a short period of time.
Let us rewrite the above Eq. (6.3) in terms of impulse and momentum:
~p + ~J = ~p 0 (Impulse-momentum theorem) (6.5)
This result is called the Impulse-momentum theorem. We thus obtain a simple

expression between the initial and final velocities of the object without having to
know its acceleration. This is equivalent to Newton’s second law.
Unlike work and energy, impulse and momentum are vector quantities. Their
directions should be taken into consideration when used in formulas.
6.2. LAW OF CONSERVATION OF MOMENTUM 97
Example 6.1
Answer
We should be careful to treat velocities as vectors when us-
ing the impulse-momentum theorem (Eq. 6.3). Choosing the
direction of the outgoing ball as the positive x -direction, we
write:
mv + F ∆t = mv0
A tennis ball incoming with a speed of 20 m/s is hit by a racket 0.056 × (−20) + F × 0.05 = 0.056 × (+30)
and is sent back at a speed of 30 m/s . The mass of the ball is
56 g and its contact with the racket lasts 0.05 seconds. Calcu- Solving for F , we find that
late the average force acting on the ball. F = 56 N .
Example 6.2
dimensions:
~F ∆t = m~v 0 − m~v → ~F = m ∆~v
∆t
Hence, ~F will have the same direction as the vector ∆~v .
As seen in the figure, when we calculate the vector ∆~v =
~v 0 − ~v by the triangle rule as the sum [~v 0 + (−~v)] , it is found
to be perpendicular to the wall.
Therefore, the vector ~F that is proportional to ∆~v will also
be perpendicular to the wall and in the +x direction.
A ball with a mass of 1 kg and a speed of v=10 m/s hits a wall As the initial and final velocities have equal magnitude,
at an angle of 53◦ and is reflected back by the same angle and we get ∆v = 2v cos 37◦ . From this, we can calculate the
at the same speed. If the ball is in contact with the wall for magnitude of the force F :
0.02 s , find the direction and magnitude of the force applied by m ∆v 2mv cos 37◦ 2 × 10 × 0.8
the wall on the ball. F= = = = 800 N
∆t ∆t 0.02
Answer We write the impulse-momentum theorem in 2-
6.2 LAW OF CONSERVATION OF MOMENTUM

Rewriting Newton’s second law in terms of momentum is not very interesting
for a single object. Its real impact can be seen when expressed for two interacting
objects.
Let us consider two masses with initial momentums ~p1 =m1~v1 and ~p2 =m2~v2 .
Let us further consider that no external force acts on this system and that the
masses only interact with each other (Figure 6.2). Let the momentums become
~p1 0 = m1~v1 0 and ~p2 0 = m2~v2 0 after this interaction.
Let us write Newton’s law for the mass m1 in its Eq. (6.3) form:
Figure 6.2: Two interacting
~p1 + ~F21 ∆t = ~p1 0 masses.
In this equation, ~F21 denotes the force applied by the second object on the first.
Let us write the same law for the second object:
~p2 + ~F12 ∆t = ~p2 0
If we add these two equations, we get:

~p1 + ~p2 + (~F21 + ~F12 ) ∆t = ~p1 0 + ~p2 0
The sum inside of the brackets is zero, because ~F21 = −~F12 according to the third
law. From this, we get the law of conservation of momentum:
~p1 + ~p2 = ~p1 0 + ~p2 0 = constant (law of conservation of momentum) (6.6)
Notice that there is no need to know the details of the interaction between the
two objects.
This is a very general result, and is also valid for more than two objects. In
such a case, the total momentum of a system is defined as follows:
~P = ~p1 + ~p2 + · · · + ~pN (total momentum) (6.7)
Therefore, the general law of conservation of momentum can be expressed as:
Law of Conservation of Momentum

If the external forces acting on a system are zero, the total mo-
mentum of the system will remain constant in time:
Fiext = 0 =⇒ ~P = ~P 0
X
i (6.8)
~p1 + ~p2 + · · · + ~pN = ~p1 0 + ~p2 0 + · · · + ~pN 0
This vector equation is true for each component. Let us write them for two
objects:
m1 v1x + m2 v2x = m1 v01x + m2 v02x
(6.9)
m1 v1y + m2 v2y = m1 v01y + m2 v02y
Important notice: The law of conservation of momentum is actually more
general than the one that we stated above. Let us say briefly that the external
forces need not be zero; it is sufficient that their impulse be zero for the the law of
conservation of momentum to hold. For example, there may be external forces,
but the time interval may be so short that their impulse can be neglected. We will
not go into these fine points.
Example 6.3
A block with mass m1 = 1 kg travels at a speed of 8 m/s on fric- Answer

tionless rails and collides into a block with a mass of m2 = 10 kg Let us write the formulas (6.9) that we found for the conser-
traveling in the same direction at a speed of 5 m/s . After the vation of momentum in one dimension (the subscripts are
collision, the block m2 is observed to travel at a speed of 6 m/s not necessary):
in the same direction. Find the magnitude and direction of the m1 v1 + m2 v2 = m1 v01 + m2 v02
final velocity of block m1 . Choosing +x in the direction of motion of the two blocks,
we get
1 × 8 + 10 × 5 = 1 × v01 + 10 × 6
And, from here, we get v01 = −2 m/s . The negative sign
shows that the block m1 is going backwards.
Example 6.4
A mass M = 10 kg traveling at a speed of 8 m/s on a fric-

tionless horizontal road suddenly explodes and splits into two
pieces. The piece with mass m2 =7 kg travels in the same direc- Answer There is a single mass M at the start. If we indicate
tion with a speed of 14 m/s . Find the velocity of the piece with its velocity with V , conservation of momentum is expressed
mass 3 kg . as follows:
6.3. COLLISIONS IN ONE DIMENSION 99
MV = m1 v01 + m2 v02 From here, we get v01 = −6 m/s . The velocity of mass m1 is
Substitute m1 =3, m2 =7 kg : in the opposite direction.
10 × 8 = 3v01 + 7 × 14
6.3 COLLISIONS IN ONE DIMENSION

In this section, we will examine collision problems in more detail. This is
a very important class of problems that can arise, from billiard balls to shells
hitting targets, from atomic particles smashing into each other to celestial bodies
interacting over immense distances via the gravitational force.
We will first consider collisions in one dimension to minimize mathematical
difficulties. Collisions are classified into two groups in this analysis.
Elastic Collisions
We have established that the total momentum of the system is conserved in
all collisions in which the net external force is zero. Let us write this only for the
x component:
m1 v1 + m2 v2 = m1 v01 + m2 v02 (6.10)
(Here, we do not write the x -subscripts, but you must remember that these
velocities are vectors, in other words, they can have positive or negative values.)
Collisions in which the total kinetic energy of the colliding objects is conserved
are called elastic collisions. For example, in collisions between steel or glass balls,
when the balls come into contact with each other, their surfaces stretch like a
spring and store the energy. After the collision, as they return to their previous
forms, this potential energy converts into kinetic energy. This property is not
observed in collisions between objects made of clay or dough; they get deformed
and lose the kinetic energy as heat.
According to this definition, kinetic energy is also conserved in elastic colli- Figure 6.3: Collision in one di-
sions: mension.
2 m1 v1 + 2 m2 v2 = 2 m1 v1 + 2 m2 v2 (6.11)
1 2 1 2 1 02 1 02
Therefore, by solving the two Eqs.(6.10 and 6.11), we can find the velocities v01
and v02 after the collision. Explicit expressions of the general solution exist, but
they are too complicated to reproduce here.
Let us seek the solution for a special case here. If mass m2 is at rest before
collision, we can set v2 = 0 in the two equations above and simplify as follows:
m1 v1 = m1 v01 + m2 v02
m1 v21 = m1 v02
1 + m2 v2
02
Let us isolate the unknown v02 in both equations:
m1 (v1 − v01 ) = m2 v02

m1 (v21 − v02
1 ) = m2 v2
02
If we write the difference of two squares as (v1 − v01 )(v1 + v01 ) and divide both
sides, we get
v1 + v01 = v02
and we can use this equation instead of the energy equation with squares. Now, if
we consider this last expression together with the equation for the conservation
of momentum (6.10), we get the following solution:
m1 − m2
v01 = v1
m1 + m2
(6.12)
2m1
v02 = v1
m1 + m2
Now, let us draw some important conclusions from these results:
1. If the masses are equal ( m1 =m2 ): We get v01 = 0 and v02 = v1 . In other words,
they exchange their velocities; the incoming object stops and the other one
starts with exactly the same velocity. We observe this in billiard balls in
head-on collisions. The elegant toy called Newton’s cradle uses this property
(Figure 6.4): When the ball at one end is pulled and released, the balls in the
middle remain motionless and the ball at the other end bounces off with the
Figure 6.4: Newton’s Cradle. same velocity. Because the masses are equal, each ball in the middle transfers
Because the masses are equal, all of its energy and momentum to the next one and remains motionless.
the momentum of the incoming 2. If m1 > m2 , in other words, if the incoming mass is heavier, Eqs. (6.12) tell us
ball is fully transferred to the that the final velocities of both masses will be positive and both will continue
last ball.
in the same direction. A small mass at rest will never bounce a heavier
mass back. When a truck crashes into a motionless automobile, it drags it
along. (The truck-automobile collision is not elastic, but we may expect this
behavior as an approximation.)
3. If m1 < m2 , in other words, if the incoming mass is lighter, the final velocity
of the incoming mass will be negative, which means that it will bounce back.
When an automobile crashes into a truck, it may sometimes bounce back.
4. If the incoming mass is much much lighter ( m1 /m2 ≈ 0 ), then the final
velocities will be v01 ≈ −v1 and v02 ≈ 0 . In other words, the incoming mass
will bounce back with the same velocity and the heavy mass will not move.
It is as if it is bouncing back from a wall. Rutherford discovered the structure
of the atom using this feature. When he bombarded heavy gold (Au) atoms
with lighter alpha particles, he observed that some alpha particles bounced
back with the same velocity. This was a clue to the existence of a heavy
nucleus at the center of the atom.
Inelastic Collision
Collisions in which kinetic energy is not conserved are called inelastic colli-
sions. In this case, the colliding objects can become deformed and kinetic energy
can be transformed into heat. Thus, the total kinetic energy of the system will
have decreased. The collisions of plastic balls used in sports or the collision of
cars in traffic are inelastic collisions.
Conservation of momentum is still valid in inelastic collisions:
m1 v1 + m2 v2 = m1 v01 + m2 v02 (6.13)
However, we can no longer write conservation of kinetic energy. Therefore, it is

difficult to find both of the final velocities. Additional information may be given
in certain cases, so that final velocities can be calculated in such cases.
6.3. COLLISIONS IN ONE DIMENSION 101
Totally Inelastic Collision

In this special case of inelastic collision, the two objects stick together after
the collision. A meteor striking Earth or a bullet or an arrow hitting a target are
examples of such types of collisions. In this case, the final velocities of the objects
are equal and the conservation of momentum gives the solution:
v01 = v02 = v0
m1 v1 + m2 v2 = (m1 + m2 )v0 Figure 6.5: Cars are manufac-
tured in such a way that their
From here, we can find the common final speed: front and rear parts collapse
readily in a collision. Thus, less
m1 v1 + m2 v2 energy and momentum will be
v0 = (6.14)
m1 + m2 transferred to the passengers.
Example 6.5 m1 − m2 3−5

v01 = v1 = × 10 = −2.5 m/s
m1 + m2 3+5
A ball with mass 3 kg travels at a speed of 10 m/s and collides The negative sign means that the ball m1 bounces back. The
into a ball with mass 5 kg at rest. As the collision is elastic, velocity of the second ball is calculated using the same for-
find the final velocities of the balls. mula:
Answer 2m1 2×3
v02 = v1 = × 10 = 7.5 m/s
We can use Eqs. (6.12) derived for elastic collision, because m1 + m2 8
one of the masses is at rest:
Example 6.6
Answer
We first find the velocity of block m2 on the ground level.
According to conservation of energy, its potential energy at
height h is supplied by the kinetic energy on the ground:
√
2 m2 v2 = m2 gh → v2 = 2gh = 2 × 10 × 0.45 = 3 m/s
1 02 0
p
We then write conservation of momentum for the collision:

A block with a mass of m1 =1 kg and a speed of 5 m/s
m1 v1 + m2 v2 = m1 v01 + m2 v02
hits another block with a mass of m2 =2 kg at rest. After the
collision, the block m2 climbs up to a height of 45 cm on a fric- We calculate v01 by substituting the numerical values:
tionless inclined plane. Calculate the velocities of these blocks 1 × 5 + 0 = 1 × v01 + 2 × 3
right after the collision. v01 = −1 m/s
Example 6.7
terms of height h . Numerical example: m = 50 g , M = 5 kg ,
L = 2 m and h = 80 cm .
Answer
We first find the speed of the (bullet+block) system right after
the collision, in other words, when the pendulum is in the
vertical position. We write conservation of momentum for
this inelastic collision:
m1 v1 + m2 v2 = (m1 + m2 )v0
Here, we have m1 =m and m2 =M , and if we show the final
velocity of the system with V , we get:
A ballistic pendulum is a device used to calculate the muzzle m
mv + M.0 = (m + M) V → V = v
velocities of bullets. It is a pendulum constructed by hanging a m+M
wooden block with a very high mass at the end of a rope with The (bullet+block) system rises to a height h due to its ki-
length L . When a bullet with mass m traveling at a speed v netic energy. According to conservation of energy, all of the
embeds itself into the wooden block with mass M , the (bul- kinetic energy at the start will be converted into gravitational
let+block) system rises by h . Calculate the speed of the bullet in potential energy:
1
+ M) V 2 = (m + M) gh m+M p
2 (m v= 2gh
m
Substituting V , we solve for v and get With the given numerical values, we get v = 400 m/s .
Example 6.8
Conservation of energy and momentum is thus sufficient for
A spring with spring constant k = 100 N/m is mounted onto the solution.
the front of a block with a mass of 2 kg at rest on a frictionless (a) If we use v0 to show the common velocities of the blocks
horizontal plane. Another block with a mass of 1 kg approaches when the spring is compressed to a maximum, conservation
at a speed of 6 m/s and collides with the spring. (a) How much of momentum is expressed as follows:
will the spring get compressed and what will the velocities of m1 v1 + 0 = (m1 + m2 ) v0
m1 v1 1×6
the blocks be at that instant? (b) What will the velocities of the v0 = = = 2 m/s
blocks be after they are separated? m1 + m2 1 + 2
Spring potential energy is also taken into consideration when
writing conservation of energy:
2 m1 v1 = 2 (m1 + m2 ) v + 2 kx
1 2 1 02 1 2
From here, we can calculate the amount of compression of

the spring:
s
m1 v21 − (m1 + m2 )v02
Answer x= = 0.49 m .
k
This is also an elastic collision, because energy is stored by (b) After the blocks are separated, the initial kinetic energy
the spring during the collision and later given back. Accord- is regained, in other words, the collision is elastic. We use
ingly, we can immediately solve item (b). However, let us first the formulas (6.12), because the block m2 is at rest before
look into item (a). It seems that there are three variables (the collision:
final velocities of the blocks and the amount of compression m1 − m2 1−2
v01 = v1 = × 6 = −2 m/s
of the spring) and we only have two equations. However, m1 + m2 1+2
consider this point: at the instant of maximum compression, 2m1 2
the velocities of the blocks will be equal for a short instant. v02 = v1 = × 6 = 4 m/s
m1 + m2 3
6.4 COLLISIONS IN TWO DIMENSIONS

Collisions in one dimension are rarely observed. Two colliding objects usually
go off into different directions. In the most general case, collisions in three
dimensional space can also be examined in two dimensions. This is because, the
momentum vectors of the two objects intersect at the collision point and form a
plane that is called the collision plane. Since total momentum is conserved, the
final momentum vectors should also lie in this plane. Hence, it will be sufficient
to study collisions in two dimensions.
Figure 6.6: Coordinates for a

collision in two dimensions.
Eq. (6.6) which gives the law of conservation of momentum was found as a
vector equation. Therefore, it should be valid separately for both components in
a collision in two dimensions.
 m1 v1x + m2 v2x = m1 v1x + m2 v2x


 0 0
~p1 + ~p2 = ~p1 + ~p2
0 0
=⇒  (6.15)

 m1 v1y + m2 v2y = m1 v0 + m2 v0

1y 2y
6.4. COLLISIONS IN TWO DIMENSIONS 103
There are four unknown variables in this two-equation system, because the
final velocities have two components and the problem cannot be solved without
additional information. Another equation is provided by the conservation of
energy for elastic collisions:
2 m1 v1 + 12 m2 v22 = 21 m1 v02
1 + 2 m2 v2 (for elastic collision) (6.16)
1 2 1 02
Hence, an additional equation is needed in order to solve the problem. In general,

one of the variables is measured experimentally.
There is no such problem for totally inelastic collisions. If we write the final
velocities as equal, we get
~v1 0 = ~v2 0 = ~v 0
m1 v1x + m2 v2x = (m1 + m2 )v0x
(6.17)
m1 v1y + m2 v2y = (m1 + m2 )v0y
Two equations with two unknowns will give the components of the final velocity.
Example 6.9
of 30◦ with the eastern direction. Using only this information,
calculate the ratio v2 /v1 and determine which car was going
faster.
Answer
Choosing the axes as shown in the figure, we use Eqs. (6.17)
for a totally inelastic collision in two dimensions:
m1 v1x +m2 v2x = (m1 +m2 )v0x → m1 v1 +0 = (m1 +m2 )v0 cos 30◦
m1 v1y +m2 v2y = (m1 +m2 )v0y → 0+m2 v2 = (m1 +m2 )v0 sin 30◦
Dividing both sides of the equation, we find the ratio v2 /v1 :
A car with a mass of m1 =800 kg traveling East and a pickup v2 m1 800
= tan 30◦ = × 0.58 = 0.38
truck with a mass of m2 =1200 kg traveling North collide head v1 m2 1200
on in an intersection and stick together. Both drivers claim According to this result, the car with mass m1 was traveling
that the other entered the intersection with excessive speed. You approximately 3 times faster than the pickup truck. (Do not
arrive at the site as the expert. You measure the black tire tracks underestimate this example. Conservation of momentum and
left on the tar surface and find that the tracks make an angle energy are the most important tools of traffic experts.)
Example 6.10
found from conservation of energy ( m1 =m2 =m ):
1 2
2 mv1 q+ 0 = 12 mv02
1 + 2 mv2
1 02
√
v02 = v21 − v021 = 52 − 32 × 106 = 4 × 106 m/s
In order to find the angles, we must write conservation of
momentum in two dimensions with axes as shown in the
figure:
m1 v1x + m2 v2x = m1 v01x + m2 v02x → v1 = v01 cos θ1 + v02 cos θ2
m1 v1y + m2 v2y = m1 v01y + m2 v02y → 0 = v01 sin θ1 − v02 sin θ2
Proton-proton collisions are an important source of infor- Substituting known velocities and simplifying, we get:
mation in nuclear physics. A proton with a speed of 5×106 m/s 5 = 3 cos θ1 + 4 cos θ2 → 4 cos θ2 = 5 − 3 cos θ1
has an elastic collision with another proton at rest. After the 0 = 3 sin θ1 − 4 sin θ2 → 4 sin θ2 = 3 sin θ1
collision, the incoming proton is observed to scatter off at a We add the squares of both sides and use the identity (sin2 θ2 +
speed of 3 × 106 m/s . Calculate the scattering angles of the cos2 θ2 = 1) , finding that
protons and the speed of the second proton.
16 = 9 sin2 θ1 + (9 cos θ12 − 30 cos θ1 + 25)
Answer We again use the identity (sin2 θ1 + cos2 θ1 = 1) and simplify:
The velocity v2 of the second proton can immediately be
0
cos θ1 = 0.6 → θ1 = 53◦
Substituting this θ1 value in the equation for the y - 3 3

sin θ2 = sin θ1 = → θ2 = 37◦
component of momentum, we find θ2 : 4 5
6.5 CENTER OF MASS

When we try to carry a piece of timber, we carry it by lifting it from its center.
If one end of a stick is heavier than the other, we hold it somewhere near the
heavier end (Figure 6.7). People on a boat will not gather at the same spot, but will
Figure 6.7: The point that bal- sit at different places so that the boat will not capsize. These examples indicate
ances a stick with a heavier end. that there is a mean point that represents the mass of extensive objects.
Definition: The “weighted” average of mass positions constituting a system
is called the center of mass (CM).
By the weighted average, we understand that each position is included in the
average as multiplied by the value of the mass at that position.
According to this definition, the expression for the center of mass of
m1 , m2 . . . mN located at x1 , x2 . . . xN along the x -axis will be as follows:
PN
m1 x1 + m2 x2 + · · · + mN xN m i xi
xcm = = Pi=1 (6.18)
m1 + m2 + · · · + mN N
i=1 mi
The y - and z -coordinates of the center of mass can be similarly defined in three-
dimensional space:
i m i yi
P P
m i zi
ycm = , zcm = i (6.19)
M M
Here, M = m1 + · · · + mN is the total mass. If we write these three formulas as a
single vector formula, we get:
i mi ~
P
ri
~rcm = (center of mass) (6.20)
M
In the simplest system, let us consider two point masses m1 and m2 located
at positions x1 and x2 along the x -axis (Figure 6.8). Applying the formula for
center of mass, we get
m1 x1 + m2 x2
xcm = (6.21)
m1 + m2
If the masses are equal (m1 =m2 ) , the center of mass is the midpoint xcm =
Figure 6.8: The center of mass (x1 + x2 )/2 . This expression conforms to our daily experience. We hold a straight
of two objects. stick from the center, but if a weight is put on one end, we hold it from a point
nearer to that side.
Center of Mass of a Continuous Mass Distribution
For objects that have a continuous mass distribution, we can calculate their
center of mass using either one of the following two methods:
1. Symmetry. As seen in Figure 6.9, objects may have certain geometrical
shapes, such as a triangle, square, rectangle, disc or circle, that have a center
of symmetry. In addition, if they are homogeneous, in other words, if their
density is distributed evenly, then we can assume that the center of mass will
be located at that center of symmetry. The summing expression for center of
mass that we found for pointlike objects can thus also be applied to whole
symmetrical parts.
6.5. CENTER OF MASS 105
Figure 6.9: The centers of mass

of symmetric objects whose mass
is distributed homogeneously.
2. Integration. If the mass distribution in space is known as a function of

position, integration technique can be used to find the center of mass. Let us
see how the integral form is set up.
Let us divide the object with mass M into N number of small ∆mi masses
located at the position (xi , yi , zi ) . Applying the formula for center of mass for
these small parts, for example, for the x -component we get
i xi ∆mi xi ∆mi
P P
xcm ≈ P = i
i ∆mi M
The sum in the denominator is written as the total mass M . Similar formulas can
be written for the y - and z -components.
Now, when we take the limit ∆mi → 0 , the sum in the numerator will become
an integral. Therefore, the integral expressions of all three components can be
written as follows:
Figure 6.10: The dm mass
y dm
R R R
x dm z dm element for continuously dis-
xcm = , ycm = , zcm = (6.22)
M M M tributed mass.
In these integrals, mass should be expressed as dm = ρ(r) dr in terms of a density
that depends on position. The details will be clear in the worked examples below.
Example 6.11 P
mi xi 1 × 1 + 4 × 3 + 5 × 0
xcm = Pi = = 1.3 m
i mi 1+4+5
mi yi 1 × 0 + 4 × 1 + 5 × 2
P
ycm = Pi = = 1.4 m
i mi 10
The position of the center of mass is indicated on the

figure below.
Determine the center of mass of the 3-mass system whose
masses and positions on the xy plane are given in the figure.
Answer
We write Eqs. (6.20), which we found for the coordinates of
the center of mass, for the x - and y -components and calcu-
late as follows:
Example 6.12
Find the center of mass of a plate shaped like the letter L with
dimensions shown in the figure and with homogeneous density.
Answer Let us first separate the plate into two simpler pieces,
as shown in the figure. The mass of each plate will be propor-
tional to its surface area, because the same density coefficients
will cancel each other out in the numerator and denominator.
The centers of mass of these pieces will be their symmetry m1 x1 + m2 x2 4 × 0.5 + 2 × 2

xcm = = = 1 units
center points, because their densities are homogeneous. m1 + m2 4+2
Accordingly, the coordinates of the center of mass are calcu- m1 y1 + m2 y2 4 × 2 + 2 × 0.5
ycm = = = 1.5 units
lated as follows: m1 + m2 6
Example 6.13
and consider a thin piece with thickness dx between x and
Use the integration method to calculate the coordinates of the x + dx along the x -axis. The area of this piece is y dx and
center of mass of a plate shaped like a right triangle, in which its center of mass will be its own x coordinate.
the perpendicular sides have the lengths a, b and the density The equation for a line passing through the origin with a
is homogeneous. slope of (b/a) is y = (b/a) x . Accordingly, Eq. (6.22) can be
written as:
Z Z a
1 1
xcm = x dm = x (y dx)
M ab/2 0 a
Z a Z a
2 2 2 x3
= x (b/a)x dx = 2 x dx = 2
2
ab 0 a 0 a 3 0
Answer
Since the plate is homogeneous, its mass is proportional to xcm = 23 a
its surface area. Let us choose the axes as shown in the figure Using a similar calculation, we find that ycm = 13 b .
Dynamics of the Center of Mass

Consider a system consisting of two particles. In Chapter 4, we learned that
forces acting on interacting masses could be separated into two groups: The
external forces and the internal forces that they exert on each other. Accordingly,
ext
let us use ~F1 to show the external force exerted on m1 and ~F21 for the force
exerted by the second mass. When writing Newton’s law separately for each
particle, let us write accelerations as the second derivative of position:
2
for m1 : ~F1ext + ~F21 = m1~a1 = d (m1~r1 )
dt2
2
for m2 : ~F2ext + ~F12 = m2~a2 = d (m2~r2 )
dt2
Adding these two equations, we get:
2
~F1ext + ~F2ext + ~F21 + ~F12 = d (m1~r1 + m2~r2 )
dt2
According to the third law, the action and reaction forces between the masses will
be equal and opposite, ~F21 = −~F12 . Hence, the content of the second brackets on
the left side is zero. Dividing and multiplying with the same total mass M , we
obtain:
~F1ext + ~F2ext = M d m1~r1 + m2~r2 = M d ~rcm
2 2
dt2 M dt2
The second derivative on the right-hand side of the equation is just the accelera-
Figure 6.11: The center of mass tion ~acm of the center of mass. Consequently, we obtain Newton’s law, which
of your body moves up and determines the motion of the center of mass:
down as you walk, and you thus
perform work against gravity. X
~Fiext = M ~acm
You will get less tired if you take (motion of center of mass) (6.23)
smaller steps. i
6.6. ROCKET MOTION 107
This result is important: In many-particle systems, there is one point that moves
with the resultant force of external forces, such that it obeys Newton’s law as if the
whole mass was concentrated at one point. The expression (6.23) is actually the
real definition of the center of mass.
Figure 6.12: The arms and legs

of this boy jumping into the
sea can follow many different
paths, but he will always have a
point that moves as if the whole
mass was centered at that point.
That point is the center of mass.
Note that the center of mass of
the boy, which is around his
belly, draws a perfect parabola.
(Photo: Erik Forsberg).
6.6 ROCKET MOTION

Airplanes fly by using the pressure differences of the air. For this reason, for
many years, it was believed that it would be impossible to build crafts capable of
flying in outer space, because there is no air to push. However, today, we have
rockets that can fly in space using the principle of conservation of momentum
and the reaction force of Newton’s third law. A rocket moves with the forward
momentum that it gains by ejecting its fuel backwards.
Let us examine the motion of the rocket with respect to an observer on the
ground (Figure 6.13). In a gravity-free environment, let the total mass of the
rocket and its fuel be m and velocity v in the positive direction at a given initial
time t . Let a small mass ∆m be ejected backwards as fuel during a time interval
∆t with a speed vexh with respect to the rocket.
Let the mass of the rocket be m − ∆m and its velocity v + ∆v at a later time
t + ∆t . We write the conservation of momentum for the (rocket+fuel) system:
Figure 6.13: Initial and final ve-

locities of the rocket-fuel system.
mv = (m − ∆m)(v + ∆v) + ∆m(v − vexh )

Notice that we have written the velocity of the fuel with respect to the ground as
( v − vexh ). Simplifying and neglecting the very small term ∆m ∆v , we find:
vexh ∆m = m ∆v (6.24)
We divide by ∆t in order to form derivatives:
∆m ∆v
vexh =m
| {z∆t} ∆t
Fthr
The term on the left-hand side has the dimensions of a force and it is called the
thrust of the fuel. And, in the limit ∆t → 0 , the derivative on the right-hand
side gives the acceleration of the rocket. In conclusion, we have Newton’s law
for rocket motion:
∆m
Fthr = vexh = ma (rocket motion) (6.25)
∆t
The ratio ∆m/∆t in the expression of the thrust on the right-hand side is the
mass of fuel exhausted per unit time. We can see that, the higher the exhaust
speed of the fuel, the higher the thrust will be.
During launch from Earth’s surface, the weight −mg should also be added to
the left-hand side of this expression, in other words, the thrust should be greater
than the weight: Fthr − mg > 0 .
We now want to calculate the rocket speed as a function of its mass. Let us
return to the expression (6.24):
vexh ∆m = m ∆v
Here, ∆m was the positive change in exhausted fuel mass. If we want to write the
change in the rocket’s mass, it will be just the opposite of this quantity. Hence, by
taking −∆m and dividing by m , we can write the equation governing the change
in rocket mass:
(−∆m)
vexh = ∆v
m
Let the rocket’s initial mass be m0 and its initial velocity v0 , and let them reach
the final values m and v . Taking the limits (∆m, ∆v) → (dm, dv) and integrating
both sides, we get:
Z m Z v
dm
−vexh = dv
m0 m v0
m
−vexh ln = v − v0
m0
Hence, the final velocity of the rocket in terms of mass loss is:
m0
v = v0 + vexh ln (6.26)
m
where the property ln(a/b)=− ln(b/a) of the logarithm was used to remove the
negative sign and to obtain a positive logarithm.
Example 6.14
using the formula (6.26):
A rocket traveling in space at a constant speed of 80 m/s ejects m0
v = v0 + vexh ln
an amount of fuel backwards with a ratio of 1/20 to its own m
mass at a speed of 200 m/s with respect to the rocket. What Here, we have m 0 /m = 20/19 . Also, the numerical values of
will the final velocity of the rocket be? v exh and v0 are given in accordance with the definitions in
the formula. From here, we can find the final velocity:
Answer
As the mass loss m0 /m is given, the velocity is calculated v = 80 + 200 × ln(20/19) = 80 + 200 × 0.51 = 90 m/s
Example 6.15
velocity, to find that
A rocket with a mass of 1000 kg is ejecting fuel at a rate of Fexh = 1200 × 10 = 12000 N
10 kg per second and with a speed of 1200 m/s during its (b) In addition to thrust, there is now the weight of the rocket
launch from the surface of the earth. (a) What is the thrust? (b) opposing motion. We write the second law for the net force
What is the initial acceleration? acting on the rocket:
Fexh − mg
Answer Fexh − mg = ma → a =
m
(a) We use Eq. (6.25) for the thrust: 12000 − 10000
∆m a= = 2 m/s2
Fthr = vexh 1000
∆t This is only the initial acceleration; it will change later as the
We substitute the given ratio ∆m/∆t = 10 and the exhaust mass changes.
1. Which quantity is preserved during all types of colli-
sions? 6. Which is correct for an elastic collision?
(a) Potential energy. (a) Only momentum is conserved.
(b) Kinetic energy. (b) Only kinetic energy is conserved.
(c) Momentum magnitude. (c) Both momentum and kinetic energy are conserved.
(d) Momentum vector. (d) None of the above.
2. Which is incorrect? 7. A block with a velocity of 3 m/s elastically collides with

(a) Momentum is zero if the kinetic energy is zero. another block with equal mass at rest. What will be the
(b) Momentum is zero if the velocity is zero. final velocity of the incoming block?
(c) Momentum is zero if the potential energy is zero. (a) 0 (b) −3 m/s (c) +3 m/s (d) 6 m/s
(d) Momentum is constant if the net force is zero.
8. Which one of the following is the expression of kinetic
3. Two objects with equal masses of m collide with equal energy in terms of momentum?
and opposite velocities v . What is the total momentum (a) K = p2 /2
of the system? (b) K = p2 /2m
(a) 0 (b) mv (c) 2mv (d) −mv (c) K = 2mp2
(d) K = 2p2
4. A particle’s speed is doubled and its mass is tripled. By
what factor will its momentum increase? 9. A boy with mass 2m standing on a frictionless horizon-
(a) 2 (b) 6 (c) 12 (d) 18 tal plane throws a package of mass m with a velocity
v in the +x direction. What will be the velocity of the
5. Which is correct for two objects? boy?
(a) If their momentum is equal, their kinetic energies (a) 0 (b) −v (c) −2v (d) −v/2
are also equal.
(b) If their momentums are equal, their velocities are 10. Which of the following is correct if two masses collide
also equal. and stick to each other?
(c) If their momentums are equal, they will move in the (a) Momentum is conserved.
same directions. (b) Kinetic energy is conserved.
(d) If their momentums are equal, their masses are also (c) Both momentum and kinetic energy are conserved.
equal. (d) Potential energy is conserved.
11. An object at rest gains a velocity v under the action of 16. Which of the following is correct if the net force acting
a force F that lasts t seconds. How much velocity will on an object is zero?
the object gain by a force 3F acting for 3t seconds? (a) The impulse is zero.
(a) v (b) 3v (c) 9v (d) v/3 (b) The momentum will remain constant.
(c) Its center of mass will be in uniform linear motion.
12. A spring is placed between two blocks with unequal
masses and compressed. What will happen when the
blocks are released?
(a) They will move away with equal and opposite ve- 17. If the position of mass m is x1 = 0 and the position of
locities. mass 2m is x2 = 6 , then what will the position of the
(b) The one with the larger mass will also have higher center of mass xcm be?
velocity.
(c) The one with the lower mass will have lower veloc- (a) 2 (b) 2.5 (c) 3 (d) 4
ity.
(d) The one with the lower mass will have higher ve- 18. Two equal masses are at positions x and x . What will
1 2
locity. the coordinate of the center of mass be?
13. A spring is placed between blocks with unequal masses (a) x2 − x1
and compressed. Which one will be true after the blocks (b) x1 + x2
are released? (c) (x1 + x2 )/2
(a) The kinetic energies of the blocks will be equal. (d) (x2 − x1 )/2
(b) Their momentums will be equal and opposite.
(c) The one with the higher mass will have higher mo-
mentum. 19. Which of the following is correct when a rifle is fired?
(d) The one with the lower mass will have higher mo- (a) The momentum of the rifle is greater.
mentum. (b) The momentums of the rifle and the bullet are equal
and opposite.
14. A car and a truck moving towards each other at the same
(c) The kinetic energies of the rifle and bullet are equal.
speed collide and stick together. Which of the following
(d) The kinetic energy of the rifle is greater than that
is correct?
of the bullet.
(a) The force on the car is greater.
(b) The force on the truck is greater.
(c) The car will receive the kinetic energy lost by the 20. A spring is placed between two blocks at rest and com-
truck. pressed. What will happen when the blocks are released?
(d) The automobile will receive the momentum lost by
the truck. (a) The center of mass will travel at constant velocity.
(b) The center of mass will move towards the heavier
15. Which of the following is incorrect if the total momen- block.
tum of a system is zero? (c) The center of mass will move towards the lighter
(a) Its kinetic energy will also be zero. block.
(b) Its center of mass will be at rest. (d) The center of mass will remain at rest.
(c) The net force acting on the system is zero.
Problems
6.1 Impulse and Momentum
6.1 A football player hits a ball of mass 400 g with an incom-

ing speed of 30 m/s and sends it back in the same direction at
a speed of 40 m/s . Since the contact of the ball with his foot Problem 6.2
took 0.04 s , what will the average force be on the football
player’s foot? [A: 700 N .] 6.2 A football player hits a ball of mass 400 g coming from
PROBLEMS 111
the right corner with a speed of 30 m/s and sends it towards ground at a horizontal distance of 3 m from the edge. Find
the goal perpendicular to its incoming direction at a speed of the initial velocity of the bullet. [A: 150 m/s .]
40 m/s . Since the contact of the ball with his foot took 0.05 s ,
find the average force acting on the ball and its direction.
[A: 400 N and 53◦ with the corner.]
Problem 6.9
6.3 A ball with a mass of 0.5 kg is dropped from a height 6.9 A spring with a constant of k=100 N/m is attached to
of h1 =5 m . The ball bounces back to a height of h2 = 3.2 m the back of a block with a mass of 2 kg traveling at a speed
after hitting the ground. What is the impulse given to the ball of 3 m/s on a frictionless horizontal plane. Another block
during the collision with the ground? [A: J = 9 kg·m/s .] with a mass of 1 kg comes with a speed of 6 m/s and collides
with the spring. What will the maximum compression of the
6.2-3 Conservation of Momentum-Collisions in spring be and what will the velocities of the blocks be at that
One Dimension instant? [A: 24 cm and 4 m/s .]
6.4 A block with a mass of 1 kg traveling at a speed of 10 m/s
on a frictionless horizontal road collides into another block
with a mass of 4 kg that is at rest. If the collision is elastic,
find the final velocities of the blocks. [A: −6, +4 m/s .]

on a frictionless horizontal road collides into another block
with a mass of 3 kg that is coming from the opposite direc-
tion with a speed of 6 m/s . After collision the blocks stick Problem 6.10
together. Find the velocity of the blocks after the collision.
6.10 A bullet with a mass of m=50 g travels at a speed v
[A: −0.4 m/s .]
and hits a block with a mass of M=950 g tied to the end of
a rope of length L = 2 m . The bullet is embedded in the
block. What should the initial velocity v of the bullet be such
that the (bullet+block) system will make one complete loop
Problem 6.6
without the rope becoming slack at point B? [A: 200 m/s .]
6.6 A wooden block with a mass of 950 g is attached to the
free end of a spring with a spring constant of k = 100 N/m
on a frictionless horizontal plane. The spring is initially at
normal length. A bullet with a mass of 50 g and a speed v hits
the block and is embedded in it. If the spring is compressed
by 30 cm , what was the initial velocity of the bullet?
[A: 60 m/s .]

on a frictionless horizontal road collides with another block
with a mass of 4 kg incoming from the opposite direction. Problem 6.11
Both blocks stay at rest after the collision. What was the 6.11 A ball with mass m1 =3 kg released from rest from point
initial velocity of the second block? [A: 7.5 m/s .] A at a height of 3.2 m travels along the frictionless path ABC ,
collides with and sticks to another ball with mass m2 =1 kg
standing at rest at point B. After the collision, the (m1 + m2 )
system climbs up to point C at a height of 1 m and compresses
the spring with a constant of k=400 N/m , fixed to the plane.
Calculate (a) the velocity right after collision, (b) the velocity
at point C and (c) the amount of compression of the spring.
[A: (a) 6 m/s , (b) 4 m/s , (c) 40 cm .]

6.8 A bullet with a mass of m = 50 g is fired horizontally 6.12 Blocks with masses M1 =5 kg and M2 =0.95 kg are at
towards a wooden block with a mass of M = 950 g that is at rest on a frictionless horizontal surface. A bullet with mass
rest near the edge of a tabletop 80 cm in height. The bullet is m=50 g fired at a speed of v=400 m/s passes through the first
embedded in the block. The (block+bullet) system falls to the block and embeds in the second block. After the collision, the
velocity of the first block is measured to be 3 m/s . Find the

velocity of the second block. [A: 5 m/s .]
Problem 6.13
6.13 Two balls with masses m1 =3 kg and m2 =2 kg are hang- Problem 6.16
ing side by side on the ends of two strings of equal lengths L 6.16 A ball with mass m1 =1 kg is traveling in the +x direc-
to form two pendulums side by side. The first ball is pulled tion at a velocity of v1 =10 m/s . It collides with another ball
aside and released at an angle of 53◦ from the vertical. If the with a mass of m2 =2 kg incoming with a velocity of v2 =5 m/s
balls collide elastically, what will be the maximum angle that that makes an angle of −53◦ with the x axis. After the col-
the second ball makes with the vertical? [A: 65◦ .] lision, m1 is scattered at a velocity of 15 m/s in a direction
that makes an angle of +45◦ with the x -axis. Calculate the
6.4 Collisions in Two Dimensions velocity components of m2 . [A: 2.7, −1.3 m/s .]
6.5 Center of Mass

6.14 A ball with m1 =1 kg travels on a frictionless horizon-
6.17 In a sulfur dioxide ( S O2 ) molecule, the S-O bonds have
tal plane at velocity v0 =10 m/s and collides with another
a length of 0.14 nm and the angle between them is 120◦ .
ball with mass m2 =2 kg at rest. After the collision, the balls
Since the mass of sulfur is approximately twice that of oxy-
scatter, m1 at an angle of 60◦ and m2 at an angle of −37◦ .
gen, find the distance of the center of mass of this molecule
Calculate the final velocities of the balls.
to the S atom. [A: ycm = 0.035 nm .]
[A: v1 = 6.0, v2 = 4.4 m/s .]
Problem 6.18
6.18 Find the center of mass of a T shaped homogeneous
plate with the dimensions given in the figure above.
[A: ycm = 2.6 ]
6.6 Rocket Motion

6.19 A rocket in uniform linear motion in outer space with
Problem 6.15 a constant speed of 120 m/s fires its engine and ejects an
6.15 A car with a mass of 800 kg traveling East and a pickup amount of fuel backwards at a ratio of 1/30 to its own mass
truck with a mass of 1200 kg traveling North collide at a at a speed of 300 m/s with respect to the rocket. What will
junction and stick to each other. According to the tire tracks the final velocity of the rocket be? [A: 130 m/s .]
on the road, the vehicles had drifted off by 9 m in a direction
that makes an angle of 37◦ Given that the coefficient of fric- 6.20 A rocket with mass 800 kg is ejecting fuel at an amount
tion of the tarmac road is µ=0.8 , determine the velocities of of 500 kg per minute at a speed of 1500 m/s during its launch
both vehicles before the collision. from the surface of the earth. (a) What is the thrust? (b) What
[A: Car: 24 m/s , pickup truck: 12 m/s .] is the initial acceleration? [A: (a) 12500 N , (b) 5.6 m/s2 .]
7
ROTATIONAL MOTION
Night sky in a time-lapse pic-

ture. The Earth rotates around
its own axis, and we perceive it
as the rotational motion of the
stars.
Although each of these stars
seems to have a different rota-
tion velocity, there is one quan-
tity that is common for all. In
this chapter, we will learn what
it is.
Up until now, we have considered an object as having a speed and acceleration

as a whole. Now, consider the rotating wheel of a car. Each point of this wheel
rotates at a different speed and acceleration. Therefore, we cannot examine the
rotation of extended objects using the methods that we have learned so far.
In previous chapters, we have made calculations based on the assumption
that all forces on an object acted on a single point. This is called a pointlike object.
Although we drew objects such as cars, blocks and spheres, we always examined
them as pointlike objects. However, there are extended objects in real life; they
have volumes, and forces act on different parts of them. In this chapter, we will
take this property of rigid bodies into consideration.
A body that takes up volume in space and does not change its shape is called
a rigid body. (They are also called solid bodies.) Wheels, doors, automobiles, the
Earth, the Moon, etc., are all examples of rigid bodies.
The motion of a rigid body is much richer than that of a pointlike object. Let

114 7. ROTATIONAL MOTION
us consider the kinds of motions that rigid bodies can perform:
Figure 7.1: The most general

motion of an object: Translation,
rotation and vibration.
1. The motion of the rigid body’s center of mass is called the translational
motion. Both ends of a baton thrown up in the air by a majorette can rise by
rotating, but the center of mass of the stick travels on a parabolic trajectory,
just like a pointlike stone.
2. Even if its center of mass remains at rest, a rigid body can perform rotational
motion by rotating around its own center of mass. Let us apply a force ~F
perpendicular to one end of a ruler stick and a force −~F to the other end
(Figure 7.2). As the net force acting on the object is zero, according to the
first law, its center of mass will either remain at rest or perform uniform
linear motion. However, there is nothing preventing the stick from rotating
around its own center of mass.
Figure 7.2: The net force acting 3. Bodies can also perform vibrational motion. If we push two masses con-
on this ruler stick is zero. But nected by a string towards the center, the object will vibrate, even though the
it is not in equilibrium and will center of mass remains at rest. However, vibrational motion is not included
perform rotational motion. in the definition of a rigid body, because the body changes shape and will be
examined separately later in Chapter 9.
In Chapter 6, we saw that the center of mass moved as if all of the external
forces were acting at the center of mass. Hence, the translational motion of the
rigid body is the motion of its center of mass. It will thus be sufficient to examine
the rotational motion of rigid bodies here.
7.1 ANGULAR KINEMATICS

The rotational motion of rigid bodies may seem complex at first. Each point
of the body has a different position, velocity and acceleration, and you may guess
the mathematical difficulty of examining all of these together.
However, we notice something here: Although each point of a wheel rotates
at a different velocity, when the wheel makes one full revolution, each point will
have made the same revolution. If one point rotates by a certain angle, all points
will rotate by that angle. Therefore, rigid bodies should naturally be analyzed
using angular coordinates. The topic that deals with concepts of angular velocity
and angular acceleration is called angular kinematics.
Angular Position (θ)
Let us consider a point P rotating on a circular trajectory with radius r . Let
us choose a reference line from which the angles are measured. If a point P
travels some length of arc s from the reference line, the subtended angle,
s
θ= (angular position) (7.1)
r
is called the angular position of P (Figure 7.3). According to the most commonly
Figure 7.3: Angular position θ . accepted convention, counterclockwise angles are taken as positive, and clockwise
as negative.
7.1. ANGULAR KINEMATICS 115
The unit of angle is the radian in angular kinematics. Using other units, like
degrees ( ◦ ) or grades, will give incorrect results. The formula θ=s/r actually
indicates that the radian is the natural unit of angle. Indeed, if we substitute
the circumference of a circle as s=2πr , we get the angle 2π radians. Therefore,
angles should be taken as radians in angular kinematics calculations. Remember
the following rule for converting degrees into radians:
1 revolution = 360◦ = 2π radians
Accordingly, 180◦ =π radians , 60◦ =π/3 radians , 45◦ =π/4 radians , etc.
Angular Velocity (ω)
Let us again consider a point P rotating on a circle with radius r . If the
angular position of this point at t1 is θ1 , and at a later time t2 is θ2 , the ratio
θ2 − θ1 ∆θ
ωav = = (7.2)
t2 − t1 ∆t
is the average angular velocity during this time interval.
The unit for angular velocity is radians/second (rad/s). Another unit that is Figure 7.4:
used in industry is revolutions per minute (rpm):
1 revolution 2π rad
1 rpm = = ≈ 0.10 rad/s
minute 60 s
As in linear motion, instantaneous angular velocity is defined by taking
the derivative at the limit ∆t → 0 :
∆θ dθ
ω = lim = (angular velocity) (7.3)
∆t→0 ∆t dt
Angular Acceleration (α)
Angular acceleration is defined as the change of angular velocity in unit time
and the ratio
ω2 − ω1 ∆ω
αav = = (7.4)
t2 − t1 ∆t
is the average angular acceleration during this time interval.
The ∆t → 0 limit of average acceleration is instantaneous angular accel-
eration:
∆ω dω
α = lim = (angular acceleration) (7.5)
∆t→0 ∆t dt
The unit of angular acceleration is rad/s2 . Acceleration is positive if the object
gets faster as it rotates in the counterclockwise direction and negative if it gets
slower.
Let us emphasize once more: Angular velocity and angular acceleration have
the same value at all points of the rigid body.
Motion With Constant Angular Acceleration
The angular acceleration α of a rigid body is constant if its angular velocity
changes uniformly. In such a case, we get formulas for angular position and
angular velocity, just as we did for linear motion in Chapter 2. There is no need to
repeat this calculation. The results are displayed in the following table, together
with the linear motion formulas:
Circular and linear motions with constant acceleration

Circular Linear
ω = ω0 + αt v = v0 + at
θ = θ0 + ω0 t + 21 αt2 x = x0 + v0 t + 12 at2
ω2 − ω20 = 2α(θ − θ0 ) v2 − v20 = 2a(x − x0 )
Example 7.1
with correct units rad/s :
2π rad
A wheel at rest starts to rotate with constant angular accelera- 90 × = 0 + α × 10 → α = 0.3π rad/s2
tion and reaches an angular velocity of 90 rpm after 10 s . 60 s
(b) We use the angular position formula:
(a) What is the angular acceleration of the wheel?
θ = θ0 + ω0 t + 21 αt2
(b) How many revolutions has it performed during this period?
Here, we take the point θ0 = 0 of the wheel:
Answer θ = 0 + 0 + 12 × 0.3π × 102 = 15π radians
(a) We use the velocity formula for constant acceleration: The angle is requested in terms of revolutions. One revolution
ω = ω0 + α t is taken as 2π :
The radian unit must be used in all calculations. The initial 15π
θ = 15π radians = = 7.5 revolutions .
velocity is zero: ω0 = 0 . Let us substitute the final velocity 2π
Example 7.2 ω − ω0 (150 − 600) × π/30

α= =
t 3
An electrical motor rotating with an angular velocity of
α = −5π rad/s2
600 rpm starts to decelerate when the power is disconnected
(b) We use the angular position formula:
and its angular velocity decreases to 150 rpm in 3 s .
600 × π
(a) What is the angular acceleration? θ = θ0 + ω0 t + 21 αt2 = 0 + × 3 + 12 × (−5π) × 32
(b) How many revolutions will it perform during this time 30
θ = 37.5π radians = 18.8 revolutions .
interval?
(c) The final velocity is zero when the motor stops. We use
(c) How many revolutions will it perform from the start until
the velocity formula without time:
it stops?
ω2 − ω20
Answer ω2 − ω20 = 2αθ → θ =
2α
(a) We substitute the given velocities in the angular velocity 0 − (600π/30)2
formula in terms of rad/s and find the angular acceleration: θ= = 40π radians = 20 revolutions
2 × (−5π)
Relation Between Angular and Linear Kinematics

When a rigid body performs rotational motion, any point on it makes a
circular motion with a certain radius. That point has some well-defined linear
speed, tangential and centripetal accelerations. It is important to know how these
quantities are related to angular velocity and acceleration.
Let us rewrite Eq.(7.1) which gives us the relation between angular and linear
positions:
s = rθ
As the angle θ changes over time, so will the arc s . Taking the derivative of both
sides with respect to time t , we find that
Figure 7.5: Angular and linear
coordinates. ds d(rθ) dθ
= =r
dt dt dt
7.1. ANGULAR KINEMATICS 117
where the constant radius r was moved outside of the derivative. The derivative
on the right-hand side is the angular velocity ω and the derivative on the left-hand
side is just the linear velocity v of point P :
v = rω (7.6)
Acceleration has two components in circular motion (Figure 7.6). Let us

remember the formula for the centripetal acceleration:
v2
ar =
r
Substituting the expression v = r ω that we found above for linear velocity, we
can write centripetal acceleration in terms of ω :
Figure 7.6: Tangential and cen-
(r ω)2 tripetal accelerations.
ar = = r ω2 (7.7)
r
For tangential acceleration, let us rewrite the expression (3.19) that we
found in Chapter 3:
dv
at =
dt
If we substitute the expression v = rω that we found for linear velocity and take
the derivative, as the radius r is constant, we get
dω
at = r = rα (7.8)
dt
These formulas completely establish the relation between the rotational mo-
tion of the solid body and the linear motion of any point on it. Let us summarize
these formulas in a table:
The relation between angular and linear kinematics
Angular Linear
Position θ s = rθ
Velocity ω v = rω

 ar = r ω2 (centripetal acceleration)

Acceleration α

 at = r α (tangential acceleration)


Example 7.3
π/30=2π . We find angular acceleration using the velocity
formula without time: ω2 − ω20 = 2α(θ − θ0 )
A wheel starts from rest and accelerates at a constant rate to
ω2 − 0 (2π)2
reach a velocity of 60 rpm after 5 revolutions . α= = = 0.4π rad/s2
(a) What is the angular acceleration of the wheel? 2(θ − 0) 2 × 5π
It is a good practice to leave π in angular calculations.
(b) What are the linear velocity and tangential and centripetal
They are either canceled out later or the value π2 ≈ 10 can
accelerations of a point located at a distance of 2 m from
be substituted.
the center of the wheel?
(b) We use the formulas for conversion to linear kinematics
Answer (a) The angular velocity of the wheel is 60 rpm=60× for r = 2 m :
Linear velocity: v = r ω = 2 × 2π = 4π m/s Centripetal acceleration: ar =r ω2 =2×(2π)2 =8π2 ≈80 m/s2 .

Tangential acceleration: at = r α = 2 × 0.4π = 0.8π m/s2
Example 7.4
eration. In particular, the linear speeds of points on the rim
of both disks must be equal:
v = v1 = v2
We use the relation v = rω for each wheel:
r1
r1 ω1 = r2 ω2 → ω2 = ω1
r2
There is no need to change units here, because we are working
with ratios:
10 cm
ω2 = × 40 rpm = 80 rpm
5 cm
Now, in addition to linear velocities, the linear acceleration
On a bicycle the crankset (toothed disk) fixed to the pedal has of the chain is also the same everywhere. In particular, at the
a radius of 10 cm and the sprocket (smaller toothed disk) at points on the rim, they become the tangential acceleration
the center of the rear wheel has a radius of 5 cm . The two are for each circular motion:
connected by a chain. a = a1t = a2t
(a) If the angular velocity of the crankset is ω1 =40 rpm and Using the relation at =rα that connects the tangential and
its angular acceleration is α1 =7 rad/s2 , what will be theangular acceleration for each wheel,
angular velocity and angular acceleration of the rear
r1 10 cm
wheel? α2 = α1 = × 7 rad/s2 = 14 rad/s2
r2 5 cm
(b) Calculate the centripetal accelerations of points on the rim
(b) The formula ar = rω2 giving the centripetal acceleration
of the crankset and of the sprocket.
is used for both disks:
Answer
(a) The chain is the connection between the two wheels. All a1r = r1 ω21 = 0.10 × (40 × π/30)2 = 1.8 m/s2
of the points on this chain have equal linear speed and accel- a2r = r2 ω22 = 0.05 × (80 × π/30)2 = 3.6 m/s2
7.2 TORQUE (MOMENT OF A FORCE)

How do we open a door? The door will open (Figure 7.7) if we push or pull
with a perpendicular force ( F1 ) from the free side of the door. However, it will
be difficult to open it if we apply the same force on a point near the hinges ( F2 ).
Likewise, it will not move if we pull with a force along the door ( F3 ). The ability
Figure 7.7: Which of these of the same force to rotate an object is different in all of these cases.
forces will rotate the door more The ability of a force to rotate an object is called its torque (or, its moment).
easily? We shall introduce the torque in a simple way, then generalize.
Definition: Let ~F be a force acting at a position ~r (Figure 7.8). The magnitude
of the torque about the origin O is,
τ = F r sin θ (7.9)
where θ is the angle between ~F and ~r .

The sign of the torque depends on the direction of rotation. According to the
most commonly accepted convention, the torque is positive if the force is trying
to rotate the stick r in a counterclockwise direction. The torque is negative if the
force is trying to rotate in a clockwise direction.
Figure 7.8: The torque of a The unit of torque is newton × meters (N·m) and does not have a special
force. name.
We can write the definition above in two different ways: In the first, the factor
F sin θ is the component F⊥ that is perpendicular to the direction r of the force
7.2. TORQUE (MOMENT OF A FORCE) 119
(Figure 7.9a):
τ=| sin}θ r = F⊥ r
F {z (7.10)
F⊥
Figure 7.9: Two methods of cal-

culating torque.
On the other hand, considering that the product r sin θ is just the opposite
perpendicular side of the triangle (Figure 7.9b), we can write
τ = F |{z}
r sin θ = F d (7.11)
d
The distance d here is called the lever arm. Whichever expression is suitable
for the problem at hand should be used. Usually, the expression F⊥ r is more
practical.
Expression of Torque As a Vector Product
The factor sin θ in the defining expression (7.9) of the torque suggests that it
can be expressed as a vector product. Indeed, if the position vector ~r is measured
from the rotation axis to the application point of the force ~F , the torque vector
can be expressed as follows:
~τ = ~r × ~F (7.12)
Figure 7.10 shows that this expression satisfies the definition given above: Let
us assume that the vectors ~r and ~F are on the xy -plane. The direction of the Figure 7.10: Expression of
vector product is found using the right-hand rule. According to this definition, torque as vector product.
the direction of the torque vector will be along z , which is the rotation axis. If
the rotation of the force about the z -axis is counterclockwise, then the torque
will be in the +z direction, in other words, positive. If it rotates in the opposite
direction, the torque will be negative.
As the forces that we will take into consideration in this chapter are all on
a plane, the torques will always be along the z -axis, with positive or negative
components. We can thus work without referring to the vector nature of torque.
Example 7.5
τ1 = r F1⊥ = 0 , because the component of the force perpen-
dicular to r is zero. Torque is always zero if the line of the
force passes through the center of rotation.
τ2 = r F2⊥ = 2 × 10 = +20 N·m . Torque is positive, because
the direction of rotation of the force F2 is counterclockwise.
τ3 = −rF3⊥ = −1 × 10 = −10 N·m . Torque is negative,
All of the forces shown in the figure have the same magnitude because the direction of rotation of the force F is clockwise.
3
of 10 N . Calculate the torques of these forces with respect to τ = F d = 0 . Torque is zero, because the lever arm d=0 .
4 4
the point O.
τ5 = −rF5 sin 37◦ = −1 × 10 × 0.6 = −6 N·m . Torque is
Answer We use one of the expressions τ=rF⊥ or τ=F d for negative, because the direction of rotation of the force F5 is
torque: clockwise.
7.3 ROTATIONAL DYNAMICS

Rotational Dynamics of a Pointlike Object
Before expressing Newton’s law for a rigid body, let us first demonstrate our
method on a simpler motion.
Let a pointlike mass m rotate on a circular trajectory with radius r and
with a net force ~F acting on it (Figure 7.11). Let us separate this force into two
components: The component Fr acts towards the center O, and thus causes the
centripetal acceleration ar . The other component Ft acts along the tangent and
thus causes the tangential acceleration at .
Let us write the Newton’s law for these two forces and convert linear acceler-
ations into angular quantities:
Figure 7.11: Tangential and ra-
dial components of forces acting Fr = mar = mrω2
on a point body in circular mo- Ft = mat = mr α
tion.
The first of these equations specifies the centripetal force required to rotate the
object at the angular velocity ω . We set it aside to concentrate on the second
equation which gives the tangential acceleration. Let us write it as multiplied by
r:
Ft r = mr2 α
The left-hand side of the equation is the torque of the force F with respect to the
center O:
τ = (mr2 ) α (7.13)
This expression resembles the Newton law F=ma . However, force F is replaced
with the torque τ , acceleration a is replaced with angular acceleration α and the
mass is replaced with product (mr2 ) . This different expression of Newton’s law
is actually nothing new. But it gives us a clue as to how to write the rotational
motion of the rigid body. The fact that all of the particles of the rigid body are
rotating with the same angular acceleration α is explicit in the right-hand side of
the equation.
Rotational Dynamics of a Rigid Body
Now, we can consider the dynamical law that determines the rotation of a
solid object. Let us assume that a rigid body rotating about an axis (Figure 7.12) is
made of N number of pointlike masses ∆m1 , ∆m2 . . . ∆mN . Let the external forces
acting on each of these masses be ~F1 , ~F2 . . . ~FN . (There is no need to consider the
internal forces because they will cancel each other out in the sum.)
For each small mass we write the equation (7.13) that we developed for a
pointlike object, to get a system of equations:
Figure 7.12: External forces act-
ing on a rigid body. τ1 = F1t r1 = (∆m1 r12 ) α
τ2 = F2t r2 = (∆m2 r22 ) α
··· = ···
τN = F Nt rN = (∆mN rN2 ) α
7.3. ROTATIONAL DYNAMICS 121
Adding these equations on each side, the angular acceleration α on the right-hand
side becomes the common factor:
τ1 + τ2 + · · · + τN = ∆m1 r12 + ∆m2 r22 + · · · + ∆mN rN2 α

X X
τi = ∆mi ri2 α
i i
The left-hand side of this equation is the total torque of the forces acting on the
rigid body. The sum in the brackets on the right-hand side becomes a new term in
which each point mass is multiplied by the square of its own radius ri and added.
This is called the moment of inertia of the rigid body and is shown with I :
X
I= ∆mi ri2 (moment of inertia) (7.14)
i
Moment of inertia serves the function of mass in rotational motion; in other

words, the larger I is, the more difficult it becomes to give angular acceleration
to the object.
As the distances ri in this definition vary depending on the chosen axis of
rotation, the moment of inertia will depend on the chosen axis of rotation.
For a continuously distributed mass, the sum is replaced by an integral in the
limit ∆mi → 0 : Z
I= dm r2 (moment of intertia) (7.15)
As a result, the equation for the rotational motion of a solid object is written as
follows:
τnet = I α (Rotational dynamics of a rigid body) (7.16)
We must emphasize one small point here: Previously, we had specified the
positive direction of the torque as being counterclockwise. However, in calcula-
tions the positive direction of the torque can be chosen in the direction of motion,
and not the clock direction. In other words, the torques in the direction that
the rigid body is assumed to rotate are taken as positive and the torques in the
opposite direction are taken as negative.
Example 7.6
wheel.
(b) A block with mass m=1 kg is attached at the end of the
rope and released. Calculate the accelerations and the
Answer
The reaction forces that may be generated on the axis of the
wheel are not taken into account, because their torques will
A wheel with radius R=2 m and mass M=5 kg can rotate with- be zero as they pass through the axis of rotation. Accordingly,
out friction about its center. Its moment of inertia is I= 12 MR2 . only the torque of force F will cause motion. We write the
(a) The wheel is pulled with a force F=10 N from the end of equation of rotational dynamics (7.16):

a rope tied around it. Find the angular acceleration of the τnet = I α → F.R = 12 MR2 α
2F 2 × 10 Therefore, an extra equation is required. This is obtained
α= = = 2 rad/s2
MR 5×2 from the relation between linear and angular kinematics.
(b) This time, there are two moving objects: The rotating
Since each point of the rope will have the same acceleration
wheel and the mass m in linear motion. The forces acting on
a , it will be the tangential acceleration of the wheel at the
each one are shown below as free-body diagrams:
point of contact with the wheel. Therefore, the third equation
is as follows:
a = at = R α (3)
Taking the value T from equation (1) and using it with the
a = Rα in equation (2), we get
mg 10
Taking the direction of motion as positive for both objects, α= R= = 1.4 rad/s2
(m + M/2) (1 + 5/2) × 2
the equations of motion are as follows:
From here, we find the other unknowns:
For M : τnet = T R = I α = 12 MR2 α (1)
For m : Fnet = mg − T = ma (2) a = Rα = 2.8 m/s2
These are two equations with three unknowns T, α, a . T = m(g − a) = 7 N .
Example 7.7
us calculate their torques separately in order to determine
which is greater:
τ1 = F1 L = 6 × 5 = 30 N·m
τ2 = r F2⊥ = r F2 sin 37◦ = 2 × 15 × 0.6 = 18 N·m
The rod will rotate in the direction of F1 , because the value
A rod of length L = 5 m can rotate without friction around its τ1 is greater.
end O. The moment of inertia of the rod with respect to point We choose the positive rotation direction for F1 and write
O is I0 = 5 kg·m2 . The two forces shown in the figure are the rotational dynamics equation:
acting on this rod. Find the direction of rotation and angular τnet = τ1 − τ2 = I0 α
acceleration of the rod. From here, we find the angular acceleration:
Answer 30 − 18
α= = 2.4 rad/s2
The torques of the two forces are in different directions. Let 5
Example 7.8
We write the equations of motion for each object (the weight
of the mass on the plane and its reaction force N cancel each
other out and therefore are not taken into consideration.):
For m1 : m1 g − T 1 = m1 a (1)
For m2 : T 2 = m2 a (2)
A block of mass m1 = 1 kg is attached to one end of a rope For the pulley: T1R − T2R = I α (3)
going around a pulley. Another block of mass m2 = 2 kg at rest
on a frictionless horizontal plane is tied to the other end of the Additionally, we write the relation between the accelerations
rope. The pulley has a mass m2 = 2 kg , a radius R = 20 cm as follows:
and a moment of inertia I = 3 kg·m2 . Calculate the linear and a = Rα (4)
angular accelerations and the tensions in the ropes.
We find the accelerations and tensions from these four equa-
Answer tions. Taking the tensions from (1) and (2) and substituting
The important point to note here is that the tensions on two in equation (3), we get
ends of the rope will be different. For, if the tensions were
equal, there would be no net torque to rotate the pulley. The m1 gR 10 × 0.2
α= = = 0.64 rad/s2
forces acting on each object can thus be shown as follows: I + (m1 + m2 )R 2 3 + 3 × 0.22
Using these, we calculate the other unknowns:
a = Rα = 0.2 × 0.64 = 0.13 m/s2
T 1 = m1 (g − a) = 9.9 N
T 2 = m2 a = 0.3 N .
7.4. MOMENT OF INERTIA CALCULATIONS 123
Example 7.9
Since m2 R2 > m1 R1 , the direction of motion is obvious. We
take the direction of motion of each object as positive and
write the equations as follows:
For m1 : T 1 − m1 g sin 37◦ = m1 a1 (1)
For m2 : m2 g − T 2 = m2 a2 (2)
For the pulley: T 2 R2 − T 1 R1 = I α (3)
A rope is wrapped around the smaller radius R1 = 10 cm As the linear accelerations will again be the tangential accel-
of a pulley with moment of inertia I = 3 kg·m2 and the other eration of the pulley, we write the relations between linear
end of the rope is tied to a block with mass m1 = 1 kg on a and angular accelerations:
frictionless inclined plane with 37◦ slope. A block with mass a1 = R1 α (4)
m2 = 2 kg is tied to the end of a rope going around the outer a2 = R2 α (5)
radius R2 = 20 cm of the pulley. Calculate the accelerations We find the accelerations and tensions from these five equa-
and the tensions in the ropes when this system is released. tions. Taking T 1 and T 2 from (1) and (2) and substituting in
Answer equation (3), we find the angular acceleration:
The linear accelerations of the blocks are different in this m2 gR2 − m1 gR1 sin 37◦
α= = 1.1 rad/s2
problem because they are tied at different radii. Let us show I + m1 R21 + m2 R22
the forces acting on each object: We then calculate the accelerations and tensions using this
value:
a1 = R1 α = 0.1 × 1.1 = 0.11 m/s2
a2 = R2 α = 0.22 m/s2
T 1 = m1 g sin 37◦ + m1 a1 = 6.1 N
T 2 = m2 g − m2 a2 = 19.6 N .
7.4 MOMENT OF INERTIA CALCULATIONS

The moment of inertia, behaving like a mass in rotational motion, depends
not only on the mass, but also on how the mass is distributed around the axis of
rotation.
Moment of inertia can be calculated using two methods:
• If the rigid body consists of pointlike objects, the moment of inertia of each
one is added using Eq. (7.14):
X
I= mi ri2
i
• If the distribution of mass is continuous, we take the integral Eq. (7.15):

Z
I= dm r2
Let us summarize the results of the calculation before delving into moment
of inertia calculations:
Moments of inertia of various rigid objects
Ring : I = MR2 Rod : I= 1

ML2
12
Disk or cylinder Rod : I= 1

3 ML2
I = 12 MR2
Rectangular plate Sphere : I= 2

5 MR2
I= 1
12 M(a2 + b2 )
Parallel Axis Theorem

The value of moment of inertia depends on the chosen axis. The table above
gives the moments of inertia Icm of rigid bodies about an axis passing through
their centers of mass. If the rigid body rotates about another axis, the moment of
inertia can still be calculated using a result known as the parallel axis theorem
(or, Steiner theorem):
If the moment of inertia of an object is Icm with respect to an axis passing
through its center of mass, then its moment of inertia with respect to another axis
Figure 7.13: Parallel axis theo- that is parallel at a distance d will be
rem.
I = Icm + M d2 (Parallel axis theorem) (7.17)
Here, M is the mass of the rigid body. We give this without a proof here. Accord-
ing to this theorem, considering that the term Md2 is positive, the axis where the
moment of inertia is the lowest, in other words, where it can rotate most easily,
is the axis passing through its center of mass.
Example 7.10
three masses shown in the figure with respect to each of the x -,
y - and z -axes.
Answer The moment of inertia will be i mi ri2 for point
P
masses. The mass m1 will have no contribution, because it is
on the axis, and therefore r1 = 0 .
We substitute the given values of m and ri :
I x = m2 r22 + m3 r32 = 0 + 5 × 22 = 20 kg·m2
Iy = m2 r22 + m3 r32 = 3 × 42 + 0 = 48 kg·m2
Calculate the moment of inertia of a system consisting of the Iz = m2 r22 + m3 r32 = 48 + 20 = 68 kg·m2
7.4. MOMENT OF INERTIA CALCULATIONS 125
Example 7.11
(a) Find the moment of inertia of a ring with mass m and
radius R about an axis passing through its center O and
perpendicular to the plane of the ring.
(b) Using the previous result, find the moment of inertia of a
disk with mass M and radius R about a perpendicular
axis passing through its center.
(b) Let us consider a small ring between the radii r and r + dr
on the disk. Its surface area will be approximately 2πr dr . If
the total mass is M , we find the mass of this small disk using
proportion:
dm 2πr dr M
= → dm = 2 2r dr
M πR2 R
We had found in item (a) that the contribution of this small
Answer disk is dm r2 . Therefore, their contributions are added, in
(a) We choose a small mass dm on the ring and use the integ- other words, integrated from r = 0 to r = R :
ral formula
Z (7.15):
Z Z R
M 2M R 3
Z
= 2
= 2
=
Z
I dm r 2r dr r r dr
I= dmR2 = R2 dm = MR2 0 R2 R2 0
As each mass dm is located at a constant distance R , it is The value of the integral is R /4 , and after simplification, we
4
taken outside the integral. The sum of dm is equal to the find the moment of inertia of the disk:
total mass M . I = 1
2 MR 2
Example 7.12 2
Solid sphere: Icm = MR2
5
The moment of inertia of each object with respect to the y -
axis is found using the parallel axis theorem, Eq. (7.17). The
distance of the parallel axis is L/2 for the rod and (L + R)
A solid sphere with mass M and radius R is welded to the end for the sphere:
of a rod with mass m and length L . Calculate the moment of L 2 1 1 1
inertia of this system with respect to the y -axis on the other Rod: Iy = Icm + m = mL2 + mL2 = mL2
2 12 4 3
end. 2
Solid sphere: Iy = Icm + M(L + R)2 = MR2 + M(L + R)2
Answer 5
The moments of inertia of the rod and sphere with respect to The total moment of inertia of the system is the sum of these
their own centers of mass are given in the table on Page 124: two:
1 1 2
Rod: Icm = mL2 Iy,total = mL2 + MR2 + M(L + R)2
12 3 5
Rotational Kinetic Energy

We cannot calculate the kinetic energy of a rotating rigid body using the
formula 12 mv2 because the linear velocity of each point is different. Instead, we
have to write an expression in terms of angular velocity.
Let as assume that the rigid body consists of N number of small masses
m1 , m2 . . . mN . Let the linear velocity of each of these masses be v1 , v2 . . . vN . The
kinetic energy of the rigid body will be the sum of the kinetic energies of these
pointlike masses:
K = 12 m1 v21 + 21 m2 v22 + · · · + 21 mN v2N
Let each of these masses rotate at distances r1 , r2 . . . rn from the axis. If we write
the relation v = r ω between the linear and angular velocity for each one, we get
K = 2 m1 (r1 ω) + 2 m2 (r2 ω) + · · · + 2 mN (rN ω)
1 2 1 2 1 2
h i
= 1
2 m 1 r1 + m2 r2 + · · · + mn rN ω
2 2 2 2
| {z }
I
Therefore, rotational kinetic energy can be expressed as follows:
K = 12 I ω2 (rotational kinetic energy ) (7.18)
Note the similarity of this expression with 12 mv2 for translational kinetic energy.
Therefore, the law of conservation of energy can still be valid using this expression
for rigid bodies.
Example 7.13
ley and the mass. The potential energy of the mass m will
A mass m=1 kg is attached to the end of a rope that is wrapped be completely converted into kinetic energy when the mass
around a pulley with radius R=20 cm and with moment of in- drops by h :
ertia I=0.5 kg·m2 . The mass is released at a height of h=2.7 m mgh = 12 mv2 + 12 Iω2
from the ground. With what velocity will it hit the ground? We write the relation between the linear velocity v of mass
m and the angular velocity of the pulley:
v = Rω
Substituting these values, we find the angular velocity:
r r
2mgh 20 × 2.7
ω= = = 10 rad/s
I + mR2 0.5 + 1 × 0.22
From this, we get the linear velocity:
Answer We write conservation of energy both for the pul- v = Rω = 2 m/s
Example 7.14
Answer
The potential energy of the rod will turn into kinetic energy
once it reaches the vertical position and its center of mass
descends by L/2 :
mg (L/2) = 12 I0 ω2
The moment of inertia here is with respect to the axis of
rotation O. We can calculate it from Icm with the help of the
parallel axis theorem:
I0 = Icm + m(L/2)2 = 12 1
mL2 + mL2 /4 = 13 mL2
A rod with mass m and length L = 1.2 m is hinged from one Substituting these values, we find the angular velocity:
end to a wall. The rod is released in horizontal position. What
r r
3g 30
will its angular velocity be once it reaches vertical position? ω= = = 5 rad/s .
L 1.2
7.5 ROLLING MOTION

In the most general motion of rigid bodies, translation and rotation can occur
simultaneously. This general motion may be very complex. But still, it can always
be examined by separating it into two components:
1. Translational motion of the center of mass: This motion is determined
with Newton’s law:
~Fi = m ~acm
X
(for translation) (7.19)
i
2. Rotational motion about the center of mass: This is determined with

the rotational dynamics law:
X
τi,cm = Icm α (for rotation) (7.20)
i
7.5. ROLLING MOTION 127
The torques here should be taken with respect to the center of mass.
In the most general motion, there is no relation between linear acceleration
acm and angular acceleration α , and each may have a different value. However,
these two accelerations are related for an object that rolls on a surface without
slipping.
Let us consider a cylinder that rolls on a horizontal plane without slipping.
The velocity of each point of this cylinder is different, but there is one point that
is at rest, in other words, has zero velocity for one instant. This is point P, where
the object is in contact with the surface (Figure 7.14). Indeed, if the object is not
slipping, the two common points where the surface and the object meet must
have equal velocity. And this means that the velocity of the point on the object is
zero.
This is true only for one instant. Immediately afterwards, this contacting Figure 7.14: Rolling motion.
point of the object will detach from the surface and gain velocity. However,
another point will take its place and its velocity will also become zero at the
moment that it contacts the surface. This point of contact is the instantaneous
axis of rotation. Therefore, we can write the linear velocity and acceleration of
the center of mass, which is at a distance R from this axis, in terms of angular
quantities:
vcm = R ω and acm = R α (Rolling conditions) (7.21)
Rolling motion can be easily calculated by adding these conditions to the afore-
mentioned translation and rotation equations.
Important note: At the point of contact between the rolling object and the
surface, there is also a friction force f . However, the formula f = µN cannot be
used here, because the object is not slipping and that friction has not reached its
maximum value.
In rolling motion, as translation and rotation take place together, the kinetic
energy is the sum of both types of energy:
K = 12 mv2cm + 12 Icm ω2 (rolling kinetic energy) (7.22)
This kinetic energy expression is used in energy conservation problems that

include rolling motion. (Note: It may seem surprising that it is possible to write
the conservation of energy despite the fact that there is a friction force in rolling
rolling. However, the friction force performs no work, as the point of the object
contacting the surface is motionless.)
Example 7.15
angle of 37◦ .
Answer
The forces acting on the cylinder are shown in the figure. We
can explain the direction of the friction force f as follows:
If this was an inclined plane covered with ice, the cylinder
would have slipped down without rolling. Therefore, friction
would be upwards.
Let us write Eqs. (7.19 and 7.20) that we found for rolling
Find the angular acceleration of a cylinder with radius R = motion, with the indicated axes (we denote acm = a ). For
50 cm and mass M = 4 kg released on a plane inclined at an translational motion, we have:
In the x -direction: Mg sin θ − f = Ma between the linear and angular accelerations:

In the y -direction: N − Mg cos θ = 0 a = Rα
(The y -equation will not be necessary, because the formula f is eliminated between these equations:
f = µN does not apply.) MgR sin 37◦
α=
For rotational motion: The friction force f is what ensures I + MR2
rotation around CM, and the torques due to other forces are Substituting the moment of inertia of the cylinder I = 12 MR2
zero: and the other numerical values,
τ= f R= Iα MgR sin 37◦ 2g sin 37◦ 20 × 0.6
α= 3
= = = 8 rad/s2 .
Also, since the cylinder is not slipping, we have the relation 2 MR 2 3R 3 × 0.5
Example 7.16
Translation in the x -direction: F − f = Ma
Find the angular acceleration and the acceleration of the cen- Rotation around CM: FR1 + f R2 = I α
ter of mass of a cylinder with radius R2 = 40 cm and mass The relation between the accelerations is: a = R2 α
M = 1 kg pulled with force F = 12 N through a rope wrapped Eliminating the force f between these two equations, we
around the cylinder at R1 = 30 cm from the center. find the acceleration α :
F(R1 + R2 )
α=
I + MR22
Substituting the moment of inertia of the cylinder I= 12 MR22
and the other numerical values,
F(R1 + R2 ) 12 × (0.3 + 0.4)
α= 3 2
= 3
= 35 rad/s2
MR × 0.42
2 2 2
We find the linear acceleration of the center of mass, which
Answer is located at a distance of R2 from the instantaneous center
The forces acting on the cylinder are shown in the figure. We of rotation:
again apply the formulas (7.19 and 7.20): a = R2 α = 0.4 × 35 = 14 m/s2 .
Example 7.17
Mgh = 21 mv2 + 21 I ω2
Since there is no slipping, we can use the relation v = Rω
A sphere with radius R and mass M released on an inclined
between the linear and angular velocities. Substituting the
planed at height h from the ground rolls down without slipping.
expression I = 52 MR2 in this equation as well, we get:
What will the velocity of the center of mass be once it reaches r
the ground? (For sphere: Icm = 25 MR2 ) v=
10gh p
= 1.4gh
7
Answer Considering that a pointlike object would have a velocity
v = 2gh after descending by h , we see that the sphere goes
p
According to conservation of energy, the initial potential en- √ √
ergy of the sphere at height h is converted into translational down slower, as the factor 1.4 is less than 2 . The reason
and rotational kinetic energies at the bottom: for this is that some of the energy is used for rotation.
7.6 ANGULAR MOMENTUM AND ITS CONSERVATION

In translational motion, momentum was defined as the product of mass and
velocity: ~p = m~v . Likewise, an angular momentum is also defined for rotational
motion. By using moment of inertia I instead of mass and angular velocity ω
instead of velocity, we get the expression
L= Iω (angular momentum) (7.23)
defining the angular momentum of a rigid body. Its unit is kg·m2 /s and does
not have a particular name.
7.6. ANGULAR MOMENTUM AND ITS CONSERVATION 129
Recall how we expressed Newton’s law in terms of momentum in linear

motion:
dv d(mv) d p
F = ma = m = =
dt dt dt
The rotational motion equation can likewise be expressed as follows:
dω d(Iω)
τ = Iα= I =
dt dt
dL
τ = (7.24)
dt
In the simplest case, the angular momentum of a point object traveling at
velocity v , with respect to an axis of rotation at a distance of r , is as follows:
L = Iω = (mr2 ) ω = mr (rω)
L = mvr (angular momentum of a pointlike object) (7.25)
Angular momentum is actually a vector quantity. Its most general definition

is given as the torque of the linear momentum vector ~p (Figure 7.15):
~L = ~r × ~p
Its direction is perpendicular to the object’s plane of rotation. We will not use the
vector properties of angular momentum here and provide it merely as information.
If the net torque of external forces acting on a rigid body is zero, then its Figure 7.15: The angular mo-
angular momentum will remain constant, according to Eq. 7.24. This expression mentum vector ~L is perpendic-
is the law of conservation of angular momentum: ular to the rotation plane.
dL
τ = 0 =⇒ =0
dt
L1 = L2 = constant (conservation of angular momentum) (7.26)
When an object is rotating, even if the torque of external forces is zero, its
moment of inertia I may change with the impact of internal forces. For example,
a man on a rotating platform may increase his moment of inertia by spreading his
arms. As the final situation can again be considered as a rigid body, the angular
velocity will decrease to conserve angular momentum.
Example 7.18
Answer
We write the conservation of angular momentum for a point-
like object:
mvr = mv0 r0
We write the linear velocities in terms of angular velocity
using the formula v = rω :
A mass m is rotating with angular velocity ω = 3 rad/s at
r2
the end of a rope with length r = 50 cm tied to an axis on a r2 ω = r02 ω0 → ω0 = 02 ω
r
frictionless table. As the rope starts to wrap around the axis, By substituting the value r0 = r/2 , we find the final angular
what will the angular velocity of the object be when the length velocity:
of the rope shortens to 25 cm ? ω0 = 4ω = 4 × 3 = 12 rad/s .
Example 7.19
We write the conservation of angular momentum:
A bullet with mass m = 50 g travels at a speed of v = 200 m/s Lbullet + Ldisk = Lbullet+disk
0
and embeds itself into the rim of a disk at rest that can rotate Initially, there is only the angular momentum mvR of the
freely, at a distance of R = 60 cm from the center. As the disk’s point mass m . After the collision, this mass will rotate with
mass is M = 900 g and the moment of inertia is I = 12 MR2 , an angular velocity ω together with the disk at a distance R .
find the angular momentum of the system (M + m) after the Therefore,
collision. mvR + 0 = (Idisk + Ibullet ) ω
Substituting the moments of inertia of the disk I = 12 MR2
and of the bullet I = mR2 , we find that:
mvR
ω=
(M/2 + m)R2
We find the angular velocity by substituting the numerical
Answer values:
The forces generated during the collision are the internal mv 0.05 × 200
ω= = = 33 rad/s
forces of the system (bullet+disk) and their net torque is zero. (M/2 + m)R (0.9/2 + 0.05) × 0.6
Example 7.20
direction with angular velocity ω2 =100 rpm . What will be the
final angular velocity when the disks are suddenly clamped
together?
Answer We write the conservation of angular momentum:
I1 ω1 + I2 ω2 = (I1 + I2 ) ω
I1 ω1 + I2 ω2
In the transmission box of a car, two disks are rotating, one ω=
connected to the engine and the other connected to the trans- I1 + I2
If we take ω1 in the positive rotation direction, then ω2 =
mission shaft. The moment of inertia of the disk connected
−100 rpm . From here, we find the final angular velocity (there
to the engine is I1 =2 kg·m2 and it rotates with angular veloc-
is no need to change units, as the ratios of I are used):
ity ω1 =2000 rpm . The moment of inertia of the disk on the 2 × 2000 + 5 × (−100)
transmission shaft is I2 =5 kg·m2 and it rotates in the opposite ω= = 500 rpm
2+5
1. Which of the following formulas is incorrect?
5. Which of the following is correct?
(a) v = rω (b) at = rα (c) ar = rω2 (d) L = Iω2
(a) An object will not rotate if the net force acting on it
2. Which of the following is incorrect for the definition of is zero.
the torque of a force? (b) An object will not rotate if the net torque acting on
(a) Force×lever arm it is zero.
(b) Distance×perpendicular component of force (c) An object will rotate with a constant angular veloc-
(c) Force×distance ity if the net torque acting on it is zero.
(d) Force×distance2 (d) None of the above.
3. A disk’s mass is doubled and its radius tripled. By what

factor will its moment of inertia increase?
(a) 5 (b) 6 (c) 12 (d) 18
4. Three forces act on an object perpendicular to the axis

of rotation. The force F acts at a distance of 4R , the
6. The rod with four equal parts marked in the figure can
force 2F acts at a distance of 3R and the force 3F acts
rotate around the O-axis. Which of the forces shown
at a distance of 2R from the axis. Which one applies the
will exert the largest torque?
lowest torque?
(a) F (b) 2F (c) 3F (d) F & 3F (a) 4 N (b) 6 N (c) 7 N (d) 8 N
7. Which of the following is incorrect about the moment (a) Centripetal accelerations are equal.
of inertia of an object? (b) Linear velocities are equal.
(c) Tangential accelerations are equal.
(a) It increases with mass.
(d) Angular velocities are equal.
(b) It is directly proportional to the square of the dis-
tance from the axis.
15. The moment of inertia of a 3 kg object with respect to its
(c) Its value depends on the chosen axis.
center of mass is Icm = 5 kg·m2 . What will its moment
(d) It increases with velocity.
of inertia be in units of kg·m2 with respect to a parallel
axis that is 2 m away?
8. Which of the following is incorrect if the angular accel-
eration of a rotating object is zero? (a) 10 (b) 15 (c) 17 (d) 20
I. Net torque is zero.
II. Angular velocity is zero.
III. Centripetal acceleration is zero.
IV. Tangential acceleration is zero.
(a) I (b) I & II (c) II & III (d) IV
16. Which of the objects in the figure above has the highest
9. A mass m is rotating around a circle with radius R . An- moment of inertia?
other mass 2m is rotating around a circle with radius (a) A (b) B (c) C (d) D
R/2 . And a third mass 4m is rotating around a circle
with radius R/2 . Which one has highest moment of 17. Which of the following is incorrect for a rolling objects?
inertia? (a) The linear velocity is different at each point.
(a) m (b) 2m (c) 4m (d) m & 4m (b) The velocity of the point in contact with the ground
is zero.
10. The same tangential force F is applied on a ring, on a (c) The angular velocity is the same everywhere.
disk, and on a sphere with the same masses and radii. (d) No friction force acts on the object.
Which one will accelerate faster?
(a) Ring (b) Disk (c) Sphere (d) Equal 18. The moment of inertia of a solid cylinder is 12 MR2 and
that of a hollow cylinder is MR2 . Two cylinders, one
solid and one hollow, with equal masses and radii are
11. A rotating object’s angular velocity is doubled and its
released at the same height on an inclined plane. Which
moment of inertia is tripled. By what factor will its
one will roll down and reach the bottom first?
kinetic energy increase?
(a) The solid cylinder.
(a) 5 (b) 6 (c) 12 (d) 18 (b) The hollow cylinder.
(c) They will arrive at the same time.
12. A rotating object’s angular velocity is tripled and its (d) It is impossible to tell.
moment of inertia is doubled. By what factor will its
angular momentum increase? 19. Which of the following is incorrect if the net torque on
(a) 5 (b) 6 (c) 12 (d) 18 an object is zero?
(a) The angular acceleration is zero.
13. What will happen to the Earth’s rotation velocity if the (b) The kinetic energy remains constant.
icebergs melt as a result of “global warming”? (c) The angular momentum increases uniformly.
(a) It will increase. (d) The angular momentum remains constant.
(b) It will decrease.
20. Which is correct for a rolling object?
(c) It will remain constant.
(d) It is impossible to tell. (a) The axis of rotation is the center of mass.
(b) The axis of rotation is the point in contact with the
14. Which of the following is correct for a point A located surface.
at a distance of 1 cm and a point B at a distance of 2 (c) The translational kinetic energy is zero.
cm from the center of a wheel rotating with constant (d) The rotational kinetic energy is zero.
angular velocity?
Problems
7.1 Angular Kinematics rod around the center of rotation O is I=5 kg·m2 , calculate
the direction of rotation and angular acceleration of the rod.
7.1 A wheel at rest starts to rotate and reaches an angular
[A: 0.8 rad/s2 , counterclockwise.]
velocity of 54 rpm in 3 s . Calculate the angular acceleration
of the wheel and the number of revolutions it makes during
this interval. [A: 0.6π rad/s2 and 1.4 revolutions .]
7.2 A motor is rotating at angular velocity 90 rpm and, due

to a power failure, stops after making 6 revolutions . Calcu-
late its angular acceleration and the time that it takes to stop.
[A: −0.38 rad/s2 and 8 s .] Problem 7.7
7.3 A wheel initially rotating at a certain angular velocity

suddenly starts to accelerate and reaches an angular velocity
of 120 rpm after making 3 revolutions in 2 seconds. Find the
angular acceleration of the wheel and its initial velocity.
[A: π rad/s2 and 2π rad/s .]
7.4 A wheel with a radius of 30 cm accelerates with angu-

Problem 7.8
lar acceleration 20 rad/s2 to reach angular velocity 45 rpm .
What will be the linear velocity and the tangential and cen- 7.8 The pulley in the figure has a radius of R=10 cm and
tripetal accelerations of a point on the rim of the wheel? moment of inertia I=0.1 kg·m2 . The masses m1 =1 kg and
[A: v = 0.45π m/s , at = 6 m/s2 , ar = 6.8 m/s2 .] m =2 kg are attached to the ends of a rope wrapped around
2
the pulley. Calculate the accelerations and the tensions in the
ropes when the system is released.
[A: α = 7.7 rad/s2 , a = 0.77 m/s2 , T 1 = 11 N , T 2 = 19 N .]
Problem 7.5
7.5 On a bicycle, the sprocket (small disk connected to the
rear wheel) has a radius of r1 =5 cm and the crankset (larger
disk connected to the pedal) has a radius of r2 =10 cm . The
Problem 7.9
biker starts from rest and increases the angular velocity of
the crankset to ω2 =30 rpm in 2 seconds. (a) What is the an- 7.9 The square plate seen in the figure has a side length
gular acceleration of the crankset? (b) What will the angular of 1 m and a moment of inertia I0 =7 kg·m2 with respect to
velocity and acceleration of the sprocket be at the end of this the center of rotation O. The forces F1 =10 N , F2 =20 N and
time? (c) What are the final centripetal accelerations on the F3 =30 N are applied on the corners of this square in the di-
rims of both disks? rections shown in the figure. What is the angular acceleration
[A: (a) α2 = 1.57 rad/s2 , (b) ω1 = 6.28 rad/s , of the plate? [A: 0.22 rad/s2 .]
α1 = 3.14 rad/s2 , (c) a1r = 2 m/s2 , a2r = 1 m/s2 .]
7.6 A rotating wheel with a radius of 80 cm starts to slow

down when the linear velocity of a point on the rim is 24 m/s
and stops in 5 seconds. (a) What are the angular acceleration
and the number of revolutions that it will make until it stops?
(b) What are the initial tangential and centripetal accelera- Problem 7.10
tions on the rim? 7.10 The pulley fixed to the upper end of the inclined plane
[A: (a) α = −6 rad/s2 , 12 revolutions , (b) at = −4.8 m/s2 , shown in the figure has a radius of R=20 cm and moment
ar = 720 m/s2 .] of inertia I=0.4 kg·m2 . The slope angle of the plane is 37◦
and the coefficient of friction is µ=0.2 . A block with mass
7.2-3 Torque - Rotational Dynamics
m=1 kg is attached to the end of the rope and released. Find
7.7 All of the forces shown in the figure below have the the accelerations and the tension in the rope.
same magnitude of 10 N . As the moment of inertia of the [A: 2 rad/s2 , 0.4 m/s2 , 4 N .]
PROBLEMS 133
a rope attached to its center. The rope makes an angle of 37◦

with the wall. The coefficient of friction of the wall is 0.3 .
A tangential downward force F=10 N is applied on the roll.
Find the angular acceleration of the roll. (Hint: Use the equa-
tions (7.19–7.20) of general motion.) [A: α = 4.1 rad/s2 .]
Problem 7.11
7.11 The moment of inertia of the pulley shown in the fig- 7.5 Rolling Motion
ure is I=0.45 kg·m2 . A rope attached to a block with mass
m1 =1 kg at rest on a horizontal plane is wrapped around
the pulley at an inner radius of R1 =10 cm . Another mass
m2 =2 kg hanging freely is attached to another rope that is
wrapped around the pulley at a radius of R2 =20 . The coeffi-
cient of friction on the plane is µ=0.4 . Calculate the accel-
erations and the tensions in the ropes when the system is
released. Problem 7.15
[A: α=7.2 rad/s2 , a1 =0.7, a2 =1.4 m/s2 , T 1 =4.7 , T 2 =17 N .] 7.15 A wheel at rest on a horizontal plane has a mass 1 kg , a
moment of inertia I=0.2 kg·m2 , an inner radius of R1 =10 cm
and an outer radius of R2 =20 cm . The wheel is slowly pulled
vertically upwards with a force F=8 N by means of a rope
wrapped around at radius R1 . If the wheel is rolling without
slipping, calculate the angular acceleration of the rolling mo-
tion and the linear acceleration of the center of mass.
Problem 7.12 [A: α = 3.3 rad/s2 , acm = 0.67 m/s2 .]
7.12 Two frictionless planes are inclined at angles of 30◦
and 53◦ . A block with mass m1 =1 kg is placed on the first
and another block with mass m2 =2 kg on the second. The
two blocks are tied to the two ends of a rope passing through
a pulley with a radius of R=50 cm and moment of inertia
I=2 kg·m2 . Find the accelerations and the tensions in the
rope. [A: 2 rad/s2 , 1 m/s2 , T 1 = 6, T 2 = 14 N .]
Problem 7.16
7.16 A sphere with mass M and radius R is rolling on a
horizontal plane with velocity v=7 m/s and then climbs to a
plane at height h=2.8 m . The moment of inertia of the sphere
is I= 25 MR2 . Find the final linear velocity of the sphere. (Note:
Problem 7.13 The mass and radius will cancel out in the final steps of cal-
7.13 A rope is wrapped around two pulleys, as shown in the culation.) [A: 3 m/s .]
figure, and its free end is attached to a block of mass m=5 kg
hanging freely. The radii of the pulleys are R1 =10 cm and
R2 =20 cm , and their moments of inertia I1 =0.1 kg·m2 and
I2 =0.2 kg·m2 , respectively. Calculate the accelerations and
the tensions in the ropes when the system is released.
[A: a=2.5 m/s2 , α1 =25, α2 =12.5 rad/s2 ,
T 1 =25 , T 2 =37.5 N .] Problem 7.17
7.17 A wheel at rest on a plane inclined at 37◦ has a mass
1 kg , a moment of inertia I=0.4 kg·m2 , an inner radius of
R1 =10 cm and an outer radius of R2 =20 cm . The wheel is
slowly pulled upwards with a force F=15 N parallel to the
plane, by means of a rope wrapped around it at radius R1 . If
the wheel is rolling without slipping, calculate the angular
acceleration of the rolling motion and the linear acceleration
of the center of mass. [A: α = 7.5 rad/s2 , acm = 1.5 m/s2 .]
Problem 7.14 7.6 Angular Momentum and Its Conservation
7.14 A cylindrical roll of paper with a radius of R=10 cm and 7.18 A disk with a moment of inertia I1 =2 kg·m2 and rotat-
moment of inertia 0.2 kg·m2 rests against a vertical wall by ing with angular velocity ω1 =8 rad/s clamps together with
another disk with a moment of inertia I2 =3 kg·m2 and rotat- When the bullet rises by h , the center of mass of the rod will
ing in the opposite direction on the same axis with angular rise by h/2 .) [A: (a) ω = 2.9 rad/s , (b) h = 0.29 m .]
velocity ω2 =6 rad/s . What will be the common angular ve-
locity of the two disks? [A: −0.4 rad/s .]
Problem 7.20
Problem 7.19 7.20 A rope tied to a mass m=1 kg on a frictionless table is
7.19 A rod with mass M=10 kg and length L=1 m can rotate passed through a hole in the center of the table and is allowed
freely about its end hinged to the ceiling. A bullet with mass to rotate in a circle when a force F is applied to the other end.
m=100 g and a horizontal velocity v=100 m/s hits and sticks (a) What should the force F be so that the mass m can rotate
to the free end of the rod. The moment of inertia of the rod with angular velocity ω = 5 rad/s on a circular trajectory
with respect to its center of mass is Icm = ML2 /12 . (a) What with radius 60 cm ? (b) What will the angular velocity be if
will be the angular velocity of the (rod+bullet) system right the rope is pulled down and the rotation radius is brought to
after the collision? (b) How high will the bullet rise? (Note: 30 cm ? [A: (a) F = 15 N , (b) ω0 = 4ω = 20 rad/s .]
8
STATIC EQUILIBRIUM
Mostar Bridge in Bosnia and

Herzegovina was commissioned
by the famous Ottoman Grand
Vizier Sokollu Mehmet Pasha
near his birthplace in 1566. Its
beauty and historical value has
been recognized by UNESCO in
its selection as a World Heritage
Site.
Is it sufficient to have the resul-
tant force be equal to zero to
keep this bridge in equilibrium?
Also, don’t the forces need to
have a certain distribution?
Stationary rigid structures are a part of our daily lives. The houses that we
live in, the roads and bridges that we pass by in traffic, dams, etc. All of these
structures are able to maintain their statical status because they can actually
keep the forces acting on them in equilibrium, in other words, they remain in
static equilibrium. The science that is concerned with objects and structures
in equilibrium is called Statics; it is an important area of study in civil and
mechanical engineering, architecture and many other branches of science and
technology.
In this chapter, we will discuss the conditions of static equilibrium of rigid
bodies.

136 8. STATIC EQUILIBRIUM
8.1 TWO CONDITIONS OF STATIC EQUILIBRIUM

In examining rolling motion in Chapter 7, we saw that the most general
motion of rigid bodies can be separated into two components:
Figure 8.1: The most general
motion of a rigid body is sep-
arated into two as the transla-
tional motion of the center of
mass and the rotational motion
around the center of mass.
The first one is the translational motion of the center of mass and was
determined by Newton’s law (Equation 7.19):
~Fi = m ~acm
X
(for translation) (8.1)
i
The second type of motion, rotational motion about the center of mass,
is determined by the law of rotation dynamics (Equation 7.20):
X
τi,cm = Icm α (for rotation) (8.2)
i
Therefore, these two types of motion should be prevented for a rigid body to
remain in static equilibrium. The condition required to prevent translational
motion can be written directly as follows:
~Fi = 0
X
(8.3)
i
Writing this vector equation for the components, we have:
First Condition of Equilibrium

The net external force acting on an object should be zero:
X X
Fi,x = 0 and Fi,y = 0 (8.4)
i i
Having the center of mass at rest, let us now examine the rotational motion
of the object. To prevent rotation, we must have
X
τi,cm = 0
i
For a rotating body, this torque was calculated with respect to the center of mass.
However, in the static condition the axis of rotation is irrelevant. This is because,
if the object is not rotating, the net torque is zero regardless of the chosen axis.
(Otherwise, it would rotate about that axis.) Therefore, we write the second
condition of equilibrium as follows:
Second Condition of Equilibrium

The net torque of forces acting on an object about an arbitrary
axis should be zero:
X
τi = 0 (8.5)
i
8.2. APPLICATIONS 137
These two conditions are, in principle, sufficient for static equilibrium. How-
ever, in real life, there are other parameters to consider, and our conditions may
not be enough to calculate the static equilibrium of complex objects that consist
of many parts (bridges, buildings, etc.) Other parameters are added to achieve the
full solution of the problem. We will discuss these through worked examples.
8.2 APPLICATIONS
Now let us see how static problems can be solved using conditions of equilib-
rium, and other information particular to the problem at hand.
Example 8.1
Answer
There will be no rotational motion, as the tensions in the
ropes and the weight W meet at the same point and their
torques are zero, hence the first condition will be sufficient.
We write it with the chosen axes:
F =0 → −T 1 cos 30◦ + T 2 cos 53◦ = 0
P
Pi i,x
i F i,y = 0 → +T 1 sin 30 + T 2 sin 53 − W = 0
◦ ◦
We substitute the numerical values:

−0.87T 1 + 0.6T 2 = 0
A block with weight W=10 N is in equilibrium hanging from 0.5T 1 + 0.8T 2 = 10
two ropes as in the figure. Calculate the tensions in the ropes. From here, we find that T 1 = 6 N and T 2 = 8.7 N .
Example 8.2
The weight W1 =20 N is at rest on a horizontal plane with a Answer

coefficient of friction µ=0.4 . It is tied to a horizontal rope that The free-body diagrams are shown above. The three forces
connects to another weight W hanging freely. Both ropes are W2 , the tension T and the friction force f meet at point O,
connected to a third rope attached to the wall at an angle of hence only the first condition is needed. The friction force
45◦ What is the maximum weight W2 that can be attached can have any value, but it will be proportional to the normal
without having the weight W1 slide? force N1 when the weight W1 starts to slide:
f = µN1 = µW1
Therefore, we write the first condition of equilibrium at point
O for these three forces:
F = −µW1 + T cos 45◦ = 0
P
Pi i,x
i F i,y = +T sin 45 − W2 = 0
◦
By taking the unknown T from the second equation and

substituting it in the first one, we get W2 :
W2 = µW1 tan 45◦ = 0.4 × 20 × 1 = 8 N
Example 8.3
cable and the reaction force on the hinge.
Answer
There are three forces acting on the beam: The weight W ,
the tension T along the cable and the reaction force R on
the hinge that can be in any direction.
We first write the first condition of equilibrium according to
the axes shown in the figure:
A beam with weight W = 10 N is hinged to a wall from one F = −T cos 37◦ + R x = 0
P
Pi i,x
i F i,y = +T sin 37 + Ry − W = 0
◦
end and tied to the wall from the other end with a cable, making
a 37 angle with the horizontal. Calculate the tension in the For the second condition, we are free to calculate the torque
◦
with respect to any arbitrary axis. Hence, it is convenient to We can immediately find T from this last equation:
choose an axis that gives a simpler equation. Here, it will be W 10
T= = = 8.3 N
smart to choose the axis at point A, because two unknowns 2 sin 37◦ 2 × 0.6
( R x , Ry ) will both have zero torque and not be included in
the equation: Substituting this value in the other two equations, we find
the components of the reaction force R x , Ry :
L
i τi,A = 0 → −W + T sin 37 L = 0 R x = 6.7 N and Ry = 5 N
P
2
Example 8.4
much force F should the biceps muscle apply to lift a weight
of W2 = 10 N ?
Answer
The forces on the arm are shown in the figure. We can take
the torque with respect to the elbow, as we do not need the
reaction force R :
i τi = 0 → F × 0.05 − W1 × 0.15 − W2 × 0.35 = 0
P
From here, we calculate the force F :
0.15W1 + 0.35W2
F= = 150 N .
The dimensions of an average human arm are shown in the 0.05
figure. The biceps muscle acts at a distance of 5 cm from the It may seem surprising that the force on the biceps is 15 times
elbow. The lower half of the arm has a weight of W1 = 20 N greater, but it is true. It is difficult for the arm to carry a load
and its center of mass is located at 15 cm distance from the in the horizontal position, and it therefore gets tired very
elbow. The distance of the hand to the elbow is 35 cm . How quickly.
Example 8.5
The friction force f can have any value when the ladder
A ladder with weight W and length L is leaned against a wall is in equilibrium, but when the ladder starts to slide, it will
with angle θ . The wall is frictionless, while the coefficient of reach its maximum value f =µN1 . Therefore, we will make
friction is µ=0.5 on the horizontal plane. What is the minimum the calculations for this limit value.
angle θ at which the ladder can stand without sliding? We write the first condition for equilibrium:
i F i,x = µN1 − N2 = 0
P
i F i,y = +N1 − W = 0
P
For the second condition, it is smart to choose point A to

calculate the torque, then two unknowns ( f and N1 ) will not
be included in the equation:
L
i τi,A = 0 → −W cos θ + N2 L sin θ = 0
P
Answer 2
The forces acting on the ladder are shown in the figure. As By eliminating the unknowns N1 , N2 from these equations,
the wall is frictionless, it will only have the perpendicular we find θ :
reaction force N2 . The ground has both the normal reaction 2 2
tan θ = = = 4 → θ = 76◦
force N1 and the friction force f . µ 0.5
Example 8.6
At what maximum height h can we apply this minimum force
F without toppling the chest?
Answer
(a) It is sufficient to apply a force greater than the friction
force in order to move the chest. As the friction force will
reach the value f = µN once the chest starts to slide, we get
F = f = µN = µW
We want to push a chest without toppling it. The chest’s weight . We substitute the numerical values to find F :
is W=500 N and its length is L=2 m and it is at rest on a hor- F = 0.8 × 500 = 400 N
izontal plane with a coefficient of friction µ=0.8 . (a) What is
the minimum horizontal force F that can move the chest? (b) (b) The chest, when it is about to roll over, will only be in
8.2. APPLICATIONS 139
contact at point A. The chest will topple when the torque L

i τi,A = 0 → −F h + W =0
P
of the force F about point A is greater than or equal to the 2
Substituting the value of F that we found in item (a),
torque of the weight W . Calculating this limit condition, we L 2
get h= = = 1.25 m
2µ 2 × 0.8
Example 8.7
Therefore, one of the arms of the ladder should be isolated

The two arms of a ladder shaped like an ‘A’, standing motionless from the other (the figure on the right). According to the law
on a frictionless horizontal plane, are hinged from the top and of action-reaction, these forces will be equal and opposite on
tied to each other with a horizontal rope near the bottom. Each each arm. (According to symmetry, the force R must also
of the ladder arms has weight W=30 N and length L=4 m . be horizontal, otherwise it would be downwards in one and
The points at which the rope is tied are 1 m from the lower end upwards in the other.)
of the ladder. Find the tension in the rope and the reaction force We can now write the conditions of equilibrium for one arm.
on the hinge. The first condition of equilibrium for the arm AC is:
Answer Fi,x = +T − R = 0
P
This problem is a good example of the isolation technique in Pi
i F i,y = +N − W = 0
the equilibrium problem of two objects.
From here, we get the results T = R and N = W .
First, let us consider the ladder as a whole (the figure below).
We write the second condition with respect to point A:
The forces on the hinge and the rope need not be shown,
i τi,A = R (4 sin 53 )−T (1 sin 53 )−W (2 cos 53 ) = 0
P ◦ ◦ ◦
because they are internal forces of this system. We can only
write 2N = 2W for the reaction forces on the ground. Hence, Using the fact that R = T in this equation, we find the result:
we will never be able to find T and the reaction force at the 2W 2 × 30
T =R= = = 15 N
hinge by considering the whole ladder. 3 tan 53◦ 3 × 1.33
Example 8.8
directions of the forces are chosen such that the top hinge
will pull and the bottom hinge will push the door.
We write the first condition of equilibrium:
F = R1x − R2x = 0 R1x = R2x
P
→
Pi i,x
i F i,y = R1y + R2y − W = 0 → R1y + R2y = W
Let us take the torque with respect to point A:
L
i τi,A = R2x H − W =0
P
2
From here, we find the horizontal components of the reaction
forces:
A barn door with height H=2 m ,width L=1 m , and weight L 1
R1x = R2x = W = 50 = 12.5 N
50 N is hinged to the wall at its top and bottom edge corners. 2H 4
Calculate the horizontal and vertical components of the reaction
We have no information for the vertical components of the
forces on the top and bottom hinges.
reaction force other than the equation R1y + R2y =W found
Answer above. Even if we took torque with respect to another point,
This problem is interesting in that it shows how the num- only the sum of these two vertical components would appear
ber of equations may sometimes be insufficient for the full in the equation. Therefore, it is not possible to calculate the
solution of static equilibrium problems. vertical components separately; we can only know their sum:
The forces acting on the door are shown in the figure. The R1y + R2y = W = 50 N .
Problems
Problem 8.6
Problem 8.1 8.6 A stick with negligible mass and a length of 5 m is hinged
8.1 Two weights hanging from ropes as shown in the figure to a wall from one and, with its other end freely resting on
above, are in equilibrium. If W1 =10 N , find the weight W2 a frictionless horizontal plane. A horizontal force F=10 N
and the tensions in the ropes. is applied to the bottom end of the stick. Calculate the hori-
[A: W2 = 13, T 1 = 20, T 2 = 22, T 3 = 17 N .] zontal and vertical components of the reaction force on the
hinge. [A: R x = 10, Ry = 7.5 N .]
Problem 8.7
Problem 8.2 8.7 A horizontal beam with length L=1 m and weight
8.2 A block with weight W1 =10 N is placed on a plane in- W1 =10 N is hinged to a wall from one and tied to the wall
clined at an angle of 30◦ and with a coefficient of friction from the other end with a rope making a 37◦ angle with the
µ=0.9 . Another weight W2 is tied to the weight W1 with a horizontal. A block with weight W2 =40 N is placed on this
rope parallel to the inclined plane and with another rope at beam at distance x . The rope is able to endure a maximum
an angle of 37◦ to the wall. What is the largest weight W2 tension of 50 N . Find the maximum distance x that the block
that can be hung without having the weight W1 slide? can go and the components of the reaction forces on the hinge.
[A: 1.8 N .] [A: x = 0.63 m , R x = 40, Ry = 20 N .]
8.3 A weight W1 =5 N is hung at the 10 cm mark of a uni-

form meter stick. The stick is balanced horizontally if a knife
edge is placed under the 40 cm mark. What is the weight of
the meter stick? [A: 15 N .]
Problem 8.8
8.4 A horizontal beam of weight W1 =10 N and length L=1 m 8.8 The beam in the figure above with length L and weight
is supported by two vertical ropes attached to each end. A W1 =10 N is hinged to the ground from one end at 53◦ angle
weight W2 =30 N is hung at a distance d = 30 cm from the with the horizontal and tied to the ground with a rope making
right end. Find the tensions in the ropes. a 30◦ angle with the horizontal. A weight W2 =20 N is hung
[A: 14N and 26 N .] on the free end of the stick. Calculate the tension in the rope
and the components of the reaction force on the hinge.
[A: T = 38, R x = 33, Ry = 49 N .]
Problem 8.5
8.5 In the figure, one end of a beam with weight W2 =20 N
and length L is hinged to the wall and a weight W1 =10 N is
hung on the other end. The beam is kept at an angle of 37◦ Problem 8.9
with the horizontal by means of a horizontal rope tied at a 8.9 A ladder with weight W1 =50 N and length L=3 m is
distance 3L/4 from the hinged end A. Find the tension in leaned against a frictionless wall at an angle of 53◦ with the
the rope and the horizontal and vertical components of the horizontal. The coefficient of friction on the ground is 0.4 . A
reaction force at the hinge. [A: T = R x =36, Ry =30 N .] man with weight W2 =700 N is climbing on the ladder. How
PROBLEMS 141
far from the lower end can the man climb before the ladder of W2 /2 to each of the hanging points.)
slides? [A: x=1.61 m .] [A: T = 31, R x = 24, Ry = 12 N .]
Problem 8.13
Problem 8.10 8.13 A uniform rod with two ropes tied at each end is in
8.10 The dimensions of an average human arm are shown equilibrium, as shown in the figure. The rope attached to the
in the figure. The biceps muscle acts at a distance of 5 cm wall is horizontal. The rope attached to the ceiling makes
from the elbow. The lower half of the arm has a weight of an angle of α=53◦ with the horizontal. What is the angle β
W1 =20 N and its center of mass is located at 15 cm distance of the rod? (Note: The weight and length of the rod will not
from the elbow. The distance of the hand to the elbow is appear in the final expression.) [A: β=34◦ .]
35 cm . How much force F should the biceps muscle apply
to lift a weight of W2 = 50 N ? [A: 470 N .]
Problem 8.14
Problem 8.11 8.14 A cylinder with radius R=50 cm and a mass of 1 kg
is pulled at its center by a force F along a horizontal plane.
8.11 A rod with weight W=10 N and length L=1 m is freely
However, a step with height h=10 cm on the plane is pre-
placed vertically on a horizontal plane with a coefficient of
venting motion. What should the minimum value of the force
friction µ=0.5 . The top end of the rod is tied to the ground
F be for the cylinder to overcome the step? (Hint: What
using a rope making a 53◦ angle with the vertical. The rod is
will the normal force be when the cylinder is just leaving the
being pulled by a horizontal force F=20 N acting at height
ground?) [A: F = 7.5 N .]
h from the ground. At what height h will the lower end of
the rod start to slide? [A: h = 0.55 m .]
Problem 8.15
Problem 8.12 8.15 The two arms of a ladder shaped like a ‘slanted A’,
8.12 A beam with length L and weight W1 = 10 N is hinged standing on a frictionless horizontal plane, are hinged from
to a wall from one end, and tied to the same wall from the the top and tied to each other with a horizontal rope near
other end with a rope making a 37◦ angle with the horizontal. the bottom. The ladder arms have weights 30 N and 40 N ,
A signboard with mass W2 = 20 N is hung on this beam. One and lengths 3 m and L=4 m , respectively. The rope is at a
end of the signboard is tied to the outer end of the beam and height of 1 m from the ground. Find the tension in the rope,
the other end at a distance of L/3 from the hinge. Calculate the normal forces on the ground, and the components of the
the tension in the rope and the components of the reaction reaction force on the hinge.
force on the hinge. (Hint: The signboard applies equal forces [A: T = R x = 40, Ry = 7, N1 = 37, N2 = 33 N .]
9
HARMONIC MOTION
The colibri (hummingbird), with

a size of 2 cm, is the smallest
bird that is not extinct. It can
flap its wings as fast as 80 beats
per second.
Which concepts should be used
to characterize the periodic mo-
tion? How can we associate
these concepts with the laws of
physics?
A motion that repeats itself regularly is called a periodic motion. It is the

most frequently encountered type of motion in nature, in our daily lives and in
technology. The oscillation of a pendulum, the vibration of a guitar string, our
heartbeats or the vibration of our vocal cords, the ebb and tide of the sea, etc.
Even an object at rest is made of vibrating atoms.
Oscillatory motion occurs when an object is disturbed from its equilibrium
position and a restoring force exists that is trying to bring it back. In real life, there
are also other dissipative forces that quickly dampen the oscillations.
We will examine the basic concepts and properties of this motion that repeats
itself within a certain time interval.
9.1 SIMPLE HARMONIC MOTION

The oscillations of a mass connected to a spring represent the simplest motion
in which we can observe the basic features of vibrational motion. Let us fix one
end of a spring with spring constant k to a wall and attach a mass m to its other

144 9. HARMONIC MOTION
end on a frictionless horizontal plane (Figure 9.1).

The mass starts to vibrate when we pull the spring by x from its normal
length and release. The weight and the normal reaction force perpendicular the
to motion cancel each other out, and are thus not taken into consideration. Since
the spring force causes the motion of the mass, let us write Newton’s law:
Figure 9.1: Spring force is al-
ways in opposite direction to ex- F = ma
tension. d2 x
−kx = m 2 = m x00
dt
Here, we used x00 to indicate the second derivative of position x . We have thus
obtained an equation that determines the relation between the position and its
second derivative:
k
x00 + x = 0 (9.1)
m
This expression is called the differential equation of harmonic motion. We
will be able to determine the motion if we can find its solution in the form x=x(t) ,
in other words, the position as a function of time.
Differential equations are part of advanced mathematics, and their solution
methods are known. However, we can find the solution here without using
advanced techniques. Let us write the equation as follows:
k
x00 = − x
m
What is the function x(t) such that its second derivative will be proportional
to its negative? We know these functions: They are sine and cosine functions.
Therefore, we can predict the form of the solution:
x = A cos ωt (9.2)
Here, A and ω are two constants to be determined later. We could have chosen
the solution as a sine function as well (We will discuss when to choose sine or
cosine below.)
Let us first make sure that this function satisfies the differential equation. Let
us take the derivative twice:
x0 = −ωA sin ωt
x00 = −ω2 A cos ωt
We substitute the expressions for x and x00 in Eq. (9.1) and simplify to get
k
−ω2 A cos ωt + A cos ωt = 0
m
h k i
− ω2 + A cos ωt = 0
m
This equation must be true at every time t . The cosine is not always zero, therefore
the expression in the brackets should be zero:
h ki
− ω2 + =0
m
9.1. SIMPLE HARMONIC MOTION 145
From here, we find the constant ω , called the angular frequency, in terms of
the mass and the spring constant:
r
k
ω= (angular frequency) (9.3)
m
We will discuss the physical meaning of the angular frequency ω below. At this
stage, the equation of motion of the mass m can be written as follows:
x = A cos ωt (simple harmonic motion) (9.4)
This motion is called simple harmonic motion or sinusoidal motion.

Figure 9.2 shows how position x varies with respect to time t .
Selection of the cosine or sine function: We chose the cosine function for
vibration in Eq. (9.4) above. We could have also chosen this as the sine function,
because the sine also satisfies the same equation of motion. The initial position
of the object determines which one is to be chosen (Figure 9.2).
We take x = A cos ωt if the object is pulled to a certain distance and released
from that distance at time t = 0 . Indeed, we get cos 0◦ = 1 and x = A at t = 0 ,
as it should be. On the other hand, if the object is thrown from its equilibrium
position at the start, then we should have (x = 0) at t = 0 . Indeed, x = A sin ωt
fulfills this condition, because the sine function gives x = 0 at time t = 0 .
(Actually, the cosine and sine are just the same curve. The name varies only Figure 9.2: The sine and cosine
with respect to the place where you draw the vertical axis. You get the cosine if functions.
you have the vertical axis pass through the maximum of the curve and the sine if
you pass it through the zero value.)
In conclusion, our function selection becomes as follows:
 A cos ωt (if the object starts at maximum distance)



x= (9.5)

 A sin ωt
 (if the object starts at x = 0)
Now, let us learn the basic concepts of the harmonic motion.

Amplitude ( A )
When the cosine function takes values within the range [−1, +1] , the position
x=A cos ωt will also vary between the values [−A, +A] (Figure 9.3). When we
pull the mass attached to the spring by a distance A and the release it, it does
indeed start to go back and forth between [−A, +A] , and never venturing outside
of this range. This quantity A , which is the absolute value of the maximum
extension, is called the amplitude.
Period ( T )
Harmonic motion is periodic, in other words, it repeats itself within a certain Figure 9.3: Amplitude and pe-
time interval. This complete motion in which the object passes through the same riod.
point in the same direction is called a cycle, and the time to complete one cycle
is called the period (Figure 9.3). Let us find the period formula from the equation
of motion.
Let us rewrite the position of the object at any time t :
x = A cos ωt
There should pass a time interval T such that the object passes through the same
x again:
x(t + T ) = x(t)
A cos ω(t + T ) = A cos ωt
cos ω(t + T ) = cos ωt
The cosine function is periodic in the interval [0, 2π] , in other words, it repeats
itself after 2π . Therefore, the difference between the arguments of the cosines on
both sides of the equation must be 2π :
ω(t + T ) − ωt = 2π
ωT = 2π
From here, we find the period of the harmonic motion:
2π
T= (period) (9.6)
ω
This expression is true for all kinds of harmonic motions. In particular, if we use
Eq. (9.3), which we found for the angular frequency of the mass-spring system,
the period formula becomes:
r
m
T = 2π (Period of spring-mass system) (9.7)
k
It can be checked that this expression gives the period unit in terms of seconds
(s). According to this formula, when we attach an object with a known mass m
to a spring with a known constant k and vibrate it, its period T is determined.
Frequency ( f )
The number of cycles that the objects performs per unit time is called the
frequency and is denoted by f . Accordingly, frequency is the reciprocal of
period. For example, if the period is 3 seconds, then the number of oscillations in
1 second will be 1/3. Therefore, the expression for frequency is:
1 ω
f = = (frequency) (9.8)
T 2π
This expression is true for all kinds of periodic motions. In particular, the fre-
quency of the spring-mass system is
r
k
f = 2π (9.9)
m
The unit of frequency is 1/second = s−1 and is also called the Hertz (Hz) in
technology:
1 s−1 = 1 Hz (9.10)
If we write the ω in the formula (9.8) in terms of frequency, we get
ω = 2π f (9.11)
This expression explains why ω is called the angular frequency. When the
frequency is 1, in other words, when a cycle is completed in 1 second, ω takes
the value of 360◦ = 2π as if it were rotating in one complete circle.
As a conclusion, the equations of simple harmonic motion can be written in
one of the three following forms:
A cos ωt





 2π
x= (9.12)


A cos t



 T
 A cos 2π f t

Example 9.1 2π 2π
T= = = 2.5 s
ω 0.8π
The position of an object with mass m=3 kg attached to a spring
(b) We use Eq. (9.3), which gives the relation between spring
varies with time as
constant and
p angular frequency:
x = 6 cos 0.8πt (meters)
ω = k/m → k = mω2
(a) What are the amplitude, angular frequency and period of
k = 3 × (0.8π)2 = 1.92π2 ≈ 19 N/m .
the motion?
(c) We use t = 5/12 s in the equation of motion:
(b) What is the spring constant? 5 π
(c) What is the position of the object at time t=5/12 s ? x = 6 cos 0.8π × = 6 cos
12 3
(d) How many seconds does it take for the object to reach the Substituting cos π/3= cos 60◦ = 0.5 , we find x=3 m .
positions x=0 m and x=2 m ?
(d) The distance from inital position x=A to x=0 is one fourth
Answer of a cycle. Therefore, it will reach x=0 in T/4=2.5/4=0.63 s .
(a) Amplitude means maximum extension. As the cosine In order to find the time to reach the other position, we substi-
function varies within the range [−1, +1] , the object will tute the value x=2 m in the equation of motion and calculate
go back and forth within the range [−6, +6] . Therefore, the the t in the cosine:
amplitude will be A=6 m . x = 6 cos 0.8πt → 2 = 6 cos 0.8πt
Angular frequency is the coefficient in front of the variable cos 0.8πt = 1/3 = 0.33
t . Therefore, we get ω=0.8π s−1 =0.8π Hz . (It is convenient We use the table in Appendix C to find the angle whose cosine
to leave the constant π in these calculations. It is sometimes is 0.33 as 71◦ = 1.23 radian . It is necessary to use radian units
canceled out and sometimes taken as π2 ≈ 10 .) in these calculations. From here, we find time t :
We use Eq. (9.6), which relates period to angular frequency: 0.8πt = 1.23 → t = 0.49 s
Example 9.2
T = 4/5 = 0.8 s .
A mass m attached to a spring with a constant k=125 N/m From here, we can calculate the angular frequency and the
completes 5 cycles in 4 seconds. mass of the object:
2π
(a) What are the period, angular frequency and m ? ω= = 2.5π Hz
(b) The mass is stretched by 60 cm from its equilibrium po- 0.8
k 125
sition and released. The time t = 0 is started on the m= 2 = = 2 kg .
ω (2.5π)2
stopwatch as it passes through the equilibrium position.
(b) The object’s position at t = 0 is x = 0 . Therefore, we
Write the equation of the simple harmonic motion.
must use the sine function, which is zero at t = 0 . Amplitude
Answer is given as A = 0.6 m . The equation becomes:
(a) Period is the time for one cycle: x = 0.6 sin 2.5πt .
Velocity and Acceleration in Simple Harmonic Motion

We have learned how the position x of a mass attached to a spring varies
with time t . What are the velocity and acceleration of this object at time t ? In
order to determine these, let us start off by using the derivative definitions of
velocity and acceleration:
x = A cos ωt
dx
v = = −ω A sin ωt (9.13)
dt
dv
a = = −ω2 A cos ωt (9.14)
dt
These variations are shown in Figure (9.4). Let us emphasize the important points
about the position, velocity and acceleration of harmonic motion.
• The velocity and acceleration are also harmonic; they oscillate as sine or
Figure 9.4: Position, velocity cosine with the same frequency.
and acceleration. • The variations of position and velocity are opposite to each other: When
position reaches its maximum value (x=A) , velocity is zero. In contrast,
as the object passes through the equilibrium point (x=0) , its velocity is
maximum.
• Velocity and acceleration also act in opposition to each other: The accelera-
tion of the object is zero when its velocity is maximum (passing through the
equilibrium point). In contrast, the object has maximum acceleration when
it reaches maximum distance, where it has zero velocity.
• There is a relation between position and velocity that is always true regardless
of what time t is equal to. We use the identity sin2 a + cos2 a = 1 to eliminate
t between x and v :
x2 v2
cos2 ωt = ; sin2 ωt =
A2 ω 2 A2
x2 v2
sin2 ωt + cos2 ωt = + =1
A2 ω2 A2
Simplifying this expression, we get
p
v = ω A2 − x 2 (9.15)
Hence, if either position or velocity is known, the other can be calculated

without any need to know time.
• There is also a relation between position and acceleration:
a = −ω2 A cos ωt
a = −ω2 x (9.16)
Therefore, if either position or acceleration of the object is known, the other

can be calculated without reference to time t .
Example 9.3 r r
k 72
ω= = = 6 Hz
A mass m=2 kg attached to a spring with spring constant m 2
k=72 N/m is stretched by 50 cm from its equilibrium posi- In Eqs. (9.13–9.14), which we found for velocity and accelera-
tion and released. (a) What will be its maximum velocity and tion, we set sine and cosine to their maximum values of 1 to
acceleration? (b) What will its velocity and acceleration be as calculate the maximum values of velocity and acceleration:
it passes through position x = 30 cm ? vmax = ωA = 6 × 0.5 = 3 m/s
Answer (a) We first calculate the angular frequency: amax = ω2 A = 62 × 0.5 = 18 m/s2 .
(b) We use Eq. (9.15), which gives the relation between posi- We use Eq. (9.16), which gives the relation between position
tion and velocity:
√ √ and acceleration:
v = ω A − x = 6 0.5 − 0.3 = 6 × 0.4 = 2.4 m/s .
2 2 2 2 a = −ω2 x = −62 × 0.3 = −11 m/s2 .
Example 9.4

v2 = ω2 A2 − x2
The angular frequency of an object performing simple harmonic We take the amplitude A from this formula:
motion is 5 Hz . This object passes through position x = 3 m A = x + (v/ω)2
p
2
with velocity v = 20 m/s . (a) What is the amplitude of the

p
A = 32 + (20/5)2 = 5 m .
motion? (b) What is its maximum velocity? (b) Velocity will be maximum when x = 0 in the afore-
Answer mentioned formula, in other words, when the object passes
(a) We use Eq. (9.15), which gives the relation between posi- through the equilibrium point. Accordingly,
tion and velocity: vmax = ωA = 5 × 5 = 25 m/s .
Energy of Harmonic Motion

Harmonic motion will have a kinetic energy dependent on the velocity of
the mass and a potential energy dependent on the extension of the spring. Both
types of energy will change with respect to time.
Using the expression for velocity from Eq. (9.13), in the definition of kinetic
energy, we get
K = 12 mv2 = 21 mω2 A2 sin2 ωt (9.17)
Using Eq. (9.4) for position x in the elastic potential energy of the spring, we get
U = 12 kx2 = 21 k A2 cos2 ωt (9.18)
The variations of the kinetic and potential energies are shown in Figure 9.5a.
Figure 9.5: (a) The variations of

kinetic and potential energies: (a)
against time t , (b) against posi-
tion x .
Let us calculate the total mechanical energy:

E = K + U= 12 |{z}
mω2 A2 sin2 ωt + 12 kA2 cos2 ωt= 21 kA2 (sin
| ωt{z
2
+ cos2 ωt
})
k 1
E = 2 k A = constant
1 2
Therefore, the total energy is constant in simple harmonic motion and has the
value of the potential energy at maximum extension. This is easy to understand,
because its velocity will be zero when it reaches maximum extension, in other
words, all of it will have turned into potential energy:
E = K+U = 1
2 kA2 (total energy of harmonic motion) (9.19)
Figure 9.5b also shows how kinetic and potential energies vary with position x
and how their sum E remains constant.
Note that the energy is proportional to the square of the amplitude A . The
energy is multiplied by a factor of 4 when the amplitude is doubled. This is why
the destructive impact of earthquakes increases with amplitude.
Example 9.5
E = K + U = 12 kA2
r r
A mass m=1 kg attached to a spring with spring constant 2E 1600
k = 400 N/m has a total energy of 800 J . A= = = 2m
k 400
(a) What is the amplitude of the oscillations?
(b) What will the velocity of m be when the potential energy (b) If the potential energy is 200 J , the kinetic energy will be
of the system is 200 J ? 800 − 200 = 600 J : r
Answer 2 × 600
2 mv = 600 → v = = 35 m/s
1 2
(a) We use Eq. (9.19), which we found for total energy: 1
Example 9.6
that they are both half of the total energy. This value is taken
In a system performing simple harmonic motion, for potential energy:
x 1
2 kx = 2 2 kA = √
(a) At which x/A ratio will the kinetic and potential energies 1 2 1 1 2
→
be equal? A 2
(b) What will the ratio of the kinetic and potential energies (b) Let us write the formula that gives velocity in terms of
be when x = A/2 ? position:
v2 = ω2 A2 − x2
Answer
We write the ratio of energies for the value x = A/2 in this
(a) We write the total energy expression:
expression:
2 mv + 2 kx = 2 kA
1 2 1 2 1 2
K mv2 H mωH2 (A2 − A2 /4)
If the kinetic and potential energies are equal, this means = = =3
U kx2 ZZk (A/2)2
Phase Angle
We had previously discussed when to use cosine or sine in the harmonic
motion equation: We use sinus if the object starts from the origin, and cosine if it
is released from the maximum distance.
What if the motion starts at any place other than these two points? The
answer to this question is seen in Figure 9.6. We can make the function start at
any point by adding another term besides (ωt) as the argument of cosine. Having
the unit of an angle, this constant argument is called the phase angle or phase
difference and is indicated with φ :
Figure 9.6: Phase angle φ .
x = A cos(ωt + φ) (9.20)
In this expression, we should find the sine function when we take φ = −90◦ =
−π/2 , in other words, when we start the cosine function 90◦ from behind. Indeed,
the trigonometric identity cos(α − 90◦ )= sin α gives us the sine equation.
We can calculate the phase angle φ in terms of the initial position x0 when
we set t = 0 in this expression:
x0
x0 = A cos φ =⇒ cos φ = (9.21)
A
The phase difference will be important when we consider the superposition of two
harmonic motions. We will return to this issue later, in the section on interference
of waves.
Example 9.7
position and released. The clock is started as it passes through
position x = 1 m . Find the phase angle and write the equation
A mass-spring system can oscillate at an angular frequency
of the harmonic motion.
ω=5 Hz . The mass is stretched by 2 m from its equilibrium
9.2. PENDULUM MOTION 151
Answer 1 = 2 cos(0 + φ) → cos φ = 0.5

We write the most general harmonic motion expression:
The angle whose cosine is 0.5 is 60◦ = π/3 . Therefore, we
x = A cos(ωt + φ)
write the equation of the motion as follows:
We require that this equation satisfy the value x = 1 m at
time t = 0 : x = 2 cos(5t + π/3)
Example 9.8
given equation, and find the position: √
The equation of a harmonic motion is given as x = 3 cos(0 − π/4) = 3 cos(π/4) = 3 × 2/2 = 2.1 m .
x = 3 cos(0.2πt − π/4) (b) Maximum position is the amplitude x = A . We substitute
the amplitude value for x :
(a) What is its position at the start?
3 = 3 cos(0.2πt − π/4) → cos(0.2πt − π/4) = 1
(b) When will it reach the maximum position?
The angles whose cosine is 1 are 0, 2π, 4π . . . . We choose
Answer the smallest one 0 and find the time t :
(a) We use the starting time, in other words, t = 0 , in the 0.2πt − π/4 = 0 → t = 1.25 s .
9.2 PENDULUM MOTION

Simple Pendulum
A mass tied to the end of a rope performs oscillatory motion. The period of
a pendulum was used to measure time in old wall clocks. This motion can be
analyzed as simple harmonic motion under certain conditions.
A mass m is tied to the end of a string with length L . Let us pull the string
by angle θ from the vertical direction and release it. The forces acting on the
rope at any instant are shown in Figure 9.7.
The tension force T has no effect on the motion, because it is along the string.
However, the weight mg has a component in the tangential direction to the circle
with radius L . The force that moves the mass along the path is the force mg sin θ .
When writing Newton’s law along the tangent, we choose the direction Figure 9.7: Simple pendulum.
towards which the angle θ increases as positive. If we also write acceleration as
the derivative of velocity, we get
Ft = mat
dv
−mg sin θ = m
dt
We write the velocity v in terms of angular velocity as the mass performs circular
motion with radius L . The angular velocity is the derivative of the angle θ :
dθ
v = Lω = L
dt
Substituting this expression and arranging the terms, we get
d2 θ g
− mg sin θ = mL → θ00 + sin θ = 0
dt2 L
This equation looks similar to Eq. (9.1), which we found for harmonic motion of
the mass-spring system, except that it is written for the angle θ . Also, we have
sin θ instead of θ , and therefore we cannot write the solution directly.
However, if we consider oscillations with small amplitudes, we can approxi-

mately take sin θ ≈ θ . Then, we obtain the simple harmonic motion equation for
the angle θ :
g
θ00 + θ=0 (9.22)
L
|{z}
ω2
Comparing this expression with the equation (9.1), the coefficient of the term θ
becomes the square of angular frequency ω2 . From here, we find the period T of
the pendulum using the relation T = 2π/ω :
s
L
T = 2π (Period of simple pendulum) (9.23)
g
The maximum angle θmax is used instead of the amplitude A when writing the
pendulum motion equation. Accordingly, the solution of pendulum motion is
θ = θmax cos ωt (9.24)
and is valid for oscillations with small angles. Notice that Eq. (9.23) for the period
of pendulum is independent of mass and amplitude. This property was used in
the designing of clocks for centuries.
Physical Pendulum
The simple pendulum is an ideal case. Actually, every rigid body can perform
pendulum motion. A rigid body that can rotate about an axis that is not through
the center of mass is called a physical pendulum. The pendulums used in wall
clocks and in the industry are physical pendulums.
Consider a rigid body with mass m and moment of inertia Icm with a rotation
axis a distance d from its center of mass (Figure 9.8). Only the mg sin θ component
of the weight causes the rotation motion.
Choosing the positive direction towards which the angle θ increases and
writing the net torque with respect to the center of rotation O , we get,
Figure 9.8: Physical pendulum.
τo = Ft d = −mg sin θ d = I α
If we write the angular acceleration α as the second derivative of the angular
position θ and arrange the terms, we get
mgd
θ00 + sin θ = 0
I
We again use the approximate value sin θ ≈ θ for small angles:
mgd
θ00 + θ=0 (9.25)
I
|{z}
ω2
We again obtain the form of a simple harmonic motion equation. According to

this equation, the coefficient of the term θ will be angular frequency ω2 . We get
the formula for period using T = 2π/ω :
s
I
T = 2π (Period of physical pendulum) (9.26)
mgd
9.3. DAMPED HARMONIC MOTION 153
Note that the moment of inertia I in this formula is with respect to the axis of
rotation O. We use the parallel axis theorem to relate it to the Icm values given in
the tables (Chapter 7):
I = Icm + m d2
Example 9.9 s
L gT
T = 2π → L= 2
We want to construct a simple pendulum with period 1 s . g 4π
(a) What length of string should be used? We use the values T = 1 s and g = 10 :
(b) What length would be needed on planet Jupiter, where 10 × 1
L= = 0.25 m
2
gravitational acceleration is 24 m/s ? 4π2
(b) We can find the length of the pendulum on Jupiter by
Answer taking g = 24 in the same formula:
We solve the formula (9.23), which we found for the period L=
24 × 1
= 0.60 m .
of a pendulum, for the length L : 4π2
Example 9.10 r
2 × 0.60 2π
T = 2π = = 1.26 s .
A physical pendulum is constructed by hanging a rod with 3 × 10 5
length L=60 cm and mass m from one of its ends. (b) If we write the value I = Icm + md2 for the axis at distance
(a) Calculate the period of the oscillations. d and then solve
s for d , we get
12 mL + md
1
(b) At what distance from its center of mass should it be hung 2 2
gT 2 L2
for the period to be 2 s ? T = 2π → d2 − 2 d + =0
mgd 4π 12
Answer Substituting T = 2 s and the other numerical values, we get
(a) The moment of inertia of the rod with respect to its center d2 − d + 0.03 = 0
of mass is Icm = 12 mL . We had previously found its mo- The two solutions to this equation are d = 0.03 and 0.97
1 2
ment of inertia with respect to one end to be 13 mL2 using the and the solution greater than L = 0.6 is not taken into con-
parallel axis theorem. Therefore, if we calculate the period sideration. Therefore, the solution is as follows:
for d = L/2 using the formula d = 0.03 m
s s (9.26), we get s
I mL2 /3 2L
T = 2π = 2π = 2π
mgd mgL/2 3g
9.3 DAMPED HARMONIC MOTION

An oscillating body left to itself will stop after some time. If a swing is not
pushed, its amplitude gradually decreases and then stops. The energy of a mass
attached to a spring decreases due to friction. However, we sometimes deliberately
want the oscillation to stop. For example, when a car travels over a bump, the
passengers are protected by oscillations resulting from the springs in the shock
absorbers that last some reasonable time. Sometimes, we do not want oscillations
to last long. The needle of a bathroom scale should stop without oscillating too
much.
Therefore, the damping effect should also be taken into consideration in a
real oscillatory motion. In the simplest case, let us reconsider a mass attached to
a spring, but this time, let us place the mass inside of a box filled with oil. The
viscosity or the friction of the oil during the motion of the mass will provide the
necessary damping.
In liquids, the friction force is proportional to the velocity of the moving
object. We can feel this when trying to walk in the sea. We can move easily with
slow steps, but the resistance of the water suddenly increases when we try to run.
Using Fd to show this friction force, we get
Fd = −b v (9.27)
and the constant b is called the damping coefficient. The negative sign in this
expression shows that the force is always in the opposite direction to the velocity.
In this new setup, there are two forces acting on the mass m (Figure 9.9): The
Figure 9.9: Damped oscillator. spring force is F and the friction force is Fd . Let us write Newton’s law:
F + Fd = ma
d2 x
−kx − bv = m 2
dt
Let us rearrange the terms and write the differential equation of motion:
d2 x b dx k
+ + x=0 (9.28)
dt2 m dt m
We see that a new term with a first derivative is added.
We can likewise guess the solution to this equation without going into the
topic of differential equations. In simple harmonic motion, we had found the
solution by guessing that cosine and sine are the functions proportional to their
own second derivative. Here, we have the additional friction term with the first
Figure 9.10: Damped harmonic derivative. Likewise, let us consider which function is proportional to its own
motion. It performs sinusoidal derivative. This is the exponential function. Therefore, we look for a solution
oscillations as it decreases expo- that includes both behaviors simultaneously, in other words, the product of
nentially. the exponential and cosine functions. We can then determine the necessary
coefficients by requiring that this solution satisfy the equation. Let us directly
write the solution here without going into too much detail of this long operation:
r
b2
x = A e−(b/2m) t cos ω20 − t with (ω20 = k/m)) (9.29)
4m2
We can check that this solution satisfies Eq. (9.28). The parameter ω20 = k/m is
just the angular frequency of the simple oscillator with no damping.
Let us examine what types of motion this expression involves:
• As shown in Figure 9.10, this is a sinusoidal motion whose amplitude de-
creases exponentially over time.
Figure 9.11: Critically damped • It is easy to see that this equation reduces to simple harmonic motion if
and overdamped oscillations. friction is zero, in other words, if b = 0 .
• Critical damping: When the argument of the cosine is zero, we get cos 0◦ =1
and the amplitude decreases exponentially without the object making any
oscillation. The following must be true for this to happen:
b2 √
ω20 − =0 =⇒ b= 4mk
4m2
No oscillation is observed for a coefficient of friction b above this limiting
value and the object exponentially comes to rest. In this case, it is said to be
overdamped.
9.4. DRIVEN HARMONIC MOTION – RESONANCE 155
Regardless of the type of damping, the mechanical energy of the system

vanishes at the end and all of the energy is converted into heat. An important
application of damped motion in technology is the shock absorber. It provides
for easy driving by preventing vibrations on cars and motorcycles (Figure 9.12).
Large shock absorbers are used to dampen vibrations caused by earthquakes in
modern buildings and bridges.
Figure 9.12: The shock ab-
9.4 DRIVEN HARMONIC MOTION – RESONANCE sorber of a motorcycle.
Did you ever think about how we push a child on a swing? We push the
swing gently at the end of each oscillation. In other words, we apply a periodic
force. And to do this, we wait for the instant when the swing is about to go back.
In other words, we apply a force in phase with the swing.
When an external is force is applied on a harmonic oscillator, the effect of
friction is partly compensated and much richer behaviors are observed. Let us
again use the setup of the damped motion. This time, we apply an external force
Fa cos ωt with angular frequency ω (Figure 9.13).
We write Newton’s law for the sum of the three forces acting on the system: Figure 9.13: Oscillator driven
by a periodic force.
F + Fd + Fa cos ωt = ma
dx d2 x
−kx − b + Fa cos ωt = m 2
dt dt
Let us rearrange the equation using the expression k/m = ω20 :
d2 x b dx
+ + ω20 x = Fa cos ωt (9.30)
dt2 m dt
This time, we have a new term that is not dependent on the unknown x , but is
rather a function of time.
Again, we shall only write the solution of this equation without going into
its explicit solution:
x = A cos(ωt + φ) (9.31)
Fa /m
A = q (9.32)
(ω2 − ω20 )2 + b2 ω2 /m2
Let us emphasize the important properties of the driven harmonic motion:

• The motion is again simple harmonic motion, due to the cos ωt term. Its
angular frequency is equal to the angular frequency of the driving force and
follows it with a phase difference.
• The most important property is that the amplitude A is dependent on the
frequency of the applied force. This amplitude is shown in Figure 9.14 as a
function A(ω) .
√
• Resonance. ω0 = k/m is determined by the properties of the spring and
mass that constitute the oscillator and is called the natural frequency. The
frequency ω of the external force can be arbitrarily varied. Figure 9.14: Variation of ampli-
As seen in the figure, the amplitude of the oscillator increases enormously as tude with respect to external fre-
ω → ω0 . This is called resonance. It is surprising that it has a large impact quency in driven harmonic mo-
even if the amplitude Fa of the external force is small. tion.
Resonance is a very important phenomenon in everyday life and in technology.

In some cases, it is completely intentional:
• In musical instruments, each string is tuned such as to resonate at a different
frequency.
• When searching for a radio station, the receiver is made to resonate with the
broadcast frequency of the transmitter that we want.
• A microwave oven emits electromagnetic waves that resonate the water
molecules inside of food.
• Magnetic Resonance. The neutrons and protons in the nucleus of all atoms
have a magnetic property. They receive and return energy only at certain
frequencies to and from external magnetic fields. Certain atoms are made
to resonate by applying a magnetic field close to their frequencies, and the
location of the radiation that they emit can be measured to map the profile
of organs.
Figure 9.15: Magnetic reso- Sometimes, resonance is unwanted:
nance imaging (MRI) device.
• Soldiers are not marched in unison when crossing a bridge. In this phe-
nomenon, discovered during World War I, the periodic steps of soldiers may
cause a bridge to resonate and collapse.
• Some accessories and equipment made of crystal glass may resonate and
break due to a surrounding vibration.
• Attention is paid to ensure that the natural vibration frequencies of buildings
are not close to earthquake or wind gust frequencies, because, even if the
building is strong, it may resonate and collapse. One example of this was
the collapse of the Tacoma River Bridge in the United States due to the
periodic wind gusts. In order to prevent this hazard, the natural frequencies
of buildings can be determined and their amplitude can be reduced using
shock absorbers.
1. Which is incorrect for simple harmonic motion?
4. If the amplitude of a harmonic oscillator with total en-
(a) The number of oscillations per unit time is the fre-
ergy E is doubled, what will be its total energy?
quency.
(b) The time for one oscillation is the period. (a) E/2 (b) 2E (c) 4E (d) E/4
(c) The maximum oscillation distance is the amplitude.
5. Which is incorrect for simple harmonic motion?
(d) Acceleration is constant.
(a) Velocity is zero when the position is maximum.
2. The angular frequency ω is equal to which of the fol- (b) Position is zero when the velocity is maximum.
lowing? (c) Acceleration is zero when the velocity is maximum.
(a) 2π/ f (b) 2π/T (c) 2π/A (d) 2πA (d) Acceleration is zero when the position is maximum.
3. If a harmonic oscillator passes through the equilibrium 6. Which of the following are incorrect for the period of a
position at time t = 0 , which of the following describes mass+spring system?
its motion? I. It depends on the amplitude.
(a) A cos ωt II. It depends on the mass.
(b) A sin ωt III. It depends on the spring constant.
(c) A tan ωt IV. It depends on the phase angle.
(d) A cos(ωt + π/3) (a) I & III (b) I & IV (c) II & III (d) II & IV
7. By what factor will the period of a pendulum increase if (a) Maximum velocity will double.
its length is increased by a factor of 4? (b) Maximum acceleration will double.
(c) Energy will increase by a factor of 4.
(a) 2 (b) 4 (c) 6 (d) 8
(d) Period will increase by a factor of 4.
8. Which is incorrect for the energy of a harmonic oscilla- 15. Which of the following will not help to increase the
tor? maximum velocity of a mass+spring system?
(a) Potential energy is maximum at the maximum posi-
(a) Increasing its amplitude.
tion.
(b) Increasing its angular frequency.
(b) Kinetic energy is maximum at the maximum veloc-
(c) Increasing its energy.
ity.
(d) Increasing its phase angle.
(c) Kinetic+potential energy is constant.
(d) Kinetic energy is zero at the equilibrium position. 16. How will a pendulum clock behave when taken to the
surface of the Moon?
9. In a mass+spring system, the spring constant is doubled (a) It will lag.
and the mass is halved. By what factor will the period (b) It will run too fast.
decrease? (c) It will not change.
(a) 2 (b) 4 (c) 6 (d) 8 (d) It will stop.
10. The oscillation period of a simple pendulum depends on 17. What is the source of damping in damped harmonic
which of the following? motion?
(a) Mass (a) Spring force.
(b) Amplitude (b) Friction force.
(c) Length (c) Amplitude.
(d) Kinetic energy (d) Potential energy.
18. The period of a physical pendulum does not depend on

11. A simple pendulum used as a clock is lagging. How can
which of the following?
you fix it?
(a) Moment of inertia.
(a) Increase its length. (b) Mass.
(b) Decrease its length. (c) Distance between the axis and the center of mass.
(c) Increase its mass. (d) Amplitude.
(d) Decrease its mass.
19. Which will change when we choose another starting
12. While a simple pendulum is oscillating, half of the mass time t = 0 in a harmonic oscillator?
on its end breaks off and falls. How will the motion (a) Amplitude
change? (b) Frequency
(a) It will get slower. (c) Energy
(b) It will get faster. (d) Phase angle
(c) It will not change.
(d) It will stop. 20. An oscillator consisting of a mass+spring system is sup-
plied with a periodic external force. Which of the fol-
13. The period of a harmonic oscillator is doubled. Which lowing is correct?
of the following is correct? (a) It will approach resonance as the amplitude of the
(a) Its frequency will double. external force increases.
(b) Its angular frequency will be halved. (b) It will approach resonance as the amplitude of the
(c) Its amplitude will double. external force decreases.
(d) Its amplitude will be halved. (c) It will approach resonance as the external frequency
approaches the natural frequency.
14. Which is incorrect when the amplitude of a harmonic (d) No resonance will be observed.
oscillator is doubled?
Problems
9.1 Simple Harmonic Motion 9.7 A system with m=2 kg attached to a spring with spring
9.1 The position of an object with mass m=2 kg undergoing constant k=72 N/m has a total energy of 900 J . (a) What
simple harmonic motion varies with time as, is the amplitude of the oscillation motion? (b) At a certain
x = 3 cos 4π t (meters) . moment, the kinetic energy is 500 J . What is the extension
of the spring at that moment? [A: (a) 5 m , (b) 3.3 m .]
(a) What are the amplitude, angular frequency and period
of the motion? (b) What is the spring constant? (c) What is
9.8 When the body of a car with mass 1200 kg is pressed
the position of the object at time t=1/24 s ? (d) How many
from above and released, it vibrates at a frequency of 3 Hz .
seconds does it take for the object to reach the positions
What will the vibration frequency of the car be if 5 people,
x = −3 m and x = 1.5 m ?
each weighing 70 kg , get into the car? [A: 2.6 Hz .]
[A: (a) A = 3 m , ω = 4π Hz , T = 0.5 s , (b) k = 320 N/m ,
(c) 2.6 m , (d) 1/4 and 1/12 s .]
9.9 A system performs simple harmonic motion. (a) At which
9.2 A mass m=0.5 kg attached to a spring completes 25 cy- x/A ratio will the kinetic energy be twice the potential en-
cles in 10 s . (a) What are its period, angular frequency and ergy? (b) What will the ratios of the kinetic and potential
spring constant k ? (b) This object is stretched by 40 cm from energies be when x = A/5 ? √
its equilibrium position and released at the time t=0 . Write [A: (a) x/A = 1/ 3 , (b) K/U = 24 .]
the equation for the simple harmonic motion equation.
[A: (a) T =0.4 s , ω=5π Hz , k=125 N/m , (b) x=0.4 cos 5π t .] 9.10 A harmonic oscillator with angular frequency 3 Hz
is pulled by 5 m from its equilibrium position and released.
9.3 A mass m=3 kg attached to a spring with spring constant The time t=0 is initiated as it passes through position x=4 m .
k=75 N/m is pulled by 50 cm from its equilibrium point and Find the phase angle and write the equation of the harmonic
released. (a) What will be its maximum velocity and accel- motion. [A: x = 5 cos(3t + 37◦ ) .]
eration? (b) What will its velocity and acceleration be as it
passes through position x = 20 cm ? 9.11 The equation of a harmonic motion is given as
[A: (a) 2.5 m/s , 12.5 m/s2 , (b) 2.3 m/s , −5 m/s2 .] x = 4 cos(πt − π/3)
(a) What is its position at the start? (b) When will it reach
9.4 The angular frequency of an object performing simple the maximum position? [A: (a) 2 m , (b) 0.33 s .]
harmonic motion is 2 Hz . This object passes through position
x = 1 m with velocity v = 4.8 m/s . (a) What is the amplitude
of the motion? (b) What is its maximum velocity?
[A: (a) 2.6 m , (b) 5.2 m/s .]
Problem 9.12
9.12 A bullet with mass m=50 g and velocity v = 200 m/s
hits and embeds in a wooden block with mass M=950 g ,
which is attached to a spring with constant k=64 N/m . Cal-
culate the angular frequency and amplitude of the resulting
Problem 9.5 harmonic motion. [A: ω = 8 Hz , A = 1.25 m .]
9.5 The figure above shows the motion of a mass m = 2 kg
attached to a spring. (a) Find the angular frequency and the
spring constant. (b) Write the equation of this motion.
[A: (a) ω = π/4 , k = 1.25 N/m , (b) x = 0.8 sin πt/4 .]
Problem 9.13
9.13 A mass m=1 kg is placed on the plate of a spring with
constant k=50 N/m that can vibrate in the vertical direction.
How large can the amplitude of the harmonic motion be such
Problem 9.6 that the mass does not leave the plate? The mass of the plate
9.6 Write the equation of the simple harmonic motion shown can be neglected. (Hint: Consider the maximum acceleration
in the figure. [A: x = 0.05 cos(πt−37◦ ) .] of the harmonic motion.) [A: A 6 0.2 m .]
PROBLEMS 159
Problem 9.14
9.14 A mass m1 =1 kg is placed on another mass m2 =2 kg
lying on a frictionless horizontal plane. The friction coef-
ficient between the two blocks is µ=0.5 . The mass m2 is Problem 9.18
attached to a spring with spring constant k=60 N/m and set 9.18 A ring with mass M and radius R is hung on a nail, as
into harmonic motion. What is the maximum amplitude such shown in the figure. Show that the period of the harmonic
that both masses can oscillate together without the top mass motion about the axis O is,s
sliding? (Hint: Compare the maximum acceleration that can 2R
be given to the mass m1 solely by the friction force and the T = 2π
g
maximum acceleration of the harmonic motion.)
[A: A 6 0.25 m .] 9.19 When a rigid body with mass m=1 kg is oscillating
about an axis located at a distance of 20 cm from its center
9.2 Pendulum Motion
of mass, its period is 2 s . Calculate the moment of inertia of
9.15 The combined effect of the centripetal force at the equa- the rigid body with respect to its center of mass.
tor and the oblate shape of the Earth result in a g=9.78 m/s2 [A: 0.16 kg·m2 .]
at the equator and a g=9.83 m/s2 at the poles. By how much
will a simple pendulum clock with a period of 1 s on the
equator lag behind on the poles in 1 day? [A: 220 s .]
9.16 A simple pendulum with length 2.5 m is pulled by 5◦
from its equilibrium position and released. (a) In how many
seconds will it reach its equilibrium position? (b) In how
many seconds will it reach the angle θ = 2.5◦ ?
Problem 9.20
[A: (a) T/4 = π/4 = 0.79 s , (b) π/6 s .]
9.20 Calculate the period of oscillation of the solid sphere
with radius R=20 cm attached to the end of a rope with length
L=1 m , as shown in the figure. The moment of inertia of the
solid sphere with respect to its center of mass is Icm = 25 MR2 .
[A: 2.2 s .]
Problem 9.17
9.17 A stick with mass m and length L=1.20 m can rotate
about an axis O located at distance L/4 from one end. Calcu-
late the period of the pendulum motion. [A: 1.67 s .]
10
WAVES
Surfing is the art of riding a

wave. The size of the swell
(ocean surface wave) depends
on the strength of the wind and
the area of open water.
What really moves forward in a
wave motion? And what does a
wave carry?
We all know how waves are formed by a pebble thrown into a still pond. The
ripples move in ever growing circles. Likewise, waves propagate along a rope
when we shake it from one end. The propagation of any change, vibration or
perturbation in a medium is called a wave. There is a broad range of wave motions
in our everyday life and in nature: Sound waves, water waves, electromagnetic
waves that form light and radio-TV signals, etc.
The source of all of these waves is a vibration: we shake a rope, and a wave
propagates along the rope; a tuning fork is set to vibrate, and sound waves appear,
etc.
Notice that it is not the matter itself but the disturbance that propagates. If
you examine a bottle floating on wavy water, you will notice that the bottle goes
up and down as the wave passes, but does not move forward with the wave.
What is it that moves forward in wave? It is energy and momentum that prop- Figure 10.1: A bottle does not
agate. Each water molecule transfers the energy and momentum that it receives move forward as a wave does, it
to the next one as it goes up and down. Energy and momentum transferred in a only moves up and down.
certain harmony mediate the propagation of waves.

162 10. WAVES
10.1 GENERAL PROPERTIES OF WAVES

Waves can be classified according to their various features. According to the
type of perturbation, they can be separated into two distinct groups (Figure 10.2):
Figure 10.2: Transverse and lon-

gitudinal waves.
• Transverse waves: The perturbation is perpendicular to the wave’s direc-

tion of propagation. The waves on a string are a good example of this type.
• Longitudinal waves: The perturbation is along the wave’s direction of
propagation. When we compress and release a spiral spring (Figure 10.2), a
wave is observed to propagate along the spring.
Sound waves are good examples of longitudinal waves. Sound waves consist
of the back and forth motion of air molecules.
Figure 10.3: Sound wave is a Another classification is according to the medium in which the wave propagates:
longitudinal wave.
• Mechanical waves: A flexible or elastic medium is needed for the wave
to propagate. The wave on a string propagates by moving the string. A
sound wave is produced by the density fluctuations in the surrounding air.
Mechanical waves must have a medium in which to travel.
• Electromagnetic waves: They can propagate in empty space and need no
medium. In the modern view, an electromagnetic field fills all space, and
perturbations in this electromagnetic field form the wave.
We will discuss only mechanical waves in this chapter. Electromagnetic waves
will be mentioned in the discussion on light.
Another classification of waves is in terms of periodicity:
• Periodic wave: The perturbation propagates by repeating itself in time. For
example, if one end of a string vibrates, the wave propagating along the
string will be periodic.
• Wave pulse: This represents the propagation of a single disturbance. For
example, if we shake the end of the string once, we will only observe the
progress of this perturbation.
Figure 10.4: Impulse wave and Regardless of type, every wave has a propagation speed, which is the rate at
periodic wave. which the perturbation propagates. This speed, denoted by v , will be constant if
the medium is homogeneous, and is independent of the cause of the waves.
Wave Speed
The wave speed in a medium depends on the various properties of said
medium, such as density, heat, pressure, tension in the rope, etc. It can be
calculated for a particular type of wave and medium. Without considering how
these speeds are calculated, each of which requires a different method, let us only
give the results:
10.1. GENERAL PROPERTIES OF WAVES 163
• Wave speed on string: The linear mass density of a string is defined as

µ = m/` (kg/m). Accordingly, if a tension F is applied between the two ends
of a string, the wave speed is given with the formula,
s
F
v= (Wave speed on a string) (10.1)
µ
(The tension is indicated here as F to prevent confusion with the period T .)

The more tautly the string is stretched, the faster the wave speed will be.
Wave speed is lower in strings made of denser materials.
• Speed of sound in air: If a gas with mass density ρ (kg/m3 ) has pressure
P , the speed of sound waves in this gas is
s
γP
v= (Speed of sound in air) (10.2)
ρ
Here, γ is a dimensionless constant that describes the thermodynamic prop-

erties of the gas and is approximately γ = 1.4 for air.
• Liquid waves: Wave speed depends on the wavelength λ and the depth h
of the liquid:
 p



 gh (In shallow waters : h λ)
v= (10.3)

 r
gλ
(In deep waters : h λ)




2π

• Electromagnetic waves: The speed of electromagnetic waves in vacuum is

equal to the speed of light c :
v = c = 2.997 × 108 m/s
The speed of electromagnetic waves decreases when they enter a material

medium. Electromagnetic wave speed in a medium with refractive index n
is v = c/n .
Wave Function
What is the mathematical expression of a propagating wave? Figure 10.5
shows the images of a pulse wave propagating with speed v in the +x direction
at two different times. Let the profile of the string be the function f (x) at time
t = 0 . What will be the mathematical expression of this wave at a later time t ?
Figure 10.5: The image of a

wave propagating at speed v at
various times.
Note that the profile given with the function f is only displaced without
changing its form. The wave at time t = 0 is displaced by a distance +vt at time t .
Therefore, it is sufficient to write the same function by displacing it by +vt . And
this becomes the function f (x − vt) . (You may check this with a simple function:
The function y = x crosses the x -axis at 0 , but the function y = x − a crosses it
at +a and the function y=x−2a crosses it at +2a . . . )
164 10. WAVES
Therefore, every function in the form f (x − vt) represents a wave propagating

in the +x direction with speed v .
The reverse is also true: Every function in the form f (x + vt) represents a
wave propagating in the −x direction with speed v .
Sinusoidal Wave
When we vibrate the free end of a string in a periodic fashion, a sinusoidal
wave is produced that propagates along the string. Figure 10.6 shows the dis-
placement along the y -axis of a string extended along the x -axis. This profile
can be shown as a sine or cosine wave. Now let us examine the main parameters
of sinusoidal waves.
Figure 10.6: The distance be-
tween two consecutive crests is
the wavelength λ in a sinusoidal
wave.
The distance between two consecutive crests (maximum points) of a wave is

called the wavelength and is indicated with λ (lambda).
If this wave is generated from a source that vibrates with a period T and the
resulting wave propagates with speed v , the distance λ between the two crests
wavelength = speed × (duration of one vibration)

λ = vT (10.4)
The maximum displacement of a wave is called the amplitude A . Let us

write the profile of the string at t = 0 as a sine function. This function with
wavelength λ and amplitude A is
x
y = A sin 2π (10.5)
λ
The factor 2π was added to ensure that the wave is periodic at multiples of λ .
Now, let us turn this string profile at t = 0 into a wave that propagates at
speed v . We have seen how to do this above: Replacing the function f (x) with
f (x−vt) produces a function that moves in the +x direction at speed v . Therefore,
a sinusoidal function propagating at speed v can be written as:
2π(x − vt)
y = A sin
λ
We can substitute v as v/λ = 1/T to make this expression more symmetrical and
find the sinusoidal wave expression as follows:
x t
y(x, t) = A sin 2π − (Sinusoidal wave function) (10.6)
λ T
The displacement y(x, t) is a function with two variables, depending on both

position x and time t . This expression clearly shows that the wave is periodic in
both space and time: As sin(α + 2π) = sin α , the displacement y of the positions
x and x + λ will always be equal; likewise, the displacement at times t and t + T
will be equal.
10.1. GENERAL PROPERTIES OF WAVES 165
For another form of the wave equation, a quantity called a wavenumber and
indicated with k is defined as follows:
2π
k= (wavenumber) (10.7)
λ
The wavenumber is a measure of the number of wavelengths per unit length, but
gives it in terms of 2π . For example, if the wave length is 1 m, , 12 m, 13 m, . . . ,
then the wavenumber will be k = 2π, 4π, 6π, . . . .
On the other hand, let us remember the angular frequency ω that we know
from Chapter 9:
2π
ω= = 2π f (10.8)
T
The sinusoidal wave is written as follows in terms of these two quantities:
y(x, t) = A sin(kx − ωt) (Sinusoidal wave functions) (10.9)
If the wave is propagating in the −x -direction, we take A sin(kx + ωt) .

This expression is very convenient in calculations, because the coefficients of
x and t can be immediately identified as k and ω . Then, other quantities can be
calculated in terms of ω and k as:
ω 2π 2π
v= , λ= , T= (10.10)
k k ω
If these wave functions also have a phase angle φ , then the equation becomes
y = A sin(kx − ωt + φ) . We will add the phase angle where necessary.
Example 10.1
Answer
The speed of sound in air is 340 m/s . The note A (La) com- We write the relation between frequency and wavelength:
ing out of a musical instrument has a frequency of 440 Hz . v 340
λ = vT = = = 0.77 m = 77 cm .
Calculate the wavelength of the note A (La). f 440
Example 10.2
Answer
We first calculate the wave speed for each string with
Eq. (10.1):
p
v1 = F/µ1 = 40/0.1 = 20 m/s
p
p
v2 = F/µ2 = 40/0.2 = 14 m/s
p
Two strings, each with length L=1 m , are attached end to end.
The linear density of the first is µ1 =0.1 kg/m and that of the A wave with speed v travels the distance L in time L/v .
second is µ2 =0.2 kg/m . The tension F=40 N is applied to the Accordingly, we calculate the total time as follows:
outer ends of the strings. How much time does it take for a wave L L 1 1
t = t1 + t2 = + = + = 0.12 s
to travel from end A to end B on this composite string? v1 v2 20 14
Example 10.3
Answer
The sinusoidal wave on a string with linear density µ=0.1 kg/m (a) We can directly identify the following quantities, as the
is given as wave is given in the form A sin(kx − ωt) :
y(x, t) = 0.03 sin(5πx − 40t) (meters) A = 0.03 m, k = 5π m−1 , ω = 40 Hz
(a) What are the wave amplitude, wavelength, period and The others are found using Eqs. (10.10):
wave speed? ω 40
(b) What is the tension in the string? v= = = 2.5 m/s
k 5π
166 10. WAVES
2π 2π v = F/µ → F = µ v2
p
λ= = = 0.4 m
k 5π We substitute the numerical values:
2π 2π
T= = = 0.05π s F = 0.1 × (2.5)2 = 0.63 N .
ω 40
(b) We use Eq. (10.1) for the wave speed on the string:
Example 10.4
is f (x + vt) , then it propagates in the −x -direction.
Determine the direction of propagation and wave speed for the We then convert the given functions into this form (without
following wave functions. considering the amplitudes or the signs):
(a) y = 2 sin(3x − 18t) (a) sin(3x − 18t) = sin[3(x − 6t)]
(b) y = 3 cos(4t − 10x) → v = 6 m/s, in the + x direction
(c) y = 4 sin(3x + 21t) (b) cos(4t−10x) = cos[−(4t−10x)] = cos[10(x− 25 t)]
Answer → v = 25 m/s, in the + x direction
If a function has the form f (x − vt) , then it is a wave propa- (c) sin(3x + 21t) = sin[3(x + 7t)]
gating in the +x -direction with speed v . However, if its form → v = 7 m/s, in the − x direction
Example 10.5
A = 0.20 m , λ = 0.60 m
To find the wave speed, let us note that the point at the origin
at time t = 0 was displaced by 1.80 m at time t = 0.6 s .
Accordingly, we calculate the wave speed as follows:
1.80
v= = 3 m/s
0.6
From here, we calculate the period:
The figure shows the profile of a sinusoidal wave on a string at λ 0.6
times t = 0 and t = 0.6 s . Find the wave function. T= = = 0.2 s .
v 3
Answer We find the wave function based on this information:
The amplitude and wavelength can be immediately identified
x t x t
y = A sin 2π − = 0.2 sin 2π −
by examining the figure: λ T 0.6 0.2
Example 10.6 s r
F 100
v= = = 20 m/s .
Two ends of a string with linear density µ=0.25 kg are stretched µ 0.25
with a pair of forces F=100 N . The string is transversely pulled We calculate the period and the wavelength:
from one end by 3 cm and released. It vibrates with a frequency 1 1
T = = = 0.125 s
of 8 Hz . f 8
(a) Find the wave speed, period and wavelength. λ = v T = 20 × 0.125 = 2.5 m .
(b) Write the equation of the sinusoidal wave produced. (b) We use Eq. (10.6), which expresses the wave function in
terms of T and λ:
Answer x t x t
y = A sin 2π − = 0.03 sin 2π −
λ T 2.5 0.125
(a) We use Eq.(10.1) for speed:
10.2 INTERFERENCE, REFLECTION AND TRANSMISSION OF WAVES

Superposition Principle
What will be the combined effect of two or more waves in the same medium?
For example, the sounds from many musical instruments in an orchestra arrive at
our ears simultaneously. Likewise, electromagnetic waves from many TV stations
affect the television antenna simultaneously.
The superposition principle determines the combined effect of waves:
10.2. INTERFERENCE, REFLECTION AND TRANSMISSION OF WAVES 167
Superposition Principle
The total effect of two or more waves propagating in the same

medium is equal to the algebraic sum of their wave functions:
y(x, t) = y1 (x, t) + y2 (x.t) (10.11)
We can explain the source of this principle in mathematical terms as follows:
Functions in the form f (x−vt) satisfy a differential equation that has the following
property: If y1 and y2 are two separate solutions, then the function y1 + y2 is
also a new solution. Experiments show that the principle is true for waves with
small amplitudes.
Let us emphasize the most important conclusions to be drawn from the
superposition principle:
• If two waves reach a point with the same sign, the displacement of that point
increases. This is called constructive interference (Figure 10.7a).
Figure 10.7: (a) Constructive in-

terference and (b) Destructive in-
terference in the superposition of
two waves.
• If the algebraic sum of two waves at a point is zero, then that point will
remain motionless. This is called destructive interference (Figure 10.7b).
• Waves continue along their paths with their former shapes after overlapping
and separating.
Interference
Let us examine in detail the constructive and destructive interference that
we mentioned above. Consider two sinusoidal waves propagating in the same
medium and in the same direction (Figure 10.8). In the simplest case, let the
amplitudes, wavelengths and frequencies of both waves be equal. Let the only
difference between them be a phase difference φ on one of the waves:
y1 = A sin(kx − ωt)
y2 = A sin(kx − ωt + φ)
According to the superposition principle, the total wave in the medium will be
the algebraic sum of these two: Figure 10.8: The interference
of two identical waves with a
y = y1 + y2 = A sin(kx − ωt) + A sin(kx − ωt + φ) small phase difference.
If we apply an identity from trigonometry on this sum, we get
a+b
! !
a−b
sin a + sin b = 2 sin cos
2 2
sin(kx − ωt) + sin(kx − ωt + φ) = 2 sin(kx − ωt + φ/2) cos(−φ/2)
168 10. WAVES
Since cos(−a) = cos a , the total wave function can ultimately be written as
follows: h i
y = 2A cos(φ/2) sin(kx − ωt + φ/2) (10.12)
| {z }
A0
The term sin(kx − ωt + φ/2) here shows that this expression is again a wave
propagating in the +x direction with the same wavelength and same frequency,
but with a phase difference of φ/2 . The factor 2A cos φ/2 at the front of this
expression is independent of x and t and is considered the amplitude of the new
wave:
y = A0 sin(kx − ωt + φ/2) with A0 = 2A cos(φ/2) (10.13)
Let us consider two special cases of the phase angle:
• If φ = 0◦ , in other words, if the two waves are in equal phase, we get
cos 0◦ = 1 . In this case, the wave function y becomes the same function
with amplitude 2A . This is called constructive interference.
• If φ = 180◦ , in other words, if the two waves are in opposite phase, we get
cos 90◦ = 0 . In this case, we get y = 0 for every value of x and t . And this
is called destructive interference.
We have considered here the interference of two identical waves as the sim-
plest case. In more general cases, interference between two waves with different
wavelengths and frequencies exhibits more interesting features, as the following
example shows.
Standing Waves
Another important case of interference is that of two waves propagating
in opposite directions. Let us consider two waves with equal amplitude, equal
wavelength and equal frequency, one traveling in the +x -direction and the other
in the −x -direction:
y1 = A sin(kx − ωt) (wave traveling in the +x-direction)
y2 = A sin(kx + ωt) (wave traveling in the −x-direction)
According to the superposition principle, the total wave function will be
y = y1 + y2 = A sin(kx − ωt) + A sin(kx + ωt)
Figure 10.9: Standing wave pro-

duced by two waves propagating
in opposite directions.
If we again use the trigonometric identity above for sin a+ sin b ,

y = 2A sin kx sin ωt (10.14)
This result shown in Figure 10.9 is called a standing wave. Let us emphasize the
important features of standing waves:
• This wave is no longer a propagating wave, because it does not have the
form f (x ± vt) . If you examine the wave produced on a guitar string, you will
notice that it does not propagate (Figure 10.10). Each point makes a vibration
motion with a different amplitude. This is the vibration motion cos ωt with
amplitude A0 = 2A sin kx :
y = [2A sin kx] sin ωt

| {z }
A0
Figure 10.10: A standing wave

produced in laboratory (Harvard
Natural Sciences Lecture Demon-
strations).
• The points at certain x values do not move. These points, called nodes,
occur at the values sin kx = 0 :
sin kx = 0 −→ kx = 0, π, 2π, 2π, 3π . . .
Using the definition k = 2π/λ for the wavenumber, we get

λ 3λ λ
x = 0, , λ, ... = n (n = 0, 1, 2, . . .) (Nodes) (10.15)
2 2 2
• The points at certain x values vibrate at maximum amplitude. These points,

called antinodes, occur at the values sin kx = 1 :
π 3π
sin kx = 1 −→ kx = , ,...
2 2
λ 3λ 5λ λ
x= , , . . . = (2n + 1) (n = 0, 1, 2, . . .) (Antinodes) (10.16)
4 4 4 4
Harmonics of a Musical Instrument
Waves are generated on stringed instruments like the violin or the guitar,
upon which a string has two fixed ends. According to our results above, the two
fixed ends of such a string must be a node. Therefore, on a string with length L ,
the points x = 0 and x = L should fulfill the following condition:
sin k.0 = sin kL = 0 =⇒ kL = 0, π, 2π, 2π, 3π . . .
Writing this for wavelength λ = 2π/k we have,

λ 2L
L=n =⇒ λn = (n = 1, 2, 3 . . .)
2 n
170 10. WAVES
The length L of the string should accommodate half-wavelengths. According to

this result, only waves with wavelength λn can be generated in a string with two
fixed ends (Figure 10.11).
We use the formula f = v/λ to convert these into frequencies:
v
fn = n (n = 1, 2, 3 . . .) (10.17)
2L
These frequencies that can be generated by a string with a certain length are
called natural frequencies or resonance frequencies. In musical terms, they are
called harmonics. The one with n = 1 is called the fundamental frequency
or the first harmonic; and fn = 2 f1 , 3 f1 . . . are called the nth harmonics:
v
f1 = (The fundamental frequency or the first harmonic) (10.18)
2L
Here, let us also use Eq. (10.1) v = F/µ , which we previously gave for the wave
p
speed on a string:
s
n F
fn = n f1 = n (n = 1, 2, 3 . . .) (10.19)
2L µ
When “tuning” a musical instrument, we vary the tension in the strings to ensure
Figure 10.11: Natural frequen- that it generates standing waves at the required frequency.
cies or harmonics on a string of Timbre: The property that distinguishes the sound of a violin from that of a
length L . guitar is called timbre. Timbre is a result of mixing harmonics. In each instrument,
together with the fundamental frequency, a few more of these harmonics are
generated. The mixing rates of these harmonics depends on the type of musical
instrument. A different harmonic mixture is generated in the violin than is
generated in the guitar.
Reflection and Transmission of Waves
What happens when a wave encounters an obstacle or reaches the boundary
between two mediums? If the second medium is also flexible, some of the wave
will be transmitted and some will be reflected back. Let us review the general
rules of reflection and transmission before going into their detailed calculation.
Figure 10.12 shows a wave on a string with one end fixed to the wall, with
the wave moving towards the wall. When the wave reaches the wall, it will be
fully reflected back, because it cannot pass to the other side. However, the point
A of the string connected to the wall must always remain motionless, due to the
superposition principle. The only way to ensure this is for the reflected wave to
have opposite phase with the incoming wave. It means that the reflected wave is
inverted.
Figure 10.12: A wave reflected

from a fixed end will have a 180◦
opposite phase.
Therefore, a wave reflected from a fixed end will be inverted. Another way
to see this is as follows: When the wave hits the wall, according to Newton’s
law, the wall will push the string back with an equal and opposite force. In other
words, the wave will be pushed downwards if it tries to get displaced upwards.
Figure 10.13: A wave reflected

from a free end will have the
same phase.
On the other hand, the end of the string could be free to move. For example,
we can consider a ring that can slide freely up or down the wall (Figure 10.13). In
this case, the reflection will be in the same phase, because this end of the string is
displaced in the same direction for the waves in both directions.
Figure 10.14: (a) A wave com-

ing from a denser medium will be
reflected in the same phase. (b)
A wave trying to enter a denser
medium will be reflected in the
opposite phase.
Apart from these two extreme cases, a wave is usually partially transmitted
into a second flexible medium and partially reflected. Whether or not there
is a phase difference during the reflection depends on the densities of the two
mediums. It gets reflected back without phase difference if it comes from a denser
medium. In contrast, the reflected wave will be inverted if transmitting into a
denser medium. Both cases are summarized in Figure 10.14.
Example 10.7
We first calculate the wavelength to find the nodes and antin-
Two waves propagating on a string are given as odes:
2π 2π
y1 = 3 sin(5πx − 4t) λ= = = 0.4 m .
k 5π
y2 = 3 sin(5πx + 4t) The nodes will be located at the integer multiples of λ/2 and
antinodes at the odd multiples of λ/4 :
(a) Determine the nodes and antinodes of the standing wave.
Nodes: x = 0, λ/2, λ, 3λ/2 · · · = 0, 0.2, 0.4 · · · m
(b) Calculate the amplitude at the point x = 0.05 m .
Antinodes: x = λ/4, 3λ/4, 5λ/4 · · · = 0.1, 0.3, 0.5 · · · m
Answer (b) The amplitude of the vibrational motion of any position
(a) The standing wave generated by the waves A sin(kx ± ωt) x will be the coefficient of cos ωt :
propagating in opposite directions was given by Eq. (10.14):
A0 = 2A sin kx = 6 sin 5πx
y = 2A sin kx cos ωt
Therefore, the values k = 5π and ω = 4 are read from this We find the amplitude by substituting the √ value x=0.05 :
expression. A = 6 sin 5π × 0.05 = 6 sin π/4 = 6/ 2 = 4.2 m .
0
Example 10.8
wavelength can fit in the string:
A tension of F = 100 N is applied on both ends of a string with λ1
L= → λ1 = 2L
length 1 m and linear density 0.1 kg/m . What are the funda- 2
mental frequency (1st harmonic) and 2nd harmonic frequency Frequency can be calculated if the wavelength and speed v
generated on this string? are known:
v 33.3
f1 = = = 17 Hz
Answer λ1 2
Let us first
p calculate
p the wave speed on the string: Likewise, two half-wavelengths should fit in the length L to
v = F/µ = 100/0.1 = 33 m/s obtain the 2nd harmonic:
We had previously found the formulas for the fundamental λ2
L=2 → λ2 = L
and harmonic frequencies. However, it is more convenient to 2
From here, we calculate the frequency:
keep in mind the wavelength condition, rather than memorize
v 33.3
these formulas. f2 = = = 33 Hz
The fundamental frequency should be such that a half- λ 2 1
172 10. WAVES
Example 10.9
We take n=1 for the fundamental frequency:
v 240
The wave speed on a string with length 80 cm is 240 m/s . f1 = = = 150 Hz .
2L 2 × 0.80
(a) What is the fundamental frequency? (b) We write the harmonic frequencies in terms of the funda-
(b) Considering that the human ear can hear sounds with a mental frequency:
maximum frequency of 20 000 Hz , what is the highest fn = n f1
harmonic that can be heard on this string? In order to be audible by the human ear, the frequency fn
Answer should be as close as possible to the value 20 000 Hz :
20 000
(a) We use the formula (10.17), which gives the harmonic n f1 6 20 000 → n 6 = 133.3
frequencies: 150
The closest value to this is n = 133 :
v
fn = n (n = 1, 2, 3 . . .) f133 = 133 × 150 = 19 950 Hz .
2L
Example 10.10
v = F/µ → F = µv2
p
A guitar string with linear density µ = 0.001 kg/m and length F = 0.001 × 3142 = 99 N
60 cm was tuned so that its fundamental frequency produces (b) The wave speed v is the same, because the tension of the
the note C (Do, frequency 262 Hz ). string has not changed. However, this time, the fundamental
(a) What is the tension in the string? frequency is changed to f 0 , because the length of the string
(b) Where should the string be pressed so that it can produce is changed to L0 :
the note D (Re, frequency 294 Hz )? v
v = λ0 f 0 = 2L0 f 0 → L0 = 0
Answer 2f
We first calculate the wave speed: We use the frequency of the note D and the wave speed found
v=λf in item (a):
314
The wavelength of the fundamental frequency is such that L0 = = 0.53 m .
λ = 2L . We can calculate the wave speed from this: 2 × 294
v = 2L f = 2 × 0.60 × 262 = 314 m/s Accordingly, one must press at 60 − 53 = 7 cm from one end
We calculate the tension F giving this speed: of the string.
10.3 THE DOPPLER EFFECT AND SHOCK WAVES

In this section, we will discuss two interesting effects caused by waves: The
Doppler effect is observed when the wave source or the observer is moving. A
shock wave is observed when the wave source moves faster than the wave.
Both effects have very significant technological applications.
The Doppler Effect
Have you ever stood by a highway and listened to the sound of a car passing
by at top speed? The sound that you hear as the car approaches and the one
that you hear as it moves away have a distinguishable difference to the ear. The
sound has a higher pitch as the car approaches and a lower pitch as it moves
away. Expressed in terms of frequencies, the frequency of the approaching sound
is greater and the frequency of the receding sound is lesser. This is called the
Doppler effect.
The Doppler effect is observed when both the wave source and the observer
are moving. Let us first calculate both cases separately, then combine them
together.
Standing source, moving observer
Let us clearly designate our parameters: v is the speed of sound. f s and λ s
are the original frequency and wavelength emitted when the source is at rest. v s
and vo are the speeds of the source and the observer, respectively.
10.3. DOPPLER EFFECT - SHOCK WAVES 173
As seen in Figure 10.15, a source at rest is emitting sound at frequency f s and

an observer approaching the source is traveling at speed vo .
Let us remember the relation between the frequency f s and wavelength λ s
emitted by a source at rest:
v
fs = Figure 10.15: Sound source at
λs rest and observer moving to-
In other words, the distance between two consecutive wavefronts is λ s . The wards the source at speed vo .
speed of this wave with respect to the ground is v , whereas the observer running
at speed vo observes the wave to be approaching at speed v + vo . In this case, the
frequency heard by the observer is
v + vo
fo =
λs
If we eliminate the wavelength λ s between these two equations and rearrange
the terms, we get
v + vo
fo = fs (10.20)
v
The observer will hear the sound at a higher frequency.
If the observer is moving away from the source, we take the velocity of the
observer as negative. In this case, v + vo will be smaller and the sound heard by
the observer will have a lower frequency.
Moving source, stationary observer
As seen in Figure 10.16b, let the source travel with a speed v s and approach a
stationary observer. If the emitted sound has frequency f s and period T s = 1/ f s
when the source is stationary, the wavelength of the sound emitted from the
stationary source will be
v Figure 10.16: (a) The source at
λs = v T s = rest generates sound with wave-
fs
length λ s . (b) The distance be-
Now, let us consider that the first of the two consecutive wavefronts is emitted. tween two fronts (wavelength
The source will have traveled a distance of v s T s = v s / f s during the time that λ0 ) decreases when the source
it takes for the second wavefront to be emitted. Therefore, the new distance is moving.
between two consecutive fronts, in other words, the wavelength λo measured by
the observer, will be, vs
λo = λ s − v s T s = λ s −
fs
The stationary observer will perceive the frequency of the sound approaching
with wavelength λo , as follows:
v
fo =
λo
If we substitute the value of λo and rearrange, we find that:
v
fo = fs (10.21)
v − vs
The observer will again observe the sound at a higher frequency. If the source
is moving away from the observer, we will take the value v s as negative in the
formula. In that case, the sound heard by the observer will have a lower frequency.
Let us note one point here: The respective speeds of the source and observer
do not appear as symmetrical in these formulas. In other words, the Doppler
174 10. WAVES
effect gives different results depending on whether the object or the source is
approaching with the same speed. The reason for this is that the wavelength of
the sound propagating in the air is different in both cases. Relative velocity does
not change this fact.
Both source and observer moving
We can easily calculate this case by using the two formulas found above for
only one of the observer or the source moving.
The frequency fo heard by a stationary observer from a moving source was
given by Eq. (10.21). This fo value replaces f s in Eq. (10.20), which we found for
a moving observer. Therefore, if we apply both formulas successively, we get
v + vo v
fo = fs
v v − vs
v + vo
fo = fs (Doppler formula) (10.22)
v − vs
Note the signs of the velocities v s and vo when using this formula: Velocities in
the approaching direction are taken as positive for both the source and the observer.
In the “away” direction, they are taken as negative.
The Doppler formula for electromagnetic waves is slightly different from
this. This is because the relative velocity addition formula of the speed of light is
different from that of classical physics.
The Doppler effect has a wide range of applications in science and technology.
The most important one is determining the velocities of celestial bodies. The
approaching or receding speed of a star moving relative to the Earth can be
calculated by measuring the change in the frequency of a known color in the
Figure 10.17: The speed con-
light emitted by that star. Measurements indeed show that stars mostly move
trol device called radar, operates away from each other. This means that the universe is expanding.
with the Doppler effect in traffic. Another application is the device known as “radar” which is used by highway
patrols to monitor the speed of vehicles. In this device, an electromagnetic
wave (microwave) sent forward is reflected from a moving vehicle back into the
device. The speed of the vehicle can be calculated from the difference between
the frequencies of the outgoing and incoming waves.
Shock Wave
What happens if a wave source travels faster than the wave it generates?
We can again answer this question with wavefronts. Figure 10.18 shows the
wavefronts generated at various times by a source traveling at speed v s . These
are spherical surfaces in three-dimensional space. If v s <v , in other words, if the
source velocity is less than the wave velocity (Figure 10.18a), the distance between
consecutive fronts will decrease in the direction of motion and increase in the rear
direction, as seen in the figure. In this case, the Doppler effect will be observed.
Figure 10.18: Wavefronts

formed when the source velocity
v s is (a) less than, (b) equal to
or (c) greater than the wave
velocity v .
10.3. DOPPLER EFFECT - SHOCK WAVES 175
However, if v s >v , in other words, if the source velocity is greater than the
wave velocity (Figure 10.18c), the source moves by leaving behind the wavefronts
that it previously emitted. Let us examine the surface of the cone tangent to all
of these wavefronts. When this cone reaches a stationary observer, the observer
will hear the sum total of the magnitudes of each wavefronts at once. He/she will
hear a booming explosion (sonic boom). Thus, we have what is called a shock
wave.
When modern fighter planes fly faster than the speed of sound, this conic
wavefront produces a sound that resembles an explosion everywhere it passes. It
is said that the “plane broke through the sound barrier.” This is actually incorrect,
because a shock wave is not an instantaneous phenomena. The plane is already
flying faster than the speed of sound, but the conic wavefront passes through an Figure 10.19: The shock wave
observer only once. As it applies the energy of all of the wavefronts at once, its formed when a fighter plane ex-
impact can be large enough to break window glasses. ceeds the sound barrier causes
Shock waves are used to break kidney stones in modern medicine. Sound the air to concentrate where it
passes, and a conic wavefront is
waves at frequencies called ultrasound, which are sent from outside of the body
observed.
transfer their energy into dense stones and break them into small parts.
Example 10.11 v + vo
fo = fs
v − vs
The siren of an ambulance can emit sound at the 440 Hz fre- The velocity of the stationary observer will be vo =0 . We
quency. The ambulance is approaching a junction at a speed of convert the velocity v s of the ambulance into m/s units:
144 km/hour . 144 × 1000
vs = = 40 m/s
3600
(a) What frequency is heard by an observer standing at the The velocity v s of the ambulance (source) is taken as posi-
junction? tive when it is approaching. Accordingly, we calculate the
(b) What frequency is heard by the observer when the am- frequency heard by the observer:
bulance is moving away? (The speed of sound in air is 340 + 0
v = 340 m/s .) fo = × 440 = 499 Hz
340 − 40
(b) We take the velocity v s as negative as the ambulance
Answer moves away:
(a) We calculate fo , the frequency heard by the observer, by 340 + 0
f = × 440 = 394 Hz .
using Eq. (10.22), which we found for the Doppler effect: 340 + 40
Example 10.12
and the speed of sound:
An orchestra in an open-top train car is playing the note A (La) 340
466 = × 440 → v s = 19 m/s .
(frequency 440 Hz ) as it travels. 340 − v s
(a) An observer sitting in a stationary car beside the railroad (b) We use the Doppler formula in which both the source and
tracks hears this sound as the note B-flat (Si-bemol) (fre- the observer are moving:
quency 466 Hz ). Calculate the speed of the train. v + vo
fo = fs
(b) In what direction and at what speed should the observer v − vs
drive his/her car to hear this sound as the note B (Si) (fre- The source velocity was found in item (a). We find the ob-
quency 494 Hz )? server velocity by also substituting the frequencies:
340 + vo
Answer 494 = 440 → vo = +20 m/s
(a) We set vo =0 in The Doppler formula (Eq. 10.22) as the 340 − 19
The observer should drive the car towards the train, because
observer is stationary:
v the velocity is positive.
fo = fs (Note: The experimental proof of the Doppler effect was pro-
v − vs
We find the source velocity by substituting the frequencies vided by a similar orchestra playing on a train in 1845.)
176 10. WAVES
1. The speed of a wave on a string is not dependent on
which of the following? 8. When will a shock wave occur?
(a) If the source velocity is less than the speed of sound.
(a) The force stretching the string.
(b) If the source velocity is equal to the speed of sound.
(b) The density of the string.
(c) If the source velocity is greater than the speed of
(c) The length of the string.
sound.
(d) None of the above.
(d) If the amplitude of the sound is high.
2. A string is vibrated from one end. Which of the follow-
ing is incorrect if the vibration frequency is doubled? 9. Which is incorrect for two waves propagating on the
same string?
(a) The wave speed will double.
(b) The angular frequency will double. (a) Their frequencies may be different.
(c) The period will halve. (b) Their velocities may be different.
(d) The wavelength will halve. (c) Their wavelengths may be different.
(d) Their amplitudes may be different.
3. Two strings of equal lengths are stretched with equal
forces. Which of the following is correct? 10. The tension in a string is increased by a factor of 4. What
(a) The one with higher density will have a higher wave will happen to the wave speed?
speed. (a) It will decrease by a factor of 4.
(b) The one with the lower density will have a lower (b) It will be halved.
wave speed. (c) It will be doubled.
(c) The one with the higher density will have a lower (d) It will increase by a factor of 4.
wave speed.
(d) Their wave speeds will be equal. 11. The density of a string is increased by a factor of 4. What
will happen to the wave speed?
4. How would you hear the frequency of a sound source
(a) It will decrease by a factor of 4.
approaching you?
(b) It will be halved.
(a) At a higher frequency. (c) It will be doubled.
(b) At a lower frequency. (d) It will increase by a factor of 4.
(c) At the same frequency.
(d) At double the frequency.
12. Two strings with different densities are attached end to
5. Which is correct if the frequency of a wave is doubled end. Which will remain constant when a periodic wave
on a string with constant tension? is transmitted from one string to the other?
(a) The velocity will double. (a) Speed
(b) The wavelength will double. (b) Wavelength
(c) The wavelength will halve. (c) Frequency
(d) The period will double. (d) Amplitude
6. Which is correct when two waves reach the same point 13. Which is incorrect when a wave propagating on a light
at the same time? string reaches a dense string?
(a) Destructive interference will occur if they arrive (a) It will be reflected with the same phase.
with same phase. (b) It will be reflected with the opposite phase.
(b) Constructive interference will occur if they arrive (c) It will be transmitted with the same phase.
with opposite phases. (d) It is impossible to tell.
(c) Destructive interference will occur if they arrive
with opposite phases. 14. Which is correct for two waves propagating towards
(d) There will be no interference. each other on the same string?
7. Which of the following is correct? (a) They will collide like billiard balls.
(b) They will pass through each other.
(a) Sound waves are transverse waves.
(c) They will merge and become a single wave.
(b) Water waves are longitudinal waves.
(d) It is impossible to tell.
(c) Sound waves are longitudinal waves.
(d) The wave on a string is a longitudinal wave.
PROBLEMS 177
15. Which of the following is a wave propagating in the (a) They exchange energy.
−x -direction? (b) They will affect each other’s propagation.
(a) sin 3x cos 5t (c) They lose energy during the interference.
(b) sin(3x − 5t) (d) They are added algebraically.
(c) sin(3x + 5t)
(d) cos(3t − 5x) 19. Which wave property distinguishes the sound of a violin
from that of a guitar?
(a) Amplitude
16. Which is a standing wave?
(b) Frequency
(a) sin 3x cos 5t (c) Wavelength
(b) sin(3x − 5t) (d) Harmonics
(c) sin(3x + 5t)
(d) cos(3t − 5x) 20. Which of the following situations will decrease the fre-
quency heard by the observer due to the Doppler effect?
17. A wave propagating on a string does not transmit which (a) When the observer approaches the source.
of the following? (b) When the source approaches the observer.
(a) Material (b) Energy (c) Momentum (d) Work (c) When the source and observer approach each other
simultaneously.
18. Which of the following is correct when there is interfer- (d) When the source moves away from the observer.
ence between two waves?
Problems
(The speed of sound in air is to be used as 340 m/s wherever
end to end. If a tension of F=50 N is applied from the two
necessary.)
outer ends, how much time will it take for a wave to travel
10.1 General Properties of Waves from one end to the other? [A: 0.11 s .]
10.1 The note C (Do) coming out of a musical instrument 10.6 Tsunami waves in oceans have very long wavelengths.
has a frequency of 262 Hz . Calculate its wavelength. The wavelength of a tsunami wave that occurred after an
[A: 1.3 m .] earthquake in Japan was observed to be 200 km . This wave
was observed to reach the coasts of Australia, which are 7000
10.2 The range of visible light, which is a type of electromag- km away, in 9 hours. (a) What is the wave speed? (b) As
netic wave, is between red light at the frequency 4.3×1014 Hz the wavelength is very large with respect to the depth of the
and violet light at frequency 7.5 × 1014 Hz . Taking the speed ocean, what is the average ocean depth between Japan and
of light as c = 3 × 108 m/s , calculate the wavelength range Australia? [A: (a) 216 m/s , (b) 4.7 km .]
of visible light in units of nanometers (nm).
[A: 400 − 700 nm .] 10.7 The sinusoidal wave on a string with linear density
µ=0.2 kg/m is given as
10.3 Sound waves above 20 kHz , which is the highest fre-
quency audible by the human ear, are called ultrasound. They y(x, t) = 0.7 sin(0.4πx − 12t) (meters)
are used to obtain images by passing them through human
skin, after which they are reflected by the internal organs. The (a) What are the wave amplitude, wavelength, period and
speed of ultrasound in the body is 1540 m/s . The wavelength wave speed? (b) What is the tension on the string?
should be around 1 mm for high-quality imaging. What [A: (a) A = 0.7 m , λ = 5 m , T = π/6 s , v = 9.5 m/s ,
should the frequency of the ultrasound be that gives 1 mm (b) T = 18 N .]
wavelength inside of the body? [A: 1.54 MHz .]
10.8 Determine the direction of propagation and speed of
10.4 Wave speed is 30 m/s when a tension of 9 N is applied the wave for the following wave functions:
between two ends of a string. How much tension should be (a) y = 2 sin(2x − 7t)
exerted to obtain a wave speed of 40 m/s ? [A: 16 N .] (b) y = 3 cos(3t − 8x)
(c) y = 4 cos(7x + 21t)
10.5 Two strings with linear densities µ1 =0.05 kg/m and [A: (a) 3.5 m/s in the +x -direction, (b) 0.38 m/s in the +x -
µ2 =0.2 kg/m and with equal length L=1.20 m are attached direction, (c) 3 m/s in the −x -direction.)
178 10. WAVES
10.9 Two ends of a string with linear density µ=0.5 kg are frequency of 20 000 Hz , what is the highest harmonic that
stretched with a force F=8 N . When the string is transversely can be heard on this string?
pulled from one end by 5 cm and released, it oscillates with [A: (a) 450 Hz , (b) n = 44 .]
a frequency of 20 Hz . (a) Find the wave speed, period and
wavelength. (b) Write the equation of the sinusoidal wave 10.15 On a string with length 40 cm , the frequencies of two
that is produced. successive harmonics are 440 Hz and 500 Hz . (a) What is
[A:(a) 4 m/s , 0.05 s , 0.2 m , (b) y = 0.05 sin(10πx − 40πt) .] the fundamental frequency? (b) What is the wave speed on
the string? [A: (a) 60 Hz , (b) 48 m/s .]
10.16 A guitar string with linear density µ=0.001 kg/m and

length 30 cm is tuned such that its fundamental frequency
is the note E (Mi, frequency 330 Hz ). (a) How much is the
string stretched? (b) Where should the string be pressed so
that it can produce the note F (Fa, frequency 349 Hz )?
Problem 10.10
[A: (a) 39 N , (b) 2 cm .]
10.10 The figure shows the shape of a sinusoidal wave on a
string at times t = 0 and t = 2 s . Write the expression of the
10.17 A sound wave sent from a speaker at one end of a
wave function. [A: y = 0.05 sin(5πx − 6πt) .]
conference hall hits the opposite wall and is reflected back.
10.11 The distance between two successive maximum points An observer walking between these two walls notices that
of water waves in a pool is measured as 1.5 m . 10 maximum no sound is heard at 2 m intervals. Calculate the frequency
points pass by an observer in 5 s . (a) What is the wave speed? of the sound. [A: 85 Hz .]
(b) Considering that the pool is shallow, what is the average
depth of the pool? [A: (a) 3 m/s , (b) 0.9 m .] 10.3 The Doppler effect and Shock Waves
10.2 Interference, Reflection and Transmission 10.18 The whistle of a train traveling at a speed of 50 m/s is
of Waves able to emit sound at a frequency of 330 Hz . What frequency
10.12 Two waves propagating on a string are given as will a stationary observer beside the railroad tracks hear, (a)
2π when the train is approaching, (b) when the train is moving
away? [A: (a) 387 Hz , (b) 288 Hz .]

y1 = 2 sin x − 7t
3
2π
10.19 A train is approaching as its whistle blows the note C

y2 = 2 sin x + 7t
3 (Do, frequency 262 Hz ). (a) An observer sitting in a station-
(a) Determine the nodes and antinodes of the standing wave ary car beside the tracks hears this sound as the note D-flat
produced. (b) Calculate the vibration amplitude of the point (Re bemol, frequency 277 Hz ). Calculate the speed of the
x=0.25 m . [A: (a) Nodes: x = 0, 0.75, 1.5, 2.25 . . . m ,
train. (b) In what direction and at what velocity should the
Antinodes: x = 0.38, 1.13, 1.88 . . . m . (b) A0 = 3.5 m .] observer drive his/her car to hear this sound as the note D
(Re, frequency 294 Hz )?
10.13 A tension F=90 N is applied to both ends of a string
[A: (a) 18 m/s , (b) 21 m/s towards the train.]
with length 1.20 m and linear density 0.1 kg/m . What are
the fundamental frequency (1st harmonic) and 3rd harmonic
10.20 A stationary observer beside a railroad track mea-
frequency generated on this string?
sures the frequency of the train’s whistle. He/she finds the
[A: 12.5 Hz, 37.5 Hz .]
frequency to be 400 Hz while the train is approaching and
10.14 The wave speed on a string with length 40 cm is 240 Hz while it is moving away. Calculate the velocity of the
360 m/s . (a) What is the fundamental frequency? (b) Consid- train and the actual frequency of the whistle.
ering that the human ear can hear sounds with a maximum [A: 85 m/s and 300 Hz .]
11
FLUIDS
The Niagara Falls is located on

the border between Canada and
USA and, with its three water-
falls, has the highest flow rate in
the world.
The properties of liquids are
much richer and more complex
than those of solids. What
are the macroscopic quantities
that define a liquid without
looking into the movement of
each molecule? Which laws of
physics specify the relations be-
tween them?
We can distinguish the three states of matter known as solid, liquid and gas
by looking at their physical properties. Solids are hard, liquids flow and gases are
volatile. The reason for such behavior can only be understood by looking into their
microscopic structures and by examining the forces between atoms and molecules.
Solid atoms are regularly positioned and tightly bonded with atomic bonds; they
may only vibrate around their equilibrium positions. Liquid molecules, on the
other hand, do not have a very regular structure and the intermolecular force is
very weak. Their bonds are repeatedly broken and renewed. And, in gases, the
molecules are located too far apart to interact with each other and, they move
freely. The only force keeping gases together is the walls of the container. Liquids
and gases are jointly known as fluids.
Can we apply the dynamic and static methods that we have learned thus
far to fluids? The motion of fluids is much more complex, and it is practically
impossible to work with Newton’s law for so many molecules. Instead, we can
reach many conclusions using energy concepts.

180 11. FLUIDS
11.1 GENERAL PROPERTIES OF FLUIDS

Density
It is not practical to work with total mass in liquids and gases, the results
obtained could be mistaken to be valid only for that specific amount of liquid.
Instead, it is more useful to work with the mass of fluid per unit volume, in other
words, with density.
If the mass inside of a small volume ∆V around a certain point of the fluid is
∆m , then the density of the fluid at that point is
∆m
ρ= (11.1)
∆V
but the density may be different at another place in the fluid. The unit of density
is kg/m3 , but it is more common to use g/cm3 in daily life. The table below lists
the densities of some fluids.
Densities of some liquids and gases

(The values are for 0◦ C and 1 atm pressure, unless specified otherwise.)
material density (kg/m3 ) material density (kg/m3 )
Water 1 000 Air 1.20
Sea water 1 030 Carbon dioxide 1.98
Olive oil 920 Hydrogen 0.09
Ethyl alcohol 790 Oxygen 1.33
Mercury (Hg) 13 595 Water vapor (100◦ C) 0.80
You may find the densities of of the other elements in the periodic table
provided in Appendix D.
Pressure
A bicycle or automobile tire can carry very heavy loads, despite containing
air. Likewise, we feel a discomforting pain in our ears when we dive deep into
the sea. All fluids exert a force on the walls of their containers and on the objects
contained inside of them. This force is perpendicular to the surface; fluids cannot
exert a force parallel to the surface.
Figure 11.1: Pressure in daily

life: Automobile tire, pressure
cooker, blood pressure gauge.
If a fluid applies a force F on a surface area A , the pressure it exerts on that

surface is,
F
p= (11.2)
A
The unit of pressure is N/m2 and is called the Pascal (Pa):
1 pascal = 1 Pa = 1 N/m2
11.1. GENERAL PROPERTIES OF FLUIDS 181
Other units, such as atmosphere, the bar, the millibar, and the height of a
column of mercury are used in technology and meteorology:
1 atm = 76 cm Hg = 1.013 bar = 1.013 × 105 Pa

1 bar = 1000 millibar = 105 Pa
Hydrostatic Pressure
Pressure increases as you go deeper into a liquid. To calculate this, let us
consider a thin layer of liquid with surface area A and a small thickness dy at a
depth y , inside a liquid with density ρ (Figure 11.2). The mass of the liquid in
this layer is the product of volume and density:
m = ρ V = ρ A dy
If the pressure at the top surface of this mass of liquid is P , the pressure at its
bottom surface will be higher by dP . We may consider only the forces in the
vertical direction, as the forces on the lateral surfaces will balance each other:
Figure 11.2: Forces acting on
Ftop + mg = Fbottom the top and bottom surfaces of
P A + ρA dy g = (P + dP) A a layer with thickness dy inside
a liquid.
By simplifying we get
dP = ρg dy
Let us take the level y = 0 at the surface of the liquid. If there is an external
pressure P0 as well, acting on the surface from the outside, we take the integral
from that point down to depth y :
Z P Z y
dP = ρg dy
P0 0
P − P0 = ρg y
Therefore, the hydrostatic pressure inside of a liquid varies with depth as

follows:
P = P0 + ρg y (Hydrostatic pressure) (11.3)
y should be replaced with −y for upward cases, like calculating the air pressure
in the atmosphere. In obtaining this formula, we considered density and gravity
to be constants and took them outside of the integral. However, the change in
density and the gravitational acceleration must also be taken into consideration
if the change in height is very high.
Very high pressures are exerted on submarines and divers when they dive
very deep below the sea level. The surfaces of submarines are reinforced with
special steel cages. Likewise, special suits prevent a diver’s thorax from collapsing
in.
182 11. FLUIDS
Pascal’s Principle
In the expression (11.3), which we found for the variation of pressure with
depth, notice that the surface pressure P0 is effective at every depth. When the
surface pressure changes, its impact will be felt at every depth. This is called the
Pascal’s principle:
The external pressure applied to a liquid is equally transmitted to
each point of the liquid.
Hydraulic cranes used in lifting jobs in industry operate on Pascal’s principle
(Figure 11.3). As the force exerted on a small surface of the liquid is transmitted
as pressure to another large surface, we can write Pascal’s principle for both
surfaces:
F1 F2
P1 = P2 =⇒ = (11.4)
A1 A2
Thus, the force on the large surface will also be larger. The hydraulic brake system
of automobiles also uses this principle. With the small force we exert on the brake
Figure 11.3: Hydraulic crane. pedal, the brake pads compress a disk on the wheel with a much greater force.
Measuring Pressure
The general name for an apparatus that measures pressure is pressure gauge,
also called a manometer; however, gauges used to measure atmospheric pressure
are called barometers. There are two main types of pressure gauge: closed-tube
and open-tube.
A closed-tube gauge is shown in Figure 11.4a. A glass tube with one closed-
end is fully filled with mercury and its open end is dipped inside of a tank of
mercury. The external atmospheric pressure forces some mercury to remain
inside of the column. The height of the mercury column is equal to the external
atmospheric pressure P0 :
P0 = ρHg gh
Figure 11.4: (a) Closed-tube

manometer, (b) Open-tube
manometer.
An open-tube manometer compares the pressures in two columns of a U-tube.

As seen in Figure 11.4b, one end of a U-tube containing liquid is connected to
the container whose pressure is to be measured. The other end is open. Let us
examine the difference in height between the two columns. For example, let
the height of the liquid on the open column be higher by h . As the hydrostatic
pressure will be equal at equal heights, the pressures of points A and B should
be equal:
PA = P B
The end A has the pressure P , which is to be measured. On the other hand, the
end B is higher by h , and above it, there is atmospheric pressure P0 . Therefore,
writing this equality as
P = P0 + ρg h (11.5)
11.1. GENERAL PROPERTIES OF FLUIDS 183
we get the pressure P in terms of the liquid height in the column and density.
Gauge pressure
Let us return to the pressure formula in the U-tube:
P = P0 + ρg h
The air pressure P0 of 1 atmosphere is present in all measurements on the surface

of the Earth. We find the absolute pressure by adding the pressure that results
from the level difference in the U-tube to this value.
However, in industry, the difference with respect to the atmospheric pressure
is more useful than the absolute pressure. This is called the gauge pressure.
Accordingly, the gauge pressure is found by only calculating the ρgh pressure
value of the mercury column in the U-tube above:
Pg = ρgh (11.6)
Recommended pressure values for automobiles, blood pressure values, city water
and gas pressure values, etc. should all be considered to be gauge pressure.
The Force and Torque on a Dam Wall
As a good application of hydrostatic pressure, let us calculate the total force
acting on a dam wall and the torque that is trying to topple it.
Let us consider a dam with width L and a height of water H enclosed by it
(Figure 11.5). As the pressure exerted by the water on the wall is different at each
height, we can calculate the total force only by integration. For this purpose, let
us consider the small force dF exerted by a strip of water of length L and height
dy located between y and y + dy as measured from a point O at the bottom.
As the origin is selected at the bottom of the water, the depth of this strip from
the surface is (H − y) . Accordingly, the force acting on the strip with surface
dA = L dy due to hydrostatic pressure is
Figure 11.5: Force exerted on a
dF = P dA = [ρg(H − y)] (L dy) = ρgL (H − y) dy strip with thickness dy on the
dam’s surface.
We can find the total force by summing, in other words, integrating the contribu-
tions of these small strips from the value y = 0 to the value y = H :
y2 H
Z H
F = ρgL (H − y) dy = ρgL Hy −

0 2 0
We find the total force by substituting the integral limits:
F= 1
2 ρgLH 2 (11.7)
Note that the force is proportional to the square of the water height.
The torque of the force trying to topple the dam around the point O is likewise
calculated by integration. We again write the small torque dτ of the force dF in
the figure about the point O:
dτ = dF y = ρgL (H − y) dy y
The total torque will be the integral of these small dτ contributions:

Z Z H y2 y3 H
τ= dF y = ρgL (Hy − y ) dy = ρgL H −
2
0 2 3 0
184 11. FLUIDS
Substituting the integral limits, we find the total torque to be:
τ= 1
6 ρgLH 3 (11.8)
Note that the total torque is proportional to the cube of the height of water. In
dam constructions, the toppling torque of water is more dangerous that the force
exerted by the water.
Example 11.1
volume of the classroom:
(a) What is the total mass of air in a classroom with dimen- m = ρ V = 1.2 × (10 × 20 × 4) = 960 kg
sions 10 × 20 × ×4 m3 ? It is surprising to see that the air in a typical room is so heavy.
(b) A gold crown weighing 1.5 kg is submerged in water and (b) We take the density of gold from Appendix D as
it displaces an 80 cm3 volume of water. Is the crown made 19.3 g/cm3 . We then calculate what the volume of 1.5 kg
of pure gold? pure gold would be:
Answer V = m/ρ = 1500/19.3 = 77.7 cm3
(a) We first take the density of air from the table on page 180 The volume to be displaced is lower, therefore the crown is
as 1.2 kg/m3 . We then calculate the mass of air using the not pure gold.
Example 11.2
with m1 , the mass of ethyl alcohol will be (132 − m1 ) . We
3
The density of water is 1.0 g/cm and the density of ethyl alco- write the volumes as V = m/ρ and add:
m1 132 − m1
hol is 0.8 g/cm3 . A water-ethyl alcohol mixture with volume V = V1 + V2 = +
150 cm3 has a mass of 132 g . Calculate the amount of water ρ1 ρ2
m1 132 − m1
and ethyl alcohol in this mixture. 150 = +
1.0 0.8
Answer From this equation, we first find the mass of water and then
Total volume will be the sum of the volume of water V1 and the mass of alcohol:
the volume of alcohol V2 . If we indicate the mass of water m1 = 60 g and m2 = 132 − 60 = 72 g .
Example 11.3
umn, in other words, 38 cm Hg .
The absolute pressure is the mercury column plus the at-
The pressure of the gas inside of a balloon is to be measured us-
ing an open-ended manometer in a place where the atmosphericmospheric pressure acting on it, in other words, 38 + 72 =
pressure is 72 cm Hg . If the open-ended mercury column is 110 cm Hg .
We use the definition 1 atm = 76 cm Hg if we wish to find
38 cm higher, what is the absolute pressure and gauge pressure
of the gas? these pressures in terms of atmospheres:
Answer Gauge pressure: P = 38/76 = 0.5 atm
The gauge pressure is only the pressure of the mercury col- Absolute pressure: P = 110/76 = 1.4 atm
Example 11.4
calculate the density of the oil.
Answer
The hydrostatic pressure of the water in two columns is equal
at points 1 and 2, which have the same height:
P0 + ρ1 gh1 = P0 + ρ2 gh2
Using this, we find the relation between the densities:
h1
ρ2 = ρ1
h2
An unknown type of oil is added to the water column in a U-tube We take the density of water as 1 g/cm3 and calculate:
with two open ends. Since the height of the water is h1 = 36 cm 36
ρ2 = 1 × = 0.72 g/cm3
and the height of the oil is h2 = 50 cm as seen in the figure, 50
11.2. ARCHIMEDES’ PRINCIPLE 185
Example 11.5
to Pascal’s principle. Accordingly, the relation between the
forces is
In a hydraulic crane used to lift cars, the circular piston pushing F1 F2
the large platform has a radius of 20 cm . The other piston =
A1 A2
pushing the liquid has a radius of 4 cm . How much force is We calculate the force by taking the area of the circles πr12
required to lift a car with a mass of 1500 kg ? and the weight of the car W = mg :
Answer πr2 42
F1 = 12 mg = 2 × 15000 = 600 N
The pressures on the piston and platform are equal according πr2 20
Example 11.6
the force by integration and found Eq. 11.7. We calculate by
substituting the numerical values:
The wall of a dam has a height of 170 m and length of 1800 m .
F = 12 ρgLH 2 = 12 ×1000×10×1800×1702 = 2.6×1011 N
Find the total force acting on the dam wall and the torque trying
to topple it over. Likewise, we had found the torque with respect to a point at
the bottom of the wall (Eq. 11.8):
Answer τ = 16 ρgLH 3 = 16 × 1000 × 10 × 1800 × 1703
As the pressure varies with the height, we had calculated τ = 1.5 × 1013 N·m
11.2 BUOYANCY AND ARCHIMEDES’ PRINCIPLE

Objects immersed in a fluid feel lighter. A balloon filled with helium gas floats.
Fish go up or down in water by compressing their swim bladders. Submarines
surface or dive by pumping air or water into their diving tanks. Likewise, when
whales wash up on a shore, their lungs get crushed under their own weight and
they die. This property is called buoyancy.
Fluids always exert an upward force on the submerged objects, called the
buoyant force. Discovered by Archimedes, its value can be expressed as follows:
Archimedes’ Principle
Every object immersed in a liquid will be pushed upwards by
a force equivalent to the weight of the liquid displaced by that
object.
Attention should be paid to the expression “liquid displaced” in this principle.

The whole volume of the object will be taken into consideration if it is fully
immersed in the liquid. However, as in the case of a piece of wood, if it floats
while partially submerged, then only the volume inside of the liquid will be taken
into consideration.
The source of buoyant force is the variation of hydrostatic pressure with
depth. We can easily see this by calculating the forces exerted on the surfaces of
an object shaped like a cylinder (Figure 11.6).
Let us consider a cylinder with base area A and height h , immersed inside
a liquid with density ρ0 . There is no need to consider the forces on the lateral
surfaces of the cylinder, because they mutually balance each other. Let the top Figure 11.6: Forces acting on
base of the cylinder be at depth y and the bottom base at depth y+h . Accordingly, the top and bottom surface in-
the buoyant force F B will be the difference of the forces at the bottom and top side of a liquid.
186 11. FLUIDS
surfaces. If we write the force on each surface as the product of the pressure at
that depth and the surface area, we get
F B = Fbottom − Ftop = Pbottom A − Ptop A

= ρ0 g(y + h) A − ρ0 gy A
= ρ0 g hA = ρ0 g V
The value Ah here is the volume V of the object. Therefore, the expression is
found for Archimedes’ buoyant force F B :
F B = ρ0 g V (Archimedes’ buoyant force) (11.9)
Archimedes’ principle also applies for partially immersed bodies, but in this
case, only the immersed volume is taken into consideration. Likewise, if we write
the hydrostatic pressure difference between the top and bottom surfaces of the
object and repeat the calculations, the volume V of the object is replaced with
the volume V 0 of the immersed part:
F B = ρ0 g V 0 (for partially immersed bodies) (11.10)
The expression for the apparent weight of an object inside of a liquid can also be
found from this formula. If the object is made of a material with density ρ , its
real weight will be mg = ρVg . If we subtract the lightening effect of the buoyant
force, we get
W 0 = W − F B = mg − F B = ρVg − ρ0 Vg
W 0 = (ρ − ρ0 )Vg (apparent weight) (11.11)
Example 11.7
volume V 0 , the buoyant force F B will be equal to the weight
What percent of the volume of an iceberg floating on water is of the displaced water:
underwater? The density of sea water is ρ0 = 1.03 and the mg = F B
density of ice is ρ = 0.92 g/cm3 . ρVg = ρ0 V 0 g
Answer From here, we find the volume:
The buoyancy of the water balances the weight of the ice- V0 ρ 0.92
= = = 0.89
berg. If the submerged part of an iceberg with volume V has V ρ0 1.03
Example 11.8
mg = F B → ρ1 gV = ρ2 gV 0
A wooden block is observed to float half-immersed when placed We find the density of the wood by taking the density of the
in water. 60% of its volume is immersed when it is placed in water as ρ2 = 1 g/cm3 :
V0 1
oil. Find the densities of the wood and the oil. ρ1 = ρ2 = × 1 = 0.5 g/cm3
V 2
Answer Likewise, if the part submerged in oil has volume V 00 ,
Let us denote the densities of the wood, water and oil with the weight of the displaced oil will be equal to the weight of
ρ1 , ρ2 and ρ3 , respectively. If the submerged part of the wood the wood:
with volume V has volume V 0 , this means that a buoyant ρ1 gV = ρ3 gV 00
force F B equal to the weight of water with volume V 0 can V
ρ3 = 00 ρ1 =
1
× 0.5 = 0.83 g/cm3
balance the weight mg of the wood: V 0.6
11.3. SURFACE TENSION AND CAPILLARITY 187
Example 11.9
Answer
The downward weight of the stone is jointly balanced by the
upwards buoyant force and the tension in the rope. We write
the balance of the forces:
F B + T = mg
The buoyant force is equal to the weight ρ0 Vg of the mass
of water displaced by the volume V . From here, we calculate
3 the tension T :
A stone block with volume 0.01 m and density ρ=2.7 g/cm
3
is hanging in water as attached to a rope, as seen in the figure. T = mg − F B = ρVg − ρ0 Vg = (ρ − ρ0 ) Vg

Calculate the tension in the rope. T = (2.7 − 1) × 103 kg/m3 × 0.01 × 10 = 170 N .
11.3 SURFACE TENSION AND CAPILLARITY

Liquid surfaces that we encounter in daily life can exhibit very interesting
behavior. Raindrops that fall on leaves or car windows do not spread, but rather
form spherical bubbles. Likewise, liquid mercury does not stick to a surface, but
collects in spherical drops (Figure 11.7).
Figure 11.7: Examples of sur-

face tension: Raindrops, mercury
bubbles and an insect walking on
water without sinking.
In all of these examples, the liquid surface acts like an elastic membrane.
Liquid molecules bond together and try to form a stretched surface. For example,
a sewing needle carefully dropped on water may float. Mosquitoes and certain
insects can walk on the surface of water without sinking.
The property of liquid surfaces to resist external forces is called surface
tension. Water molecules, although they are neutral, can still attract each other
(we shall examine this in Chapter 14 as dipol interaction.) Briefly, if a water
molecule is considered as a rod with one positive end and one negative end, a
weak but attractive force called the van der Waals force occurs between the
two molecules. This force generates surface tension.
Let us consider a molecule inside of a liquid and another molecule on the
surface (Figure 11.8). As the forces inside of the liquid balance each other, this
molecule can circulate freely throughout the liquid. However, the forces are not Figure 11.8: Forces acting on
in equilibrium for a molecule on the surface; a net force is exerted that pulls this water molecules inside and on
molecule towards the liquid and prevents it from leaving the surface. the surface of a liquid.
Consider an imaginary line with length L drawn on the surface of the liquid
(Figure 11.9). If the force required to separate the molecules on both sides of this
line is F , then the coefficient
Figure 11.9: Surface tension co-
F
γ= (surface tension coefficient) (11.12) efficient.
L
is called the surface tension coefficient. The surface tension coefficients of
some liquids are as follows:
188 11. FLUIDS
Surface tension coefficients of some liquids

liquid γ (N/m) liquid γ (N/m)
Water (0◦ ) 0.076 Mercury 0.487
Water (20◦ ) 0.072 Ethyl alcohol 0.023
Water (100◦ ) 0.059 Acetone 0.024
Soapy water (20◦ ) 0.025 Glycerin 0.064
In laundry, water should be able to reach the stains between the fibers of the
clothing material. Pure water cannot seep into these small gaps, because the
surface tension of water is high. However, the surface tension of water greatly
decreases when it is heated and soap is added, and it thus becomes able to pass
through fine gaps.
As the surface tension of soapy water is lower, it can spread over a larger
surface without breaking and larger foams can be formed. The surface of soapy
water can be stretched over a wire frame. The areas of these surfaces are minimum.
Figure 11.10: A minimal sur-
Various minimal surfaces can thus be formed (Figure 11.10).
face formed by soapy water.
Capillarity
When a tube with two open ends is submerged in water, the water inside of
the tube is observed to rise. The smaller the diameter of the tube is, the higher
the liquid rises. Water can be observed to rise up to a meter high in tubes with
very small diameters. This effect is called the capillarity. However, certain other
liquids do not rise, but in contrast, sink in capillary tubes.
A liquid exhibits two different behaviors when it comes into contact with a
solid surface, depending on the forces between the molecules. For example, some
liquid surfaces are observed to curve upwards where they touch a glass tube. On
the other hand, the surface of mercury curves downward (Figure 11.11).
This behavior can be explained with the forces between molecules. The
attractive force between two of the same type of molecule in a liquid is called the
cohesion force. On the other hand, the attractive force between molecules of
Figure 11.11: In a glass tube, different types is an adhesion force.
the surface curves upward for In the case of a water-glass surface, the adhesion force is greater and the water
water, and downward for mer- is attracted to the glass; the level of water rises and these molecules attracted
cury.. to the surface are replaced with other molecules from the inside. As the contact
angle changes during the rise, the adhesion force steps in again and attracts
new liquid molecules. This process in which the liquid rises along the glass tube
continues until it is balanced by gravitational force.
The angle θ that the liquid surface makes with the solid wall is called the
contact angle. Measured from solid surfaces, this angle is acute (less than 90◦ )
for liquids curved upward, such as water, and obtuse for liquids curved downward,
such as mercury (Figure 11.12).
In a capillary tube, the height h is expressed in terms of the contact angle θ ,
the surface tension coefficient γ and the tube radius r as follows:
Figure 11.12: Contact angles of 2γ cos θ
mercury and water with a glass h= (11.13)
tube.
ρgr
Capillarity is perhaps the most vital application of surface tension. It plays a
very significant role in the survival of living beings. The sap of a tree can reach
11.4. VISCOSITY 189
very high branches above; the blood in humans and animals can reach the finest
veins in the circulation system, due to capillarity. Capillarity is used in many
applications in medicine and technology. For example, a capillary glass tube is
used to draw blood sample; paper towels absorb water through capillary action.
11.4 VISCOSITY
Every liquid has a consistency. Liquids such as water and alcohol flow easily,
while liquids such as glycerin, tar and honey are difficult to stir. As the latter
shows, moving a solid object inside of a liquid is difficult. It is relatively easy to
walk in water but becomes more difficult when you try to move faster.
The internal friction effect of liquids, one that resists its own fluidity and
the motion of bodies inside of it, is called viscosity. Liquids with high consis-
tency have higher viscosity. Gases also have internal friction, but it is much less
compared to that of liquids.
The sources of viscosity are different in liquids and gases: Internal friction Figure 11.13: The higher con-
is caused by attractive van der Waals forces in liquids and by the collision of sistency a liquid has, the higher
molecules in gases. Because of this difference, the viscosities of liquids decrease, is its viscosity.
whereas those of gases increase with temperature.
A viscosity coefficient is defined to specify the effect of internal friction: Let
us consider a layer of liquid of width L between two solid horizontal plates as
seen in Figure 11.14. Let the top plate be pulled with speed v as the bottom plate
is kept fixed. The speed of the liquid layer will vary from one plate to the other:
The liquid surface in contact with the bottom plate will remain motionless, while
the surface in contact with the top plate will have speed v due to adhesion forces.
If the speed v is not too high, the speed of the liquid will increase linearly from
one layer to the other. This layered structure is called laminar flow. If the surface
area of each plate is A , the exerted force is
v Figure 11.14: Velocity profile
F = ηA (11.14) in laminar flow.
L
and the proportionality constant is called the viscosity coefficient, indicated
with the Greek letter η (eta). Its unit is (N/m2 )·s = Pascal × s = (Pa · s). Another
unit used in industry is the poise, denoted with P, and its value is one tenth that
of the SI unit. One hundredth of a poise is called a centipoise and is indicated
with cP:
1 Pa · s = 10 poise = 1000 cP
The viscosity coefficients of some liquids are as follows:
Viscosity coefficients of some liquids

liquid η (cP) liquid η (cP)
Water (20◦ ) 1.00 Olive oil (20◦ ) 84
Water (40◦ ) 0.65 Motor oil (SAE20, 20◦ ) 125
Water (100◦ ) 0.28 Motor oil (SAE20, 150◦ ) 3
Honey (20◦ ) 10 Tar (20◦ ) 30 000
Viscosity varies rapidly with temperature. Motor oils that are viscous at room
temperature become as fluid as water when they reach the high temperature of a
running engine.
190 11. FLUIDS
Stokes’ Law
Solid bodies moving in fluids experience a resistant force caused by viscosity.
The magnitude of this drag force depends on the speed and geometric shape of
the object and the viscosity of the liquid. The result may be different for each
geometrical shape. The formula found by the British scientist G.G. Stokes for the
drag force acting on a sphere with radius r is as follows:
F = 6π η r v (Stokes’ law for drag force) (11.15)
v is the speed of the fluid (or the speed of the object going in the opposite
Figure 11.15: A sphere inside direction). The fact that drag force is proportional to velocity v was used in the
viscous liquid. damped harmonic oscillator that we examined in Chapter 9.
In gases, the relation between drag force and velocity can be much different.
Internal friction is proportional to v in low speeds, but in airplanes traveling
faster than the speed of sound (supersonic), it becomes proportional to v2 and
enormous power must be spent at such speeds.
11.5 BERNOULLI’S EQUATION

The motion of a fluid can take a wide range of appearances and is very difficult
to describe. A laminar flow in a calm river, the turbulent flow of the exhaust gas
of an airplane, vortexes in seas, etc. All of these types of flows are subject to
completely separate analyses. In this section, we shall consider an ideal fluid and
reach certain conclusions under general energy conservation laws.
First, let us give some definitions. An ideal fluid is a type of fluid that has
no viscosity and is incompressible. Of course, all fluids have internal friction and
may change volume under pressure. However, with a small margin of error, we
can take water and other liquids with low viscosity as ideal.
In a liquid medium, curves on which the velocity is considered as tangent at
every point are called streamlines. Streamlines can be made visible by adding
coloring at various points in the medium (Figure 11.16). There is a steady flow
if the streamlines do not vary with time.
Figure 11.16: Streamlines The structure of streamlines provides information about the type of motion. If
around an automobile. streamlines form smooth and soft curves, it is called a laminar flow. In contrast,
if the curves exhibit irregular behavior like whirlpools or turbulences, it is called
a turbulent flow.
Laminar flow can occur in low velocities. As velocity increases, turbulent
flow starts after a certain value. We shall only consider laminar flow here, as the
analysis of turbulent flow is complicated.
Flow Rate and Equation of Continuity
The volume of a liquid that passes through a surface with cross-section A
per unit time is called the flow rate and is indicated with Q . Its unit is m3 /s . If
the fluid passes through a cross-section A with a constant velocity v , the volume
that passes through this cross-section in a time interval ∆t is the volume of a
cylinder with length v ∆t and base area A . Accordingly, the flow rate expression
will be as follows:
∆V A v ∆t
Q = =
∆t ∆t
Q = Av (Flow rate) (11.16)
11.5. BERNOULLI’S EQUATION 191
In laminar flow, streamlines do not intersect. Streamlines that surround any

cross-section A stretch out like a tube inside of the liquid. Let us consider two
different cross-section of such a tube (Figure 11.17). Let the velocity through cross-
section A1 be v1 and that through A2 be v2 . As this fluid is ideal, in other words,
incompressible, an equal volume of liquid should pass through each cross-section
per unit time. Therefore, the flow rates should be equal in both cross-sections.
Q1 = Q2
A1 v1 = A2 v2 (Equation of continuity) (11.17)
This equation is known as the equation of continuity. We can flush water
much faster through a garden hose by squeezing its end with our fingers, because Figure 11.17: Equal amount of
the same amount of water is passing through a narrower cross-section. Therefore, material passes through each
it should flow faster to keep the flow rate constant. cross-section in unit time.
The equation of continuity states that volume is conserved, but since this is
an incompressible liquid, it actually states that mass is conserved. In other words,
the liquid undergoes no loss in mass along the streamline.
Conservation of energy
Let us consider the energy of a fluid mass in a flow tube at two different points.
The fluid mass passing through the cross-section A1 located at height y1 with
velocity v1 later passes through the cross-section A2 located at height y2 with
velocity v2 (Figure 11.18).
The work performed by the external forces acting on this tube will be equal
to the increase in total energy. Let us remember the expression (5.17) that we
found for the general conservation of energy when discussing the topic of energy
in Chapter 5:
(K1 + U1 ) + Wnc = K2 + U2
Here, K is the kinetic energy, U the gravitational potential energy and Wnc the
work performed by nonconservative forces. Let us write this law as follows for
convenience:
Wnc = (K2 − K1 ) + (U2 − U1 ) (11.18) Figure 11.18: Two cross-
sections of a flow tube.
The nonconservative external force acting on the flow tube is caused by the
pressures on both ends. The lateral forces acting on the tube do not perform any
work, as they are perpendicular to the path. The force exerted by pressure P1 on
the first cross-section is equal to P1 A1 and the force exerted by pressure P2 on
the other end is P2 A2 . Let the fluid mass travel a distance of L1 under the effect
of force in the first cross-section and let the mass in the second cross-section
travel a distance of L2 . Accordingly, the net work performed is
Wnc = F1 L1 − F2 L2 = P1 A1 L1 − P2 A2 L2
The work performed by the second force is negative, because it is opposite the
displacement. Now, let us write the gravitational potential energy of the mass in
both cross-sections. As the displaced liquid mass has the same value m in both
places, we get
U2 − U1 = mgy2 − mgy1
The difference between the kinetic energies can likewise be written directly:
K2 − K1 = 12 mv22 − 21 mv21
192 11. FLUIDS
Adding these terms in the (11.18) equation, we get
P1 A1 L1 − P2 A2 L2 = 12 mv22 − 12 mv21 + mgy2 − mgy1
Let us write the mass m in terms of density and volume. The two cross-sections
of the tube and the distances traveled may be different, but the volume will remain
the same:
V = A1 L1 = A2 L2
If we also express mass in terms of density, we get
m = ρV = ρ A1 L1 = ρ A2 L2
Therefore, if the products A1 L1 and A2 L2 are written as m/ρ , the expression

above simplifies as follows:
P1 − P2 = 12 ρv22 − 12 ρv21 + ρgy2 − ρgy1
Rearranging the terms, we find the Bernoulli’s equation:
P1 + 21 ρv21 + ρgy1 = P2 + 12 ρv22 + ρgy2 (Bernoulli’s equation) (11.19)
The Bernoulli’s equation is the version of conservation of energy for liquids. This
equation is valid for ideal, in other words, non-viscous and incompressible fluids,
but it is also approximately correct for other fluids.
Figure 11.19: Applications of

the Bernoulli’s equation: Carbu-
retor in gasoline engines, per-
fume sprays, Pitot tubes in air-
planes.
Let us emphasize the conclusions that can be drawn from the Bernoulli’s
equation:
• Hydrostatic pressure. If we take v1 = v2 = 0 in the Bernoulli’s equation
for a mass of liquid at rest, we get
P1 + ρgy1 = P2 + ρgy2
If we take one end at y1 = 0 and the other at y2 = y , we get
P = P0 + ρgy
This is the hydrostatic pressure formula that we discussed earlier.

• Venturi tube. Let us consider two cross-sections at the same height in a
liquid tube (Figure 11.20). If we set y1 = y2 in Bernoulli’s equation,
P1 + 12 ρv21 = P2 + 12 ρv22
According to the equation of continuity, the pressure will be lower in the

Figure 11.20: Venturi tube. narrow cross-section of the tube because the velocity is higher there. This
11.5. BERNOULLI’S EQUATION 193
device, called the Venturi tube, is used in many technologies: in the carbure-
tors of gasoline engines for mixing gasoline and air, in pesticide sprayers, in
perfume sprays, etc. As seen in Figure 11.19, when air is blown from a tube
passing through the open end of the liquid, the pressure will decrease there,
and the liquid in the tank will rise and mix with the passing air.
The Pitot tube, which is a similar device, is used to measure the velocity
of airplanes with respect to air (Figure 11.21). Air passes with velocity v at
one end of a U-tube filled with liquid and the velocity is ensured to be zero
at the other end. The velocity of the air can be calculated using Bernoulli’s
equation by measuring the height h of the liquid rising due to the pressure Figure 11.21: Pitot tube.
difference.
• Torricelli’s formula. Let us consider that a hole is opened at a distance h
below the surface on a closed tank containing liquid. Let us compare the
top liquid surface in the container and the point of the hole in Bernoulli’s
equation. The velocity of the liquid can be taken as zero at the top, because
the liquid surface is very large. Let the liquid surface inside of the container
be at pressure P and the pressure at the hole, which is just the atmospheric
pressure P0 , because it is open to the air. Accordingly,
P + 0 + ρgh = P0 + 12 ρv2
s
2(P − P0 ) Figure 11.22: Velocities of wa-
v= 2gh + (11.20) ter at various depths according
ρ
to Torricelli’s formula.
We have P = P0 if the top of the container is also open to atmospheric
pressure, and we get
v = 2gh (11.21)
p
Called Torricelli’s formula, this relation shows that, just like a stone
dropped at a height of h in free fall, a liquid will also gain the velocity
v= 2gh . This is not surprising, because Bernoulli’s equation is an expres-
p
sion of conservation of energy. Figure 11.22 shows the velocities of water

coming out of holes opened at various depths.
• Aeronautics technology has the most striking application of Bernoulli’s
equation. This equation can be used to understand how a lifting force is
produced on the wings of a heavy steel bird. Examining the cross-section of
the wing of an aircraft in Figure 11.23, the lower part is more straight and
the top part is more curved. When an aircraft enters a beam of streamlines,
the air passing through the top will travel a longer distance and meet with Figure 11.23: Speed and pres-
the others at the back. This means that the air at the top has a higher speed. sure on an aircraft wing.
Therefore, according to Bernoulli’s equation, the pressure at the top of the
wing will be lower than that at the bottom. This pressure difference causes a
net upward lift force. This force should not be confused with Archimedes’
buoyancy.
• The Bernoulli effect is likewise observed in curved balls in football and
baseball. When the ball spins as it is kicked forward, its effect on the air
speed is different on sides A and B due to the rotation of the ball (Figure 11.24).
Consider an observer moving with the ball: On side A, the rotation against Figure 11.24: Forces on a ball.
194 11. FLUIDS
the air decreases the air speed locally, while the effect is opposite on side B.
Hence the air speed is greater at B than at A. This air speed difference causes
a pressure difference on both sides. As a result of the Bernoulli effect, a net
force arises that pushes the ball in the direction of lower pressure B shown
in the figure.
Example 11.10 Q 24
v1 = = = 6 m/s
A1 4
The main pipe of the city water network shown in the figure
We use the flow rate or equation of continuity (11.17) to find
has cross-section area A1 =4 m2 and another pipe connected to
the speed at the other cross-section:
it has a cross-section of A2 =2 m2 . The water flowing through
A1 4
cross-section A1 has a flow rate of Q=24 m3 /s . A1 v1 = A2 v2 → v2 = v1 = × 6 = 12 m/s
A2 2
(a) Calculate the speed of water in both cross-sections.
(b) If the pressure at the first cross-section is P1 =2 atm , what (b) We write Bernoulli’s equation (11.19):
is the pressure at the second cross-section? P1 + 12 ρv21 + ρgy1 = P2 + 12 ρv22 + ρgy2
We take y1 = y2 , because the cross-sections are at the same
height and solve for P2 :

P2 = P1 − 12 ρ v22 − v21
Answer We calculate the pressure by first converting the atmosphere
(a) The definition of flow rate is given in Eq. (11.16): unit into Pascal:
Q = A1 v1 P2 = 2 × 1.013 × 105 − 12 × 1000 × (122 − 62 )
From here, we find the speed v1 : P2 = 1.5 × 105 Pa = 1.5 atm
Example 11.11 A1 5
A1 v1 = A2 v2 → v2 = v1 = × 2 = 10 m/s
A2 1
We find the pressure using Bernoulli’s equation:
P1 + 12 ρv21 + ρgy1 = P2 + 12 ρv22 + ρgy2
We take y1 = 0 at the ground level and solve for P2 :
P2 = P1 − 12 ρ(v22 − v21 ) − ρgy2
We convert the pressure values into Pascal units, take the
density of water as ρ = 1000 kg/m3 and calculate:
P2 = 4×1.013×105 − 21 ×1000×(102 −22 )−1000×10×4
Water is pumped with a pressure of P1 =4 atm into the water
pipe at the entrance of a building. The water speed at the pipe P2 = 320 000 Pa = 3.1 atm .
with cross-section 5 cm2 is 2 m/s at the entrance. (b) The water rising up to the 10th floor means that it reaches
(a) What is the speed and pressure of the water flowing from that floor with at least zero speed. According to the equa-
a tap with cross-section 1 cm2 at a height of 4 m on the tion of continuity, the velocity at the ground should also be
second floor? zero. Therefore, taking v1 =v2 =0 and P2 =1 atm , Bernoulli’s
(b) What is the minimum pressure at the entrance such that equation simplifies as follows:
water can rise up to the 10th floor at a height of 40 m ? P1 + 0 + 0 = P2 + 0 + ρgy2
Answer Since the tap is in open air, P2 =1 atm . From here, we calcu-
(a) We find the water speed at the second floor using the late P1 :
equation of continuity: P1 = 1×1.013×105 +1000×10×40 = 501 000 Pa = 5 atm
Example 11.12
The mercury column of a Pitot tube on the wing of a flying
aircraft has risen by 6 cm . What is the speed of the aircraft
with respect to the air?
Answer
Let us compare the end points of both columns. The speed of
air will be zero on one of end and v on the other. Since both
ends are at the same height y1 =y2 =y , Bernoulli’s equation is
ρHg
r
written as follows: v = 2gh
P1 + 12 ρ × 02 + ρgy = P2 + 21 ρv2 + ρgy ρ
P1 − P2 = 2 ρv
1 2 We take the density of air as 1.2 and the density of mercury
Now, we look at the U-tube. The pressure difference will be as ρ Hg = 13600 kg/m 3
and calculate the speed:
equal to the pressure of the mercury column:
r
13600
ρHg gh = 12 ρv2 v = 20 × 0.06 × = 117 m/s .
1.2
1. In which of the containers in the figure below is the (a) Lead
pressure at depth h lower? (b) Copper
(a) A (b) B (c) C (d) Equal (c) Equal
8. Lead is denser than copper. When two equal volumes of

lead and copper are submerged in water, which one will
experience a larger buoyant force?
(a) Lead
(b) Copper
2. In which of the containers in the figure above is the (c) Equal
pressure at the base lower? (d) It is impossible to tell.
(a) A (b) B (c) C (d) Equal
9. Archimedes’ buoyancy force is equal to which of the
3. In which of the containers in the figure above is the force following?
at the base greater? (a) The weight of the submerged part of the object.
(a) A (b) B (c) C (d) Equal (b) The weight of the floating part of the object.
(c) The weight of the water that it displaces.
(d) None of the above.
4. Which of the following are correct for hydrostatic pres-
sure?
10. An iron sphere attached to a wooden block floats when
I. It increases with depth.
placed on water. In which case will the buoyancy of the
II. It decreases with depth.
water be greater?
III. It increases with the density of the liquid.
IV. It decreases with the density of the liquid. (a) When the iron is on the top and out of the water.
(b) When the iron is at the bottom and under the water.
(a) I (b) I & II (c) I & III (d) I & IV
(c) Buoyancy is the same.
5. Which of the following is incorrect for the buoyant force
of liquids? 11. A piece of ice is floating in a glass of water. What will
(a) It increases with depth. happen to the level of water when the ice fully melts?
(b) It increases with the density of the liquid. (a) It will increase.
(c) It increases with the volume of the object. (b) It will decrease.
(d) It increases with the gravitational acceleration. (c) It will remain the same.
6. Will a boat sink more in lake water or salty sea water?
(a) Lake 12. A man inside a boat in a pool steps out of the boat into
(b) Sea the pool. What will happen to the water level of the
(c) Equal pool?
(d) It is impossible to tell. (a) It will increase.
7. Lead is denser than copper. When two equal masses of (c) It will remain the same.
lead and copper are submerged in water, which one will (d) It is impossible to tell.
experience a larger buoyant force?
196 11. FLUIDS
13. A metal block floats half-submerged in a container full (a) The radius of the sphere.
of mercury. How will the level at which it floats in the (b) The speed of the sphere.
mercury be affected when it is taken to the surface of (c) The viscosity of the liquid.
the Moon? (d) All of the above.
(a) The level will be higher.
(b) The level will be lower. 17. Two spheres with equal radius, one made of lead and
(c) The level will stay the same. one made of wood, are dropped from the same height.
(d) It is impossible to tell. Assume that the drag force of the air on both is the same.
Which one reaches the ground first?
(a) The lead one.
(b) The wooden one.
(c) Both will reach the ground at the same time.
14. Two immiscible liquids (water and mercury) are at equal 18. In which cross-section of the pipe in the following figure
heights on two sides of a U-tube when the central valve is the speed greater?
is closed. On which side will the level rise when the (a) A (b) B (c) C (d) Equal
valve is opened?
(a) On the water side.
(b) On the mercury side.
(c) The two sides will remain equal.
15. A bird lands on a man sitting in a boat in a pool. What 19. In which cross-section of the pipe in the figure above is
will happen to the level of the water? the pressure greater?
(a) It will rise. (a) A (b) B (c) C (d) Equal
(b) It will fall.
(c) It will remain the same. 20. In which cross-section of the pipe in the figure above is
(d) It is impossible to tell. the flow rate greater?
(a) A (b) B (c) C (d) Equal
16. The viscosity force acting on a sphere moving at speed
v inside of a liquid depends on which of the following?
Problems
11.1 General Properties of Fluids
The density of sea water is ρ = 1030 kg/m3 .
11.1 In a place where atmospheric pressure is 1.2 atm , an [A: (a) 1.25 × 107 Pa , 123 atm , (b) 1250 N .]
open-ended gauge is used to measure the pressure of a gas
in a balloon. If the mercury column of the gauge indicates
40 cm , what are the absolute pressure and gauge pressure of
the gas? [A: 1.73 and 0.53 atm .]
11.2 Air density is 1.3 kg/m3 at sea level. What would the
thickness of the Earth’s atmosphere be if the density of air
did not decrease with height, remaining constant? (Hint: The Problem 11.4
pressure on the surface is still 1 atm.) [A: 7.8 km .] 11.4 An unknown liquid is added to the mercury column in
11.3 (a) What is the pressure at a depth of 1200 m below a U-tube with two open ends. Since the height of the mer-
the ocean? Calculate in terms of Pascal and atmosphere. (b) cury is h1 = 6 cm and the height of the unknown liquid is
What amount of force should the lens of a research camera h2 = 68 cm , as seen in the figure, calculate the density of the
with an area of 1 cm2 be able to withstand at such a depth? liquid. [A: 1.2 g/cm3 .]
PROBLEMS 197
Problem 11.12
11.12 The scale in the figure shows 2.4 kg when a container
of water is placed on it. A piece of iron with a mass of 780 g
Problem 11.5 and a density of 7.8 g/cm3 is tied to a rope and suspended
inside of the water. What will the scale show? [A: 2.5 kg .]
11.5 The pressure in veins is called blood pressure. This is the
gauge pressure that is present in addition to the atmospheric
pressure. When a patient is fed with serum, the serum pres-
sure at the entry point into the body should be greater than
the blood pressure. The serum inside of the bottle shown in
Problem 11.13
the figure has a density of 1040 kg/m3 . The liquid is able to
11.13 An empty spherical shell made of plastic has an in-
enter the blood only when the serum bottle is hung at a height
ternal radius of 5 cm and an external radius of 6 cm . This
of 80 cm . Calculate the blood pressure (gauge pressure) of
object floats half-submerged when dropped into the water.
the patient in terms of Pascal and cm Hg units.
Calculate the density of the plastic. [A: 1.2 g/cm3 .]
[A: 8320 Pa, 6.2 cm Hg .]
11.14 A block suspended on a spring scale shows 1.4 , 1.9
11.6 In a hydraulic crane used to lift vehicles, the cylin- and 2.5 kg , respectively, when suspended in water, oil and
der pushing the platform has a radius of 30 cm . The piston an unknown liquid. As the density of water is 1000 kg/m3
pushing the liquid has a radius of 3 cm . How much force is and the density of oil is 900 kg/m3 , what is the density of
required to lift a 5-ton truck? [A: 500 N .] the unknown liquid? [A: 780 kg/m3 .]
11.7 The wall of a dam has a height of 210 m and a length 11.5 Bernoulli’s Equation
of 1100 m . Find the total force acting on the dam wall and
the torque that is trying to topple it over.
[A: 2.4 × 1011 N and 1.7 × 1013 N·m .]
Problem 11.15
11.2 Archimedes’ Buoyancy
11.15 The main pipe of a city’s water network, shown in the
11.8 What percentage of the volume of a wooden block with figure, has a cross-sectional area A1 =3 m2 and another pipe
density 0.7 g/cm3 will sink when placed in an oil with a connected to it has A2 =1 m2 . The water flowing through
density of 0.92 g/cm3 ? [A: 76 % .] cross-section A1 has a flow rate of Q=15 m3 /s . (a) Calculate
the flow speed at each cross-section. (b) If the pressure is
11.9 An aluminum block with a density of 2.7 g/cm3 floats P1 =8 atm at the first cross-section, what will the pressure be
as partially submerged when placed in mercury with a density at the second? [A: 5 and 15 m/s , (b) 7 atm .]
of 13.6 g/cm3 . Water is poured on the mercury to make the
aluminum block become fully submerged. What percentage
of the block is underwater? [A: 87 % .]
Problem 11.16
11.16 The radii of the three separate cross-sections of the
pipe in the figure are r1 =3 , r2 =2 and r3 =5 cm . Considering
Problem 11.10
that the flow velocity of the water at the first cross-section
11.10 A wooden block with a mass of 2 kg and a density of is 2 m/s , calculate (a) the velocities in the other two cross-
400 kg/m3 is submerged by being tied to a rope at the bottom sections and (b) the flow rate of the water.
of a container of water, as seen in the figure. Calculate the [A: (a) v2 = 4.5 , v3 = 0.72 m/s , (b) 0.006 m3 /s .]
tension in the rope. [A: 30 N .]
11.17 The flow rate of water in a fire hose is constant at
11.11 What is the minimum volume of an ice layer that can 1 m3 /s . Water is to be sprayed on a fire located at a height
carry a car with a mass of 1000 kg on a lake? The density of of 40 m using a metal nozzle attached to the end of the hose.
ice is 0.92 g/cm3 . (Hint: Consider that the ice floats as fully What should the maximum radius of the nozzle be such that
submerged at the limit condition.) [A: 12.5 m3 .] the water can reach the fire? [A: 0.11 m .]
198 11. FLUIDS
of 3 m/s into a pipe with a cross-section area of 5 cm2 at the

entrance to a building. What are the speed and pressure of
the water flowing from a tap with a cross-section of 1.5 cm2
at a height of 12 m on the 3rd floor?
[A: 10 m/s and 3.41 × 105 Pa = 3.37 atm. ]
Problem 11.18
11.18 The Pitot tube in the wing of an aircraft contains alco-
hol. What is the speed of the aircraft with respect to the air
if the alcohol column rises by 32 cm ? The density of alcohol
is 800 kg/m3 and the density of air is 1.2 kg/m3 .
[A: 65 m/s .]
Problem 11.20
11.20 Water is flowing at a speed of 3 m/s from the cross-
section of a pipe with an area A1 =4 cm2 located at a height
of h=12 m . If the pressure at the top cross-section is 1.2 atm ,
Problem 11.19 what is the pressure at the bottom cross-section with an area
11.19 Water is pumped with pressure P1 =5 atm and a speed A2 =5 cm2 ? [A: 2.4 × 105 Pa = 2.4 atm .]
12
TEMPERATURE AND HEAT
Temperature in shade can rise

up to 50 ◦ C at the Sharqiya
Sands in Oman. Only camels
can survive such high temper-
atures.
What is the meaning of temper-
ature in terms of atoms? How
does temperature change lead to
changes in physical properties?
Up until now we have consistently examined the motion of a small number

of particles. However, this gets difficult in systems with an enormous number of
particles. Consider a glass of water. There are approximately 1024 molecules in
this glass. In principle, we can write Newton’s laws for each molecule and find
the solutions. However, solving that great number equations will take years, even
with the fastest computers.
Suppose that we solved that many equations and found the positions and
velocities of each molecule. What purpose would that much information serve?
On the other hand, 3-5 quantities are sufficient to specify the state of a glass of
water: Temperature, volume, pressure, etc. These are quantities that specify the
system at hand as a whole, in other words, macroscopic quantities.
The branch of science that macroscopically examines the interactions of a
system with its surroundings is called Thermodynamics. We will learn the
basic concepts of thermodynamics in this chapter and apply them in the next
chapter.

200 12. TEMPERATURE AND HEAT
12.1 THERMODYNAMIC EQUILIBRIUM AND TEMPERATURE

Quantities that specify a system as a whole are called state variables. These
variables include temperature T , pressure P , volume V , mass m , and internal
energy E , which we will discuss later. No matter how complex the internal
structure of a macroscopic system is, it is possible to describe the system as a
whole using a limited number of state variables.
Consider a cup of tea. If we leave the tea as it is, it will start to cool down and
reach room temperature after a while. If we mix it using a spoon, the rotational
motion of the liquid will stop after a while. Likewise, if we put some sugar in
tea and mix it, the concentration of sugar will distribute evenly in the tea after a
while.
So, if the state variables of a system do not vary over time and have the same
value throughout the system, this state is called thermodynamic equilibrium.
Temperature
When we touch an object, we feel it to be “hot” or “cold.” Also, we know that,
if we put two samples of water of different temperature into a container and wait,
their temperatures will become equal.
As the most basic thermodynamic quantity, temperature is the quantitative
measure of thermodynamic equilibrium. Let us take two objects A and B and
put them in contact with each other, insulating them from the external environ-
ment (Figure 12.1). These two objects will have equal temperature once they reach
thermodynamic equilibrium.
Figure 12.1: Two objects insu- This definition specifies temperature macroscopically. Microscopic definition,
lated from the environment will however, is the subject of the branch of physics called Statistical Physics. Briefly
have equal temperature when stated, temperature is a measure of the average kinetic energy of molecules that
they reach equilibrium. constitute a system.
Temperature Measurement – Thermometer
As the temperature of an object changes, another one of its quantities will
inevitably change as well. For example, a heated rod of iron will expand, the
pressure of a heated gas will increase and the electrical resistance of a heated
wire will increase. These changes help us to measure temperature and define a
unit for it.
The instrument used to measure temperature is called a thermometer. The
most commonly known type is the mercury thermometer. The liquid mercury
inside of a chamber located at the bottom expands in proportion with the temper-
ature and rises in a column. Temperature can be calculated from the height of
the mercury column. (Today, dyed alcohol is used instead, because mercury is a
toxic substance.)
Figure 12.2: Types of ther-
mometer: mercury thermometer,
constant volume gas thermome-
ter, thermocouple.
Another type of thermometer is the constant-volume gas thermometer.

In this instrument, a gas inside of a glass chamber is trapped with mercury filling
a U-tube (Figure 12.2). When the gas inside of the chamber is placed inside of
a liquid and heated, its volume and pressure increases. However, the mercury
12.1. THERMODYNAMIC EQUILIBRIUM AND TEMPERATURE 201
column is adjusted so that the volume remains constant and only the pressure
increases. The temperature can again be measured from the height of the mercury
column.
Yet another type of thermometer called a thermocouple, is used in technol-
ogy and laboratories. Thermocouple has two ends made of two different metals
(copper, iron or tungsten) or alloys (constantan, alumel or nisil). A voltage differ-
ence occurs between these two ends when one is kept at a standard temperature
( 0◦ ) and the other one is brought to a different temperature. The temperature is
determined by measuring this voltage, which is proportional to temperature.
Temperature Scales – Celsius, Fahrenheit, Kelvin
To define a temperature scale, we must first decide on two reference temper-
atures that can easily be realized by anyone. A temperature scale is formed by
dividing the interval between these two temperatures into equal parts. There
are three commonly used temperature scales in everyday life, technology and
science: the Celsius ( ◦ C ), the Fahrenheit ( ◦ F ), and the Kelvin (K) scales.
In the Celsius scale, the freezing point of water under 1 atmosphere pressure
is taken as zero degrees Celsius ( 0 ◦ C ) and the boiling point of water as 100 ◦ C .
The distance between these two points is divided into 100 equal parts.
In the Fahrenheit scale, the freezing point of water under 1 atmosphere
pressure is taken as 32 degrees Fahrenheit ( 32 ◦ F ) and the boiling point as 212 ◦ F .
The distance between these two points is divided into 180 equal parts.
The conversion formula between the Celsius and Fahrenheit scales is:
TC = 5
T F − 32◦ F (12.1)

9
The Celsius and Fahrenheit scales both require two constant points.
The absolute temperature scale, or the Kelvin scale, which requires only
one fixed point, was adopted in modern science. The unit of temperature in the
SI unit system is Kelvin and is indicated as K (the ◦ sign is not used). The fixed
point of this scale is the point called the triple point of water, at which all three
states (water-ice-steam) of water coexist. The property of the triple point is that
it occurs at a single value of temperature and pressure.
In the Kelvin scale, the temperature of the triple point is fixed as follows:
T K = 273.16 K (Triple point temperature) (12.2)
The relation between the Kelvin and Celsius scales is as follows:
T K = TC + 273.15 (12.3)
The difference in the last digit between these two formulas may seem odd, but
it is, in fact, correct, because the triple point of water is actually at temperature
0.01 ◦ C . We will take it as zero in practice in this course:
0 ◦ C ≈ 273 K 100 ◦ C ≈ 373 K (in this course) (12.4)
The zero point of the Kelvin scale, in other words, the temperature −273.16 ◦ C ,
is the absolute zero of the universe. At this temperature, all motion stops in the Figure 12.3: Celsius and Kelvin
universe and lower temperatures are meaningless. scales.
12.2 HEAT
A pot of water starts to get warmer when put over the fire on a gas stove.
Here, the combustion energy of gas is transferred into the water. This energy can
be given to the water because the burning gas is hotter than water. Therefore,
the main reason for this energy transfer is temperature difference.
The type of energy that a system gives or receives due to a temperature
difference with its environment is called heat. Heat is indicated with Q .
The unit of heat is the Joule (J), because it is a measure of energy. The calorie
(cal) unit used in industry has the following value in terms of joules:
1 cal = 4.186 J
The expression “calorie” used for food in daily life is actually the kilo-calorie
(kcal) unit.
We must emphasize a point here: Heat is always something that is given or
taken. A system will not have heat as one of its properties, in other words, it is
wrong to say “the heat of the water”; it is proper to say “taken heat” or “given
heat.” In terms of thermodynamic language, heat is not a state variable.
Specific Heat
Objects use the heat that they receive in various ways: The temperature of
the object may increase, it may start to boil or expand, etc. Let us consider these
respectively.
Every object has a different rate of temperature change upon heating. A
coefficient called the specific heat is defined to indicate this feature. If Q is the
amount of heat required to increase the temperature of the mass m of an object
by ∆T , then the coefficient
Q
c=
m ∆T
is called the specific heat of the object.
∆T = T 2 − T 1 is the temperature increase here. In other words, specific heat
is the amount of heat required to increase the temperature of 1 unit of mass per 1
unit of temperature. The unit of specific heat is J/kg·K .
Based on this definition, the amount of heat required for a ∆T increase in
temperature is expressed as follows:
Q = mc ∆T (12.5)
This formula is used in the technique called calorimetry. To find the final temper-
ature when different liquids are mixed, the algebraic sum of received and given
heats is set to zero.
The specific heats of various materials are given in the following table. Specific
heat varies slightly with temperature, but can be taken as constant for small ∆T
variations.
12.2. HEAT 203
Specific heats
c c c c
(J/kg·K) (cal/g·◦ C) (J/kg·K) (cal/g·◦ C)
Copper 390 0.092 Water 4186 1.000

Aluminum 900 0.215 Ice 2100 0.5
Gold 129 0.030 Water vapor 2010 0.48
Silver 234 0.056 Ethyl alcohol 2400 0.58
Iron 448 0.107 Glass 837 0.200
Note that the highest specific heat in this table belongs to water. So much
energy is required to heat or cool water that it prevents sudden temperature varia-
tions in neighboring environments. In other words, water makes the surrounding
climate temperate.
Here, we defined specific heat per mass; molar specific heat, which is more
convenient for gases, will be defined later. In defining specific heat for gases,
separate specific heats are defined depending on the type of heating. The constant-
volume specific heat cV is defined if heating takes place under constant volume,
and the constant-pressure specific heat cP is defined if it takes place under
constant pressure.
Latent Heat
If we continue to heat water at 100◦ C it will start to boil. The system
will continue to receive heat during this change of state, but surprisingly, its
temperature will not change. The received heat is used to convert water at 100◦ C
into vapor at 100◦ C . This is called a change of phase.
Another phase change is observed during the melting of ice. If we give heat
to ice at 0◦ C , it will start to melt and turn into water at 0◦ C . We say that there
is a latent heat if no temperature change occurs during a phase change.
The heat received or given by a unit mass of a system during a change of Figure 12.4: Temperature al-
phase is called the latent heat. Latent heat is indicated with L . According to ways remains constant when
this definition, the heat received or given by an object with mass m is water boils.
Q = mL ( L : Latent heat) (12.6)
L will have a positive sign if the object is vaporizing or melting, and a negative
one if it is condensing or freezing. The heat received by the system is always
considered to be positive. Although many types of latent heats are defined in
thermodynamics, here we will only consider the latent heat of fusion (melting)
and the latent heat of vaporization. Some latent heats of fusion and vaporization
are provided in the following table:
Latent heats of fusion and vaporization

Fusion Heat of fusion Boiling Heat of vaporization
point point
◦
( C) (cal/g) (kJ/kg) (◦ C) (cal/g) (kJ/kg)
Water 0 80 333 100 540 2260

Ethyl alcohol −114 25 104 78 204 850
Lead 327 6 25 1750 208 870
Silver 961 21 88 2193 558 2300
Tungsten 3410 44 184 5900 1150 4800
Example 12.1
multiply by 5, obtaining the following:
(Note: The specific heats required in the problems in this chapter TC = 59 (T F − 32◦ F)
will be taken from the tables above.) For T F = 100 ◦ F , TC = 38 ◦ C
(a) Convert the temperatures 100 ◦ F and −30 ◦ F into ◦ C . For T F = −30 ◦ F , TC = −34 ◦ C
(b) Convert the temperatures 100 ◦ C and −30 ◦ F into Kelvin (b) Celsius temperatures are converted into Kelvin by adding
units. 273:
Answer For TC = 100 ◦ C , T K = 100 + 273 = 373 K
(a) We subtract 32 from the Fahrenheit value, divide by 9 and For TC = −30 ◦ C , T K = 243 K .
Example 12.2
It is more convenient to make such calorimetry calculations
in calorie and gram units. At the end of the calculation, we
How much heat is required to increase the temperature of iron
can take 1 cal = 4.18 J and convert it into the joule unit.
with a mass of 250 g from 20 ◦ C to 80 ◦ C ?
The specific heat of iron is given as c=448 J/kg=0.11 cal/g in
Answer the table. We calculate by substituting the numerical values:
We use the formula (12.5): Q = 250 × 0.11 × (80 − 20) = 1650 cal
Q = mc ∆T Q = 1650 × 4.18 = 6900 J .
Example 12.3
by the water and the alcohol should be zero:
◦
200 g -water at a temperature of 20 C is mixed with 100 g - Q1 + Q2 = 0
alcohol at a temperature of 30 ◦ C . What will the final temper- m1 c1 (T − T 1 ) + m2 c2 (T − T 2 ) = 0
ature of the mixture be? Notice that one of the received heats ( Q2 ) will be negative.
Answer We calculate the final temperature by substituting the num-
We take the specific heat of water from the table as c1 =1 cal/g bers:
and that of alcohol as c2 =0.58 cal/g . If we show the final 200 × 1 × (T − 20) + 100 × 0.58 × (T − 30) = 0
temperature with T , the algebraic sum of the heats received T = 22 ◦ C .
Example 12.4
Let us write the stages of the whole process:
How much heat is required to turn a block of ice with a mass of Ice at −30 ◦ C → ice at 0 ◦ C → water at 0 ◦ C →
50 g at a temperature of −30 ◦ C into water at 20 ◦ C ? → water at 20 ◦ C .
We write the heats required at each stage and add them up:
Answer
In this problem, latent heat is also required during the melting Q = mc1 [0 − (−30)] + mL + mc2 (T − 0)
of the ice. From the table, we take the latent heat of fusion We calculate the result by substituting the numerical values:
as L = 80 cal/g and the specific heat as c1 = 0.5 for ice and Q = 50 × [0.5 × 30 + 80 + 1 × 20]
c2 = 1 cal/g for water. Q = 5750 cal .
12.3. THERMAL EXPANSION 205
Example 12.5
the specific heat of ice as c1 =0.5 , of water as c=1 and of
Some amount of water vapor at a temperature of 110 C is ◦ vapor as c2 =0.48 cal/g .
mixed with a block of ice with a mass of 30 g at a temperature The heat received by the ice with mass m1 is set to be equal
of −20 ◦ C in an insulated container. How much vapor was to the heat given by the vapor with mass m2 :
used, considering that the final temperature of the mixture is Q1 = Q2
20 ◦ C ? (The mass of the container can be neglected.) m1 c1 [0 − (−20)] + m1 L1 + m1 c(20 − 0) =
Answer = m2 c2 (110 − 100) + m2 L2 + m2 c(100 − 20)
As they all turn into water in the end, the ice and the water va- We find the amount of vapor by substituting the numbers:
por have undergone phase changes to reach their final states. 30 × [0.5 × 20 + 80 + 1 × 20] =
From the table, we take the values for latent heat of fusion as m2 [0.48 × 10 + 540 + 1 × 80
L1 =80 cal/g , the latent heat of vaporization as L2 =540 cal/g , m2 = 5.3 g .
Example 12.6
with mass m1 as equal to the heat given by some m2 amount
◦
100 g of ice at a temperature of −10 C is put inside 500 g of water turning into ice. (If, at the end of the calculation,
water at a temperature of 0 ◦ C in an insulated container. What this value of m2 turns out to be higher than the total amount
will the final state be? of water, it means that the final temperature will be less than
zero. In that case, we will use a different route.) There follows:
Answer m1 c1 [0 − (−10)] = m2 L
The fact that the amount of ice increases in this problem may We calculate by taking the specific heat of ice as c1 =0.5 cal/g
seem odd at first glance, but it is true. Ice (at −10 ◦ C ) will and the latent heat of fusion as 80 cal/g :
receive heat from water because it is colder. Water at 0 ◦ C 100 × 0.5 × 10 = m2 × 80 → m2 = 6.3g .
can only give heat by turning some of it into ice. According to this result, there will be 100 + 6.3 = 106.3 g
Let us assume that the final temperature will be 0 ◦ C , with of ice and 500 − 6.3 = 493.7 g of water at a temperature of
water and ice coexisting. We set the heat received by the ice 0 ◦ C in the container.
12.3 THERMAL EXPANSION

We know from our daily lives that heated objects increase their volume, or
expand. We have seen that the mercury heated inside of a mercury thermometer
expands when heated, cables in electrical transmission lines sag on hot days,
heated balloons expand, etc. Expansion may sometimes produce unintended
results (Figure 12.5). A margin is left in the construction of building, bridges and
roads in order to account for this.
If an object with length L0 at a reference temperature T 0 , has the length L
at another temperature T , its linear expansion ∆L = L − L0 will be proportional
to the increase in temperature ∆T = T − T 0 and to its initial length L0 :
∆L = αL0 ∆T
Here, α is the coefficient of linear expansion and depends on the type of

material. Its unit is (1/◦ C) or (1/K) . If we write this formula in terms of the final Figure 12.5: Railroad tracks
length L of the object, we get that have expanded due to heat
on the Delaware river (1935).
L = L0 + ∆L
L = L0 [1 + α(T − T 0 )] (12.7)
The coefficient α is taken as a constant in calculations, although it varies slightly

with temperature.
A similar behavior is observed in volume expansion. The increase in volume
∆V caused by temperature increase ∆T is
∆V = βV0 ∆T (12.8)
and the constant β is called the coefficient of volume expansion. Rigid bodies
have the same expansion property in three dimensions, therefore the relation
between the α and β coefficients is as follows:
β ≈ 3α (12.9)
The expansion coefficients of some materials are given in the following table:
Coefficients of linear and volume expansion of materials at 20 ◦ C

Solids α (◦ C)−1 Liquids and gases β (◦ C)−1
Copper 17 × 10−6 Water 2.1 × 10−4

Iron or steel 12 × 10−6 Ethyl alcohol 11 × 10−4
Aluminum 23 × 10−6 Mercury 1.8 × 10−4
Concrete 12 × 10−6 Gasoline 9.5 × 10−4
Glass 9 × 10−6 Air 34 × 10−4
The most striking feature in this table is the fact that the expansion coefficients
of iron, steel and concrete are equal. The fact that buildings made of steel and
concrete can withstand very high temperature changes without cracking is only
possible with this property of concrete.
Examining the table, the expansion coefficients of iron and concrete may
seem to be small (on the order of one ten thousandths), but this may result in
considerable expansion. In a climate with a [−20, +40 ◦ C] temperature range,
such a difference in length means 1 cm per 10 meters. This is taken into account
in the construction of buildings, bridges and railroads with margins being left.
Figure 12.6: Expansion margin Also note that, in the table, the expansion coefficient of alcohol is much
in a bridge. greater than those of water and mercury; this is why alcohol is preferred in
thermometers. Gasoline is also observed to expand as much as alcohol. For this
reason, an expansion margin should always be left in the tank of vehicles when
filling with gasoline.
Anomalous behavior of water: The expansion of water exhibits an unusual
behavior in the temperature range [0, 4 ◦ C] . The volume of water at 0 ◦ C starts
to decrease when heated. This anomalous behavior continues until 4 ◦ C . Then,
it starts to expand with temperature. In other words, water is densest at 4 ◦ C .
This property allows fish and other living beings to stay alive under ice in frozen
Figure 12.7: This is how life lakes. This unusual behavior of water is caused by the special structure of H2 O
goes on in lakes during winter. molecules.
Example 12.7
∆L = αL0 ∆T
A steel bridge with a length of 1200 m at 0 C will expand by The coefficient of expansion of steel is given as α=12 × 10−6
◦
how much in length at 50 ◦ C ? in the table. We substitute the values:

Answer ∆L = 12 × 10−6 × 1200 × (50 − 0)
We use the formula (12.3), giving the increase in length: ∆L = 0.72 m = 72 cm
Example 12.8
filled with water and heated on an stove. How much water will
spill at a temperature of 95 ◦ C ?
A copper vessel with a volume of 5 liters at 20 ◦ C is completely
12.4. HEAT TRANSFER 207
Answer The coefficient of volume expansion is β ≈ 3α in terms of

When two objects expand together, it is more convenient linear expansion. From here, the expression for the difference
to work with (α2 − α1 ) , the difference of their expansion in volumes becomes:
coefficients. ∆V2 − ∆V1 = 3(α2 − α1 )V0 ∆T
Accordingly, if we use ∆V1 to indicate the increase in the We find α coefficients for copper and water from the table
boiler’s volume and ∆V2 for the increase in the water’s vol- and substitute to calculate the result:
ume, according to the formula (12.8), we get ∆V2 − ∆V1 = 3 × (210 − 17) × 10−6 × 5 × 10−3 × (95 − 20)
∆V2 − ∆V1 = β2 V0 ∆T − β1 V0 ∆T ∆V2 − ∆V1 = 0.00022 m3 = 220 cm3
Example 12.9
Let us take the ratio of periods t0 and t at temperatures
A simple pendulum used to measure time is made of a mass 0 ◦ C and 40 ◦ C , respectively, and use the thermal expansion
attached to the end of a steel wire. The period of the pendulum formula: s
L0 (1 + α ∆T ) √
r
correctly shows 1 s when the temperature is 0 ◦ C . Calculate t L
= = = 1 + α ∆T
how much time the clock will lose or gain in 1 day in a hot t0 L0 L0
country where the average temperature is 40 ◦ C ?
We take α = 12 × 10−6 for steel and t0 = 1 s and calculate
Answer the new period for the temperature difference ∆T =40 ◦ C :
(a) We had calculated the period of a simple pendulum in p
t − t0 = 1 + 12 × 10−6 × 40 − 1 = 24 × 10−6 s
Chapter 9 (Eq. 9.23). However, let us use t here to show the
This difference is calculated for one day:
period so that it will not get confused with the temperature
T: 24 × 3600 × (t − t0 ) = 20 s
t = 2π L/g The pendulum lags by 20 s per day.
p
Example 12.10
of the aluminum rod is α2 L0 ∆T , then the difference in ex-
pansion will be (α2 −α1 )L0 ∆T . We can reach the same result
A steel ruler measures correctly at a temperature of 0 ◦ C . This
by considering that the steel ruler is not expanding, but the
ruler is used to measure the length of an aluminum rod at a
aluminum rod is expanding with a coefficient of (α2 − α1 ) .
temperature of 0 ◦ C and it reads as 2.800 m . (a) How much
However, this time, the marks of the steel ruler will remain
will the ruler measure the length of the aluminum rod when
at the same place and correctly show the relative expansion
the temperature of both the steel ruler and the aluminum rod
of the aluminum.
is increased to 400 ◦ C ? (b) How much has the length of the
Therefore, we calculate the relative expansion of aluminum
aluminum rod really expanded?
using the coefficient (α2 − α1 ) :
Answer L = L0 [1 + (α2 − α1 ) ∆T ]
(a) This problem can be solved in various ways: We find the We look up the expansion coefficients from the table and
amount of expansion of the steel and aluminum and then calculate:
find their proportion. However, the easier method is to work L = 2.800 × [1 + (23 − 12) × 10−6 × 400] = 2.812 m
with the coefficient difference (α2 − α1 ) for jointly expanding The expanded ruler shows the length of the expanded rod as
objects: 2.812 m .
Of the two objects with the same length L0 =2.8 m , if the (b) We calculate the absolute expansion of the aluminum:
coefficient of expansion of the steel ruler is α1 L0 ∆T and that L = L0 [1 + α2 ∆T ] = 2.8[1 + 23 × 10−6 × 400] = 2.826 m .
12.4 HEAT TRANSFER

The conduction of heat from one medium to another can take place through
three different mechanisms: conduction, convection and radiation. In conduction,
heat is conducted without any displacement of matter. In convection, heat is
transferred by the displacement of material. And in radiation, heat is conducted
through electromagnetic waves.
Now let us examine in turn these mechanisms:
Conduction
If we hold a copper rod at one end and place the other end on the heat of a
stove, we will soon feel heat in our hand. This is because, heat is transmitted
through conduction from one end to the other end of the rod. We have to examine
this at the atomic level to understand how it occurs.
There are two separate mechanisms for heat transmission at the atomic level:
• Lattice vibrations (Figure 12.8a): Copper atoms may only oscillate about
their fixed positions. However, their vibration amplitude, and thus their
energy, increases with the heat that they receive. They transmit some of this
energy to their neighboring atoms. A vibration wave thus reaches one end
of the copper from the other end.
• Free electrons (Figure 12.8b): As we will discuss when addressing electricity

(Section 14.1), some of the electrons in the atoms of conductors, such as copper
and steel, become free in a solid medium. They therefore carry energy by
Figure 12.8: freely traveling from one end of the copper to the other. (This is why metals
that are good electric conductors are also good heat conductors.)
The amount of heat transferred by a medium depends on many factors: the

thickness of the medium, temperature difference, etc. To examine this, let us
consider a medium with thickness ∆x and cross-section area A (Figure 12.9). Let
one of the surfaces of this medium be kept at temperature T 1 and the other at
temperature T 2 . Accordingly, the amount of heat Q transferred from one region
into the other in time t is
Q ∆T
= kA (12.10)
t ∆x
The coefficient k here depends on the type of material and is called the thermal
Figure 12.9: Heat transfer conductivity. Its unit is watt/m·◦ C . The ratio of the temperature difference
through conduction. ∆T = T 2 − T 1 to the thickness ∆x in this formula is called the temperature
gradient ∆T/∆x .
Thermal conductivities of some materials are given in the following table:
Thermal conductivities of some materials

k (W/m·◦ C) k (W/m·◦ C)
Lead 35 Air 0.024

Iron 80 Wood 0.1
Aluminum 238 Water 0.6
Copper 397 Glass 0.8
Silver 427 Concrete 0.8
The larger the coefficient k is, the better the heat is transferred. We want heat
conductivity to be low or high when selecting a material for any job in technology
or daily life. A material with good heat conductivity is required for heating food
on a stove. For this reason, we use copper and aluminum pots. On the other hand,
poor conducting materials should be used to insulate houses against the cold and
prevent heat losses. Therefore, glass (or, even better, double glazing with a layer
of air between two glass panes) is used.
12.4. HEAT TRANSFER 209
As seen in the table, metals such as copper, silver and aluminum, that are
good electrical conductors are also good heat conductors. This is because the
free electrons in the structure of metals also conduct energy while conducting
electrical charge.
The table shows that air is a poor conductor. This is true for still air. Hence, the
air trapped in double glazing or the material called styrofoam severely decreases
the heat conductivity of that material. The super insulators recently developed in
the aeronautics industry are also spreading into our daily life (Figure 12.10).
Convection
We cool off with the cool air blown by a fan or an air conditioner on a hot day. Figure 12.10: The supercon-
Here, heat is transferred by the motion of air molecules. Heat transfer through ductor aerogel developed by
NASA.
the displacement of matter is called convection.
Convection can occur in two ways. The air heated around a stove or heater
will expand to have less density, and thus, start to rise. It is replaced by the cooler
air above, which also soon gets heated and rises. A convection current is thus
formed automatically. This is called natural convection.
The second type of convection is forced convection. The cool air generated by a
fan or air conditioner is pushed in the desired direction by a fan motor. Although
still air is a poor conductor, it turns into a good conductor through convection.
The most important example of convection is the currents formed in the
atmosphere and the oceans. The natural convections of air and water are two of
the most important factors that make all kinds of life possible.
The detailed examination of convection is complex, and the transferred heat Figure 12.11: Air convection in
depends on many factors, such as type of material, velocity distribution, pressure a heated room.
and temperature conditions in the environment, etc. It is examined in detail in
the branch of science called Fluid Mechanics.
Radiation
The third heat transfer mechanism is the type of heat transferred through
electromagnetic waves. Its most important feature is its ability to transfer heat
in vacuum, without the presence of material in the medium. The solar rays that
pass through the empty space between the Earth and the Sun, which has no
conducting medium, to heat the Earth and are the source of all life.
All bodies radiate at every temperature. However, we can only see that part
that is in the visible light spectrum. Pieces of coal in a grill radiate by emitting red Figure 12.12: Radiation of hot
light. When the coals cool down, the radiation continues as infrared waves, but coal.
we cannot see them. Today, binoculars and cameras that operate with infrared
rays can be used to view the radiation of all objects.
The relation between radiated energy and heat is given by the Stefan-
Boltzmann law:
Q
= σ A T4 (Stefan-Boltzmann law) (12.11)
t
Here, A is the surface area of the object and T is the temperature in Kelvin units.
As for the other two coefficients, σ = 5.7 × 10−8 W/m·K4 , the Stefan-Boltzmann
constant is the same for all objects. The other dimensionless coefficient is called
emissivity. This is a parameter that can take values within the range [0, 1] and
depends on the type of radiating material.
The transfer of radiated energy between bodies with different temperatures
is obtained using this formula. If a body with temperature T 1 is kept in an
environment with temperature T 2 , the radiated power is

Q
= σ A T 14 − T 24 (12.12)
t
It is remarkable that the radiated energy is proportional to the fourth power of
temperature. Heat losses through radiation increase to a tremendous degree at
Figure 12.13: A cat in the night
high temperatures.
through infrared binoculars.
Heat transfer is a very extensive topic. It will be sufficient in this course to
know only the basic concepts.
12.5 IDEAL GAS

Gases have a special place in thermodynamics, because they can easily ex-
change heat and energy with their environment over a wide range of temperature
and pressure. For this reason, they play a role in many fields of industry, through
the use of gas fuels, gas turbines, steam boilers, etc. In this section, we will discuss
the basic concepts that describe the properties of gases.
Avogadro’s Number and Mole
A constant called Avogadro’s Number is used to specify the amount of
material in terms of atomic scale. The value of this number is:
NA = 6.022 × 1023 (Avogadro’s number) (12.13)
The amount of a material containing Avogadro’s number of atoms (or molecules)

is called 1 mole (symbol: mol). In other words, 1 mole of carbon contains NA
number of carbon atoms, and 3 mole of water contains 3NA number of H2 O
molecules.
The number of moles in a given amount of gas is indicated with n . Accord-
ingly, if the given number atoms (or molecules) of a substance is N , the amount
of moles n that it contains is:
N
n=
NA
The total mass of 1 mole of atoms (or molecules) of a substance is called the
molar mass and is indicated with M . If the given mass of a substance is m , the
amount of moles n that it contains is found as follows:
m
n=
M
For example, to find the number of moles in 100 grams of oxygen (O2 ) , we divide
it by the molar mass of oxygen M = 32 g :
m 100 g
n= = = 3.125 mol
M 32 g
The molar masses of some materials are given in the following table:
12.5. IDEAL GAS 211
Molar masses of some materials

Gases M (gram/mole) Solids and liquids M (gram/mole)
Hydrogen (H2 ) 2 Water (H2 O) 18

Helium (He) 4 Aluminum (Al) 27
Oxygen (O2 ) 32 Iron (Fe) 56
Nitrogen (N2 ) 28 Copper (Cu) 64
Argon (Ar) 40 Silver (Ag) 108
Carbon dioxide (CO2 ) 44 Mercury (Hg) 201
Air 29 Lead (Pb) 207
Ideal Gas Law

It is impossible to exactly determine the microscopic behavior of a gas, be-
cause it contains approximately Avogadro’s number of molecules. However,
working with the ideal gas assumption gives us very good results that are
close to experimental reality. As a definition, the ideal gas is one in which the
intermolecular interactions are neglected. For this, the gas must have low pressure
and high temperature.
The equation that determines the relation between the mass, temperature,
volume and pressure of a gas in any state is called the equation of state. The
equation of state for ideal gas, also called the ideal gas law, is as follows:
PV = nRT (Ideal gas law) (12.14)
Here, P is the pressure, V is the volume, n is the number of moles and T is the
temperature in Kelvin units. The proportionality constant indicated with R is
called the universal gas constant and its value is
R = 8.31 J/mol·K (universal gas constant) (12.15)
The constant R is also expressed as follows for use in practical calculations in
which the pressure is given in atmosphere units and the volume in liter units:
R = 0.082 liter·atm/mol·K (12.16)
If the mass m of the gas is given, taking its mole number as n = m/M , the ideal
gas law can also be expressed as follows:
m
PV = RT (12.17)
M
If we wish to write the density ρ of the gas, we form the ratio ρ = mass/volume =
m/V in the last equation:
m MP
ρ= = (12.18)
V RT
The ideal gas law allows us to see which quantity will vary when the state of a gas
changes. For example, if the volume V of the gas is kept constant, the pressure
will increase with the temperature. Likewise, when the pressure is kept constant,
Figure 12.14: Isothermal
the volume of the gas will increase with the temperature. curves.
When the temperature is kept constant, we get what is called isothermal curves
(Figure 12.14). In isothermal changes, volume and pressure change inversely: One
of them decreases when the other increases.
Example 12.11
in the mass m :
m 100
(The table on Page 211 will be used for the molar masses required n= = = 1.6 moles .
M 64
in these problems.) (b) A mole of mass contains NA Avogadro’s number of
(a) The molar mass of copper is 64 g . How many moles does molecules. From the given mass, we calculate the number of
100 g of copper contain? moles n and then the number of molecules N . The molar
(b) How many molecules does 100 g of oxygen contain? mass of oxygen is 32 g , and we find the number of molecules
as follows:
Answer m 100
N = n NA = NA = × 6.02 × 1023 = 19 × 1023
(a) The number of moles is the number of M molar masses M 32
Example 12.12 nRT

PV = n RT → P =
V
A container with a volume of 8 liters has 10 g of hydrogen gas Taking the constant R in units of liters × atmosphere will be
( H2 ) at a temperature of 27 ◦ C . very convenient in these calculations:
(a) How many moles of hydrogen are there in the container? R = 0.082 liter·atm/mol·K
(b) What is the pressure of the gas? We substitute the number of moles and the other data we
Answer found in the previous item into the ideal gas law and calcu-
(a) We divide the given mass with the molar mass 2 g of H2 : late the pressure. Do not forget to convert temperature into
m 10 kelvins.
n= = = 5 moles . nRT 5 × 0.082 × (273 + 27)
M 2 P= = = 15 atm
(b) We solve the ideal gas law (12.14) for the pressure P : V 8
Example 12.13
(b) We use Eq. (12.17), which gives the ideal gas law in terms
(a) How much volume will 1 mole of ideal gas hold at standard of masses: m
temperature and pressure (STP)? PV = RT
M
(b) What is the density of 1 mole of oxygen ( O2 ) under STP In this equation, we form density as ρ = m/V :
conditions? m MP
ρ= =
Answer V RT
(a) The reference point called STP (standard T and P ) for
We take the STP temperature as 273 K and the pressure as
gases refers to 0 ◦ C temperature and 1 atm pressure. In such
1 atm and use the molar mass of oxygen, which is 32 g . Using
a case, we calculate the volume of 1 mole of gas using the
the constant R in units of liters × atmospheres here, the other
ideal gas law:
nRT 1 × 0.082 × 273 units simplify, and we get the density in terms of grams/liter:
V= = = 22.4 liters 32 × 1
P 1 ρ= = 1.43 g/liter
All ideal gases have the same volume of 22.4 liters at STP. 0.082 × 273
Example 12.14
terms of pascals, and we take the universal gas constant as
Air is present under STP conditions in a room with volume R=8.31 J/mol·K :
6 × 10 × 4 m3 . PV 1.013 × 105 × 240
n= = = 10 700 mol
(a) How many moles of air is present in the room? RT 8.31 × 273
(b) How many moles of air is left in the room if it is heated (b) Using ratios is very convenient in problems that involve
up to 27 ◦ C ? state changes of a gas. We write the ideal gas law for each tem-
perature and compare them. As the pressures and volumes
Answer are equal, we get
(a) We solve the ideal gas law for the number of moles n : P2X
X VX Rn2 T 2 T1
2
PV = → n2 =
S
n1
PV = nRT → n = P1X
X VX
1 Rn1 T 1 T2
RT S
Let us calculate in terms of the SI units this time. The volume We substitute the values and calculate as follows:
of the room is 6 × 10 × 4=240 m3 and, in STP conditions, 273
n2 = × 10700 = 9740 mol
1 atmosphere of pressure equals 1 atm=1.013 × 105 Pa in 300
Example 12.15
(b) If we indicate the initial variables with (P1 , V1 , T 1 ) and
The cylinder of a diesel motor with a volume of 4 lt contains air the final variables with (P2 , V2 , T 2 ) , we can write the ideal
at a temperature of 27 ◦ C and a pressure of 1 atm . The piston gas law for each state and divide on both sides:
is pushed to decrease the volume of the air by a factor of 16 and P2 V2 H nRT
H 2 T2
= =
increase the pressure by a factor of 40 . P1 V1 nRT
HH 1 T1
(a) How many moles of air does the cylinder contain? We solve this expression for the final temperature T 2 :
P2 V2
(b) What will be the final temperature of the air? T2 = T1
P1 V1
Answer We calculate the final temperature by substituting the numer-
(a) We calculate the number of moles from the ideal gas law: ical values:
40 1
PV 1×4 T2 = × × 300 = 750 K
n= = = 0.16 moles . 1 16
RT 0.082 × 300
Example 12.16 PV2 nRT 2 V2 T 2
= → =
PV1 nRT 1 V1 T 1
A cylinder contains an ideal gas at a temperature of 27 ◦ C and From here, we calculate the final temperature:
a pressure of 5 atm . The gas is subject to the following processes V2
after the same initial state each time: T2 = T 1 = 3 × T 1 = 3 × 300 = 900 K .
V1
(a) Its volume is increased by a factor of 3 by heating under (b) The ratio of the change under constant volume is found
constant pressure. What will its final temperature be? the same way:
(b) This time its pressure is increased by a factor of 4 by heat- P2 T 2
=
ing under constant volume. What will its final temperature P1 T 1
be? P2
T2 = T 1 = 4 × 300 = 1200 K .
(c) It is heated such that both its pressure and volume increase P1
by factors of 3. What will its final temperature be? (c) The ratio is as follows when both the pressure and the
(d) Its absolute temperature is doubled by heating under con- volume change:
P2 V2 T 2
stant volume. What will its final pressure be? =
P1 V1 T 1
Answer P2 V2
T2 = T 1 = 3 × 3 × 300 = 2700 K .
(a) It is sufficient to establish the ratios in this problem. If P1 V1
we write the ideal gas law twice for each process, only the (d) The ratio for heating under constant volume was found
changing variables will remain in the formulas. in item (b):
P2 T 2 T2
The ratio is as follows for heating under constant pressure: = −→ P2 = P1 = 2 × 5 = 10 atm .
P1 T 1 T1
1. How many ◦ C is a temperature of 373 K ?
4. The specific heat of iron is approximately twice the spe-
(a) 0 ◦ C (b) 73 ◦ C (c) 100 ◦ C (d) 173 ◦ C
cific heat of silver. How will their temperatures increase
if the same amount of heat is given to pieces of iron and
2. Which of the following is not a unit of temperature?
silver with the same mass?
(a) Fahrenheit (a) The iron will heat up more.
(b) Kelvin (b) The silver will heat up more.
(c) Centigrade (c) They will heat up equally.
(d) Celsius (d) It is impossible to tell.
3. How will a pendulum clock made of metal wire show 5. Two quantities of waters with equal mass at tempera-
the time when temperature increases? tures of 20 ◦ C and 50 ◦ C are mixed. What will the final
(a) It will lag. temperature be?
(b) It will run too fast. (a) 25 ◦ C (b) 30 ◦ C (c) 35 ◦ C (d) 40 ◦ C
(c) It will not change.
(d) It is impossible to tell. 6. The absolute temperature of a gas in a container is dou-
bled under constant volume. What will its pressure be?
(a) 2P (b) 4P (c) P/2 (d) P/4 15. How many atoms are there in 16 g of oxygen ( O2 ) in
terms of Avogadro’s number NA ?
7. The absolute temperature of a gas in a container is dou-
(a) NA (b) 2NA (c) 3NA (d) 8NA
bled under constant pressure. What will its volume be?
(a) 2V (b) 4V (c) V/2 (d) V/4
16. Which of the following is not the ideal gas law?
8. The volume of a gas in a container is doubled under PV nRT P nRT
constant temperature. What will its pressure be? (a) =R (b) P= (c) V= (d) V=
nT V nRT P
(a) 2P (b) 4P (c) P/2 (d) P/4
17. Which of the following is correct when ice at 0 ◦ C is
9. The pressure of a gas in a container is doubled and its heated?
volume is halved. What will its temperature be? I. Its temperature will increase.
(a) 2T (b) 4T (c) T (d) T/2 II. It will melt under constant temperature.
III. Its volume will decrease.
10. A gas at a pressure of 1 atm in a container of volume IV. Its volume will increase.
V has its temperature increased from 0 ◦ C to 273 ◦ C at
(a) I & II (b) I & IV (c) II & III (d) II & IV
constant pressure. What will its volume be?
(a) V (b) 2V (c) V/2 (d) 4V
18. Which of the following is correct if a system is in ther-
11. Which of the following is incorrect? modynamic equilibrium with its surroundings?
(a) The volume of a heated ice will decrease. (a) The temperatures are equal.
(b) Water heated at 10 ◦ C will expand. (b) The pressures are equal.
(c) Heated water vapor will expand. (c) The volumes are equal.
(d) Water heated at 2 ◦ C will expand. (d) All of the above.
12. If thermal expansion coefficients are sorted in increasing

order, which sorting will be correct? 19. Under which conditions do real gases approach the ideal
gas model?
(a) Solid→liquid→gas.
(b) Liquid→solid→gas. (a) Low pressure and low temperature.
(c) Solid→gas→liquid. (b) Low temperature and high pressure.
(d) Gas→liquid→solid. (c) High temperature and low pressure.
(d) High temperature and high pressure.
13. How many grams of substance are there in 2 moles of
hydrogen ( H2 )? 20. Which of the following are types of latent heat?
(a) 1 g (b) 2 g (c) 4 g (d) 10 g I. Specific heat.
II. Heat of fusion.
14. How many moles of hydrogen gas ( H2 ) are there in 10 g III. Heat of vaporization.
of hydrogen?
(a) I (b) I & II (c) II & III (d) I & III
(a) 1 (b) 2 (c) 5 (d) 10
Problems
12.2 Heat 12.3 An aluminum block of 200 g at a temperature of 300 ◦ C

(Note: In the problems of this section the containers are assumed is put into 300 g of water at a temperature of 20 C in a ◦con-
◦
to be insulated and have zero specific heats.) tainer. What will the final temperature be? [A: 55 C .]
12.4 Some amount of ice at a temperature of −40 ◦ C is put

12.1 How much heat is required to turn 100 g of water at
together with 10 g of steam vapor at a temperature of 120 ◦ C
30 ◦ C into steam at 120 ◦ C ? [A: 63 kcal .]
in an insulated container. How much ice was used if the final
temperature of the mixture is 30 ◦ C ? [A: 48 g .]
12.2 600 g of water at a temperature of 90 ◦ C is added to
400 g of water at a temperature of 20 ◦ C in a container. What 12.5 A 600 g piece of iron at a temperature of 500 ◦ C is put
will the final temperature be? (The container is thermally into 100 g of water at a temperature of 30 ◦ C in a container.
insulated and its specific heat is negligible.) [A: 62 ◦ C .] How much of the water will evaporate? [A: 35 g .]
PROBLEMS 215
12.6 A 200 g piece of iron at a temperature of 400 ◦ C is put grams of 5 moles of carbon dioxide ( CO2 )?
together with 50 g of ice at a temperature of −10 ◦ C in an [A: (a) 5.6 moles , (b) 220 g .]
insulated container. What will the final state be?
[A: Water and iron at 60 ◦ C .] 12.14 A container with a volume of 20 liters has 160 g of
oxygen ( O2 ) gas at 25 atm of pressure. (a) How many moles
12.7 400 g of vapor at a temperature of 150 ◦ C is mixed of oxygen are there in the container? (b) What is the temper-
with water with a mass of 100 g at a temperature of 100 ◦ C ature of the gas? [A: (a) 5 moles , (b) 1220 K .]
in an insulated container. What will the final state be?
[A: 82 g of water and 418 g of vapor at 100 ◦ C .] 12.15 An oven with a volume of 1 m contains air at STP
3
conditions ( 0 C and 1 atm ). (a) How many moles of air are

◦
12.3 Thermal Expansion present in the oven? (b) How many moles of air are left in
(Note: Expansion coefficients that may be needed in the prob- the oven if it is heated up to a 200 ◦ C temperature?
lems of this section shall be taken from the table on Page 206.) [A: (a) 45 moles , (b) 26 moles .
12.8 A steel tower with a height of 800 m at 0 ◦ C will expand 12.16 An oxygen tube used in industry contains oxygen gas
by how much at a temperature of 40 ◦ C ? [A: 38 cm .] compressed at 100 atm and at 27 ◦ C temperature in a vol-
ume of 200 liters . (a) What is the mass of the oxygen inside
12.9 A glass tube with a base area of 1 cm2 and a height of of the tube? (b) It is observed that the pressure of the tube
1 m is fully filled with mercury at 0 ◦ C . How much mercury decreases to 80 atm in the next week due to a leak in its valve.
will spill when the temperature is 150 ◦ C ? [A: 2.3 cm3 .] How much oxygen remains in the tube? (Assume that the
12.10 A physical pendulum made of an aluminum rod of temperature remains constant.) [A: (a) 14.6 kg , (b) 11 kg .]
length 1 m is used to measure time. The period of the pen- 12.17 The Loschmidt constant used in chemistry is defined as
dulum is 1 s at a temperature of 0 ◦ C . How much error will the number of gas molecules under STP conditions in 1 cm3
accrue in the pendulum’s measurement in one day when the volume. Calculate the Loschmidt constant. [A: 2.7 × 1019 .]
temperature becomes 50 ◦ C ?p(The period of a physical pen-
dulum of length L is T = 2π 2L/3g .) 12.18 The cylinder of a gasoline engine with a volume of
[A: It will lag by 50 s .] 3 lt contains air-gasoline mixture at a temperature of 27 ◦ C
and a pressure of 1 atm . The piston is pushed to decrease
the volume of the mixture by a factor of 8 and increase the
pressure by a factor of 12 . What will the final temperature
of the mixture be? [A: 447 ◦ C .]
Problem 12.11
12.19 A cylinder contains 10 liters of an ideal gas at a temper-
12.11 A steel bridge consists of two parts, each with a length ature of 27 ◦ C . The gas is subject to the following processes
of 150 m and hinged to each other at the center. The bridge each time after the same initial state:
can stretch flat in the horizontal position at 0 ◦ C . By how (a) Its pressure is increased by a factor of 5 by heating under
much will the central point rise when the temperature is constant volume. What will its final temperature be?
50 ◦ C ? [A: 5.2 m .] (b) Its volume is increased by a factor of 8 by heating under
constant pressure. What will its final temperature be?
12.12 A steel measuring tape measures correctly at 0 ◦ C .
(c) It is heated such that both its pressure and volume double.
The length of a copper pipe is measured by the measuring
What will its final temperature be?
tape as 25 m at 0 ◦ C . (a) What will the copper pipe’s mea-
(d) Its absolute temperature is increased by a factor of 4 by
surement be when both the measuring tape and the pipe are
heating under constant pressure. What will the final volume
heated up to 250 ◦ C ? (b) What is the real length of the pipe
be? [A: (a) 1500 K , (b) 2400 K , (c) 1200 K , (d) 40 liters .]
at that temperature? [A: (a) 25.03 m , (b) 25.11 m .]
12.20 On a day when air pressure is 1 atm and the temper-
12.4 Ideal Gas
ature is 27◦ , an air bubble with a volume of 1 cm3 breaks
12.13 The molar mass of the hydrogen atom (H) is 1 g , the off from the bottom of a lake with 25 m depth. What will its
molar mass of the oxygen atom (O) is 16 g and the molar volume be when it reaches the surface? (Hint: Calculate rates
mass of the carbon atom (C) is 12 g . (a) How many moles with the P = P0 + ρgh formula that we found in Chapter 11
are there in 100 g of water ( H2 O )? (b) What is the weight in for the hydrostatic pressure in liquids.) [A: 3.5 cm3 .]
13
THE LAWS OF
THERMODYNAMICS
The heat engine that started the

industrial revolution: The steam
train.
Steam expands and performs
work by pushing a piston. From
where does the steam draw the
energy to perform this work?
Is there an internal source that
can be used either as work or as
heat? (Photo: Christoph Ehlen).
In the previous chapter, we learned the basic concepts of thermodynamics.

Now, we are ready to discuss what a system can do and what changes it can
undergo with the heat that it receives. Thermodynamics is the science that
deals with systems that exchange heat and work with their environment. It is
a phenomenological (experimental) science; it makes no assumptions about the
structure of matter. It derives macroscopic results for the properties of matter
based on two general laws. It is a remarkable and complete classical theory.
The first law of thermodynamics makes a bilan of energy for both heat and
work exchanged between the system and its environment. But it does not indicate
the direction in which the system will develop. The second law of thermodynamics
establishes which processes, among those allowed by the first law, can really
occur.

218 13. THE LAWS OF THERMODYNAMICS
13.1 WORK DONE BY A GAS

We have previously discussed the fact that a system could exchange energy
with its surroundings in the form of heat. Another way to exchange energy is to
perform work.
Let us consider a gas inside of a cylinder. The top surface of this cylinder is a
movable piston, as seen in Figure 13.1. When we heat this gas, it will expand by
pushing the piston. It will thus perform work. Let us calculate the amount of this
work.
Let the gas inside of the cylinder have pressure P and volume V . The total
force exerted by the gas at this pressure on the piston with surface area A will
be F = P A . If the piston is displaced by dx as a result of this force, the small
Figure 13.1: The pressure of amount of work performed by the gas along this displacement will be
the gas will push the piston with
a force of F = PA .
dW = F dx = PA dx = P dV
Here, dV=A dx is the increase in the volume of the gas. If the gas expands ( dV>0 ),
the work it performs is positive; indeed, it expands by pushing the piston upwards.
However, if the piston is pushed downwards by an external force, the volume
decreases ( dV<0 ), hence the work performed by the gas is negative. (Of course,
the work done by the external force is positive, but we are interested in the work
performed by the gas.)
If, during a finite expansion, the volume of the gas goes from V1 to V2 , the
total work performed by the gas will be the limit sum of these small works, in
other words, its integral:
Z V2
W= P dV (Work performed by a gas) (13.1)
V1
The pressure P inside of the integral cannot be taken outside of the integral as it
is a function of both the volume V and the temperature T .
It will be instructive to examine the work performed by the gas on a P - V
diagram. As shown in Figure 13.2, recalling the definition of an integral, the area
Figure 13.2: In the P - V dia- of the region under the curve will be the work W .
gram, the area under the curve The path taken by the gas when going from V1 to V2 during expansion is
is the work W performed by the important. The work will vary depending on the path. The work performed by
gas. the gas can be calculated as follows for the processes indicated in Figure 13.3:
Figure 13.3: The work per-

formed by the gas (a) At con-
stant pressure, (b) At constant
temperature (isothermal), (c) At
constant volume.
(a) Work under constant pressure (Figure 13.3a) : It is sufficient to calculate

the area of the rectangle as the pressure P is constant along the path shown
in the figure. Then, P can be taken outside of the integral in Eq. (13.1):
Z V2
W=P dV = P (V2 − V1 ) (Work under constant pressure) (13.2)
V1
13.1. WORK DONE BY A GAS 219
(b) Work at constant temperature (isothermal) (Figure 13.3b) : If we write

the ideal gas equation as P = nRT/V , the isothermal curves in the P - V
diagram will decrease as P ∼ 1/V . Let us use the integral to calculate the
work performed on such a curve. If we write the pressure P inside of the
integral as P = nRT/V , we get
Z V2 Z V2
dV
W= P dV = nRT
V1 V1 V
The integral of this expression is the logarithm function:

V2
W = nRT ln (Work at constant temperature) (13.3)
V1
(c) Work at constant volume (Figure 13.3c) : In this case the work is zero:
W=0 . At constant volume ( dV=0 ), the area under a vertical line is zero.
Example 13.1
W = P dV = P (V2 − V1 )
A container with 3 liters of volume contains a gas at 27 C ◦ The final volume is V2 = 4V1 = 4 × 3 = 12 L . Using the SI
temperature and 2 atm pressure. Starting each time from the units (pascal and cubic meters), we get
same initial state, calculate the work performed by this gas W = 2 × 1.013 × 105 × (12 − 3) × 10−3 = 1800 J = 1.8 kJ
during the following processes: (c) We had found the isothermal work in Eq. (13.3):
V2
(a) Its temperature is increased by a factor of 5 at constant W = nRT ln
volume. V1
Taking the final volume as V2 =2V1 , we get ln 2V1 /V1 = ln 2 :
(b) Its volume is increased by a factor of 4 at constant pressure.
W = nRT ln 2
(c) Its volume is increased by a factor of 2 at constant temper-
The number of moles n here could be calculated separately,
ature.
but this is not necessary, because we use the left-hand side
Answer rather than the right-hand side of the ideal gas equation
(a) The definition of work is dW=P dV . If the volume is PV = nRT :
constant, then dV=0 and the work performed will be zero: W = P1 V1 ln 2
W=0 . We substitute ln 2 = 0.69 and the other values:
(b) Since the pressure is constant, integration is not needed: W = 2 × 1.013 × 105 × 3 × 10−3 × 0.69 = 420 J
Example 13.2
Along the path ab , work will be zero, as the volume is con-
stant : Wab =0 . The other work is found using the formula
W=P(V2 − V1 ) as it is performed under constant pressure:
Wabd = 0 + Pb (Vd − Vb ) = 2 × 105 × (8 − 5) = 600 kJ
Path acd : We again calculate in two steps and no work is
performed along the path taken at constant volume:
Wacd = Wac + Wcd = Pa (Vc − Va ) + 0
Wacd = 6 × 105 × (8 − 5) = 1800 kJ
Path ad : We know the work is the area under the curve in the
Calculate the work performed by the gas during all three pro- P - V diagram. Accordingly, the work performed by Wad is
cesses ( abd, acd, ad ) shown in the P - V diagram above. the area under the diagonal. So, we simply add the area of the
triangle to the work Wbd that we had calculated previously:
Answer (Pa − Pb ) (Vd − Vb )
Wad = Wbd +
Path abd : We calculate in two steps: 2
Wabd = Wab + Wbd Wad = 600 kJ + 12 × (6 − 2) × 105 (8 − 5) = 1200 kJ .
Example 13.3
A gas expands from volume V1 =1 m3 to volume V2 =3 m3 Answer

along the parabola curve P = 3 V 2 (Pa) shown in the figure. The work performed is found by integration if the pressure
Calculate the work performed by the gas. varies:
Z 3
W= P dV
1
We substitute the equation of the curve as P=3V 2 and inte-
grate:
Z 3 3
W= 3V 2 dV = V 3 .
1 1
W = 3 − 1 = 26 J .
3 3
13.2 INTERNAL ENERGY – FIRST LAW OF THERMODYNAMICS

Consider heating a gas in a container at constant volume. The gas will perform
no work, as the volume is constant: W=P dV=0 . So, where does the heat energy
that we give to the system go? This energy will, of course, be received by the
molecules of the gas. The kinetic energies of the molecules will increase as the
gas is heated. The total energy, and therefore, the temperature of the gas will
increase.
Let us consider another case: This time, let us insulate a container closed with
a piston such that the system cannot exchange heat with the environment. Then,
let us start to compress this gas. The gas will perform negative work, in other
words, it will receive work from the outside, because its volume is decreasing.
However, it cannot give it as heat outside, because the gas is insulated. Therefore,
where does this work go? Again, the work will be used to increase the energies
of the gas molecules. The gas molecules will thus move faster and their kinetic
energies will increase.
These two cases show that a gas can convert the heat or work that it receives
into an internal energy. This is defined as internal energy in thermodynamics:
The total energy of molecules (or atoms) constituting a gas in a stationary
reference frame is called internal energy, and it is indicated with E .
At this stage, we do not yet know how to calculate internal energy. The first
law tells us how to do it:
First Law of Thermodynamics

In any process, the difference between the heat Q received and
the work W performed by a system is independent of the pro-
cess and equal to the increase in its internal energy ∆E :
∆E = Q − W (13.4)
Figure 13.4: The First Law:
The difference between received Let us emphasize the important points of the first law:
heat and work done equals the • The first law is a kind of energy bilan: It states that the difference between the
increase in internal energy. received heat and the work done is spent on increasing the internal energy.
• Note the difference between the signs of Q and W : The heat received by the
system and the work done by the system are accepted as positive. Therefore,
the energy increase is included as their difference.
13.3. APPLICATIONS OF THE FIRST LAW 221
• Internal energy, like pressure P , temperature T or volume V , is a quantity

that characterizes a system; in other words, it is a state variable. But the
heat Q and work W are path-dependent, therefore it is wrong to say “the
heat of a system” or the “work inside of a system.” But it is correct to say the
“internal energy of a system.” Regardless of how it is obtained, this internal
energy is now independent of the path taken.
• Two special cases:
• If the system changes state without performing work ( W = 0 ), then
according to the first law, we get
∆E = Q
In other words, all of the received heat goes into increasing the internal
energy.
• If the system performs work but does not receive heat, then Q = 0
and, according to the law, we get
∆E = −W
Hence, the system performs this work by drawing it from its internal energy.
• This is a law that applies not only to gases, but to all systems. All kinds of
tools, machinery, etc., operate by obeying this law. For example, claims about
perpetual motion machines that operate without receiving any energy are Figure 13.5: The perpetual
not taken seriously in science, as they violate the first law (Figure 13.5). motion machine suggested by
• However, the first law does not tell us the direction in which the change Robert Boyle. In this machine,
water will not flow and circu-
occurs. For example, let us bring together two objects, one hot and the other
late by itself, because the point
cold. One of these objects will give heat Q to the other, but which one? The at which water exits the hose
first law says nothing and allows for heat transfer in both directions, as long should be below the surface of
as the heat received by one is Q and the other is −Q , in other words, as the water in the container.
long as the net heat is zero. However, we know that, in nature, the hot body
always gives heat to the cold one. It is the second law that that will tell us
the direction towards which the events will develop.
13.3 APPLICATIONS OF THE FIRST LAW

The first law determines the relation between heat, work and internal energy.
We can calculate many processes using this relation.
Internal Energy of an Ideal Gas
First, we must slightly change the definition of specific heat for gases. We
had defined specific heats as being per mass in section 12.2:
Q = mc ∆T
We can use the same definition in terms of the number of moles: Q = nc∆T . This
is more useful for gases, as we shall see later.
However, the temperature increase may take place under constant pressure
or constant volume, and therefore a separate specific heat is defined for each one
as follows:
Figure 13.6: 1→2 : Heating at
 ncv ∆T (Heating at constant volume)


Q= (13.5) constant volume. 1→3 : heating

 ncp ∆T
 (Heating at constant pressure) under constant pressure.
Now let us consider that we increase the temperature of a gas by ∆T at constant

volume. The work performed during this process will be zero, because the volume
is constant. If we take W = 0 in the first law,
∆E = Q = ncv ∆T
Integrating this expression from T = 0 to T , we get
E − E0 = ncv (T − 0)
If we decide that the internal energy of the gas at absolute zero temperature
T = 0 is E0 = 0 , then we find the internal energy of a gas at temperature T as
follows:
E = ncv T (Internal energy of an ideal gas) (13.6)
The increase in the internal energy would still have been ncv ∆T if we had per-
formed this heating under constant pressure. This is because internal energy is a
state variable and the value that it takes at that temperature is independent of
how it is reached.
The relation between cv and cp
Now, let us consider that we heat the same gas by ∆T under constant pressure.
It will both heat and perform work during this process. We can write the heat
directly:
Q = nc p ∆T
If the volume of the gas changes from V1 to V2 under constant pressure P , the
work can be written as follows:
W = P(V2 − V1 )
Using the ideal gas equation PV = nRT here, we get
W = PV2 − PV1 = nR T 2 − nR T 1 = nR ∆T
Now, let us use these heat and work expressions in the first law:
∆E = Q − W
ncv ∆T = nc p ∆T − nR ∆T
Simplifying, we find the relation between the specific heats:
c p = cv + R (Specific heats of an ideal gas) (13.7)
This expression is used to find one of the specific heats if the other is known. We
will see how to find the specific heat cv . Here are the results without proof:
Specific heats of gases

cv cp
Monatomic gases 3
R 5
R
2 2
(He, Ne, Ar, Kr . . . )
Diatomic gases (H2 , 5
R 7
R
2 2
O2 , N2 . . . )
Specific heat expressions are more complex for polyatomic gases.

13.3. APPLICATIONS OF THE FIRST LAW 223
Adiabatic Process
The process that a system undergoes without any heat exchange ( Q = 0 ) is
called an adiabatic process. In such a case, there is only exchange of work with
the environment. Let us consider the gas in a container (Figure 13.7). No heat
will be exchanged if the container is insulated and all changes of state will be
adiabatic. Likewise, let us compress the gas in a cylinder by suddenly compressing
the piston. The gas does not have time to exchange heat with the environment,
as it takes place in a short period of time.
We can use the first law to calculate the final temperature, pressure and
volume of an ideal gas at the end of an adiabatic process. Consider that the
temperature of a gas changes by a small amount dT during an adiabatic process. Figure 13.7: No heat is ex-
If we take Q = 0 in the first law and write P dV for work and ncv dT for the changed in an adiabatic process:
increase in internal energy, we get Q = 0.
∆E = 0 − W
ncv dT = −P dV
Let us take the pressure as P = nRT/V from the ideal gas equation:
nRT
ncv dT = − dV
V
dT R dV
+ =0
T cv V
In this expression, let us substitute R with the formula R = c p − cv that we found
above for specific heats:
!
dT cp dV
+ −1 =0
T cv V
If we define the ratio of the specific heats as a dimensionless constant γ=c p /cv ,
we always get γ > 1 :
dT dV
+ (γ − 1) =0
T V
Taking the indefinite integral of both sides of this expression, we get
ln T + (γ − 1) ln V = constant
Using the properties ln a + ln b = ln(a · b) and c ln d = ln d c of the logarithm,

the result is as follows:
T V γ−1 = constant
If we wish to find the relation between the volumes, we substitute the temperature
in this formula as T = PV/nR from the ideal gas equation and simplify:
P V γ = constant
It is meaningful to compare this relation between pressure and volume with

isothermal curves on the P - V diagram. As seen in Figure 13.8, the adiabatic Figure 13.8: Comparison of adi-
curve decreases more rapidly compared to the isothermal curve. abatic and isothermal curves.
A similar formula can be found for (P, T ) if necessary. According to these

results, these products remain constant at each stage of the adiabatic process, in
particular, for the initial and final values:
γ γ
P1 V1 = P2 V2
1−γ γ 1−γ γ
P1 T 1 = P2 T2 (Adiabatic process) (13.8)
γ−1 γ−1
T 1 V1 = T 2 V2
It is sufficient only to keep the first of these formulas in mind; the others can be
obtained from the ideal gas equation PV = nRT .
Work in an adiabatic process. As Q=0 in an adiabatic process, we get ∆E=−W
from the first law. Using Eq. (13.6) for the internal energy, we get
∆E = ncv ∆T
W = −∆E = −ncv ∆T (13.9)
The work performed by the gas is fully provided by the internal energy.
Example 13.4
under constant pressure:
Q = nc p ∆T
A diatomic gas of 5 moles is heated by giving it 600 J under
constant pressure. ( cv = 5R/2 , c p = 7R/2 .) We substitute the values and calculate ∆T as follows:
Q 600
(a) What will the temperature increase ∆T be? ∆T = = = 4.1 K
nc p 5 × 72 × 8.31
(b) What will the increase in internal energy be?
(b) The energy increase will be ∆E = ncv ∆T regardless of
(c) How much work will the gas perform?
the process being at constant pressure or volume:
Answer ∆E = n 52 R ∆T = 5 × 52 × 8.31 × 4.1 = 426 J
(a) It may seem that the information provided in this problem (c) The work is calculated from the first law when the heat
is insufficient, but it is actually sufficient for the requested and the increase in internal energy are known:
calculations. We use the formula for temperature increase ∆E = Q − W → W = Q − ∆E = 600 − 426 = 174 J .
Example 13.5
Work is zero, because the volume is constant: W = 0 . Then
A vessel contains 4 moles of a monatomic gas at 0 C . From we find the increase in internal energy using the first law:
◦
the same initial state: ∆E = Q − W = Q − 0 = 2500 J .

(a) The temperature of the gas is increased to 50 C at con- (b) The heat taken under constant pressure is likewise calcu-
◦
stant volume. Calculate the given heat, the work performed lated by taking c p = 25 R :
by the gas and the increase in internal energy. Q = nc p ∆T = 4 × 52 × 8.31 × (50 − 0) = 4200 J .
(b) The temperature of the gas is increased to 50 ◦ C under We do not have the information to calculate the work directly.
constant pressure. Calculate the given heat, the work per- However, if we first calculate the increase in internal energy,
formed by the gas and the increase in internal energy. we can then find the work by using the first law. The increase
Answer (a) The heat received by the gas at constant volume in internal energy is found using the formula (13.3):
is calculated using Eq. (13.5) and the cv = 32 R value in the ∆E = n cv ∆T = 4 × 32 R × (50 − 0) = 2500 J
table: We calculate the work from the first law using Q and ∆E :
Q = ncv ∆T = 4 × 3 × 8.31 × (50 − 0) = 2500 J . ∆E = Q − W → W = Q − E = 4200 − 2500 = 1700 J .
2
Example 13.6
into steam at 100 ◦ C in temperature. As the heat of vaporiza-
tion of water is L=540 cal/g , calculate the increase in internal
The volume of a 1 g of water at 100 ◦ C in temperature and
energy.
under 1 atm of pressure expands to 1670 cm3 when converted
13.4. KINETIC CALCULATION OF PRESSURE 225
Answer into vapor. We calculate the work under 1 atm at constant

In this problem, we will first calculate heat and work and pressure:
then find the increase in the internal energy using the first W = P (V2 − V1 ) = 1.013 × 105 × (1670 − 1) × 10−6
law. Since no temperature change is involved, there is only W = 170 J
the latent heat of vaporization during the change of phase: We find the increase in the internal energy using the values
Q = m L = 1 × 540 = 540 cal = 540 × 4.18 = 2260 J of Q and W :
The 1 cm3 volume of 1 g of water expands when it is turned ∆E = Q − W = 2260 − 170 = 2100 J .
Example 13.7
increase in internal energy will be zero: ∆E = 0 . Accord-
ingly, the net heat received during a cycle will be spent on
the net work done by the gas:
∆E = Qabca − Wabca = 0 → Qabca = Wabca
We do not have sufficient information to directly calculate
the heat, but we can calculate the work from the P - V dia-
gram. The work performed at each step of the cycle is the
area under that curve. However, the work will be negative if
V decreases in the direction moved. Accordingly, the work
A gas completes a cycle through the 3-step process ( a → b → performed along the path a → b will be positive, the work
c → a ) shown in the figure. What is the net heat received by performed along the path b → c will be negative and the work
the gas? performed along the path c → a will be zero. As a result, the
net work will be equal to the area of the red triangle shown
Answer
in the figure.
This problem may seem complicated at first glance, but it
Qabca = Wabca = Area of the triangle ABC
becomes simple when you think about it. When the gas starts
at point a and undergoes any process, it will have the same Qabca = 12 × (4 − 1) × (150 − 50) × 103 = 150 000 J
internal energy after it returns to point a, in other words, the Q = 150 kJ .
Example 13.8
equation twice and find their proportion:
A cylinder with 5 liters of volume contains a diatomic ideal gas P2 V2 T 2 P2 V2
= → T2 = T1
at 27 ◦ C in temperature and 1 atm in pressure with γ = 1.4 . P1 V1 T 1 P1 V1
The piston is suddenly pushed to compress the volume of the 9.5 × 1
T2 = × 300 = 570 K
gas to 1 L before it can exchange any heat. 1×5
(a) What will the final temperature and pressure be? (b) Adiabatic work is equal to the decrease in the internal
energy (Eqs. 13.9):
(b) How much work will the gas perform?
W = −∆E = −ncv ∆T = −ncv (T 2 − T 1 )
Answer Rather than calculating the number of moles n here, we find
(a) This is an adiabatic process, because there is no heat ex- the pressures and volumes using the formula nRT = PV :
change ( Q = 0 ). We first find the pressure using Eqs. (13.8): cv cv
W = (nRT 1 − nRT 2 ) = (P1 V1 − P2 V2 )
P1 V1γ = P2 V2γ R R
From here, we calculate the final pressure: The value cv = 52 R is taken from page 222 for diatomic gases:
V γ
1
5 1.4 W = 52 (P1 V1 − P2 V2 )
P2 = P1 = × 1 = 9.5 atm From here, we calculate the work performed by the gas:
V2 1
In order to find the final temperature, we write the ideal gas W = 52 × (1 × 5 − 9.5 × 1) × 1.013 × 105−3 = −1140 J .
13.4 KINETIC CALCULATION OF PRESSURE

What is the origin of the air pressure of a tire that can withstand the weight
of a car? The answer at the microscopic scale is that the collisions of those small
molecules with the rubber wall keep such a heavy car up.
There is a kinetic theory of gases that describes the thermodynamic properties
of a gas, such as temperature and pressure, in terms of the microscopic motions
of its atoms or molecules. Let is calculate the pressure of a gas according to the
kinetic theory in order to show how calculations are made in this theory. Figure 13.9: The pressure in-
We will make some assumptions to keep the calculation simple: side the automobile tire is a
result of the collision of the
molecules with the rubber wall.
• Gas molecules will be assumed to be monatomic. Therefore, they will act as

point particles, and thus only perform translational motion and not rotation
or vibration.
• The collisions among the molecules will be ignored.
• The molecules will be assumed to make elastic collisions with the wall.
Let us remember the concepts of impulse and momentum in mechanics: The
momentum of an object with mass m subject to force F in the time interval ∆t
changes as follows:
~F ∆t = m~v2 − m~v1
Now let us consider that a molecule with velocity ~v inside of a cubic volume of
side length L collides with a wall that is perpendicular to the x -axis (Figure 13.10).
As a result of the elastic collision, the component vy of the molecule parallel
to the wall does not change and the horizontal component v x gets reflected
back. Therefore, the force on this molecule by the wall is only in the x -direction
(Figure 13.11):
Figure 13.10: Gas molecules in F ∆t = m0 v x − m0 (−v x ) = 2m0 v x
a cubic volume.
We are using m0 here for the mass of a molecule to avoid any confusion.
This molecule, after colliding with the wall, rebounds toward the opposite
wall, subsequently returning and colliding once more. Therefore, the time ∆t
between two collisions is the time of one collision. As the length of one side is L ,
we calculate the time required to travel back and forth along the path 2L :
2L
∆t =
vx
Using ∆t from the equation above, we write the force F as follows:
2m0 v x m0 v2x
F= =
∆t L
This is the force acting on the molecule. According to the action-reaction law,
Figure 13.11: The collision of a the molecule will push the wall with an equal and opposite force. We will ignore
molecule with a wall. the sign, because we are only concerned with the magnitude of the force here.
We have found the force exerted by one molecule on the wall. If there are N
molecules in the gas, the force exerted by each one of them on the wall will have
the same form. Therefore, the sum required to find the force of N molecules can
be written as follows:
m0 2
Ftotal = v1x + v22x + · · · + v2N x
L
Let us examine the sum in the brackets separately: If we divide this sum by
the number of molecules we will find the mean (or, the average) hv2x i of the v2x
velocities:
v2 + v22x + · · · + v2N x
hv2x i = 1x
N
In this expression, we want to see the mean of the velocity itself, rather than
the mean of the component hv2x i . As each velocity vector can be written as
v2 =v2x + v2y + v2z , the mean of both sides will be
hv2 i = hv2x i + hv2y i + hv2z i

13.4. KINETIC CALCULATION OF PRESSURE 227
Ignoring the effect of gravity, the mean velocities of the molecules will be equal
in all three directions due to symmetry:
hv2x i = hv2y i = hv2z i

hv2 i = 3hv2x i
We can now write the force above in terms of this mean velocity:
Nm0 hv2 i
Ftotal =
L 3
From here, we can calculate the pressure: As the surface area of the side is A = L2
and the volume is V = L3 , the pressure is expressed as follows:
Ftotal Nm0 2 Nm0 2
P= = 3
hv i = hv i (13.10)
A 3L 3V
Many important conclusions can be drawn from this expression:
• Internal energy and specific heat: Let us move the volume to the left-hand
side of this expression, and form the kinetic energy of one molecule:
PV = 2
3 N 1 m0 hv2 i
| 2 {z }
E
The product of the mean kinetic energy of one molecule, m0 hv2 i/2 , and the
number N is the total kinetic energy of the gas. At the start, we had assumed
that the molecules did not interact and only engaged in elastic collisions with
the wall. Therefore, according to the definition, the total kinetic energy will
be the internal energy E of this gas:
PV = 2
3 E
We write the left-hand side as PV = nRT from the ideal gas law:
nRT = 2
3 E
E = 3
2 nRT (Internal energy of a monatomic gas) (13.11)
We had previously found the expression for the internal energy in terms of
the specific heat cv , in Eq. (13.6):
E = ncv T
This result shows that the temperature is a measure of the total energy of
a body. We find the specific heat cv of an ideal gas at constant volume by
comparing the last two expressions:
cv = 3
2 R (Specific heat of a monatomic gas) (13.12)
We also know that c p =cv +R = 52 R . In conclusion, the heat received by, work
performed by and the internal energy increase of ideal gases in all kinds of
processes can thus be calculated using these formulas for E, cv , c p .
• Root-mean-square speed: Let us return to the expression (13.10):

Nm0 2
PV = hv i
3
The Nm0 product on the right-hand side of the equation is the product of
the number of molecules with the mass of one molecule, in other words, the
total mass m of the gas. We can write this in terms of the number of moles n
and the molecular mass M , as m = nM . Using the ideal gas equation again
on the left-hand side, we get
nRT = nM hv2 i
1
3
3RT
hv2 i =
M
We find the root-mean-square (rms) speed formula by taking the root of
this expression:
r
p 3RT
vrms = hv i =
2 (Root-mean-square speed) (13.13)
M
This formula is also valid for polyatomic gas molecules. According to the formula,
the mean speed of gases with lower molecular mass (hydrogen, helium) will be
higher. Therefore, light gases are very scarce in the atmosphere, because they
have already escaped gravity.
Example 13.9
gases and form their ratio:
r
vrms,2
r
What will the root-mean-square speed of oxygen gas (O2 ) be 3RT M1
vrms = → =
at the temperature at which the root-mean-square speed of M vrms,1 M2
Helium gas (He) is 1500 m/s ? Molar masses are MO2 =32 g ,
MHe =4 g . We find the speed by substituting the molar masses:
r r
Answer M1 4
vrms,2 = vrms,1 = × 1500 = 530 m/s
We write Eq. (13.13) for root-mean-square speed for both M 2 32
Example 13.10
the internal energy:
A vessel contains 2 moles of neon (Ne) gas, which has molar E = 32 nRT = 32 × 2 × 8.31 × 300 = 7500 J
◦
mass 20 g , at 27 C in temperature. (b) We use Eq. (13.13) for the root-mean-square speed:
(a) What is the total kinetic energy of the gas molecules? r
3RT
(b) What is the root-mean-square speed of one molecule? vrms =
M
Answer We calculate by substituting the molar mass of neon and the
(a) The gas is monatomic, in other words, it has no rotational temperature:
or vibrational energy; its internal energy consists only of the
r
3 × 8.31 × 300
total kinetic energy. We use the formula (13.11) to calculate vrms = = 610 m/s .
0.020
13.5. HEAT ENGINES 229
13.5 HEAT ENGINES

Converting energy into work is the basis of technology. Hydroelectric power
plants convert the potential energy of water, wind turbines convert the kinetic
energy of air, and gasoline motors convert combustion energy into work.
A heat engine is the general name for machines that convert some of the
heat that they receive into work. The steam engine, the internal combustion
engine and the diesel motor are heat engines.
Regardless of their details, every heat engine performs W work with the heat
QH that it receives from a hot source at temperature T H and gives the remaining
heat QC to a cold source at temperature TC (Figure 13.12).
Heat engines operate in cycles, in other words, they return to their initial
state after each process of receiving heat and performing work. For example,
a steam engine performs W amount of work with the heat QH of the steam
that it receives from the boiler and then releases the heat QC into the external
atmosphere with the expanded steam. In the end, this cycle returns to the point Figure 13.12: Heat engine dia-
at which it starts. gram.
If we write the first law for one cycle, the change in the internal energy of
the system is zero:
∆E = Q − W = (QH + QC ) − W = 0
W = QH + QC
QC is actually a negative quantity here. If we write the absolute values of the

heats, we find that the work is always equal to the difference:
W = |QH | − |QC |
Efficiency
The efficiency of a heat engine is the ratio of the net work performed to the
spent heat:
W |QH | − |QC |
e = =
QH |QH |

QC
e = 1 − (Efficiency of a heat engine) (13.14)
QH
Efficiency is a number within the range [0, 1] , for example, if e = 0.3 , then the
efficiency is 30 %.
Now, let us review the operating principles and efficiencies of some of the
most common heat engines.
Internal Combustion Engine – Otto Cycle
A cycle consists of four processes in an internal combustion engine, also
known as the gasoline engine or the four-stroke engine. Let us examine these pro-
cesses by following them on the adjacent P - V diagram 13.13 and on Figure 13.14
below:
A gasoline-air mixture is pumped into the cylinder through the left valve
being opened when the engine is at point a . Known as intake, this process is
not included in the cycle. This gasoline-air mixture is suddenly compressed after
point a along the path ab . Known as the compression stroke, this process is
adiabatic, because the system has no time for heat exchange. When the fuel is
ignited at point b (ignition), the pressure of the system increases under constant
volume before finding the opportunity to expand, as well as increasing its pressure
along the path bc under constant volume, the gas receiving the heat QH during
this process. As a result of this pressure increase, the gas expands adiabatically
to push the piston along the path cd (power stroke). At point d , the valve on
the right is opened and the burnt fuel-air mixture is discharged from the exhaust
Figure 13.13: Otto cycle. along the path da (exhaust stroke). Once the point a is reached, the cycle starts
again with the intake stroke.
Figure 13.14: Four-stroke en-

gine. Each figure corresponds to
a different process in the P - V
diagram.
Let us calculate the efficiency of this cycle, also known as the Otto cycle, which
consists of two adiabatic and two constant-volume processes. It is sufficient to
calculate heat along the paths bc and da , as there is no heat exchange in adiabatic
processes:
bc process at constant volume: QH = ncv (T c − T b )

da process at constant volume: QC = ncv (T a − T d )
Observing from the diagram that T c > T b and T a < T d , it can be understood
that the heat QH is positive and the heat QC is negative.
Using the definition of efficiency,
QH + QC (T c − T b ) + (T a − T d )
e= =
QH Tc − Tb
The four temperatures here can actually be reduced to a single parameter. Let us
use V to indicate the value Vb = Vc at which the gas is compressed at a maximum
in the Otto cycle. If we use rV to indicate the volumes Va = Vd at which the gas
expands, r is defined as the compression ratio of the motor. Now, if we use the
formulas (13.8) that we found earlier for the adiabatic processes ab and cd , we
get
The process ab: T a (rV)γ−1 = T b V γ−1

The process cd: T d (rV)γ−1 = T c V γ−1
If we eliminate the common term V γ−1 in these equations and substitute the
temperatures T , the formula simplifies as follows:
T d rγ−1 − T a rγ−1 + T a − T d
e =
T d rγ−1 − T d rγ−1
1
e = 1− (Efficiency of the Otto cycle) (13.15)
rγ−1
13.5. HEAT ENGINES 231
Here, γ = c p /cv and is approximately γ=1.4 for air. This result is surprising. In
the gasoline engine, regardless of the type of fuel or dimensions that you use,
the efficiency depends only on the compression ratio r of the engine. Of course,
it is not possible to achieve high compression with certain fuels, as they burn
halfway through. Modern engines using high-octane gas have compression ratios
around r = 10 . If we take r = 10 and γ = 1.4 for air, the efficiency is found to
be e = 0.6 . In reality, the efficiency of a gasoline engine is around 25 %, due to
friction and other effects until the energy is transmitted to the wheels. Figure 13.15: The Jaguar V12
engine has a record compression
ratio of r=14 .
Example 13.11
Now, let us analyze each step along these principles:
Determine the signs of heat ( Q ), work ( W ) and internal en- The AB process:
ergy increase ( ∆E ) in each step of the ABC cycle shown in the
Temperature T and volume V increase along the path.
figure, and complete the table.
Therefore, Q and ∆E and W are all positive.
The BC process:
Temperature T increases, but volume V remains con-
stant along the path. Therefore, Q and ∆E are positive but
W = 0.
The CA process:
Answer Temperature T and volume V decrease along the path.
This problem constitutes the basis before examining cycles. Therefore, Q and ∆E and W are all negative. As a result,
First, let us remember some of the general principles: the table is completed as follows:
• In the P - V diagram, temperature increases with distance
to the origin.
• Internal energy always increases with T : ∆E=ncv ∆T .
• Heat is positive if the temperature increases. It can be in
either of two forms: Q=ncv ∆T or nc p ∆T
• The work dW=P dV by the gas is positive if it expands.
Example 13.12
Therefore, we will calculate the steps of the cycle at which
the gas receives heat and the steps at which it gives heat, and
use their ratio in the efficiency formula. In general, the result
will depend only on temperature differences.
In this problem, no heat exchange will take place in step
ab , as it is adiabatic. When the gas is compressed in step
bc , it is forced to give off heat in order to keep the pressure
constant. Therefore, QC =Qbc . In step ca , its pressure is in-
A gas with γ=c p /cv =1.4 undergoes a 3-step cycle, as shown in creased by heating at constant volume. Therefore, QH =Qca .
the figure. It expands adiabatically along the path ab , is com- In short, the situation is as shown in the figure below:
pressed under constant pressure along the path bc and is heated
at constant volume along the path ca . T a =500 K , T b =400 K
and T c =300 K . Calculate the efficiency of the cycle.
Answer
Calculating the efficiency of a cycle may seem complicated
at first glance, but it becomes easy when given some thought.
Efficiency was defined in Eq. (13.14) as the ratio of the net
work W to the positive heat: Now let us calculate the heats QH and QC :
W |QC | QH = Qca = ncv ∆T = ncv (T a − T c )
= =1− Q C = Qbc = nc p ∆T = nc p (T c − T b )
QH |QH |
The ratio of these heats is used in the efficiency formula: We calculate the efficiency by substituting the data:
|nc p (T c − T b )| Tb − Tc 400 − 300

=1− =1−γ = 1 − 1.4 × = 0.30 = 30 %
ncv (T a − T c ) Ta − Tc 500 − 300
Example 13.13
Calculate the efficiency of the Diesel cycle shown in the figure
in which T a =300 K , T b =1000 K , T c =1600 K and T d =800 K .
Diesel cycle. In a diesel engine, air compressed at high temper-
( γ=c p /cv =1.4 for air.)
ature is combusted by injected oil. The exploded air-oil mixture
expands and pushes the piston. The stages of the cycle are as Answer
follows: It is easy to see where the heats QH and QC are received
and given in this cycle. No heat is exchanged in the adiabatic
steps. The gas receives heat along the path bc because it is
heated, and gets cooler and gives off heat along the path da .
Therefore, we get QH = Qbc and QC = Qda . Let us calculate
these:
Under constant pressure : QH = Qbc = nc p (T c − T b )
Under constant volume : QC = Qda = ncv (T a − T d )
The ratio of these heats is used in the efficiency formula:
|ncv (T a − T d )| 1 Td − Ta
ab : adiabatic compression =1− =1−
nc p (T c − T b ) γ Tc − Tb
bc : expansion under constant pressure (explosion)
We calculate the efficiency by substituting the data:
cd : adiabatic expansion (work by pushing the gas piston)
1 800 − 300
da : cooling at constant volume (discharge from the exhaust) =1− × = 0.40 = 40 %
1.4 1600 − 1000
13.6 SECOND LAW OF THERMODYNAMICS – THE CARNOT CYCLE

When two bodies, one hot and the other cold, are brought together, heat flows
from the hot body to the cold one, and never the other way around. A drop of
ink dropped in water spreads out, but an ink that has been spread out does not
spontaneously gather at one point. Likewise, if we remove the wall between two
different types of gas in two adjacent containers, the two gases get mixed, but
they do not separate spontaneously.
Although all of these processes are possible in terms of energy, in other words,
although they comply with the first law, they are not observed in nature. There
is a direction of development preferred by nature in thermodynamic events. The
second law of thermodynamics specifies this direction.
There are many varied expressions of the second law in macroscopic and
microscopic scales. Its microscopic expression includes concepts such as entropy,
disorder, phase space, etc., and is the subject of the branch of physics called
Statistical Mechanics. We shall examine the macroscopic expression based on
heat engines here.
Second Law of Thermodynamics

It is not possible for a cyclic heat engine to convert all of the heat
that it receives from a reservoir into work.
Let us emphasize the important points of the second law:

• If a machine had converted all of the energy it received into work, according
to the definition of efficiency (Eq. 13.14), we would have QC =0 and efficiency
would be 100 %. This is possible according to the first law, because the heat
13.6. SECOND LAW OF THERMODYNAMICS – THE CARNOT CYCLE 233
received is equal to the performed work. However, the second law states that
no heat engine can operate at 100 % efficiency. In other words, it specifies
the direction towards which natural events develop.
• According to the second law, a gasoline engine cannot operate without dis-
charging the burnt gas, or a steam engine without discharging the expanded
steam. If it could not discharge this heat into the environment, the machine
would eventually heat up and become inoperable.
• If there is no way to give heat to the cold source, then work cannot be done
either. For example, the engine of a ship could, in principle, be able to take
heat from the sea water and turn some of it into work. However, the machine
would not operate, because it would not be able to return the remaining heat
to the sea after the work is done.
The Carnot Cycle
The second law states that it is impossible to build a 100 % efficient heat
engine. So, what is the most efficient engine that can be built and what is its
efficiency? Scientists and engineers have pursued ways to increase the efficiency
of the heat engines ever since the invention of the steam engine. Eventually, a
French engineer named Sadi Carnot (1796-1832) demonstrated what it takes to
make the most efficient heat engine. Carnot proved that it is not possible to build
a heat engine more efficient than the Carnot cycle, which is named after him.
The Carnot cycle is an ideal cycle, in other words, it has no application in real
life.
The Carnot cycle consists of two adiabatic and two isothermal processes. As
seen in Figure 13.16, starting from point a at the high temperature T H , expansion
at constant temperature (isothermal) takes place along the path ab, after which
cooling takes place through adiabatic expansion until the temperature TC is
reached along the path bc, and then isothermal compression takes place along the
path cd, until, finally, the cycle is completed with adiabatic compression along
the path da.
In order to calculate the efficiency of the Carnot cycle, let us examine where Figure 13.16: The Carnot cycle.
heat is received and where it is given off. As Q = 0 in adiabatic processes, heat
exchange occurs only during the isothermal processes. The gas receives heat
during the ab process, because it expands, and it gives off heat to the environment
during the cd process. Therefore, the heat received from the hot source along the
path ab will be equal to QH and the heat given off to the cold source along the
path cd will be QC .
As ∆T = 0 in isothermal processes, we get ∆E=0 . Therefore, according to
the first law ∆E=Q−W , we get Q=W in isothermal processes. We can therefore
directly calculate work using Eq. (13.3), rather than the heat:
Vb
The isothermal process ab: QH = Wab = nRT H ln
Va
Vd
The isothermal process cd: QC = Wcd = nRTC ln
Vc
We use these values in the definition of efficiency. Since Vd < Vc in the last
expression, the heat Qc will be negative. We make the logarithm positive as
ln(Vd /Vc ) = − ln(Vc /Vd ) :
|QC | TC ln(Vc /Vd )
e=1− =1−
|QH | T H ln(Vb /Va )
Now let us use the formula T V γ−1 = constant for the paths bc and da :
γ−1 γ−1
T H Vb = T C Vc
γ−1 γ−1
T H Va = T C Vd
Dividing both sides of these two equations, we get

γ−1 γ−1
Vb Vc Vb Vc
γ−1
= γ−1
−→ =
Va Vd Va Vd
The final expression for efficiency is:

TC
e=1− (Efficiency of the Carnot Cycle) (13.16)
TH
It is remarkable that efficiency depends only on the ratio of temperatures of the
hot and cold sources. In the Carnot cycle, the smaller the ratio TC /T H gets, in
other words, the higher the temperature difference gets, the higher the efficiency
will be. Carnot proved that no other cycle can have a higher efficiency.
1. Which is the first law of thermodynamics?
6. Which of the following is the efficiency of a heat engine?
(a) ∆E = Q + W
(b) ∆E = Q − W (a) Ratio of the work performed to the heat received by
(c) ∆W = E − Q the gas.
(d) ∆Q = E − W (b) Ratio of the work performed to the heat given by
the gas.
2. Which is correct for the second law of thermodynamics? (c) Ratio of the work done to the total heat.
(a) Received heat can be fully converted into work. (d) Ratio of the work performed to the internal energy.
(b) The difference between received heat and given heat
is converted into work. 7. Which of the following is correct?
(c) The sum of received heat and given heat is converted (a) A gas performs work if it is heating.
into work. (b) A gas performs work if it is cooling.
(d) It is not possible to fully convert received heat into (c) A gas performs work if it is expanding.
work. (d) A gas performs work if its pressure is increasing.
3. Which is correct for the adiabatic process?
8. The volume of a gas is increased by 2 m3 at a constant
(a) Gas performs no work.
pressure of 10 Pa . How much work will the gas per-
(b) Gas exchanges no heat.
form?
(c) Internal energy does not increase.
(d) Internal energy does not change. (a) 5 J (b) 1/5 J (c) 12 J (d) 20 J
4. Which is correct for the isothermal process? 9. A gas with 3 m3 in volume and 2 atm in pressure is
(a) The volume remains constant. heated at constant volume to increase its pressure to
(b) The pressure remains constant. 10 atm . How much work will the gas perform?
(c) The temperature remains constant. (a) 0 (b) 18 J (c) 12 J (d) 30 J
(d) Gas exchanges no heat.
10. A gas receives 50 J of heat and performs 30 J of work.
5. Which of the following is the formula for the work per-
What is the change in internal energy?
formed by a gas?
(a) P dV (b) V dP (c) V/dP (d) dP/V (a) -20 J (b) 20 J (c) 80 J (d) -80 J
PROBLEMS 235
11. The internal energy of an ideal gas changes with which II. There can be no heat engine more efficient than the
of the following? Carnot cycle.
(a) Temperature. III. The Carnot cycle has 100 % efficiency.
(b) Volume. IV. The Carnot cycle is used in industry. .
(c) Pressure. (a) I & II (b) I & III (c) II & IV (d) I & IV
12. What is the relation between cv , the specific heat at 17. The molecules of which of the three gases, hydrogen
constant volume, and c p , the specific heat at constant (H2 ) , oxygen (O2 ) and carbon dioxide (CO2 ) at the
pressure of an ideal gas? same temperature have higher root-mean-square speed?
(a) c p =cv (b) c p =2cv (c) c p =cv +R (d) cv =c p +R (a) H2 (b) O2 (c) CO2 (d) Equal
13. What is the origin of the pressure of a gas at the micro-

18. A gas is compressed at constant temperature. Which is
scopic scale?
correct?
(a) Collisions of atoms with container walls.
(a) It will receive heat.
(b) Collisions between atoms.
(b) It will give off heat.
(c) The gravitational force on atoms.
(c) Its internal energy will increase.
(d) The kinetic energy of atoms.
(d) Its internal energy will decrease.
14. The root-mean-square speed of the molecules of a gas is
independent of which of the following? 19. In which process is it necessary to give higher heat to
(a) Temperature increase the temperature of a gas by the same amount?
(b) Molar mass (a) At constant volume.
(c) The gas constant R (b) At constant pressure.
(d) Pressure (c) Adiabatic.
(d) Isothermal.
15. The efficiency of the gasoline engine is dependent on
which of the following?
(a) Temperature difference 20. When a gas expands adiabatically, what is the source of
(b) Pressure difference the work that it performs?
(c) Compression ratio (a) The heat it receives.
(d) Amount of the air-gasoline mixture (b) The heat it gives.
(c) The increase in internal energy
16. Which of the following are correct? (d) The decrease in internal energy.
I. It is not possible to build a heat engine with 100 %
efficiency.
Problems
13.1 Work Done by a Gas

13.1 A cylinder with 20 liters of volume contains 3 moles of
gas at a temperature of 27 ◦ C . Calculate the work performed
by this gas during the following consecutive processes: (a) Its
temperature is increased by a factor of 4 at constant pressure.
(b) Later, its pressure is increased by a factor of 4 at constant
volume. (c) Later, its pressure is increased by a factor of 2 at
constant temperature. [A: (a) 220 J , (b) 0, (c) −51 J .] Problem 13.3
13.2 A cylinder contains 5 moles of gas at a temperature of 13.3 Calculate the work performed by the gas during all
27 ◦ C . 2000 J of work is externally performed on this gas three processes ( abd, acd, ad ) shown in the P - V diagram
under constant pressure. What will the final temperature of above.
the gas be? [A: 348 K .] [A: Wabd = 2800 kJ , Wacd = 800 kJ , Wad = 1800 kJ .]
Problem 13.4
Problem 13.9
13.4 A gas expands from volume V1 =1 m3 to volume 13.9 We had previously seen that 1 mole of monatomic ideal
V2 =2 m3 along the curve P=4 V 3 −3 (Pa) shown in the figure gas at STP conditions ( 0 ◦ C and 1 atm ) has a volume of 22.4 L
above. Calculate the work performed by the gas. [A: 12 J .] (point a in the figure above). The volume of this gas is in-
creased by a factor of 5 in two different ways: The process ab
13.2-3-4 Internal Energy - First Law of Thermo- is isothermal ( T =constant ) and the process ac is adiabatic
dynamics and Applications ( ∆Q=0 ). Calculate and compare the work performed by the
gas in both cases. ( cv =3R/2, γ=1.4 .)
[A: Isothermal 3700 J , adiabatic 1600 J , isothermal work is
13.5 As a gas is heated by being given 3000 J of energy, it
always greater.]
simultaneously performs 500 J of work. What is the change
in the internal energy of the gas? [A: ∆E = +2500 J ] 13.10 A cylinder contains 3 moles of diatomic gas at a tem-
perature of 27 ◦ C ( cv =5R/2 ). 1000 J of work is externally
13.6 A monatomic gas of 4 moles is heated by giving it 800 J performed on this gas under constant pressure. (a) What will
under constant pressure. ( cv =3R/2 , c p =5R/2 .) (a) What will the final temperature of the gas be? (b) What will the internal
the temperature increase ∆T be? (b) What will the increase in energy be? (c) How much heat has the gas received or given
internal energy be? (c) How much work will the gas perform? off? (Notice that the work is negative.)
[A: (a) 9.6 K , (b) 480 J , (c) 320 J .] [A: (a) 260 K , (b) ∆E = −2500 J , (c) Q = −3500 J has been
given off.]
13.7 A vessel contains 3 moles of a diatomic gas at a tem-

perature of 27 ◦ C ( cv =5R/2 , c p =7R/2 ). (a) The temperature
of the gas is increased to 127 ◦ C at constant volume. Calcu-
late the given heat, the work performed by the gas and the
increase in internal energy. (b) The temperature of the gas
is increased to the same 127 ◦ C at constant pressure. Calcu-
late the given heat, the work performed by the gas and the
increase in internal energy.
[A: (a) Q = ∆E = 6.2 kJ , W = 0 , Problem 13.11
(b) Q = 8.7 kJ , ∆E = 6.2 kJ , W = 2.5 kJ .] 13.11 A gas completes a cycle through the 3-step process
( a→b→c→a ) shown in the figure. What is the net heat re-
ceived by the gas? [A: 300 kJ .]
13.12 A cylinder with 4 liters of volume contains a

monatomic ideal gas at 27 ◦ C in temperature and 3 atm in
pressure ( γ = 1.67 ). The piston is suddenly pushed to com-
Problem 13.8 press the volume of the gas to 1 L before it can exchange any
heat. (a) What will the final temperature and pressure of the
13.8 A vessel contains Va =10 L of ideal gas at pressure gas be? (b) How much work will the gas perform?
Pa =5 atm . This gas is first cooled at constant volume along [A: (a) 750 K , 30 atm , (b) −2.7 kJ .]
the path ab shown in the figure above until its pressure
reaches Pb =2 atm . Then, it is again brought to temperature
13.4 Kinetic Calculation of Pressure
T a by expansion under constant pressure. (a) What is the
volume Vc ? (b) What is the total work performed by the gas? 13.13 The molecules of an oxygen gas in a cylinder collide
(c) What is the total increase in internal energy? (d) How 3 × 1023 times in 1 s with the surface of a piston with a cross-
much heat does the gas give or receive? section area of 100 cm2 . If the root-mean-square speed of
[A: (a) 25 L , (b) W = 3000 J , (c) ∆E = 0 , (d) it receives molecules is 500 m/s , (a) What is the force exerted by each
Q = 3000 J .] molecule on the piston in 1 s ? (b) How much pressure does
PROBLEMS 237
the gas exert on the piston? (The mass of the oxygen molecule on the left in the figure above and mark them in the table on
is 5.3 × 10−26 kg .) [A: (a) 5.3 × 10−23 N , (b) 1600 Pa .) the right.
13.14 What are the root-mean-square speeds of oxygen gas
(O2 ) , which has molar a mass 32 g , and of nitrogen gas (N2 ) ,
which has a molar mass 28 g , at the temperature at which
the root-mean-square speed of argon gas (Ar) , which has a
molar mass 40 g , is 200 m/s ? [A: 224 and 239 m/s .]
13.15 A container with 5 L of volume contains 3 moles of he-
lium gas (He) at 127 ◦ C in temperature ( MH =4 g ). (a) What
is the total kinetic energy of the gas molecules? (b) What is
the root-mean-square speed of molecules?
[A: (a) 15 kJ , (b) 1580 m/s .] Problem 13.19
13.19 A gas with γ=c p /cv =1.67 undergoes a 3-step cycle
13.5 Heat Engines
as shown in the figure above. It is compressed adiabatically
13.16 (a) A heat engine receives 5000 J in heat from a hot along the path ab , then expanded at constant pressure along
source and releases 2000 J in heat to the environment. What the path bc , and finally cooled at constant volume along the
is the efficiency of the machine? (b) A heat engine oper- path ca . T a =300 K , T b =400 K and T c =600 K . Calculate the
ating with 20% efficiency performs 300 J of work in one efficiency of the cycle. [A: 10 % .]
cycle. How much heat does this machine release into the
environment? [A: (a) 60 % , (b) 1200 J .)
13.17 What is the minimum value of the compression ratio

r such that a gasoline engine operating with the Otto cycle
can have 50 % efficiency? [A: r = 5.7 .]
Problem 13.20
13.20 Diesel cycle. Calculate the efficiency of the Diesel
Problem 13.18 cycle shown in the figure in which T a =300 K , T b =900 K ,
13.18 Determine the sign of heat ( Q ), work ( W ) and internal T c =1400 K and T d =700 K . ( γ=c p /cv =1.4 for air)
energy increase ( ∆E ) in each step of the cycle ABC shown [A: 43 % .]
14
THE ELECTRIC FIELD
Lightning is the most striking

manifestation of electricity in
nature. Negatively charged elec-
trons accumulated in clouds sud-
denly flow to the Earth’s surface
and transfer large amounts of
energy.
Gravitation was the first force to be discovered among the fundamental forces
of nature. Electricity and magnetism only came to be understood later, around
the 1600s. But electric and magnetic forces shape natural events and play a major
role, perhaps the most significant role for living things and for technology. The
nervous system in the human body operates with electric currents and the oxygen
exchange in blood takes place with electric potential differences. Light bulbs,
radios, televisions and other appliances facilitate domestic life; electric motors
and electronic components in technology are included in the structure of all kinds
of instruments and machinery.
Gravitational force may hold our solar system and much larger galaxies
together. But it is the electric force that keeps atoms and molecules together to
form living bodies and organisms. With this new force, a new property of matter,
called the electric charge, arises in two forms, positive and negative.
Then, moving charges produce another fundamental force, called magnetism.
The sciences of electricity and magnetism first developed independently, and
were later united into a single theory, called electromagnetism. We will first
discuss the basic concepts and laws of electricity in this chapter and then those
of magnetism in subsequent chapters.

240 14. THE ELECTRIC FIELD
14.1 ELECTRIC CHARGE

The existence of electric forces has been observed since ancient times. Fos-
silized pieces of pine resin, known as amber, could attract small objects. The word
“electric” was derived from the word electron, which means amber in Greek. It
was later observed that other materials made of glass, rubber, plastic or fur could
gain the same property through friction.
Figure 14.1: Amber is fossilized You may perform such experiments at home too. Take a plastic comb or pen,
tree resin. rub it with a piece of cloth and then hold it close to small pieces of paper. You
will observe that the comb attracts the small pieces.
However, if you rub a second plastic comb the same way and bring it close
to the first one, you will observe that they repel each other. (You may have to
suspend the comb with a rope to observe this effect.) Likewise, if you apply
friction to a glass rod and bring it close to the pen, you will observe that, this time,
they attract each other. This attractive-repulsive property proves that electric
force has a different structure than gravity, because gravitational force can only
be attractive.
Figure 14.2: A comb rubbed Each force in nature shows its effect through an intrinsic quantity in matter.
with a piece of cloth attracts After gravitational force was observed, it was understood that it acted on a
small pieces of paper. property of matter called mass. Likewise, electric forces should also arise from
another quantity inherent in matter, this one called the electric charge. Let us
emphasize the properties that this electric charge should have:
Figure 14.3: Two different types

of electric charge.
• There are two types of electric charge with opposite signs. Only in
this way can one accommodate the attractive and repulsive forces that arise
in electricity. These were called positive and negative charges. Among the
elementary particles that were discovered later, protons (p+ ) were assumed
to be positively charged and electrons (e− ) to be negatively charged. We must
emphasize that this is only an assumption. Today, scientists consider that this
assumption was unfortunate, and that it would have been more practical to
assume electrons as positive. Indeed, in the majority of electrical phenomena,
it is usually electrons that move in matter and positively charged protons
and ions that are stationary, as they are heavier. As we shall discuss later,
the electric current is actually a flow of electrons in the inverse direction.
• Electric charge is conserved. In other words, the algebraic sum of the
electric charges in an isolated system is constant. For example, if -3 units of
charge are added to an object with +5 units of charge, the object’s net charge
will be +2 units. Objects accepted as neutral actually have both positive and
negative charges and seem neutral because they are in equal amount. If some
negative charge is taken away from such a neutral object, the object will be
positively charged.
Figure 14.4: Electron-positron In modern physics, charge conservation is not only algebraic, but also valid
pair production at Lawrence on a much more fundamental scale. Electron-positron pairs ( e− e+ ) can be
Berkeley National Lab. produced in vacuum in experiments conducted with elementary particles
14.1. ELECTRIC CHARGE 241
(Figure 14.4). As the charge of the positron is equal and opposite to the
electron, the net charge remains zero and charge conservation is respected.
• Electric charge of objects is always a multiple of an elementary
charge. It was observed that electric charge does not change continuously,
but varies as multiples of a minimum charge. In the terminology of modern
physics, it is said that electric charge is quantized. This elementary charge
is indicated with e . The electron charge (−e) and the proton charge (+e)
are exactly equal and opposite.
In today’s modern physics, this is actually not exactly correct. Particles called
quarks, considered to be the building blocks of elementary particles, are
thought to be charged as e/3 or 2e/3 , in other words, as fractions of the
electron charge. However, these quarks have not yet been observed directly. Figure 14.5: According to mod-
ern physics, a proton consists of
• Unit of electric charge. In the SI system, the unit of electric charge is the
3 quarks with fractional charges.
Coulomb (C). The value of the elementary charge e is as follows:
1 e = 1.6 × 10−19 C (elementary charge) (14.1)
A Coulomb is a very large unit, and fractions are used in practice:
1 mC (milli Coulomb) = 10−3 C

1 µC (micro Coulomb) = 10−6 C
• You should also know that the fundamental electrical unit in the SI system is
not the Coulomb. Instead, the ampere (A), the unit of current, is chosen as
the fundamental unit. The reason for this is that, in practice, it is easier to
measure current than charge. We shall define the ampere in Chapter 20.
Conductors and Insulators – Structure of the Atom
Let us take an object charged through friction and place it next to a neutral
object. When we connect these two objects with a copper rod, we observe that
some of the electric charge is transferred to the neutral object. However, when
we connect these two objects with a plastic rod, we observe that no charge is
transferred and the neutral object remains neutral.
Figure 14.6: Example of a con-
ductor and an insulator: The cop-
per rod conducts the charge and
the plastic does not.
This experiment shows that metals such as copper and iron easily conduct
electricity and substances such as glass do not conduct electricity at all. All
materials show either conductor or insulator properties in terms of electrical
conductivity. All metals are conductors. On the other hands, insulators may
become conductors under very high charge accumulations.
Other than these, there are semiconductor materials developed in labo-
ratories and used in electronic devices. Also, many elements can switch to a
superconductor state at very low temperatures. These two types of materials
cannot be explained with classical electromagnetism and must be analyzed with
modern quantum theory.
The reason why a material is a good or bad conductor can be understood by
examining the structure of the atom. As explained by modern quantum theory, the
structure of all elements consists of a positively charged nucleus with electrons

rotating around it in orbits (Figure 14.7). The nucleus contains positively charged
protons (p+ ) and neutral neutrons (n) .
In a sense, this resembles the Sun and the planets rotating around it, however,
this resemblance is not exactly correct. Electrons can rotate only at certain radii
and the number of electrons that can inhabit each orbit is limited. For example,
the innermost orbit can only hold a maximum of two electrons, while the next
one can hold 8 electrons, etc.
When these orbits become filled with the maximum number of electrons that
Figure 14.7: Atomic structure. they can hold, they are called closed shells, and the next electron is forced to
go to next outer orbit. The electrons in the innermost orbits are very strongly
bonded to the nucleus, while the outer electrons are weakly bonded.
The electrons of an atom can be separated by externally bombarding it with
particles. Observations show that it is more difficult to detach the electrons in
closed orbits and inner orbits, as they are more strongly bonded to the nucleus.
However, electrons in an outermost and incomplete orbit can be detached much
more easily.
Figure 14.8: The single elec- When atoms gather to form a rigid body, the state of these weakly-bonded
tron outside closed orbits in cop- electrons in the outermost orbit determines the electrical conductivity of that
per atom, makes it a conductor. matter. The one and two electrons in the outermost orbit of metals get released
easily without any bombardment and circulate freely in the solid medium. In
insulators, they are firmly bonded to the atoms, as the outermost orbits are
closed-shell orbits.
The source of electrical conductivity is these free electrons. Metals are
conductive because free electrons can easily move in a solid medium. Insulators
do not transfer charge because they have no free electrons.
14.2 COULOMB’S LAW

French scientist Charles Coulomb (1736–1806) was the first to examine the
nature of the force between electric charges. He had to invent new ways to
measure such forces. After charging a metal sphere, he put it in contact with an
identical second metal sphere and assumed that the charge was distributed evenly
among two spheres. He thus obtained half, quarter and one eighth charged, etc.,
spheres. He used a torsion balance to measure the very small forces among these
spheres.
And he established the law bearing his name:
Figure 14.9: The torsion bal-
ance used by Coulomb. Coulomb’s Law
The force of attraction or repulsion between two electric charges
is directly proportional to the product of the two charges and
inversely proportional to the square of the distance between
them:
q1 q2
F=k 2 (14.2)
r
Like charges repel, unlike charges attract each other.
Let us emphasize the important points of Coulomb’s law:

14.2. COULOMB’S LAW 243
• Coulomb’s law is valid for stationary charges. It is therefore also known as

the electrostatic force.
• The proportionality constant indicated with k in Coulomb’s law is called the
Coulomb’s constant:
k = 8.99 × 109 ≈ 9 × 109 N·m2 /C2 (14.3) Figure 14.10: Coulomb’s law.
Unlike charges attract, like
The approximate value will be used in the problems. charges repel.
• Later we shall see that it is more convenient to define a new constant ε0

instead of k in some of the expressions we will be developing:
1 1
k= ↔ ε0 = = 8.85 × 10−12 C2 /(N·m2 ) (14.4)
4π ε0 4π k
The constant ε0 is called the electric permittivity of free space.
Coulomb Force for a System of Charges
If there are more than two charges in an environment, the net force on any
charge is the vector sum of the forces that the other charges exert upon it. For
example, consider four charges as q1 , q2 , q3 and q4 . The net force exerted upon
one of these, for example, upon q1 , is written as follows:
Figure 14.11: System of
~F1 = ~F12 + ~F13 + ~F14 (14.5) charges.
We take the vector sum of the forces Fi j after finding their magnitudes using the
Coulomb’s law.
Example 14.1
We calculate the magnitudes of both forces with the
Coulomb’s law. The signs of the charges are ignored when
calculating the magnitudes:
q1 q3 (4 × 10−6 ) × (2 × 10−6 )
F1 = k 2 = 9 × 109 ×
r1 (0.01 + 0.04)2
Calculate the total force exerted upon the charge q3 = 2 µC by
the charges q1 = 4 µC and q2 = −3 µC shown in the figure. F1 = 29 N
q2 q3 (3 × 10−6 ) × (2 × 10−6 )
Answer F2 = k 2 = 9 × 109 ×
Charges with the same sign repel and charges with opposite r2 0.042
signs attract each other. Accordingly, the directions of the F2 = 34 N
force F1 exerted by the charge q1 and the force F2 exerted We find the total force by taking the positive x -direction
by the charge q2 are shown on the right-hand side of the towards the right:
figure. F = F1 − F2 = 29 − 34 = −5 N .
Example 14.2
charge q3 = +4 µC .
(b) Find the magnitude and direction of the total force.
Answer
(a) The figure shows the forces ~F1 and ~F2 exerted by the
charges q1 and q2 . We first calculate the magnitudes of
these forces:
q1 q3 (3 × 10−6 ) × (4 × 10−6 )
F1 = k 2 = 9 × 109 ×
For the charges q1 = − 3 µC and q2 = + 2 µC in the figure, r1 0.052
(a) Calculate the components of the total force acting on the F1 = 43 N
q2 q3 (2 × 10−6 ) × (4 × 10−6 ) (b) We calculate the magnitude and angle of the force with
F2 = k = 9 × 109 ×
r22 0.032 the known qcomponents:
F2 = 80 N F = F 2x + Fy2 = 64 N
The total force is ~F = ~F1 +~F2 , and we calculate its components
Fy −35
as follows: tan θ = = = −0.65 → θ = −33◦
F x = F2 − F1 cos 53 = 80 − 43 × 0.6 = 54 N
◦ F x 54
Fy = −F1 sin 53◦ = −43 × 0.8 = −35 N
Example 14.3
Only the forces they exert at the points in between will be in
the opposite directions. Let x be the coordinate of the point
upon which the forces cancel each other out. Accordingly,
we write the total force and set it as equal to zero:
q1 q3 q2 q3
F1 = F2 → k 2 = k
x (1 − x)2
Simplifying, we get a quadratic equation:
q1 q2
The distance between the charges q1 = + 1 mC and q2 = + 9 mC = → (q2 − q1 ) x2 + 2q1 x − q1 = 0
x2 (1 − x)2
shown in the figure is 1 m . Where should a third charge q3 be
8x2 + 2x − 1 = 0
placed such that the net force exerted on it is zero?
The roots of this equation are −0.5 and +0.25 . We will not
Answer consider the negative root, as it will be outside of the charges.
As the charges q1 and q2 have the same sign, they exert Therefore, the solution is the positive root:
force in the same direction upon the charges on the outside. x = 0.25 m .
14.3 ELECTRIC FIELD

The Coulomb law is a force exerted between two charges at a distance. New-
ton’s gravitational force is likewise exerted between two masses a distance apart
from each other. This property of “action at a distance” was disturbing for many
scientists, including Newton. Just think about it: A charge q1 examines its sur-
roundings, detects the presence of another charge q2 , determines its distance and
charge, and then exerts Coulomb’s force accordingly. Is this believable?
The English scientist Michael Faraday (1791–1867) suggested the concept of
an electric field to resolve this problem. According to Faraday, when a charge q1
is placed anywhere, it affects every point of the surrounding space and produces
Figure 14.12: First the charge an electric field (Figure 14.12). This electric field always remains there, even if no
q1 produces the electric field. other electrical charge exists. A second charge q2 that is placed later interacts
The force ~F=q2 ~E acts on charge with this electric field through Coulomb’s law. In other words, the first charge
q2 in this field. does not check the position and value of the second charge. The concept of field
was later taken much further in modern physics and was understood to be a
correct approach.
Since a charge interacts with the surrounding electric field, let us take a very
small positive q0 test charge and place it at a point in space. If the electrostatic
force acting on this charge q0 is ~F , then the electric field at that point in space
is defined as:
~
~E = F (The Electric Field) (14.6)
Figure 14.13: If a force ~F acts q0
on test charge q0 there is an The inverse is also true: The force acting on a charge q in the presence of an
electric field ~E=~F/q0 at that electric field ~E is,
point. ~F = q ~E (14.7)
Let us emphasize the main properties of an electric field:
14.3. ELECTRIC FIELD 245
• An electric field can also be considered as the “force acting on a unit charge.”
• The unit of electric field is newton/coulomb (N/C) .
• The positive test charge q0 used in the definition is selected as very small, in
order not to disturb the distribution of the electric field in the environment.
In a more correct definition, it would be necessary to take the limit q0 → 0 ,
but we will not worry about this detail.
• According to this definition, the electric field ~E and the force vector ~F on
a charge q will have the same direction if the charge q is positive, and the
opposite direction if negative. An easy-to-remember rule is derived from
this:
Positive charges always try to go along with and negative charges always try to
go opposite to the electric field.
Electric Field of a Point Charge
As the simplest case, let us calculate the electric field produced by a point
charge q located at the origin, at a point with position vector ~r . According to
the definition, we place a positive test charge q0 at position ~r and examine the
force exerted upon it (Figure 14.14). If we write Coulomb’s law for the charges q
and q0 , the magnitude of the force is
qq0
F=k 2
r
From here, according to the definition of the electric field, we find that
F kq Figure 14.14: Electric field of a
E= = 2 point charge.
q0 r
The direction of this electric field depends on the sign of charge q (Figure 14.15):
If q is positive, it will be away from the origin, in other words, in the same
direction as ~r , as it shall repel the other positive test charge q0 . If q is negative,
it will attract the test charge, q0 , in other words, the force will be towards the
origin and in the direction of −~r .
Figure 14.15: Electric field

vectors are outwards from the
+ charge (along unit vector r̂ ), in-
wards towards the – charge (op-
posite to r̂ ).
It is possible to show both cases in a single expression. Remember the concept

of unit vector that we defined in Chapter 1: We defined the vector â as the unit
vector in the same direction as the vector ~a . Here, if we use r̂ to indicate the
unit vector along the position vector ~r , the electric field vector of a point charge
~E = kq r̂ (Electric field of a point charge) (14.8)

r2
This expression gives the correct direction for both a positive and a negative
charge q . If q is positive, ~E and ~r are in the same direction. If q is negative, ~E
is in opposite direction to ~r .
Electric Field of a System of Charges

Just as the Coulomb force of several charges could be written as the sum of
vectors, the same is also true for the electric field. The net electric field at any
point P due to N charges such as q1 , q2 , . . . qN will be the vector sum of the
electric field exerted by each one:
~EP = ~E1 + ~E2 + · · · + ~EN = ~Ei

X
If ~r1 ,~r2 . . .~rN are the position vectors of point P as measured from each charge,
Figure 14.16: The electric field and if we use the unit vectors along them, we get
of multiple point charges.
~EP = kq1 r̂1 + kq2 r̂2 + · · · + kqN r̂N = k
X qi
r̂i (14.9)
r12 r22 rN2 i
ri2
Electric Field Lines

Unfortunately, the electric field is not visible, and it is difficult to imagine
certain cases. Michael Faraday, who suggested the concept of the electric field,
developed the technique of electric field lines to make it easier to visualize
them.
We rewrite the electric field ~E produced by a positive charge q at any point
~r :
~E = kq r̂
r2
We would have to draw an infinite number of small arrows if we wanted to
indicate this electric field as a separate vector at each point in space. This would
be both difficult and impractical. Instead, let us consider the following method:
Figure 14.17: Rather than draw-

ing lots of arrows, it is more
meaningful to draw rays starting
from the charge.
Let us draw the electric field of a point charge at a few points by going
further away from the charge (Figure 14.17a). The lengths of these ~E vectors will
gradually get smaller. Now, we draw a single line connecting successive vectors
in a given direction, extending from the origin to infinity (Figure 14.17b). We do
the same thing in another direction. We thus obtain a bundle that spreads out
from the origin like light beams. We put an arrow indicating the direction of the
electric field on each of these lines.
This simple example shows how to draw and interpret the field lines in the
most general case:
• At any given point, the electric field vector ~E is tangent to the field lines. The
arrow on the field lines determines the direction towards which we will draw
the tangent.
• The magnitude of the electric field at any point is proportional to the density
of the field lines around that point. For example, the electric field of a point
charge decreases with distance from the origin; therefore, the field lines also
diverge from each other. In Figure 14.18, the electric field at point A is greater
than that at point B.
• The electric field will be towards the origin at each point if the point charge
at the origin is a negative. Therefore, electric field lines start at positive charges
and end at negative charges or at infinity. Figure 14.18: At which point
• Electric field lines never intersect. If the opposite was true, it would be as if is the magnitude of the electric
field greater?
there could be two tangents, in other words, two electric fields at that point.
Example 14.4
We substitute the values and calculate q :
1.52 × 10000
(a) The magnitude of the electric field is measured as q= = 2.5 × 10−6 C = 2.5 µC
10 000 N/C at a distance of 1.5 m from a point charge. 9 × 109
What is the charge? (b) The force acting on the charge qe in the field ~E is ~F=qe ~E .
(b) What is the force exerted on an electron placed at 1 mm If we write the expression for the electric field E of a point
distance from such a charge? (Electron charge: qe = − e charge, the magnitude of the force is as follows:
= − 1.6×10−19 C ). kq
F = |qe |E = e 2
r
Answer Substituting the value q found in item (a), the electron charge
(a) We write the expression for the electric field at distance r e and the distance r = 0.001 m , we calculate the force as fol-
from a point charge q and solve for q : lows:
9 × 109 × 2.5 × 10−6
kq r2 E F = 1.6 × 10−19 × = 3.6 × 10−9 N .
E= 2 → q= 0.0012
r k
Example 14.5
opposite to the field ~E , in other words, downward. We write
Newton’s second law:
qE
F = ma → qE = ma → a =
m
This acceleration will be downward. We substitute the values
and calculate the acceleration:
1.6 × 10−19 × 10
There is a vertically upward constant electric field E=10 N/C a= = 1.8 × 1012 m/s2
9.1 × 10−31
in a region. An electron is thrown with a horizontal velocity
(b) This is a projectile motion problem with the acceleration
of v0 =106 m/s at a height of h=2 m from the ground. (For the
g replaced by acceleration a . We find the range R by finding
electron, me =9.1 × 10−31 kg and qe = − e = −1.6 × 10−19 C .)
the time t from the vertical component of the motion and
Ignoring gravitational force,
using it in the horizontal motion:
(a) What is the acceleration of the electron? p
(b) Calculate the horizontal range of the electron. h = 12 at2 → t = 2h/a
p
Answer R = v0 t = vr
0 2h/a
The force acting on a charge q in a field ~E is ~F = q~E . As the R = 106 ×

2×2
= 1.5 m
charge of the electron is q = −e , the exerted force shall be 1.8 × 1012
Example 14.6
charges and inwards towards negative charges. Accordingly,
the fields ~E1 and ~E2 at points A and B are shown below:
Therefore, we first calculate the field magnitudes E1 and E2

Calculate the total electric field at points A and B due to
at these points and take their vector sum.
charges q1 and q2 shown in the figure.
At point A:
Answer kq1 (9 × 109 ) × (5 × 10−6 )
E1 = = = 5000 N/C
The electric field of a point charge is outward from positive r12 32
kq2 (9 × 109 ) × (2.5 × 10−6 ) kq1 (9 × 109 ) × (5 × 10−6 )
E2 = = = 900 N/C E1 = = = 450 N/C
r22 52 r12 (8 + 2)2
kq2 (9 × 109 ) × (2.5 × 10−6 )
We add the parallel electric fields at point A: E2 = 2 = = 5600 N/C
r2 22
E A = E1 + E2 = 5900 N/C (towards right).
We subtract the opposite electric fields at point B:
At point B: E B = E1 − E2 = −5150 N/C (towards left).
Example 14.7
the charges q1 and q2 at point P.
We first calculate the magnitudes of these fields separately:
kq1 9 × 109 × 5 × 10−6
E1 = = = 1800 N/C
r12 52
kq2 9 × 109 × 3 × 10−6
E2 = 2 = = 1690 N/C
r2 42
We then calculate the components of the vector ~E = ~E1 + ~E2 :
E x = E1 cos 53◦ = 1800 × 0.6 = 1080 N/C
Calculate the components of the total electric field due to charges Ey = E1 sin 53◦ − E2 = 1800 × 0.8 − 1690 = −250 N/C .
q1 = 5 µC and q2 = −3 µC at point P . According to this result, as the x -component is positive and
Answer the y -component is negative, the vector ~E is in the 4th quad-
The figure shows the electric fields ~E1 and ~E2 produced by rant of the plane.
Example 14.8
we can mutually decrease the charges at diagonally opposite
The figure shows the charges q, 2q, 3q and 4q respectively sides of point P . We get the following in the end:
placed on the corners of a square with side length a . Calculate
the electric field at point P in the center of the square.
Therefore, we can solve this simpler problem. The x -

component of the total electric field of equal 2q charges
at point P is zero. For the y -component, it is sufficient to
Answer
find one of the y -components and multiply it by 2:
This problem is a good example of the use of symmetry. It is k2q 1
not necessary to calculate the electric field of four separate E = Ey = 2E1y = 2E1 sin 45◦ = 2 √ · √
charges, because there is mutual symmetry. The electric field (a/ 2)2 2
√
is zero at the center of two diagonally opposite equal charges, E=
4 2kq
because they are equal and in opposite directions. Therefore, a2
Example 14.9
electric dipole. The properties of this neutral system is important
in many applications.
The electric field of the dipole plays an important role in the
structure of matter. The attraction between neutral molecules,
the operating principle of radio and TV antennas, the behavior
of dielectric materials, etc., all result from the electric properties
of dipoles.
(a) Find the expression for the total electric field at point P
located at distance r from the perpendicular bisector of
the charges.
Electric dipole. The system consisting of two equal and oppo- (b) Find the limit of the electric field very far away from the
site ±q charges with a small distance a in between is called an dipole, when r a .
Answer Accordingly, the expression for the electric field of a dipole is

(a) If we use ~E+ and E− to indicate the electric fields of the as follows:
charges ±q , the total electric field at point P will be the vector kp
E= 2
sum of these two: (r + a2 /4)3/2
~E = ~E+ + ~E−
The magnitude ofpthe electric fields of point charges ±q is (b) At very long distances, in other words, when r a , the
equal at distance r2 + (a/2)2 : term a2 in the denominator can be neglected with respect to
kq the other term. The limit field of the electric dipole is thus as
E+ = E− = 2 follows:
r + a2 /4
If we use the components of these two vectors to calculate E≈ 3
kp
(for r a)
the components of the total vector ~E , the x -components are r
equal and in opposite directions and cancel each other out The essential property of the dipole is that the dipole field de-
due to symmetry. Therefore, we only calculate and add the creases as 1/r3 , whereas the field of a point charge decreases
y -components: as 1/r2 .
E x = E+x + E−x = 0 The electric field of a dipole has the following distribution in
Ey = E+y + E−y = 2E+ cos θ space:
E = Ey
We find the cosine of the angle in the figure and substitute it
as follows:
a/2
cos θ = p
r2 + a2 /4
kqa
E= 2
(r + a2 /4)3/2
We define a new quantity called the dipole moment here:
p = qa (dipole moment) (14.10)
Electric Field of Continuous Charge Distributions

Although an electric charge consists of point charges at the atomic scale, the
small distances between them are indistinguishable at the macroscopic scale and
appear as continuously distributed. The integration technique must be used to
calculate the electric field produced by continuous charge distributions.
Let us divide a charge continuously distributed over a region into small ele-
ments with charges ∆q1 , ∆q2 , . . . , each with a very small dimension (Figure 14.19).
We can use Eq. (14.8), which we found for a point charge, to express the contribu-
tion of any one of these items to the electric field at point P . For example, the
contribution of element i is
k ∆qi
∆~Ei = 2 r̂i Figure 14.19: The ∆~E contri-
ri bution in the electric field by a
The electric field of the whole charge distribution will be approximately the vector small item ∆q .
sum of these small contributions:
X k ∆qi
~E ≈ ∆~Ei =
X
r̂i
i i
ri2
Then, the approximate expression becomes exactly correct at the limit ∆qi → 0
and the sum turns into an integral:
X k ∆qi
~E = lim r̂i
∆qi →0
i
ri2
Z
~E = k dq
r̂ (continuous charge distribution) (14.11)
r2
This vector integral should not be confused with the one-dimensional integral
with which we are familiar. The contribution of each charge item is in a different
direction and a separate integral is required for each of their components. We
shall see how this is performed in the worked examples below.
Charge Densities
Distribution of a continuous charge over a region can be expressed more
clearly using the concept of charge density. This will make it easier to incorporate
the charge element dq into the integral expression above.
Let us review possible charge distributions and their corresponding charge
densities (Figure 14.20):
Figure 14.20: Linear charge den-

sity λ , surface charge density σ
and volume charge density ρ .
• Linear charge density (λ) : If the total charge Q is evenly distributed over
a rod with length L , the linear charge density is
Q
λ= (unit: C/m) (14.12)
L
• Surface charge density (σ) : If the total charge Q is evenly distributed
over a surface A , the surface charge density is
Q
σ= (unit: C/m2 ) (14.13)
A
• Volume charge density (ρ) : If the total charge Q is evenly distributed in
a volume V , the volume charge density is
Q
ρ= (unit: C/m3 ) (14.14)
V
With the above definitions of charge densities, we can express the charge element
dq in terms of either line, surface or volume elements ( dL, dA, dV ), depending
on the given distribution, as will be seen in the worked examples below:
dq = λ dL dq = σ dA dq = ρ dV (14.15)
Example 14.10
Infinite line of charge. An infinite linear wire has uniform

linear charge density λ . Calculate the electric field at a distance
r from the wire.
Answer
Let us take the wire as the x -axis, as shown in the figure.
We write the contribution of a small charge element with
thickness dx located at a distance x , to the electric field at
cos θ dx
Z
point P . If the unit length has charge λ , then dx will have E = kλ
charge dq = λ dx : x2 + r 2
√
k dq kλ dx Substituting the cosine term as cos θ = r/ x2 + r2 and tak-
dE = 2 =
x + r2 x2 + r 2 ing into consideration that the charges are distributed over
The components of these small d~E contributions will be the the range [−∞, +∞] , we get
components of the total field. However, the E x components Z +∞
dx
will add to zero due to symmetry, because the mutual con- E = kλr
(x + r2 )3/2
2
tributions of points on both sides of the origin are opposite |−∞ {z }
along the x direction:
Z 2/r 2
Ex = dE x = 0 We find the result of this integral to be 2/r2 from the integral

Therefore, the field E is the integral of only the y - tables. Therefore, the electric field of an infinite line of charge
components: Z is as follows:
Z 2kλ
E = Ey = dEy = dE cos θ E=
r
Example 14.11 Q
dq = ds
2πR
Ring of charge. A total charge Q is evenly distributed over a kQ ds
thin wire bent into a ring with radius R . Calculate the electric dE =
2πR h + R2
2
field at a point P located at distance h on the axis of the ring.
As the piece of arc ds rotates around the circle, the perpen-
dicular components dE⊥ of these small contributions dE
will cancel each other out due to symmetry. However, the
dE x components shall always be added in the same direction.
Therefore, the total electric field will be the integral of these
dE x components:
Z
E⊥ = dE⊥ = 0
ds cos θ
Z Z Z
kQ
E= dE x = dE cos θ =
2πR h2 + R2 √
Answer If we substitute the cosine of the angle as cos θ=h/ h2 + R2
According to the coordinate system in the figure, let us con- and take all of the constants Zout of the integral, we get
sider a piece of arc with length ds on the circle. We write the kQ h
E= · 2 ds
contribution of the small charge dq on this arc to the electric 2πR (h + R2 )3/2
field at point P: The integral, in other words, the sum of the arcs ds around
k dq k dq the circle, will be the circumference 2πR of the circle. We
dE = 2 = 2 simplify and find the electric field of the ring:
r h + R2
If the circle with circumference 2πR has total charge Q , the E= 2
kQh
amount of charge on the arc ds is calculated by proportioning (h + R2 )3/2
and then substituted: The electric field will be perpendicular to the ring.
Example 14.12
ring in the previous example. Consider a ring at a radius r
with thickness dr in the small interval [r, r + dr] .
Charged disk. Calculate the electric field at a point P located
If we use dq to show the small amount of charge on this ring,
at distance h along the axis of a disk with radius R and with
according to the previous example, its contribution to the
charge Q distributed evenly along its surface.
electric charge at point P will be as follows:
kh dq
dE = 2
(h + r2 )3/2
In order to find the amount of charge dq , let us calculate the
surface area of the piece with thickness dr : If we cut this
ring out and spread, we get a long thin rectangle with width
dr and length 2πr . Its surface area will be approximately
2πr dr . If a disk with surface area πR2 has a total charge of
Answer Q , we find the amount of charge on this small surface and
We can solve this problem using the result for the charged substitute it as follows:
Q We define the variable u = r2 + h2 and calculate du :
dq = 2πr dr
πR2 u = r2 + h2 → du = 2r dr
kQh 2r dr
dE = 2 (along the axis) The integral simplifies with this variable u :
R (h + r2 )3/2
2 Z ! R
kQh kQh 2 kQh 2
As the contributions of these small rings are always in the E= 2 −3/2
du= 2 − √ = 2 − √

u
R R R r2 + h2 0

same direction, we can directly integrate to find the total elec- u
Substituting the limit values and simplifying, we find the
tric field, without any need to separate it into its components:
Z R electric field of the charged disk:
kQh 2r dr
E= 2 2 + r 2 )3/2 2kQ
"
h
#
R 0 (h E = 2 1− √
This can be calculated using the change of variable method. R h2 + R2
Example 14.13
charge density σ . If the disk’s total charge is Q and its
Infinite plane of charge. Calculate the electric field at dis- surface area is πR2 , its surface charge density will be
Q
tance h from an infinite plane that has a constant surface σ=
charge density σ . πR2
We solve this expression for Q and substitute. Also, let us
express the constant k as k =# 1/4πε0 :
σ
"
h
E= 1− √
2ε0 h + R2
2
Now, what will happen if the radius R of the disk goes to in-
finity with the surface charge density σ remaining constant?
No difference remains between the infinite disk and the infi-
Answer nite plane. Therefore, this expression gives the electric field
of the infinite plane at the limit R → ∞ . The second term
We can solve this problem using the result of the charged
inside of the brackets becomes zero when the limit R → ∞
disk in the previous example. Let us write the electric field
is taken. We thus find the electric field of the infinite plane
expression that we found above for a charged disk with radius
of charge:
R: σ
E=
" #
2kQ h
E = 2 1− √ 2ε0
R h2 + R2 Note that the electric field is independent of h and has the
First, let us write this electric field expression in terms of constant value σ/2ε0 everywhere.
1. Which of the following are correct for the electric
(a) The force on charge A is greater.
charge?
(b) The force on charge B is greater.
I. The smallest charge is 1 Coulomb.
(c) The forces are equal.
II. There is no smallest charge, it continuously increases
or decreases.
III. The smallest charge is the electron. 5. Which of the following are correct about the relation
IV. The charge varies as multiples of the electron charge. between the structure of the atom and conductivity?
(a) I & II (b) II & III (c) III & IV (d) I & IV I. The electrons at the outermost orbit are weakly bonded
in conductors.
2. Which of the following carries current in conductors? II. The electrons at the outermost orbit are strongly
(a) Electron (b) Proton (c) Ion (d) Neutron bonded in insulators.
III. The electrons at the innermost orbit are weakly
3. What is the direction of the force acting on an electron bonded in conductors.
placed at a point at which the electric field is towards IV. The electrons at the innermost orbit are weakly
the right? bonded in insulators.
(a) Right (b) Left (c) Up (d) Down (a) I & II (b) I & III (c) II & IV (d) I & IV
4. Which of the following is true if charge qA = +3 C and 6. Which of the following are correct?
charge qB = +2 C interact? I. A neutral object has no electrical charge.
II. A neutral object has equal number of + and - (a) It is proportional to r.

charges. (b) It is inversely proportional to r.
III. There are excess electrons in a positively charged (c) It is inversely proportional to r2 .
object. (d) It is inversely proportional to r3 .
IV. There is a shortfall of electrons in a positively
charged object. 14. How does the electric field of an infinite plane of charge
vary with distance?
(a) I & II (b) I & III (c) II & III (d) II & IV
(a) It is proportional to r.
(b) It is inversely proportional to r.
7. Why do insulators not conduct electricity well?
(c) It is inversely proportional to r2 .
(a) They have no electric charges. (d) It is independent of distance.
(b) Their electric charges are strongly bonded to the
nucleus of the atoms. 15. How does the electric field of an electric dipole change
(c) Their atoms are neutral. with distance?
(d) Their electric charges are weakly bonded to the (a) It is proportional to r.
nucleus of the atoms. (b) It is inversely proportional to r.
(c) It is inversely proportional to r2 .
8. If two charges are attracting each other with a force of (d) It is inversely proportional to r3 .
100 N , what will the force of attraction be if the distance
is doubled? 16. A charge Q placed at a corner of a square produces an
(a) 10 N (b) 25 N (c) 50 N (d) 75 N electric field E at the center of the square. What will
the electric field at the center be if equal charges Q are
9. Two equal Q charges are repelling each other with force placed at four corners of the square?
F . What will the force be if both charges are increased (a) 0 (b) E/4 (c) 4E (d) 16E
to 3Q ?
(a) 3F (b) 6F (c) 9F (d) 12F 17. An upward force is exerted upon an electron placed at a
point in space. In which direction is the electric field at
that point?
10. Which of the following is the definition of the electric
field? (a) Up (b) Down (c) Right (d) Left
(a) The force acting on a 1 C charge.
18. A metal sphere with charge +Q is connected to an-
(b) The force acting on one electron.
other neutral identical metal sphere by a conducting
(c) The energy of a unit charge.
wire. What will the charge of the second sphere be?
(d) The momentum of one electron.
(a) −Q/2 (b) −Q (c) +Q/2 (d) +Q
11. If the magnitude of the electric field is E0 at distance
R from a point charge, at what distance will it have the 19. Where is the positive (+) charge in the following figure?
value E0 /4 ? (a) A (b) B (c) C (d) Nowhere
(a) 2R (b) 4R (c) R/2 (d) R/4
12. What is the force exerted upon a 5 C charge placed at a

point at which the electric field is 3 N/C ?
(a) 3 N (b) 5 N (c) 8 N (d) 15 N
20. Where is the negative (−) charge in the above figure?
13. How does the electric field of an infinite line of charge (a) A (b) B (c) C (d) Nowhere
vary with the distance r from the wire?
Problems
14.2 Coulomb’s Law

14.1 Two small conducting spheres are placed with 30 cm
of distance in between. One has a +12 µC charge and the
other −6 µC . (a) What is force on one of the spheres? (b) The
Problem 14.7
spheres are connected with a conducting wire. What will the
14.7 A ball with a mass of 3 g and a charge of 4 µC is sus-
new force be? [A: (a) 7.2 N , (b) 0.9 N .]
pended from the ceiling with a rope as shown in the fig-
ure, in a region with a uniform horizontal electric field of
E=10 000 N/C . Calculate the angle of the rope with respect
to the vertical. [A: 53◦ .]
Problem 14.2
14.2 Calculate the total force exerted upon the charge
q3 =3 µC by the charges q1 =5 µC and q2 = − 4 µC shown
Problem 14.8
in the figure. [A: 420 N .]
14.8 The charges q1 =3 µC and q2 = − 8 µC in the figure
are tied to each other with a rope of length 1 m . These two
charges are placed in a region with a uniform electric field of
E=5 × 106 N/C and the charge q1 is nailed and fixed to its
location. Calculate the tension in the rope. [A: T = 40 N .]
Problem 14.3
14.3 (a) Calculate the components of the total force exerted
upon the charge q3 = + 3 µC by the charges q1 = + 5 µC and
q2 = − 2 µC in the figure. (b) Find its magnitude and direction.
[A: (a) F x =43, Fy = − 27 N , (b) F=51 N and −33◦ ]
Problem 14.9
14.9 There is a vertically downward constant electric field
E = 1000 N/C in a region. A proton is thrown with a hori-
Problem 14.4 zontal velocity of v0 =106 m/s at a height of h=1 m from the
14.4 The distance between the charges q1 = + 9 mC and ground. (a) What is the acceleration of the proton? (b) Cal-
q2 = − 4 mC shown in the figure is 2 m . Where should a third culate the horizontal range of the proton. ( m=1.7 × 10−27 kg
charge q3 be placed such that the net force exerted upon it and q = e = 1.6 × 10−19 C for the proton and ignore gravity.)
is zero? [A: 4 m to the right of q2 .] [A: (a) 9.4 × 1010 m/s2 , (b) R = 4.6 m .]
Problem 14.10
14.10 Calculate the total electric field of the charges q1 and
Problem 14.5
q2 shown in the figure at points A and B .
14.5 The two identical spheres shown in the figure with mass [A: (a) −1380 N/C , (b) +4430 N/C .]
m=30 g and charge q are suspended from the ceiling with
two ropes of length L=1 m . What is the charge of the spheres
if the ropes each have an angle of 37◦ with the vertical in
equilibrium? [A: |q| = 6 µC .]
14.3 Electric Field

14.6 A 3.2 × 10−15 N force acts on an electron placed at a Problem 14.11
point in space. What is the magnitude of the electric field at 14.11 Calculate the components of the total electric fields at
that point? (Electron charge: e = −1.6 × 10−19 C .) point P due to charges q1 = 5 µC and q2 = −2 µC .
[A: 20 000 N/C .] [A: E x = 315, Ey = 1080 N/C .]
PROBLEMS 255
As E=0 for h=0 and for h→∞ , the electric field must be at
a maximum at a point in between. At what distance h will
the electric field be at a maximum? (Hint: The first derivative
√
is zero at the maximum.) [A: h = R/ 2 .]
Problem 14.12
14.12 Two equal charges +q , with a small distance a in be-
tween are shown in the figure. (a) Find the expression for the
electric field at distance of r from the perpendicular bisector
of the charges. (b) What will the limit of the electric field be Problem 14.17
when r a ? [A: (a) 2kqr/(r2 + a2 /4)3/2 , (b) 2kq/r2 .]
14.17 Find the total electric field in the region between the
two parallel planes shown in the figure above with surface
density ±σ . (Hint: Use the result from Example 14.13.)
[A: σ/ε0 .]
Problem 14.13
14.13 What is the electric field at point P of the isosceles Problem 14.18
triangle shown in the figure? [A: 2kq/a2 .] 14.18 The figure shows two infinite wires placed in parallel
at a distance of 2a and with linear charge densities ±λ . Find
the total electric field at the point P in the center. (Hint: Use
the result from Example 14.10.) [A: 4kλ/a .]
Problem 14.14
14.14 The figure shows the charges at the corners of a regu- Problem 14.19
lar hexagon with side length a . Calculate the electric field at 14.19 The figure shows a finite line of charge with length a
point P in the center of the hexagon. and uniform linear charge density λ . Use integration to cal-
[A: E = 2kq/a2 to the right.] culate the electric field at point P located along the extension
of the wire at distance a . [A: E = kλ/2a .]
Problem 14.15
14.15 The regular pentagon shown in the figure with side
length a has equal charges q on all but one corner. Calculate Problem 14.20
the electric field at point P in the center of the pentagon.
14.20 The inner radius of the hollow disk shown in the figure
(Hint: Use symmetry.) [A: E = 1.38kq/a2 upwards.]
is a and its outer radius is b . The total charge Q is evenly
14.16 In example 14.11, the expression for the electric field distributed over the surface of the disk. Calculate the electric
of a charged ring with radius R , at distance h along its axis field at distance h along the disk axis. (Hint: Use the method
was found to be in example 14.12.)
kQh 2kQ h h
E= 2 [A: E = 2 √ − √ .]
(h + R2 )3/2 b − a2 a2 + h2 b2 + h2
15
GAUSS’S LAW
This natural phenomenon in the

north and south polar regions,
known as the northern lights or
aurora borealis, occurs as a re-
sult of the collision of charged
particles incoming from the sun
with the atoms in the high atmo-
sphere.
In the previous chapter, we learned how to calculate electric fields produced

by systems of point charges and by the continuous distribution of charges. These
involved vector sums and integrals. These integrals can sometimes be very
complex and difficult. However, there is a very powerful and elegant technique
for calculating electric field. Known as Gauss’s law, this method can be used in
symmetrical charge distributions and it yields results very quickly. Although it
may seem a bit abstract, learning this method will develop your ability to think
mathematically.
15.1 ELECTRIC FLUX (Φ)

Let us place a surface A on the path of the electric field lines in a region. If the
field ~E is parallel to the surface (Figure 15.1a), then no line will pass through this
surface. And, if we wish to have the maximum amount of field lines pass through,
we should orient the surface perpendicular to the field lines (Figure 15.1b).

258 15. GAUSS’S LAW
Figure 15.1: (a) Parallel surface,

(b) perpendicular surface and (c)
tilted surface to the electrical
field lines.
The number of these field lines that pass through the surface A will be propor-
tional to the electric field, the surface area and the angle all at once. Accordingly,
the electric flux indicated with the symbol Φ is defined as follows (Figure 15.2):
Φ = E A cos θ = E⊥ A (15.1)
Here, θ is the angle between the vector ~E and the normal vector n̂ . The normal
vector is the unit vector perpendicular to the surface at that point.
Figure 15.2: Electric flux. This definition effectively shows that: (a) When θ=0 (or, the field lines are
perpendicular to the surface), we have maximum flux Φ = EA , and (b) When
θ=90◦ (or, the field lines are parallel to the surface), we have Φ=0 . The coefficient
cos θ in this definition also shows that, if the flux of a vector passing through the
surface in one direction is positive, the flux of a vector passing in the opposite
direction ( θ=180◦ ) will be negative. In general, flux flowing outward from a
closed surface is considered to be positive and that flowing inward to be negative.
The definition above is for a constant field ~E . The flux of a variable electric
field passing through a surface of any shape is defined with an integral. In order
to set it up, we divide the surface A into infinitely small ∆Ai elements and use
the flux definition above for each element. The total flux is the limit sum of these
fluxes as ∆Ai → 0 , in other words, the integral:
X Z
Φ = lim Ei ∆Ai cos θi = E dA cos θ
∆Ai →0 surface
i
This is usually a double integral, as it must be taken over the entire surface. If the
surface A is closed, a circle is added to the integral sign to indicate it:
I
Φ= E dA cos θ (electric flux) (15.2)
surface
The field E and the angle θ inside of the integral are usually variable, and
therefore not taken outside of the integral.
15.2 GAUSS’S LAW

In order to understand the technique of Gauss’s law, let us consider a positive
point charge q at the origin (Figure 15.3). Let us draw electric field lines that
spread out from this charge. Let us calculate the total flux of these electric
field lines on an imaginary spherical surface that we draw with radius r . (This
imaginary surface is called a Gaussian surface.)
On this Gaussian surface, the field ~E has the same value everywhere and is
perpendicular to it (θ = 0) . Therefore, we can calculate the total flux without
Figure 15.3: A closed Gaussian having to integrate:
surface around a point charge. Φ = E A cos 0◦ = E A
15.2. GAUSS’S LAW 259
If we substitute E = kq/r2 for the electric field of a point charge r and A = 4πr2
for the surface area of a sphere, we get
kq
Φ=EA = 4π rS
S
2
rS2
q
E A = 4πk q =
ε0
In this last expression, we replaced the constant k = 1/4πε0 with the permittivity
of free space ε0 .
This result is rather remarkable. While both the electric field and the surface
area on the left-hand side are dependent on the distance r , on the right-hand
side, we get a result that is independent of the distance r and only proportional
to the charge q inside! Of course, we can see how this happens: As the field E is
inversely proportional to r2 and the surface area A is directly proportional to r2 ,
they cancel each other out in the product.
We can argue the case that this result will be valid for all closed surfaces and
for all charge distributions as follows:
Figure 15.4: (a) The charge is

not in the center of the Gaussian
surface, (b) the Gaussian surface
is not spherical, (c) the charge is
outside the Gaussian surface.
• The result would be the same if the charge q was not located at the center
of the sphere (Figure 15.4a). Although the field lines that pass through the
surface are more frequent in one place and sparser in another, the number of
lines that cross the surface is still the same.
• The result would be the same if there was any other closed surface around
q instead of a sphere (Figure 15.4b). The total number of lines crossing this
surface will still be equal to that crossing the sphere.
• The situation is different if the charge q is outside of the Gaussian surface
(Figure 15.4c). In this case, each field line entering the spherical surface will
come out somewhere else. Therefore, the sum of positive and negative fluxes
will be zero.
Z
Φ= E dA cos θ = 0 (Charge outside Gaussian surface)
In the light of this analysis, we can express Gauss’s law as follows without
proof:
Gauss’s Law
The net electric flux over any closed surface is proportional to
the net charge inside of the surface:
I
qinside
E dA cos θ = (15.3)
surface ε0
The charge qinside on the right-hand side of the equation is the algebraic sum
of the charges inside the Gaussian surface.
Let us emphasize the important points of Gauss’s law:
• Although the law is expressed as an integral, we will actually avoid it through
symmetry considerations. If the given charges are symmetrically distributed,
we choose such a Gaussian surface that the field E remains the same on this
surface, and can thus be taken outside of the integral. The remaining integral
will usually be much easier to calculate.
• Regardless of how much charge is present outside of the Gaussian surface, in
the end, only the net charge inside of the surface is included in the formula.
It may seem surprising that the left-hand side of the formula has the total
electric field and the right-hand side has the sum of the charges inside, but it
is correct.
• Selection of the Gaussian surface is arbitrary and any surface can be chosen.
However, the surface must be closed for the law to be valid. Otherwise, the
flux passing through the open part will not be counted.
15.3 APPLICATIONS OF THE GAUSS’S LAW

We now show the power of Gauss’s law in electric field calculations of various
charge distributions. In the following examples, we shall find formulas rather
than numerical answers. Students will thus gain the ability to make abstract
calculations.
Example 15.1
the surface: I
qinside
Calculate the total electric flux passing through the closed sur- Φ= E dA cos θ =
faces S 1 and S 2 shown in the figure, with q = 1 µC . surface ε0
Therefore, it is sufficient to calculate flux from the right-hand
side of this equation. The surface S 1 contains only the charge
q:
q 1 × 10−6
Φ1 = = = 1.1 × 105 N·m2 /C
ε0 8.85 × 10−12
Answer The surface S 2 contains both charges:
According to Gauss’s law, the electric flux passing through q + 3q
Φ2 = = 4.4 × 105 N·m2 /C
a closed surface is proportional only to the charge inside of ε0
Example 15.2
of the cube by integration, because the electric field varies
everywhere on this face. However, the problem becomes easy
if we consider symmetry. All six faces of the cube are equiv-
alent with respect to charge q . Therefore, the flux passing
through one face will be 1/6 of the total flux:
Φtotal
Φ1 =
6
Now, as the total surface is closed, we can find the flux using
Gauss’s law:
q q
The cube with side length a = 1 m shown in the figure contains Φtotal = inside =
ε0 ε0
the charge q = 3 µC . What is the electric flux passing through From here, we calculate the flux passing through one face:
one face of the cube? Φtop qinside 10−6
Φ1 = = =
Answer 6 6ε0 6 × 8.85 × 10−12
It would be difficult to calculate the electric flux on one face Φ1 = 1.8 × 10 N·m /C
4 2
15.3. APPLICATIONS OF THE GAUSS’S LAW 261
Example 15.3
The sum of small dA surface elements over the whole spher-
Charged spherical shell. A total of charge Q is evenly dis- ical surface will be the total area A of the Gaussian sphere:
tributed over a spherical surface with radius R . E A = qinside /ε0
The surface area of a sphere with radius r is A=4πr2 . The
(a) Find the electric field at distance r outside of the sphere.
qinside is the total charge Q . We substitute these values:
(b) Find the electric field at distance r inside of the sphere.
E 4πr2 = Q/ε0
Answer We find the solution by solving this expression for E :
(a) Let us draw an imaginary Gaussian surface with radius r Q kQ
outside of the sphere (r > R) (the blue surface in the figure E= = 2 (for r > R)
4πε0 r2 r
below): (b) This time, we choose the Gaussian surface with radius r
inside of the sphere ( r < R ) (the blue surface in the figure
below):
The electric field ~E at each point on this surface should be We again use the same thinking as in item (a): According to
perpendicular to the surface due to spherical symmetry. Like- symmetry, the field ~E should be perpendicular to the surface
wise, the electric field at each point on the surface should and have the same magnitude at very point of the surface.
have the same value because, again, according to spherical Accordingly, we can directly write Gauss’s law as the product
symmetry, we go from one point to another when we rotate E A instead of the integral:
the sphere. E A = qinside /ε0
Consequently, the electric field ~E should be equal and perpen- Now, what is the charge qinside inside of the Gaussian sur-
dicular to the surface at every point on the Gaussian surface. face? As all of the charges are left outside of the Gaussian
Let us write Eq.¨(15.3) for Gauss’s law: surface,
qinside = 0 → EA=0
I
q
E dA cos θ = inside Since the surface A is not equal to zero, we find that
surface ε0
E=0 (for r < R)
The field E inside of the integral can be taken outside, as it is Let us summarize the result:
constant over the surface. Also, as the angle with the normal
kQ/r2 (for r > R)
(
vector is θ = 0 , we get cos 0 = 1 :
◦ ◦
E =
I
q 0 (for r < R)
E dA = inside This result will be used later for conductors.
surface ε0
Example 15.4
sian surface with radius r such that r < R . Due to symmetry,
the electric field will be the same and perpendicular to the
surface at each point on this sphere.
A point charge +Q is placed at the center of a spherical shell

with radius R , containing evenly distributed charge −Q . Find
the electric field inside and outside of the sphere. We write the expression for Gauss’s law: E A = qinside /ε0
Answer Inside of the sphere: Let us draw a spherical Gaus- Only the point charge +Q is located inside of the Gaussian
sphere; the charge −Q on the sphere with radius R is not surface is the same and perpendicular to the surface at each
taken into account, as it is outside of the surface: point. (We do not need to draw the figure again.) We write
qinside = +Q the expression for Gauss’s law:
E A = qinside /ε0
E 4πr2 = +Q/ε0
This time, both the point charge +Q and the charge −Q on
From here, we find the electric field inside of the sphere:
the sphere with radius R are located inside of the Gaussian
E = kQ/r2 (for r < R) surface. Therefore, we get
Outside of the sphere: We draw a Gaussian surface with ra- qinside = +Q − Q = 0
dius r > R . Again, due to symmetry, the field ~E on this E 4πr2 = 0 → E=0 (for r > R)
Example 15.5
the wire with radius r and length L . The field ~E on the lat-
eral surface of this cylinder must be the same everywhere
Infinite line of charge. Calculate the electric field of an infi-
and perpendicular to the surface due to symmetry.
nite linear wire carrying uniform linear charge density λ .
On the base surfaces, the field ~E will be parallel to the surface,
in other words, do not cross it. Therefore, the flux is zero on
the base surfaces and not taken into consideration. Let us
write the expression for Gauss’s law:
E A = qinside /ε0
A is the lateral surface here and its area is
A = 2πr L
The charge inside of the cylinder with length L is
qinside = λ L
Answer We substitute these values:
λL
We solved this problem in Example 14.10 with a lengthy inte- E (2πrL) =
ε0
gration. Now let us see how easily it can be solved by Gauss’s We simplify and find the electric field as follows:
law. λ 2kλ
E= =
As the Gaussian surface, let us chose a coaxial cylinder around 2πε0 r r
Example 15.6
everywhere, due to symmetry, because a certain angle in any
Infinite plane of charge. Calculate the electric field of an direction would violate symmetry. Also, it will have the same
infinite plane carrying a uniform surface charge density σ . value everywhere on the base surface, because, as the plane is
infinite, the charge distribution does not change if any point
Answer at the base of the cylinder is shifted onto another point.
Again, we previously solved this problem in Example 14.13 ~E will be parallel to the lateral surface of the cylinder, hence
using integration. Now, let us see how easily it can be solved
the flux is zero on the lateral surface and only the flux on
using Gauss’s law.
the base surfaces are taken into consideration. The flux is
positive, as E is outwards in both bases.
Accordingly, we write the expression for Gauss’s law for both
bases as follows:
E A + E A = qinside /ε0
The amount of charge inside of the cylinder is found by using
the surface charge density σ :
qinside = σ A
We substitute these values and solve for the electric field:
σS A
2ES A=
ε0
σ
E= (Outward from the plane)
As the Gaussian surface, let us chose a cylinder extending 2ε0
equally to both sides of the surface by L and having a base ~E would have been towards the plane if its charge had been
area of A . The field ~E should be perpendicular to the plane negative.
15.3. APPLICATIONS OF THE GAUSS’S LAW 263
Example 15.7
negatively charged plate, the electric fields in between and
outside of the plates will be as follows:
Two infinite plates with charge densities +σ and −σ are placed

in parallel. Calculate the electric field in the three regions. As shown in the figure, we find the sum and difference of two
vectors with equal magnitudes:
Answer  σ σ
After finding the electric field of a single plate in the previous 

 − =0 (Outside of the plates)
 2ε0 2ε0
example, the electric field of two plates can easily be found E=


σ σ σ
as their vector sum. If we use ~E+ and E− to indicate the  2ε + 2ε = ε (In between the plates)




0 0 0
electric fields of the positive and negative plates, the electric
field magnitude of both plates will be The electric field between two equal and opposite charged
σ plates is σ/ε0 and no electric field flows over outside of the
E± = plates. This set up is frequently used in technology (tele-
2ε0
and will have the same value. However, as the field is outward visions, capacitors, etc.) to produce a region with uniform
from the positively charged plate and inward towards the electric field.
Example 15.8 Q kQ
E= 2
= 2 (r > R)
4πε0 r r
Charged solid sphere. A total charge Q is uniformly dis- This result shows that, outside of the sphere, the electric field
tributed in a spherical volume with radius R . Find the electric behaves as if all of the charges were located at the center.
field outside and inside of the sphere.
Inside of the sphere: Again, let us take a spherical Gaussian
Answer surface with radius r such that r < R . The field ~E on this
Outside of the sphere: Let us take a spherical Gaussian surface sphere will be the same and perpendicular to the surface:
with radius r such that r > R . By symmetry, the field ~E on
this sphere will be the same everywhere and perpendicular
to the surface. We write the expression for Gauss’s law:
E A = qinside /ε0
The charge left inside of the Gaussian surface is the whole
charge Q :
We write the expression for Gauss’s law:

E A = qinside /ε0
This time, some of the charges remain outside of the Gaus-
sian surface, and are thus not taken into consideration. The
charge inside the Gaussian surface is found using proportion.
If a sphere with radius R contains Q , a sphere with radius r
will contain
Q 4πr3 Qr3
qinside = = 3
4πR /3 3
3 R
We substitute these expressions and solve for E :
qinside = Q Qr3
E 4πr2 =
The surface area of a sphere with radius r is 4πr . We sub-
2
ε0 R3
stitute these expressions and solve for E : Qr kQ
E= = 3 r (r < R)
E 4πr2 = Q/ε0 4πε0 R3 R
Example 15.9
E A = qinside /ε0
Infinite cylinder. A cylinder with infinite length and radius Here, A=2πr L is the area of the lateral surface. The amount
R has uniform volume charge density ρ in its volume. Find the of charge inside of the cylinder is the charge inside of a vol-
electric field in the regions outside and inside of the cylinder. ume with length L and base radius R :
qinside = ρV = ρ (πR2 L)
We substitute these values:
ρ (πR2 L)
E (2πrL) =
ε0
We simplify and find the electric field as follows:
ρR2
Answer E= (r > R)
2ε0 r
Outside of the cylinder (r>R) , the problem is the same as the
infinite wire problem in Example 15.5. We take a cylinder Inside of the cylinder (r < R) , we again take a cylindrical
with length L and radius r as the Gaussian surface. Gaussian surface with radius r :
E A = qinside /ε0
This time, some charge is left outside of this surface and is
not taken into consideration. We calculate the charge inside
of the cylinder with length L and radius r :
qinside = ρV 0 = ρ (πr2 L)
We find the electric field using this expression for charge:
E (2πrL) = ρ πr2 L/ε0
ρr
We write the expression for Gauss’s law: E= (r < R)
2ε0
15.4 ELECTRIC FIELD IN CONDUCTORS

Electric fields inside of and around conductors have different properties due to
the free electrons in the conductors’ structures. Gauss’s law helps us understand
these properties.
Now let us review these properties:
1. Electric field is zero everywhere inside of a conductor in equilibrium.
Remember the atomic structure of conductors: A neutral piece of copper
contains an equal number of positive and negative charges. The positively
charged ions are stationary, but some of the electrons can freely move inside
of the conductor. Now, suppose that there exists a nonzero field inside of the
conductor ( ~E , 0 ). In such a case, a force ~F = q~E would be exerted upon the
electrons. The free electrons would thus start moving and would continue
moving until the opposite electric field produced by the electrons in their
new position makes ~E = 0 inside of the conductor.
Figure 15.5: The free charges A conductor placed in an external electric field will again ensure that ~E = 0
in a conductor placed in an ex- inside. As shown in Figure 15.5, electrons that are randomly distributed in
ternal electric field get reposi- the conductor start to gather in the direction opposite to the electric field
tioned to make ~E = 0 inside. under the action of the force ~F = q~E , and the opposite electric field that they
produce will cancel the external electric field inside of the conductor.
2. Any excess charge given to a conductor distributes itself at the sur-
face.
Let us remember Gauss’s law:
I
qinside
E dA cos θ =
surface ε0
15.4. ELECTRIC FIELD IN CONDUCTORS 265
As ~E=0 is always true inside of the conductor, the left-hand side of this
equation will be zero. Accordingly the inner charge on the right-hand side
should also be zero:
qinside = 0
We can expand this Gaussian surface and make it encompass the conductor’s
surface from the inside (Figure 15.6). Even in this case, qinside =0 must be Figure 15.6: The widest Gaus-
true, as the electric field is zero inside. Therefore, the only place where the sian surface inside of a conduc-
excess charge can reside is the surface of the conductor. tor.
But, what if there were a cavity inside of the conductor and a charge +q
was placed there (Figure 15.7)? What sort of equilibrium would occur in that
case? The conductor cannot move that charge to the surface. In order to
fulfill the condition ~E = 0 inside, an amount of −q charge will have to move
from the outer surface to its inner surface. In this way, we will get qinside = 0
for all Gaussian surfaces drawn inside of the conductor.
3. The electric field just outside of the surface of a conductor is always Figure 15.7: +q charge in a
cavity inside of a conductor.
perpendicular to the surface.
If the electric field was not perpendicular to the surface, it would have a
component tangent to the surface. The tangential component would exert
a force ~F=q~E and move the free electrons. As there is static equilibrium, it
means that there is no tangential field component acting on the electrons.
A conductor placed inside an external electric field (Figure 15.8) will position
its charges such that the electric field lines are perpendicular to the surface.
We will take these properties into consideration when applying Gauss’s law Figure 15.8: Electric field lines
to conductors. will always be perpendicular to
the surface.
Example 15.10
total of ~E = 0 inside of the conductor:
A thick infinite conducting slab is placed in parallel against a
thin infinite insulator plate carrying uniform surface charge
density σ1 . The conducting slab is neutral. (a) Calculate the
surface charge densities formed on the surfaces of the slab. (b)
Calculate the electric field between the plate and the slab.
We set the total electric field to zero by taking into considera-

tion the directions of these two vectors inside the conductor:
σ1 σ2
E = E1 + E2 = −
2ε0 ε0
From here, we find the charge σ2 :
σ1
σ2 =
2
Answer (b) In Example 15.6, we showed that the electric field is zero
(a) Equal and opposite ±σ2 surface densities occur on two at the outer region of two equal and opposite charge planes.
surfaces of the conducting slab placed opposite to the charged Therefore, in this problem, we shall only find the electric field
thin plate. In Examples 15.5 and 15.6, we showed that the elec- of the insulator plate at the region in between the conductor
tric field of a single plate is E1 =σ/2ε0 and the electric field and the insulator:
between two inversely charged plates is E2 =σ/ε0 . Therefore, σ1
E=
the charge density σ2 should be such that there should be a 2ε0
Example 15.11
According to this distribution, we apply Gauss’s law in each

A point charge +Q is placed at the center of a spherical neu-
region:
tral conducting shell with inner radius a and outer radius b .
Calculate the electric field in the three regions. We take the area of the chosen spherical Gaussian spheres as
A = 4πr2 and calculate:
Answer +Q kQ
For r < a : E (4πr2 ) = → E= 2
~
We have previously learned that E = 0 is always true inside ε0 r
of a conductor. Accordingly, it transfers −Q of its free elec- +Q − Q
For a < r < b : E (4πr2 ) = → E=0
trons from its outer surface to its inner surface in order to ε0
cancel out the effect of the central point charge inside of the +Q−Q+Q kQ
conductor and the following distribution occurs: For r > b : E (4πr2 ) = → E= 2
ε0 r
Problems
15.3 Applications of Gauss’s Law radius b . Calculate the electric field in the three regions.
[A: E=0 for r<a , 3kq/r2 for a>r>b , kq/r2 for r>b ]
Problem 15.1
15.1 Calculate the total electric fluxes passing through the
closed surfaces S 1 , S 2 and S 3 shown in the figure.
[A: Φ1 = 0 , Φ2 = 2q/ε0 , Φ3 = 6q/ε0 .]
Problem 15.4
15.4 There are two coaxial infinite cylindrical shells, as

shown in the figure. The one with radius a has a linear
charge density 3λ and the one with radius b has a charge
density −λ . Calculate the electric field in the three regions.
[A: E = 0 for r<a , 6kλ/r for a>r>b , 4kλ/r for r>b ]
Problem 15.2
15.2 The regular tetrahedron shown in the figure has side
length a , and a charge +q is placed at its center. What is the
electric flux passing through one face of the tetrahedron?
[A: q/4ε0 .]
Problem 15.5
15.5 Two infinite plates with charge densities +3σ and −2σ
Problem 15.3 are placed in parallel. Calculate the electric field in the three
15.3 There are two concentric spherical shells, as shown in regions.
the figure. A total charge 3q is distributed over the one with [A: σ/2ε0 on the left, 5σ/2ε0 in the middle, σ/2ε0 on the
radius a and a charge −2q is distributed over the one with right.]
PROBLEMS 267
faces of a thick conductor slab placed perpendicularly to a

uniform electric field ~E0 ? [A: σ = ±ε0 E0 .]
Problem 15.6
15.6 The spherical volume with radius a in the figure has a
uniform volume charge density +3ρ and the spherical shell
with inner radius a and outer radius b has a uniform volume
charge density −2ρ . Calculate the electric field in the three Problem 15.10
regions.
[A: E = ρr/ε0 for r < a , 5ρa3 /(3ε0 r2 ) − 2ρr/3ε0 for 15.10 A +3Q charge is given to a conducting spherical shell
a < r b .] with inner radius a and outer radius b . A point charge +2Q
is placed at the center of the cavity inside of the conductor.
(a) What is the charge at the inner and outer surfaces of the
conductor? (b) Calculate the electric field in the three regions.
[A: (a) −2Q on the inside, +5Q on the outside,
(b) 2kQ/r2 for r < a , 0 for a < r b .]
Problem 15.7
15.7 The infinite cylindrical shell with inner radius a and
outer radius b shown in the figure has a volume charge den-
sity ρ . Calculate the electric field in the three regions.
[A: E=0 for r < a , ρ(r2 − a2 )/2ε0 r for a < r b .]
2 2
Problem 15.11
15.11 An infinite wire carrying a linear charge density λ is
surrounded by an infinite cylindrical conducting shell with
inner radius a and outer radius b , as shown in the figure.
Calculate the electric field in the three regions.
[A: 2kλ/r for r < a , 0 for a < r b ]
Problem 15.8
15.8 An infinite wire carrying a linear charge density λ is
surrounded by an infinite conducting cylindrical shell with
inner radius a and outer radius b and carrying a volume
charge density ρ , as shown in the figure. Calculate the elec-
tric field in the three regions.
[A: 2kλ/r for, 2kλ/r+ρ(r2 −a2 )/(2ε0 r) for a < r b .]
15.4 Electric Field in Conductors
Problem 15.12
15.12 The thin planar layer shown in the figure has 3σ1
in surface charge density. A neutral thick conductor slab
is placed in parallel to it. (a) What will the surface charge
density σ2 of the conductor slab be? (b) What is the electric
Problem 15.9 field in the region in between?
15.9 What are the surface charge densities produced on both [A: (a) σ2 = ±3σ1 /2 , (b) 3σ1 /2ε0 .]
16
ELECTRIC POTENTIAL
A “hair raising” experience. This

metal sphere, called the van de
Graaff generator, is used to ob-
tain very high potentials up to 1
million volts with only a small
amount of charge. When you
touch the sphere, your body will
be charged and your hair will
stand on end.
What is the relation between
electric potential and potential
energy? These concepts will
help us better understand the na-
ture of electricity.
Up until now we have dealt with Coulomb’s force and electric field, both of
which are difficult to work with because they are vector quantities. Now, it is
time to adapt the concepts of work and energy that we developed in mechanics
into their electrical equivalents. Indeed, like the gravitational force, the electric
force is also conservative, hence we can define a potential energy for electric
forces as well.
The electric potential that we shall define is a scalar quantity, and thus much
easier to deal with and to solve related problems. Also, the electric potential will
help us to better understand conductors, capacitors, electric circuits, etc., and
other technological applications.

270 16. ELECTRIC POTENTIAL
16.1 ELECTRIC POTENTIAL

Electric Potential Energy
In Chapter 5, we defined the work performed by a variable force:
Z 2
W= ~F · d~r
1
Let us also remember the force on a charge q placed in an electric field ~E :

~F = q ~E
Therefore, the work performed by the electrical force when this charge q is
moved from one point to another will be
Z 2
W=q ~E · d~r
1
Again, as we discussed in Chapter 5, the work performed against conservative

forces is equal to the potential energy difference between the two points (Equation
5.12):
Z 2
− ~Fc · d~r = U2 − U1 (16.1)
1
It can easily be observed that the Coulomb force between two charges will also be
conserved: In Chapter 5, we showed that a potential energy could be defined for
the gravitational force F=GmME /r2 . Likewise, the Coulomb force F=kq1 q2 /r2
will also be conservative, as it is also inversely proportional to r2 .
Therefore, an electric potential energy can also be defined for the electrical
force as the work performed against the force ~F = q~E :
Z 2
U2 − U1 = −q ~E · d~r (Electric potential energy) (16.2)
1
We shall directly proceed to the concept of potential without discussing potential

Figure 16.1: The work per- energy, as we already examined it in detail in Mechanics.
formed against the electrical Electric Potential
force exerted upon a charge q
The electric potential energy of the point charge q defined above in Eq. (16.2)
in an electric field.
is proportional to q . Therefore, there is no use in keeping the charge q always in
the formulas. If we know the potential energy of the unit charge, we can easily
calculate the potential energy of any charge q at that point by just multiplying
them.
Definition: If a charge q has a potential energy U at a point in space, the
electric potential at that point will be:
U
V= (electric potential) (16.3)
q
Although the potential looks like the potential energy per unit charge, this simple
definition is somehow misleading: The potential at a point exists even when no
charge q is present there. Hence, we should stress that the potential is a property
of the electric field, and not of the charge q .
16.1. ELECTRIC POTENTIAL 271
Using this definition in reverse, the potential energy of a charge q placed at a

point with potential V will be
U = qV (16.4)
Therefore, if we divide Eq. (16.2) for potential energy difference by q , we will
obtain the expression for the potential difference between two points:
Z 2
V2 − V1 = − ~E · d~r (Electric potential difference) (16.5)
1
More precisely, the work performed against electric forces when moving the unit
charge from one point to another, is the potential difference between these two
points.
It is useful to remember the following two simple rules according to this
definition:
• Potential decreases when moving in the direction of electric field lines.
• Potential increases when approaching positive charges and decreases when
approaching negative charges.
For a small displacement ∆r , the small potential difference ∆V can be ex-
pressed without integration as:
∆V = −~E · ∆~r (16.6)
Unit of Potential
According to the definition above, the unit of potential is joule/coulomb (J/C)
and is called the volt, indicated with the symbol V:
1 joule/coulomb = 1 volt = 1 V
The unit of electric field can also be re-expressed in terms of Volts. We had
previously defined the unit of electric field as (N/C). However, if we compare the
units in the formula ∆V = E ∆r , the electric field unit will be,
newton
1 = 1 V/m
coulomb
This V/m unit is used in technology, as it is more practical. The term voltage is
also used instead of potential difference.
Potential of Constant Electric Field
In Chapter 15, we learned that a constant electric field forms between two
oppositely charged parallel plates. Let us place two such plates along the x -axis,
as shown in Figure 16.2. Let the electric field be in the −x direction. (The reason
that we place them so is that the potential will increase in the opposite direction
to the electric field and we want it to increase in the +x direction.) We use
Eq. (16.5) to find the potential difference between position x1 and position x2 in
this electric field. As E is constant and in the −x direction, we get
Z x2 Z x2 Figure 16.2: Potential differ-
V(x2 ) − V(x1 ) = − (−E) dx = E dx = E (x2 − x1 ) ence in a constant electric field.
x1 x1
If we take this integral from the origin x1 = 0 to the point x2 = x , we get
V(x) − V(0) = Ex
The zero reference point of potential is always arbitrary. If we choose the negative
plate at x = 0 as the zero potential, the potential change between the plates is
found to be as follows:
V(x) = Ex (potential of a constant electric field) (16.7)
If the distance between the plates is d , the potential difference between the plates
will be V = Ed .
Example 16.1
energy U :
(a) A 2 µC charge gains 0.006 J potential energy when placed V = U/q = 0.006/(2 × 10−6 ) = 3000 V
in an electric field. What is the potential at that point? (b) The potential energy of the proton at a point with potential
(b) What will the velocity of a proton be if you convert all of its V will be eV . This is fully converted into kinetic energy:
potential energy at a point with potential 1000 V into kinetic q
1
m v 2
= eV → v = 2eV/m p
energy? (The proton’s mass is m p = 1.7 × 10−27 kg and the 2 p
charge is e = 1.6 × 10−19 C ) We substitute the given values and calculate the velocity:
r
Answer 2 × 1.6 × 10−19 × 1000
v= = 4.3 × 105 m/s
(a) We write the definition of potential in terms of potential 1.7 × 10−27
Example 16.2
V = Ex
Two parallel conducting plates with 5 mm of distance in be- We substitute the position x = 0.005 m of the positive
tween are connected to the terminals of a 20 V battery. (a) charged plate and find the electric field:
What is the magnitude of the electric field between the plates? E = V/x = 20/0.005 = 4 000V/m
(b) How much charge accumulates on one of the plates if its (b) We can assume the plates to be infinite if their dimensions
surface area is 100 cm2 ? are large with respect to the distance in between. Then, the
electric field is given by:
σ
E=
ε0
We write the surface charge density of a plate with charge Q
and surface area A :
Q
σ=
A
(Q/A)
E= → Q = ε0 EA
ε0
Answer We find the charge of a plate by substituting the numerical
(a) We know that a constant electric field E is formed be- data:
tween two oppositely charged plates. Also, we had found Q = 8.85 × 10−12 × 4000 × 100 × 10−4
Eq. (16.7) for the potential change in a constant electric field: Q = 35 × 10−11 C = 0.35 nC
16.2 POTENTIAL OF A SYSTEM OF POINT CHARGES

Potential of a Point Charge
Let us rewrite the electric field expression of a positive charge Q located at
the origin:
kQ
E= 2
r
The electric field will be outwards, as the charge Q is positive (Figure 16.3). We
use Eq. 16.5 to calculate the potential difference between position r1 and position
16.2. SYSTEM OF POINT CHARGES 273
r2 in this electric field. As the field E and the displacement dr are in the same
direction here, the scalar product can be written as the product E dr :
Z r2
V(r2 ) − V(r1 ) = − E dr
r1
Z r2 Z r2 r Figure 16.3: Potential differ-

V(r2 ) − V(r1 ) = −
kQ
dr = −kQ
dr
= −kQ − 1 2 ence of a point charge.
2 r2 r r
r1 r r1 1
!
1 1
= kQ −
r2 r1
If we take this integral from r1 = ∞ to r2 = r , we get
kQ
V(r) − V(∞) =
r
We choose the zero reference point of the potential at infinity, V(∞) = 0 . With
this choice, the potential for a point charge is as follows:
kQ
V(r) = (Potential of point charge) (16.8)
r
In other words, the potential of a point charge is the work performed to bring it
from infinity to that point.
Figure 16.4: Potential increases

towards a positive charge and
decreases towards a negative
charge.
According to this result, the potential of a positive charge is positive every-

where and the potential of a negative charge is negative. Potential increases
towards a positive charge and decreases towards a negative charge.
Potential of a System of Point Charges
When calculating the electric field of several charges, we have seen that the
electric fields of the charges were added as vectors. However, the potential of a
system of point charges at a point P will be the algebraic sum of the potential of
each charge.
We can write the potential produced by charges q1 , q2 . . . qN at a point P
using Eq. (16.8), which we found above for a point charge:
Figure 16.5: Potential of multi-
kq1 kq2 kqN X kqi ple charges at point P .
V= + + ··· + = (16.9)
r1 r2 rN i
ri
The distance ri here is the distance of charge qi to the point P (Figure 16.5).
Potential Energy of a System of Point Charges

Let us consider a system in which many charges are located at certain fixed
positions, as shown in Figure 16.6. What is the potential energy of this system?
In other words, how much work was done to bring the charges from infinity to
these positions?
Let us find the answer to this question step-by-step. Let us first bring the
charge q1 from infinity. No work needs to be done for this, therefore its potential
energy is zero:
U1 = 0
Then, we bring charge q2 . It will be subject to the potential of charge q1 . As the
Figure 16.6: Potential energy potential is V = kq1 /r , if we use the formula U = qV to calculate the potential
of three charges. energy of the charge q2 placed at distance r12 , we get
kq1 q1 q2
U2 = q2 V1 = q2 =k
r12 r12
and this potential energy is the energy of the (q1 , q2 ) charge system. It would
be incorrect to consider this as belonging to q2 , as we could have first brought
charge q2 and then charge q1 .
Now let us bring charge q3 to its position. This charge will be subject to the
potential of the charges (q1 , q2 ) . Its potential energy at its position is
! !
kq1 kq2 q1 q3 q2 q3
U 3 = q3 + =k +
r13 r23 r13 r23
Now let us stop here and examine the potential energy of the whole system:
!
q1 q2 q1 q3 q2 q3
U = U1 + U2 + U3 = k + + (16.10)
r12 r13 r23
This result can easily be generalized for more than three charges: We find the
interaction of each charge with the others and add them. The total energy of N
charges is expressed as follows:
X qi q j
U=k (16.11)
i< j
ri j
This potential energy is due to the work performed in bringing these charges
from infinity to these positions.
Law of Energy Conservation in Electrostatics
The law of energy conservation that we developed in Chapter 5 is also valid
in electrostatics. Since the potential energy of a point charge q is U = qV in a
given potential V , Eq. (5.17) for energy conservation can be written as follows:
1 2
2 mv1 + qV1 = 12 mv22 + qV2 (16.12)
However, unlike gravitational potential energy, the charge q can have positive
or negative contribution to potential energy depending on its sign, as we shall
see in the examples below.
16.2. SYSTEM OF POINT CHARGES 275
Example 16.3
potentials of all three charges at that point:
kq1 kq2 kq3
VA = + +
r1 r2 r3
−6
5 × 10−6 −2 × 10−6
!
1 × 10
VA = 9 × 10 9
+ √ +
1 2 1
VA = 22 800 V .
(b) Another definition of potential energy is the work per-
The charges q1 =1 µC , q2 =5 µC and q3 = − 2 µC are placed on
formed to move a charge q from infinity to point A. Therefore,
three corners of a square with side length 1 m . we can write the potential energy of charge q4 at point A in
(a) What will the potential at point A be? terms of the potential at that point as follows:
(b) How much work needs to be done to bring a new charge U A = q4 VA
q4 = +4 µC from infinity to point A ? We substitute the values and calculate as follows:
Answer (a) The potential of point A is the algebraic sum of U A = 4 × 10−6 × 22800 = 0.09 J
Example 16.4
As charge q3 is released from rest at point A , we take vA = 0
and solve for vr
B:
2q3 (VA − VB )
vB =
m
Let us calculate the potential differences separately:
1 × 10−6 2 × 10−6
!
The charges q1 =1 µC and q2 = − 2 µC shown in the figure with VA =
kq1 kq2
+ = 9 × 109 × −
4 m of distance in between are fixed in their current locations. r1 r2 1 3
The charge q3 =3 µC with a mass of 1g is released from rest = 3000 V
1 × 10−6 2 × 10−6
!
between these two charges at point A . What will its speed be kq1 kq2
VB = + = 9 × 10 ×
9
−
when it reaches point B ? r1 r2 3 1
= −15 000 V
Answer
From here, we calculate the speed vB :
A charge q at a point with potential V has a potential energy r
U = qV . We thus write the conservation of energy for points 2 × 3 × 10−6 (3000 + 15000)
vB =
A and B : 0.001
vB = 10 m/s .
2 mvA + q3 VA = 2 mv B + q3 V B
1 2 1 2
Example 16.5
Answer
The charge q will have zero speed at some point B that it
approaches most: vB =0 . Let us use b to show the distance of
this point to the origin. We write the conservation of energy
between points A and B :
Charges q1 = − 5 µC and q2 = + 2 µC are fixed with 1 m of dis- ˜ 2 mvA + qVA = 2 mv B + qV B
1 2 1 2
q1 q2 q1 q2
tance in between. A third charge q=1 µC with a mass of 1 g is

1
2 mv 2
A + kq + = 0 + kq +
thrown with an initial velocity of 3 m/s towards q2 from point 4 4−1 b b−1
A located along the extension of these charges at a distance of If we substitute the given values and solve for b , we get the
3 m from q2 . Calculate the maximum distance that charge q following result:
can approach. b = 1.7 m
Example 16.6
Electric dipole. Calculate the potential of an electric dipole at
any point in space and find its limit value for r a .
Answer
We had found the electric field of an electric dipole in Example
14.8. Now, let us calculate its potential.
The charges ±q in the figure are placed along the y -axis with
distance a in between. We can immediately write the electric
potential of this dipole at a point P located at distance r from

the charge +q and with angle θ with the y -axis:
kq kq kq ∆r
V= − =
r r + ∆r r (r + ∆r)
Here, we indicate the distance of point P to charge +q with
r and to charge −q with r + ∆r . The potential can be found
by calculating this expression for any given point P .
We use the following approximate expressions for r a : The equipotential surfaces of the electric dipole on the
∆r ≈ a cos θ xy -plane are shown in the figure above. The most important
r(r + ∆r) ≈ r2 feature of this potential is that it decreases with r2 . The po-
Also, the product p = q a was defined as the electric dipole tential of a point charge decreases with r and the potential of
moment. Accordingly, the expression for potential at dis- a dipole decreases much faster. The electric field lines of the
tances very far from the dipole is as follows: dipole are shown in the same figure from the positive charge
p cos θ to the negative charge, as perpendicularly intersecting the
V=k equipotential surfaces.
r2
16.3 POTENTIAL OF CONTINUOUS CHARGE DISTRIBUTIONS

Integration is used to calculate the potential of charges distributed over a
volume, surface or line. As shown in Figure 16.7, if we consider a small charge
element dq in the region where charge is distributed, its small contribution dV
to the total potential will be like that of a point charge:
k dq
dV =
r
The potential of the whole charge distribution will be the sum of these small
contributions at the limit dq → 0 , in other words, their integral:
Figure 16.7: The contribution Z
of a small charge dq on the po- dq
V=k (Potential of a continuously distributed charge) (16.13)
tential at point P . r
The charge element dq here is to be expressed in terms of charge density, by
examining the given charge distribution. It can be one of the following three
expressions if distributed over a line, surface or volume:
dq = λ dL dq = σ dA dq = ρ dV (16.14)
Also, note that the definition of potential as the integral of electric field ~E (Eq. 16.5)
can also be used to calculate V :
Z 2
V2 − V1 = − ~E · d~r (16.15)
1
Example 16.7
with linear charge density λ .
Answer
It is easier to use the definition in terms of electric field:
Z b Z b
Vb − Va = − ~E · d~r = − E dr
a a
Since ~E and d~r are parallel, we have ~E·d~r=E dr . The electric
Infinite line of charge. Find the potential difference between field of an infinite line of charge was found to be E=2kλ/r
two points located at distances a and b from an infinite wire in Example 15.5. We can use that result and integrate:
16.4. CONDUCTORS AND EQUIPOTENTIAL SURFACES 277
Z b
2kλ dr As b > a , this result shows that the potential decreases when
b
Vb − Va = − = −2kλ ln r = −2kλ ln(b/a)
a r a moving away from a positively charged wire.
Vb − Va = −2kλ ln(b/a)
Example 16.8
ds on the ring. We write its contribution to the potential at
point P and add the contributions of ds all over the ring, in
other words, take its integral:
Z Z
dq dq
V=k =k √
r R2 + h2
The denominator inside of the integral can be taken outside,
as it is constant throughout
Z the ring:
Charged ring. Calculate the potential along the axis at a dis- k
V= √ dq
tance h from the center of a ring with radius R carrying a total R2 + h2
charge Q . The remaining integral is the sum of all of the charges Q :
kQ
Answer V= √
Let us choose a small charge dq corresponding to a small arc R2 + h2
Example 16.9
In order to find the charge dq , we have to multiply the area
Charged disk. Calculate the potential of a disk with radius R of the small ring with the surface charge density:
and uniform surface charge density σ , at point P located at dq = σdA = σ (2πr dr)
distance h on the axis of the disk. We substitute this value and find the contribution of all the
rings by integrating:
Z R
2r dr
V = πkσ √
0 r 2 + h2
We make a change of variable by defining a new variable
u=r2 + h2 . Then, du=2r dr . The integral is simplified with
Answer this variable uZ:
√
Consider a ring with thickness dr with a radius in the inter- R
V = πkσ u−1/2 du = πk 2u1/2 = πkσ 2 r2 + h2
val [r, r + dr] . If we use dq to indicate the small amount of 0
charge on this ring, according to the previous example, its Substituting the limit values and simplifying, we find the
contribution to the potential at point P will be as follows: expression for the potential of the disk:
k dq h√
dV = √ V = 2πkσ h2 + R2 − h
i
r 2 + h2
16.4 CONDUCTORS AND EQUIPOTENTIAL SURFACES

Surfaces on which the potential has the same value are called equipotential
surfaces. This is similar to the isotherms used in meteorology. For example, let
us examine the expression for the potential of a charge q at the origin:
kq
V=
r
As shown in Figure 16.8, the equipotential surfaces of a point charge are spherical
surfaces centered around the charge q . Likewise, the equipotential surfaces of an Figure 16.8: The equipotential
infinite line of charge are cylindrical surfaces with the line as their axis. surfaces around a point charge
Let us emphasize the important features of the equipotential surfaces. are spherical.
• Conductor surfaces are equipotential surfaces. In order to show this,
let us write Eq. (16.5) for potential difference:
Z 2
V2 − V1 = − ~E · d~r
1
Let points 1 and 2 be located on the conductor’s surface (Figure 16.9). As

potential difference is independent of the path taken, let us choose the path
that connects these two points through the inside of the conductor. However,
we had previously seen that ~E = 0 is always true inside of a conductor.
Therefore, the right hand-side of this integral is zero:
Figure 16.9: Two points on a
conductor surface. ~E = 0 =⇒ V2 = V1
In other words, no work is performed to take a charge q from one point to

another on a conductor’s surface.
• Electric field lines are always perpendicular to equipotential sur-
faces. For a small displacement d~r the potential difference was given by
Eq. (16.6): dV= − ~E·d~r . As two points very close to each other by d~r will
have the same potential on the equipotential surface (Figure 16.10), dV=0
and we get
dV = 0 =⇒ −~E · d~r = 0 =⇒ ~E ⊥ d~r
We had previously proven with Gauss’s law that electric field is perpendicular
Figure 16.10: A small displace- to a conductor surface.
ment d~r on a conductor surface.
• Electric field as the gradient of potential. Let us consider that we take a
small step d~r from an equipotential surface with potential V , in the perpen-
dicular direction (in other words, towards ~E ). As the vectors ~E and d~r are
in the same direction, the scalar product turns into a simple product:
dV = −E dr
From here we get the relation between electric field and potential:
dV
E=− (16.16)
dr
The rate of potential increase perpendicular to the equipotential surface
is called the potential gradient. Temperature gradient is mentioned in
meteorology, and its meaning is likewise the temperature increase per unit
length in the direction perpendicular to the isotherm. Therefore, electric
field is the negative potential gradient. The negative sign tells us that
the potential decreases in the direction of the electric field.
Example 16.10
of the electric field. We write the formula (16.5):
Z 2
V2 − V1 = − ~E · d~r
1
If we take r1 = ∞ in this formula, we get V1 = 0 , and the
potential at point r is written as V2 = V(r) :
Z r Z ∞ Z ∞
V(r) = − ~E · d~r = ~E · d~r = E dr
∞ r r
Conducting sphere. Calculate the potential outside and inside (We reversed the limits of the integral and used ~E · d~r=E dr ,
0f a conducting sphere with radius R and charge Q . as ~E and d~r are in the same direction.)
Answer Outside of the conducting sphere (r > R) : the electric field
It is easier in this problem to calculate potential as the integral is the same as the field E = kQ/r2 of a point charge:
Z ∞
kQ dr kQ ∞ kQ

V(r) = = −
r r V(r) = (r < R)
r r2 R
The potential inside of the conductor is constant and equal
kQ to its value at the surface.
V(r) = (r > R)
r Let us look at a graphic of the potential of a conducting
Inside of the conducting sphere (r < R) : Let us write the sphere:
integral in two parts for a point r :
Z ∞ Z R Z ∞
V(r) = E dr = E dr + E dr
r r R
As E = 0 is always true inside of the conductor, the first

integral will be zero. The other is the value that the potential
outside of the conductor takes at the surface r = R :
Example 16.11
will be equal.
As the spheres are far away from each other, the potential on
the surface of each sphere only results from its own charge.
Let us denote q01 and q02 as the new charges on the spheres.
These new charges should fulfill two conditions:
Two conducting spheres are located far away from each other. Charge conservation: q01 + q02 = q1 + q2 = 1 + 8
The one with a radius of 1 m is charged with q1 =1 µC and q01 + q02 = 9 µC (1)
the other with a radius of 2 m is charged with q2 =8 µC . These 0
kq1 kq2 0
q0 q0
two spheres are connected with a metal wire. What is the new Equipotential: = → 1 = 2 (2)
r1 r2 1 2
charge of each sphere?
Answer After the spheres are connected with a wire, they Solving (1) and (2), we find the new charges:
all constitute one single conductor, and thus their potential q01 = 3 µC , q02 = 6 µC
Example 16.12
Answer (a) The potential due to the outer conducting sphere
will be constant inside of it and equal to the value on its
surface, and it will not contribute to the potential difference
between a and b . Hence, the potential difference between
the spheres will be the potential difference due to the inner
sphere charged with +Q at distances a and b :
1 1
Va − Vb = kQ −
a b
1 1
Two concentric conducting spheres have radii a=1 m and Va − Vb = 9 × 109−6 − = 4500 V
1 2
b=2 m , with charges Q= + 1 µC and −Q , respectively. (b) We write the energy conservation law for charge q :
(a) Calculate the potential difference between the two spheres.
1
2 mv
2
+ qVA = 21 mv2B + qVB
rA
(b) A q=2 µC point charge with a mass of 1 g is released
r
2q(VA − VB ) 2 × 2 × 10−6 × 4500
from rest near the positively charged sphere. What will its v= =
m 0.001
speed be when it reaches the negatively charged sphere? v = 4.2 m/s
1. If a 2 C charge gains 10 J in potential energy when
placed at a point, what is the potential of that point? 3. Of three conducting spheres, sphere A has a radius of
1 m and is charged with 1 µC , sphere B has a radius
(a) 5 V (b) 10 V (c) 20 V (d) 40 V
of 2 m and is charged with 3 µC and sphere C has a
2. If the distance between two parallel conducting plates is radius of 3 m and is charged with 6 µC . Which sphere
2 m and their potential difference is 10 V , what is the has greater potential?
electric field between the plates? (a) A (b) B (c) C (d) Equal
(a) 5 V/m (b) 10 V/m (c) 20 V/m (d) 40 V/m
4. Of two conducting spheres with equal potentials, if the
one with a radius of 1 m has a charge of 4 µC , then (a) A (b) B (c) C (d) Equal
what is the charge on the one with a radius of 2 m ?
(a) 1 µC (b) 2 µC (c) 4 µC (d) 8 µC 13. What is the electric field in a region where potential is
constant?
5. Which of the following are correct? (a) 0
I. Electrons are attracted towards higher potential. (b) Constant.
II. Protons are attracted towards higher potential. (c) Increasing linearly.
III. Electrons are attracted towards lower potential. (d) Decreasing linearly.
IV. Protons are attracted towards lower potential.
14. What is the potential in a region where electric field is
(a) I & II (b) I & III (c) II & III (d) I & IV
constant?
6. Which of the following is true at the middle point be- (a) 0
tween point charges +Q and −Q with distance r in (b) Constant
between? (c) Increasing linearly.
(d) Infinite.
(a) E = 0 (b) V = 0 (c) E=kQ/r2 (d) V = 2kQ/r
15. If the radius of a conducting sphere is doubled and its
7. Which of the following complies with the definition of
charge is increased by a factor of 4, by what factor will
potential?
its potential increase?
I. The potential energy of a unit charge.
II. The work performed to bring a unit charge from (a) 2 (b) 4 (c) 1/2 (d) Equal
infinity to that point.
III. The product of electric field and charge. 16. What is the total potential produced at point P located
IV. The electric field of a unit charge. at the center of the square by the charges shown on the
corners of the square in the figure below?
(a) I & II (b) I & III (c) II & III (d) I & IV
8. Which of the following are correct?

I. The potential is zero inside of a conductor.
II. The potential is constant inside of a conductor.
III. The potential inside of a conductor is equal to that
(a) 0 (b) 3kQ/a (c) 5kQ/a (d) 10kQ/a
on the surface.
IV. The potential is infinity inside of a conductor.
17. Which regions of two concentric conducting spheres
(a) I & II (b) I & III (c) II & III (d) I & IV charged with ±Q have constant potential?
9. Which of the following is the potential energy of a

charge q at a place where potential is V ?
(a) V/q (b) qV (c) q2 V (d) qV/2

I. Potential increases towards a positive charge. (a) A & C (b) A & B (c) Only A (d) Only C
II. Potential increases towards a negative charge.
18. Which of the following are correct for equipotential
III. Potential decreases towards a positive charge.
surfaces?
IV. Potential decreases towards a negative charge.
I. Conductors are equipotential surfaces.
(a) I & II (b) I & III (c) II & IV (d) I & IV II. Insulators are equipotential surfaces.
III. The electric field is perpendicular to the equipoten-
11. When a +1 C charge moves freely from a point at which tial surface.
the potential is 50 V to a point at which the potential is IV. The electric field is tangent to the equipotential
20 V , how much does its kinetic energy change? surface.
(a) 0 (b) −30 J (c) +30 J (d) +70 J
(a) I & II (b) I & III (c) II & IV (d) I & IV
12. Which of the following systems consisting of two
19. Which is the expression for the potential of a point
charges has a greater potential energy?
charge?
A: q1 =q2 =1 C and separated by 1 m ,
B: q1 =q2 =2 C and separated by 4 m , (a) kq/r2 (b) kq/r (c) kq2 /r2 (d) kq2 /r
C: q1 =q2 =3 C and separated by 3 m .
PROBLEMS 281
20. Which of the following figures is the potential of a

charged conducting sphere?
(a) A (b) B (c) C (d) D
Problems
16.1 Electric Potential 16.8 Two identical charges with masses m1 =m2 =1 g and
charges q1 =q2 =5 µC are released from rest with 2 m of dis-
16.1 (a) How much potential energy will a 3 µC charge have tance in between. What will their speeds be when the distance
when placed at a point with a potential of 500 kV ? (b) An is doubled? [A: 7.5 m/s .]
electron at rest at infinity is accelerated towards a metal plate
and hits it with a speed of 6 × 106 m/s . What is the potential
of the plate? (The electron’s mass is me = 9.1 × 10−31 kg and
the charge is −e = −1.6 × 10−19 C .)
[A: (a) 1.5 J , (b) 102 V .] Problem 16.9
16.9 The charges q1 = − 1 µC and q2 = + 2 µC in the figure
16.2 Two identical conducting plates are placed parallel to above are fixed with 2 m of distance in between. A third
each other with 2 mm distance in between and are connected charge q=3 µC with a mass of 1 g is released from rest at
to the terminals of a 24 V battery. (a) What is the magnitude point A located along the extension of these charges at a
of the electric field between the plates? (b) If the surface area distance of 1 m from q2 . What will be its speed at infinity?
of one plate is 100 cm2 , how much charge accumulates on [A: 9.5 m/s .]
each plate? [A: (a) 12 kV/m , (b) 1.1 nC .]
16.2 Potential of a System of Point Charges

16.3 At what distance from a point charge q=5 nC will the
potential be 30 V ? [A: 1.5 m .] Problem 16.10
16.10 The charges q1 =5 µC and q2 = − 8 µC shown in the
figure with 3 m of distance in between are fixed in their cur-
16.4 The potential is 100 V and the electric field is 80 V/m
rent locations. A third charge q3 =1 µC with mass m3 =1g
at a certain distance from a point charge. Find the amount of
is thrown from point A between these two charges with a
charge and the distance. [A: r = 1.25 m , q = 140 µC .]
speed of 10 m/s towards q1 . What is the nearest distance at
which it can approach q1 ? [A: 1.1 m .]
16.3 Potential of Continuous Charge Distribu-
tions
16.11 The potential is 400 V on the surface of a raindrop
Problem 16.5 with a radius of 1 mm . (a) What is the charge of the drop?
(b) When two such drops merge, what will the potential at
16.5 The charges q1 = − 1 µC , q2 =2 µC and q3 =3 µC are
the surface of the new drop be?
placed on three corners of a square with side length of 1 m .
[A: (a) 4.4 × 10−11 C , (b) 635 V .]
(a) What will the potential be at point A ? (b) How much
work needs to be done to bring a charge q4 = +4 µC from
infinity to point A ? [A: (a) 39 kV , (b) 0.15 J .]
16.6 Six identical 2 µC charges are placed in equal inter-

vals around a circle with a radius of 1 m . How much work
Problem 16.12
is performed to bring these charges from infinity to these
16.12 The electric field outside and inside of a sphere with
positions? [A: 0.52 J .]
radius R and with a total charge Q evenly distributed over
its volume was( found as follows using Gauss’s law:
Problem 16.7 kQ/r2 (r > R)
E=
16.7 The distance between the charges q1 = 4 µC and kQr/R3 (r < R)
q2 = −1 µC shown in the figure is 2 m . At what point will Calculate the potential outside and inside of the sphere.
the potential be zero? [A: 1.6 m to the right from q1 .] [A: kQ/r for r > R , kQ(3 − r2 /R2 )/(2R) for r < R .]
no need to take the integral again. Consider that there is a

second disk with surface charge −σ h √at the center
√ of a full
i
disk.) [A: V = 2πkσ b + h − a2 + h2 .]
2 2
Problem 16.13 16.4 Conductors and Equipotential Surfaces

16.13 An infinite cylindrical conducting shell with radius R
16.17 A conducting sphere with a radius of 3 m is kept at
has linear charge density λ on its surface. Calculate the po-
a 100 V potential. At what distance from the center of this
tential difference between the surface of the cylinder (r = R)
sphere will the potential drop to 50 V ? [A: 6 m .]
and a point located outside at distance r .
[A: Vr − VR = −2kλ ln(r/R) .]
Problem 16.18
16.18 Of two concentric conducting spheres, the one with
Problem 16.14 radius a=1 m is charged with −4 µC and the one with radius
16.14 A q2 =2 µC point charge with mass m2 =1g is thrown b=2 m is charged with +4 µC . A point charge of q= − 1 µC
with v0 =10 m/s from the center and along the axis of a ring with a mass of 1 g is released from rest near the negatively
with radius R=1 m carrying charge q1 =1 µC . What will its charged sphere. What will its speed be when it reaches the
speed be at infinity? (Hint: Use the result of the Example positively charged sphere? [A: 6 m/s .]
16.8.) [A: v∞ = 12 m/s .]
Problem 16.19
16.19 Two identical conducting spheres located far away
Problem 16.15 from each other have a radius of 1 m . These two spheres
are connected with a metal wire and are together set to a
16.15 An electron is released from rest at a distance h=1 m
potential of 90 kV . (a) What is the charge on each sphere?
from a point on the axis of a disk with radius R=1 m and on
(b) While the spheres are interconnected, the radius of the sec-
which a total charge q1 =1 µC is distributed uniformly. With
ond sphere is increased to 2 m . What will the new charges of
what speed will it collide with the disk? (Hint: Use the result
the spheres be? (c) What is the new potential of the spheres?
of Example 16.9.) [A: v = 6.1 × 107 m/s .]
[A: (a) q1 = q2 = 10 µC , (b) q1 = 6.7 µC , q2 = 13.3 µC , (c)
16.16 In Example 16.9, the expression for the potential at 60 kV .]
distance h on the axis of a disk with radius r and carrying
σ surface charge
h √was found asi follows:
V = 2πkσ h +R −h
2 2
Problem 16.20
16.20 Of two conducting spheres separated by d=20 m from
their centers, one has radius R1 =1 m and a potential of
1000 V and the second has radius R2 =2 m and a potential of
−1000 V . Calculate the charge on each sphere. (Hint: The
Problem 16.16 total potential of each sphere is the sum of its own potential
Using this result, calculate the potential at distance h on the and the potential of the sphere at distance d . As the spheres
axis of a hollow disk with inner radius a and outer radius are far away, assume that the charges are evenly distributed
b and carrying surface charge density σ . (Hint: There is on their surfaces.) [A: q1 = +0.12 µC , q2 = −0.23 µC .)
?
17
CAPACITORS AND
DIELECTRICS
The Z machine at the Sandia Lab-

oratory in New Mexico. The
most powerful laboratory radi-
ation source in the world, it
works by charging a giant bank
of capacitors. When it oper-
ates, it generates more power
than 2,500 lightning bolts. It is
used in fusion energy and other
military-related researches.
What makes capacitors useful
in technology? How can we
change their properties using in-
sulating (dielectric) materials?
We discussed the basic properties of electrical forces in the previous chapters.

We can now start discussing the technological applications of electricity. We
shall first learn about a circuit component called the capacitor, which is used
to store charge and electrical energy. Capacitors are commonly used today in
electrical technology, in automobile spark plugs, radios and television, in camera
flash lights and nanotechnology.
In this chapter, we will also take a closer look at the behavior of materials
placed inside of an electric field. We will examine how these properties can
change the performance of capacitors in energy and charge storage.

284 17. CAPACITORS AND DIELECTRICS
17.1 CAPACITANCE
A system consisting of two conductors carrying equal and opposite charges
±Q in vacuum or in an insulating medium is called a capacitor. The primary
function of capacitors is to store charge under a certain potential difference.
Regardless of their geometric shapes, the charge ±Q stored by all capacitors is
proportional to the applied potential V :
Q =CV
The proportionality coefficient C is called capacitance. It is the ratio of the
charge on the conductor to the applied potential difference:
Q
C= (capacitance) (17.1)
V
Here, V is the absolute value of the potential difference V2 − V1 between the two
conductors: V = |V2 − V1 | . ∆V should actually have been used to indicate this.
However, using V will not lead to any confusion in this chapter.
Capacitance depends on the geometric shape of the conductors and the prop-
erties of the intermediate insulator medium. You should think as follows to keep
this in mind: A capacitor with larger capacitance stores higher charge.
The unit of capacitance is coulomb/volt (C/V) and was named as the farad
(F) in memory of the great scientist Michael Faraday. The farad is a very large
Figure 17.1: Various capaci- unit and its fractions are used in practice:
tors.
1 nanofarad (nF) = 10−9 F
1 picofarad (pF) = 10−12 F
Figure 17.2: The place of capac-

itors in technology: Capacitors
supplying the flash light of a cam-
era, capacitors under the keys of
a keyboard and dynamic memory
(DRAM).
Parallel-plate Capacitor
Let us consider two parallel conducting plates (Figure 17.3). Let each have
surface area A and let the distance in between them be d . Two equal and opposite
charges ±Q will accumulate when these plates are connected to the terminals of
a battery generating potential difference V .
We can assume these plates to be approximately infinite planes if the dimen-
sions of the plate are very large with respect to the distance d . Accordingly, we
had previously found the electric field between two infinite planes carrying ±σ
surface charge density:
σ
Figure 17.3: Parallel-plate ca- E= = constant
pacitor connected to a battery. ε0
In Chapter 16, we found the potential difference at distance d under a constant
electric field:
σd
V = V2 − V1 = E d =
ε0
17.1. CAPACITANCE 285
If the plate carrying charge Q has surface area A , the surface charge density σ
will be σ = Q/A . Accordingly,
(Q/A) d
V=
ε0
and, if we write this in terms of Q ,
ε0 A
Q= V
d
Comparing with the definition in Eq. (17.1), we can identify the capacitance of a
parallel-plate capacitor as:
ε0 A
C= (capacitance of a parallel-plate capacitor) (17.2)
d
As you can see, capacitance is dependent on the geometric dimensions and the
permittivity of the intermediate insulator space. Capacitance increases with the
surface area of the plates and decreases with the distance between the plates.
Cylindrical Capacitor
Let us consider two coaxial conducting cylinders, both with length L (Fig-
ure 17.4). Let the internal cylinder have radius a and the outer b . We wish to
calculate the potential difference when these cylinders are charged with equal
and opposite charge ±Q .
The cylinders can be assumed to be approximately infinite cylinders if their
length L is very large with respect to their radii a, b . In Chapter 15, we used
Gauss’s law to find the electric field of an infinite cylinder with linear charge
density λ :
2kλ Figure 17.4: Cylindrical capac-
E=
r itor.
Let us calculate the potential difference without taking into consideration whether
the field E is positive or negative:
Z b Z b
dr b
V = Vb − Va = E dr = 2kλ = 2kλ ln
a a r a
The linear charge density of a cylinder with length L carrying charge Q will be
λ = Q/L . Also, if we take k = 1/4πε0 , we find that
Q b 2πε0 L
V= ln −→ Q= V
2πε0 L a ln(b/a)
From here, we can find the capacitance C :
2πε0 L
C= (capacitance of a cylindrical capacitor) (17.3)
ln(b/a)
We likewise observe that capacitance is dependent only on the geometric proper-
ties of the conductors.
Spherical Capacitor
Let us consider two concentric conducting spherical shells with radii a and
b with a < b . Let the inner sphere with radius a be charged with +Q and the
outer with −Q . Let us use the definition to calculate the potential difference V
between these two spheres:
Z b
V = Vb − Va = − E dr
a
We had found the electric field in the region between the two spheres (a < r < b)
Figure 17.5: Spherical capaci- as E = kQ/r2 , using Gauss’s law. Accordingly, the absolute value of the potential
tor. difference is
Z b b !
V = kQ
dr
= kQ − 1 = kQ − 1 + 1 = kQ(b − a)
2 r a b a ab
a r
If we solve this expression for Q and also write the constant k in terms of ε0 ,
4πε0 ab
Q= V
b−a
The coefficient of V will be the capacitance C of the spherical capacitor:
4πε0 ab
C= (capacitance of a spherical capacitor) (17.4)
b−a
Example 17.1
ε0 A 8.85 × 10−12 × 30 × 10−4
C= =
In a parallel-plate capacitor, the plates have a surface area of d 4 × 10−3
C = 6.6 × 10 F = 6.6 picofarad = 6.6 pF
−12
30 cm2 and a distance of 4 mm between them. This capacitor
is connected to a 5000 V voltage (potential difference). (a) Find (b) We use the definition of capacitance to find the charge:
the capacitance of the capacitor. (b) What will the total charge Q = C V = 6.6 × 10−12 × 5000 = 33 × 10−9 = 33 nC
accumulated on the plates be? (c) What is the electric field (c) As E is constant between the plates, its relation with
between the plates? potential is V = Ed :
Answer V 5000
E= = = 1.25 × 106 V/m
(a) We use Eq. (17.2) we found for a parallel-plate capacitor: d 4 × 10−3
Example 17.2 2πε0 L

C=
ln(b/a)
In a coaxial cable, the inner cylinder has a radius of 2 mm and
the outer has a radius of 4 mm . A cylindrical capacitor with We solve for the length L and substitute the numerical data:
a capacitance of 1 pF is to be made from this cable. How long C ln(b/a)
L=
should the cable be? 2πε0
Answer 10−12 × ln(4/2)
L= = 0.012 m = 1.2 cm
We use Eq. (17.3) we found for cylindrical capacitor: 2 × 3.14 × 8.85 × 10−12
Example 17.3
(c) When it is fully charged, the capacitor is disconnected
from the battery and then the distance between the plates
A parallel-plate capacitor with capacitance C=3 pF is con-
is doubled. What will the new capacitance and the new
nected to a 1000 V battery.
potential difference between the plates be?
(a) How much charge is accumulated?
(b) While the capacitor is connected to the battery, the dis- Answer (a) We calculate from the capacitance formula:
tance between the plates is doubled. What will the new Q = C V = 3 × 10−12 × 1000 = 3 nC
capacitance and new charge be? (b) As the capacitance of a parallel-plate capacitor is inversely
17.2. COMBINATIONS OF CAPACITORS 287
proportional to the distance d in between, C will halve if d is (c) The charge will not change, as the capacitor is discon-
doubled. And V will remain constant, because it is connected nected while it has a charge Q . The potential difference V
to the battery, is again calculated from Q = CV , as C is halved:
C Q Q Q
Q0 = C 0 V = V = = 1.5 nC V0 = 0 = = 2V = 2 × 1000 = 2000 volt
2 2 C (C/2)
17.2 COMBINATIONS OF CAPACITORS

Each circuit component used in electrical circuits is indicated with a standard
symbol. Capacitors are indicated with the symbol in circuits.
A single capacitor that performs the function of multiple capacitors in a circuit
is called an equivalent capacitor. The way to calculate an equivalent capacitor
depends on whether the connections are in series or parallel.
Capacitors in Parallel
Two capacitors are said to be connected in parallel if they are connected to
the same potential difference. Let two capacitors with capacitances C1 and C2 be
connected to the same battery with potential difference V (Figure 17.6).
Then, the charges transferred to each capacitor can be calculated using the
formula Q = CV :
Q1 = C 1 V Q2 = C 2 V Figure 17.6: Capacitors in par-
allel.
The total charge drawn by the capacitors will be Q = Q1 + Q2 .
Now let us insert a single equivalent capacitor Ceq between the same points a
and b such that the same charge accumulates under the same potential difference:
Q = Ceq V
As the charge Q here will be the sum of Q1 and Q2 ,
Q = Q1 + Q2
Ceq V = C1 V + C2 V
Canceling V , we get the equivalent capacitance for the parallel case:
Ceq = C1 + C2
This proof is valid for any number of capacitors connected in parallel:
Ceq = C1 + C2 + · · · + C N (Capacitors in parallel) (17.5)
The equivalent capacitance in parallel is larger than each capacitance.

Capacitors in Series
Two capacitors are said to be connected in series if they are connected end
to end, without separating into another branch between them. When two such
capacitors C1 and C2 are connected to the terminals of a battery (Figure 17.7),
the outer plates will draw charges +Q and −Q . The inner plates, although they
draw no charge from the battery, cannot remain neutral; the charge +Q on the
left plate of C1 will attract electrons in the wire to the right plate of C1 , thus Figure 17.7: Capacitors in se-
leaving the left plate of C2 with charge +Q . ries.
We write the potential differences between the points a, b and c in the figure
and use the formula V = Q/C for each capacitor:
Vac = Vab + Vbc = V1 + V2

Q Q
V = +
C1 C2
The equivalent capacitor to be inserted between ac should accumulate the same
charge under the same potential difference:
Q Q Q
= +
Ceq C1 C2
Canceling Q , we get the equivalent capacitance for the series case:
1 1 1 C1C2
= + or Ceq =
Ceq C1 C2 C1 + C2
(Note: The expression on the right is convenient for calculations, because it is

a common mistake to forget to take the inverse at the last stage when working
with sums of 1/C .)
This expression is valid for any number of capacitors connected in series:
1 1 1 1
= + + ··· + (Capacitors in series) (17.6)
Ceq C1 C2 CN
In series connection, the equivalent capacitance is less than each capacitance. Let
us remember the following two principles (in addition to the formula Q = CV )
when solving capacitor circuits:
(1) Capacitors in parallel have equal voltage.
(2) Capacitors in series have equal charge.
Example 17.4
for innermost capacitors.
In the circuit shown in the figure, we have C1 = 1 , C2 = 2 , C1 and C5 between points c and d are parallel. We use
C3 = 3 , C4 = 4 and C5 = 5 µF . C 0 to indicate their equivalent capacitance between cd and
(a) Calculate the equivalent capacitance between the termi- calculate:
nals ab . C 0 = C1 + C5 = 1 + 5 = 6 µF
(b) The terminals ab are connected to a 12 V potential dif- Now, if we consider that this capacitance C 0 is between cd ,
ference. How much charge accumulates in the equivalent it will be in series connection with C3 . We calculate their
capacitor? equivalent capacitance:
(c) How much charge accumulates on the capacitor C4 ? C3C 0 3×6
C 00 = = = 2 µC
C3 + C 0 3 + 6
Then, C4 is in parallel with C 00 . We calculate their equivalent
capacitance:
C 000 = C4 + C 00 = 4 + 2 = 6 µF
Finally, capacitance C2 is in series with C 000 and we find the
equivalent capacitance that we are looking for:
C2C 000 2×6
Ceq = = = 1.5 µF
Answer C2 + C 000 2 + 6
(a) We carefully examine the figure to seek components that (b) As the equivalent capacitor is connected to terminals ab ,
fit the definition of parallel or series. You should always look its potential difference is Vab . From here, we calculate the
17.2. COMBINATIONS OF CAPACITORS 289
charge Q : potential difference over C2 :

Q = Ceq Vab = 1.5 × 12 = 18 µC V2 = Vdb =
Q2 18
= = 9V
(c) We can solve this problem calculating charges and volt- C2 2
ages alternately. First, note that the charge on C2 is equal to The potential difference Vab facing the battery can be written
the charge on the equivalent capacitor, because one terminal in two parts:
of the battery directly sees the capacitor C2 without separat- Vab = Vad + Vdb → 12 = Vad + 9
ing into another branch. We therefore write Q2 directly as Vad = 3 V
follows: This potential difference is the one faced by C4 . From here,
Q2 = Q = 18 µC we calculate the charge Q4 :
From here, we use the formula Q = CV to calculate the Q4 = C4 V4 = Q4 Vad = 4 × 3 = 12 µC
Example 17.5
the definition of capacitance:
In the circuit shown below, we have C1 =1 , C2 =2 , C3 =3 and Q = Ceq V = 2 × 24 = 48 µC .
C4 =4 µF . (b) The connection between ab changes when the switch S
(a) First, the terminals ab are connected to a 24 V battery is closed. This time, the C1 and C3 are connected in parallel;
when the switch S is open. Calculate the equivalent ca- likewise, the pair C2 and C4 are connected in parallel. Then,
pacitance and the total charge drawn from the battery. these two pairs are connected in series. We calculate the
(b) The circuit is disconnected from the battery after charg- equivalent capacitance accordingly:
ing and then the switch S is closed. Find the equivalent (C1 + C3 )(C2 + C4 )
Ceq =
capacitance and the charge on each capacitor. (C1 + C3 ) + (C2 + C4 )
(1 + 3)(2 + 4)
Ceq = = 2.4 µF
1+2+3+4
The total charge Q remains the same when the circuit is
disconnected from the battery as charged. But, this time, it is
differently distributed into two branches. As the capacitors
C1 and C3 are connected to the common points aS , they
have the same voltage:
VaS = VaS
Answer Q1 Q3
=
(a) When the switch S is open, C1 and C4 are connected C1 C3
in series; likewise, C2 and C3 are also connected in series. Also, the sum of Q1 and Q3 should be equal to the initial
Then, these two branches are connected in parallel to each charge:
other. We calculate the equivalent capacitance accordingly: Q1 + Q3 = Q = 48 µC
C1C4 C2C3 We calculate Q1 and Q3 from these two equations:
Ceq = + = 12 µC , Q3 = 36 µC
C1 + C4 C2 + C3 Q 1
Ceq =
1×4 2×3 4 6
+ = + = 2 µF The charges on C2 and C4 are calculated using the same
1+4 2+3 5 5 method applied to the potential difference S b . The result is:
We calculate the charge of the equivalent capacitance from Q2 = 16 µC , Q4 = 32 µC
Energy of a Capacitor
Capacitors have energy because of the electric charge that they carry. They
can discharge this energy in a very short duration to produce high currents or
potential differences. For example, although automobile batteries have a 12 Volt
potential difference, a capacitor circuit can produce potential differences up to
800–1000 V which are required to ignite spark plugs.
Let a capacitor with capacitance C have a total charge Q . Its energy will be
equal to the amount of work required to bring it from zero to the final charge Q .
Consider a parallel-plate capacitor initially without charge. Let us first take a
small charge dq1 from one of the plates and carry it to the other (Figure 17.8).
The work is zero, because no potential exists yet:
dW1 = 0
When we try to carry the next charge dq2 , there will be a potential difference of
dq1 /C between the plates, and therefore the work to be performed will be
dq1
dW2 = dq2 V1 = dq2
C
We carry out this process repeatedly and reach the charge Q in the end. We
Figure 17.8: Gradual charging can calculate the energy of the capacitor by adding the works performed at each
of a capacitor. stage.
Let q be the charge accumulated in the plates at some stage. The potential
difference at that stage will be V = q/C . Now, the work to be performed if we
wish to carry an additional charge dq will be
q
dW = dq V = dq
C
The work to be performed to bring the charge from zero at the start to the value
Q , in other words, the increase in the potential energy of the capacitor, will be
the integral of the small works dW :
Q
2 Q
Q2
Z
1 1 q
W = U Q − U0 = q dq = =
1
2
C 0 C 2 0 C
The energy of the neutral capacitor is chosen as zero potential: U0 =0 . Using the
formula Q = CV for the charge, the energy of a capacitor can be written under
various forms:
Q2
U= 1
2 = 1
2 CV 2 = 1
2 QV (Energy of a capacitor) (17.7)
C
Example 17.6
We calculate the energy using any one of the formulas (17.7):
U = 12 QV = 12 × 12 × 10−6 × 12
U = 72 × 10−6 J = 72 µJ
(b) When the switch S is closed in the direction of C2 , the
charge Q at the start is distributed between C1 and C2 . Con-
sidering that the potential differences between the terminals
of these two capacitors are equal, we get
The capacitor shown in the figure with capacitance C1 =1 µC is Q1 Q2
charged by closing the switch S in the direction of the battery V1 = V2 → =
C1 C2
with a potential difference of V=12 V . Q1 + Q2 = Q = 12 µC
(a) What are the charge and energy of the capacitor C1 ? We calculate Q1 and Q2 from these two equations:
(b) The switch S is then closed in the direction of the capacitor Q1 = 4 µC and Q2 = 8 µC
C2 = 2 µC . What will the charges and total energies of
the two capacitors be? We use the formula U = 21 Q2 /C to calculate the energies:
(c) How can the energy difference in between be explained? Q2 1 Q22 1 42 82
!
U =2
0 1 1
+ =2 + × 10−12+6
Answer C1 2 C2 1 2
(a) C2 is not taken into consideration when the switch S is U 0 = 24 µJ
closed in the direction of the battery. We calculate the charge (c) In this problem, we have U 0 < U , and it seems as if the
as follows: energy of the system automatically decreased, in other words,
Q = C1 V = 1 × 12 = 12 µC not conserved.
17.3. DIELECTRICS 291
This problem is actually an ideal case that is not found tween gets converted into heat in the wires of the circuit.
in real life. Circuit wires connecting capacitors usually have Otherwise, if such losses did not exist, the system would not
a certain resistance. When one capacitor transfers its charge reach equilibrium and the charge Q would go back and forth
by moving electrons to the other, energy will be lost in the from one capacitor to the other.
resistance of wires. As a result, the energy difference in be-
17.3 DIELECTRICS
In our analysis so far, we have assumed that there was vacuum between
the two plates of the capacitor. However, the area between the plates is filled
with an insulating material in capacitors used in technology. These materials
include paper, glass, plastic, oil, etc. When insulators are placed in an electric
field, they try to accommodate their structure with the applied field. They are
called dielectrics to emphasize this property.
According to experimental observations:
• A capacitor accumulates more charge when a dielectric material is inserted
between the plates while it is connected to a battery. According to this
observation, the capacitance of the capacitor increases due to the formula
C = Q/V .
• When a dielectric material is inserted between the plates after a charged ca-
pacitor is disconnected, the potential difference between the plates decreases.
According to this observation, for example, if we recall the formula V = E d
for parallel-plate capacitors, the electric field between the plates decreases.
How is the material between the plates able to change the properties of the
capacitor despite the fact that it is an insulator? In order to understand this, we
need to examine the microscopic structure of dielectric materials.
Remember the concept of electric dipole from Chapter 13: We discussed
that a system consisting of two equal and opposite charges (±q) separated by
distance a is called an electric dipole. This system could produce an electric field
in space and interact with other charges even though it is neutral. This system
has an electric dipole moment defined as p = qa .
In some molecules (like H2 O , NO2 , HCl ), the positions of positive and
negative charges do not coincide and they form a electric dipole. These are called
polar molecules (Figure 17.9a). They have a permanent dipole moment, even
when there is no electric field. When they are in an external electric field, they
try to align their dipole moments with it.
Figure 17.9: Polar and nonpo-

lar molecules placed in an exter-
nal electric field gain a dipole mo-
ment in the direction of the field.
In some other molecules (like O2 , CO2 , CH4 ), the positions of positive

and negative charges coincide and they have no permanent dipole moment.
However, when placed in an external electric field, the geometric location of the
electrons changes under the forces exerted in the opposite direction to the + and
− charges and they gain a dipole moment. These are called nonpolar molecules
(Figure 17.9b).
Figure 17.10: The molecules of

dielectric materials placed in be-
tween the plates of a capacitor
form surface charges in the oppo-
site direction and cause the field
~E to decrease inside.
Regardless of their structure, dielectric molecules placed in an electric field

between the plates of a charged capacitor show a reaction (Figure 17.10). Polar
molecules rotate and nonpolar molecules deform to try to align their dipole
moments in the direction of the electric field (Figure 17.10b). Although they
cannot fully balance the electric field, they partially reduce its effect in the medium.
Induced surfaces charges are thus produced on dielectric surfaces facing the
plates (Figure 17.10c). This phenomenon is called polarization.
If we express the electric field produced by the charges ±Q on the plates in
terms of the surface charge density σ , we get
σ
E0 =
ε0
If we use σb to show the surface charge density induced in the opposite direction
by the bound charges of the dielectric material, the new electric field will result
from the net surface charge density σ − σb :
σ − σb σb
E= = E0 − (17.8)
ε0 ε0
If the external electric field is not too strong, the amount of the induced charge σb
will be proportional to the electric field E that they feel in the medium. Hence,
E and E0 will be proportional, with E being smaller:
E0
E= (17.9)
κ
The denominator of this expression is defined as the dielectric constant of the
medium. The electric field always decreases in a dielectric medium because κ>1 .
The decrease in the electric field will also decrease the potential difference
through the formula V=E d . If V0 is the potential difference in vacuum,
E0 E0 d
V = Ed = d=
κ κ
V0
V = (17.10)
κ
According to the definition C = Q/V , we find the capacitance in the presence of
a dielectric medium:
C = κ C0 (Capacitance of a capacitor with dielectric) (17.11)
All of these results can be expressed simply by defining the permittivity of the
dielectric medium as:
ε = κ ε0 (17.12)
Here, ε is the electric permittivity of the medium. Therefore, all formulas
become valid in capacitors with dielectrics by using ε instead of ε0 .
Some dielectric constants are given in the table:
Dielectric constants and dielectric strengths of some insulators

dielectric constant dielectric strength
κ Emax (106 V/m)
Vacuum 1 —
Air 1.0006 3
Paraffin 2.2 10
Paper 3.7 15
Glass 5 14
Porcelain 6 12
Figure 17.11: Use of dielectrics

in technology: Coaxial cables,
pin insulators, pure silicon layers
known as wafers in the manufac-
ture of microcircuits.
Dielectric Strength
When the magnitude of the electric field applied on a dielectric is too strong,
the positive and negative charges of the molecules are subject to extreme forces
in the opposite directions and their bond is broken. Molecules become ionized
and the dielectric turns into a conducting medium in which an electric discharge
is observed. The high current passing through the dielectric turns into heat,
damaging the material (Figure 17.12).
The maximum electric field that a dielectric medium can endure without
dielectric breakdown is called its dielectric strength. The dielectric strengths of
some materials are given in the table above.
As can be seen from the table, air cannot withstand high electric fields. But
if we insert a dielectric between the plates of a capacitor, according to Eq. 17.9, Figure 17.12: Cracks formed
E = E0 /κ , it will have a weaker electric field for the same accumulated charge. on a plexiglass plate as a result
And the capacitor can operate in higher voltages without getting damaged. of dielectric breakdown.
van de Graaff Generator
How large an electric potential can be generated on a conductor? We can now
answer this question using what we have learned up to now about conductors
and dielectrics. Let us consider a hollow conducting spherical shell with radius
R as the simplest case. Let us remember the expressions for the potential and
surface electric field of this sphere:
kQ kQ
V= , E= and V = RE
R R2
In principle, the more we can increase Q , the more the potential of the conductor
will increase, until infinity. However, this actually has a limit. The electric field
around the conductor increases with its potential.
The dielectric strength of the air in which the sphere is located is Emax =3 ×
106 V/m , in other words, at higher electric fields, the air molecules get ionized
and dielectric breakdown occurs, with charges accumulated on the conducting
sphere flowing to the ground. Therefore, the maximum potential that the sphere
can carry should fulfill the following equation:
Vmax = R Emax
For a sphere with radius R = 1 m , we find that the maximum charge that it can
carry is Qmax =1/3 mC and the maximum potential produced is Vmax = 3 × 106 V .
But how do we charge this sphere? If we bring the charges directly, as the
electric field around the sphere will increase gradually, the opposing force will also
increase and it will become more difficult to bring additional charges. However,
the following property of conductors will make it easier: Charges placed in the
internal surface of a conductor will accumulate in the outer surface. This is because
the electric field inside of the conductor must always be zero.
A simple diagram of the van de Graaff generator that uses this principle is
shown in Figure 17.13. In this mechanism, an insulating rubber belt moves by
turning a pulley on the lower end. A sharp metal brush connected to the positive
terminal of a battery stands very close to the belt near the bottom pulley. This
brush ionizes the air around it and transfers some of the positive charges to the
belt. A second metal brush near the top pulley has zero electric field around it, as
it is connected to the conducting sphere and easily collects the charges on the
belt and then transfers them to the external surface of the sphere.
Figure 17.13: Diagram of the The van de Graaf generator is used in nuclear physics researches to accelerate
van de Graaff generator. charged particles and operate X-ray tubes. High-voltage shows are also performed
in physics laboratories and science fairs. If your shoes are insulated from the
ground, when you touch the metal sphere, your hairs will repel each other and
stand up due to the high voltage formed on your body. (The van de Graaff
generator is not a dangerous device, despite this high potential. Considering that
the charge required for 3 million volts is on the order of merely millicoulombs,
the current that can flow from your feet to the ground is very low.)
Lightning
In stormy weather, we hear a peal of thunder like a big explosion after a flash
of light in the sky and understand that a lightning bolt has struck somewhere.
Lightning is the electric discharge that occurs after air becomes conductor when
the maximum value of electric field that the dielectric medium (air) can withstand
is exceeded, as explained above. There can be different types of lightning with
different structures. The most common type of lightning is explained as follows:
Water molecules that evaporate on the surface of the Earth start to rise, as
they are lighter than air. They start to condense into water droplets when they
reach the colder upper layers of the atmosphere. This mixture of microscopic
Figure 17.14: A tree struck by particles of water and ice collide with other rising water vapor molecules, and
lightning. ionize them by freeing their electrons. Negative electrons accumulate in the
bottom surface of the cloud and positive ions in the top surface.
Consequently, an electric field arises between the negative bottom surfaces
of the clouds and the Earth’s surface. As the magnitude of the electric field
increases, the molecules of the air in between generate a conducting path to
ensure a discharge. Hence, before the lightning strike, zigzagged fringes called
leaders consisting of ionized air molecules start to extend downward from the
clouds and upwards from the roofs on the Earth’s surface (Figure 17.15).
Once the conducting path produced by these ionized air molecules is com-
pleted, the electric discharge occurs instantly and a very large electron current
reaches the Earth’s surfaces. The heat generated by the current causes the sur-
rounding air molecules to heat and expand. As this expansion occurs instantly, it
produces an explosive effect and a shockwave.
As the speed of light is greater than the speed of sound, the light is observed
first and the sound later. As the speed of sound in air is 340 m/s, you can guess Figure 17.15: Leaders that pre-
the distance of the place where the lightning struck by counting the seconds after pare the path of lightning.
it strikes (1 km per 3 seconds).
Leaders that prepare the path of lightning are usually produced on high
buildings and trees. Therefore, it can be dangerous to stay in open air or to take
shelter under a tree. The best thing to do is to crouch, but without touching the
ground with your hands, in order to not increase the conducting path.
Example 17.7
κε0 in the capacitance formula (17.2) of the parallel-plate
There is a 2 mm separation between the plates of a parallel- capacitor:
plate capacitor. Its capacitance is C0 = 3 µF when there is κε0 A
C= = κ C0 = 5 × 3 = 15 µF
vacuum between the plates. d
We calculate the potential V using this capacitance C and
(a) The capacitor is connected to a battery with V0 = 12 V .
charge Q :
What will the charge Q0 and the electric field E0 between
Q Q V0 12
the plates be? V= = = = = 2.4 V
(b) The capacitor is disconnected as charged and a glass layer C κC0 κ 5
The electric field is calculated with this potential difference
with dielectric constant κ = 5 is inserted between the
V:
plates. Find the capacitance C , charge Q , potential V V (V0 /κ) E0 6000
and electric field E of the capacitor. E= = = = = 1.2 kV/m
d d κ 5
(c) A glass layer is inserted between the plates when the capac-
(c) The potential difference remains constant if the capacitor
itor is still connected to the battery. Find the capacitance
is kept connected to the battery:
C , charge Q , potential V and electric field E of the ca-
pacitor. V = V0 = 12 V
The change in capacitance is the same when a dielectric is
Answer likewise inserted in between:
(a) We calculate Q0 from the definition: C = κ C0 = 15 µF
Q0 = C0 V0 = 3 × 12 = 36 µC
The charge is calculated using this capacitance and potential
We find the electric field using the formula V = Ed in con-
difference:
stant electric field:
V0 12 Q = CV = (κC0 ) V0 = 15 × 12 = 180 µC
E0 = = = 6000 V/m = 6 kV/m
d 0.002 The electric field is calculated using the same method:
(b) As the capacitor is disconnected as charged, its charge V V0
E= = = E0 = 6 kV/m
will remain constant afterwards: d d
Q = Q0 = 36 µC As you may see, a capacitor connected to the same battery
To calculate the new capacitance when a dielectric is inserted accumulates higher charge when there is a dielectric but the
in between, it is sufficient to replace the coefficient ε0 with electric field between the plates still remains the same.
Example 17.8
is inserted. What will the new capacitance be?
Answer
We write the capacitance C0 of the parallel-plate capacitor
before the dielectric layer is inserted as C0 = ε0 A/d .
We can consider the system as two capacitors connected in
A parallel-plate capacitor has capacitance C0 when the area series when the dielectric is inserted. The first capacitor has
between the plates is empty. A dielectric layer with constant κ thickness a and dielectric constant κ1 and the second capac-
that fills a distance a of the total distance d between the plates itor has thickness d − a and dielectric constant κ2 = 1 .
1 κd − (κ − 1)a
!
1 d a d−a
= + =
C ε0 A κd d C0 κd
From here, we find the capacitance C in terms of C0 and
dimensions a and d :
κd
Accordingly, we write the equivalent capacitance formula of C= C0
κd − (κ − 1)a
the capacitors in series as follows:
1 1 1 a d−a In order to check the validity of this expression, we can see
= + = + that C = C0 at the limit a = 0 . Likewise, it correctly gives
C C1 C2 κε0 A ε0 A
We factor out C0 in this expression: C = κC0 within the limit a = d .
Example 17.9
Answer
(a) The fact that air’s dielectric strength is 3×106 V/m means
The 4 mm wide space between the plates of a parallel-plate that air will lose its dielectric property and turn into a conduc-
capacitor is filled with air. tor if the electric field exceeds this value. Therefore, the max-
imum potential difference between the plates of the parallel-
(a) What is the maximum operating voltage of the capacitor?
plate capacitor is calculated using this Emax value:
(b) The space between the plates is filled with a porcelain layer.
V = Emax d = 3 × 106 × 0.004 = 12 000 V = 12 kV
What is the maximum operating voltage this time? (Di-
electric strengths: Air: Emax = 3 × 106 V/m , porcelain: (b) A similar method is used for porcelain:
12 × 106 V/m ) V = Emax d = 12 × 106 × 0.004 = 48 000 V = 48 kV
Example 17.10
From here, we calculate the distance d :
V 30 000
A parallel-plate capacitor is to be manufactured with a ma- d= = = 0.002 m = 2 mm
terial with dielectric constant κ=30 and dielectric strength Emax 15 × 106
15 × 106 V/m such that its capacitance shall be 1 pF and be We write the capacitance formula of the parallel-plate capaci-
operating up to 30 kV voltage. What should the surface area tor with dielectric constant κ :
κε0 A
and distance between the plates of this capacitor be? C=
d
Answer From here, we calculate the surface area A :
The electric field between the plates by the maximum voltage Cd 10−9 × 0.002
A= =
should be equal to the dielectric strength: κε0 30 × 8.85 × 10−12
V = Emax d A = 0.0075 m2 = 75 cm2
1. What is the unit of capacitance?
III. The electric field doubles.
(a) joule (b) volt (c) faraday (d) farad IV. The capacitance doubles.
2. In a parallel-plate capacitor, the surface area of the plates (a) I & II (b) I & III (c) I & IV (d) II & IV
and the distance in between are both doubled. By what 5. The distance between the plates of a parallel-plate capac-
factor will capacitance increase? itor is doubled while it is connected to a battery. Which
(a) No change (b) 2 (c) 1/2 (d) 4 of the following are correct?
I. The potential difference remains the same.
3. The capacitance of a capacitor and the applied voltage II. The capacitance halves.
are both doubled. By what factor will the accumulated III. The charge halves.
charge increase? IV. The electric field remains the same.
(a) No change (b) 2 (c) 1/2 (d) 4 (a) I & II (b) I & III (c) I & IV (d) I, II & III
4. A parallel-plate capacitor is disconnected after being 6. When two capacitors with capacitance 1 µF and 2 µF
charged and the distance between its plates is doubled. are connected in parallel, how many µF will the equiva-
Which of the following are correct? lent capacitance be?
I. The charge remains the same. (a) 1 (b) 3 (c) 2/3 (d) 3/2
II. The potential doubles.
7. When two capacitors with capacitance 1 µF and 2 µF

are connected in series, how many µF will the equiva-
lent capacitance be?
(a) 1 (b) 3 (c) 2/3 (d) 3/2
16. While the capacitors are connected to the circuit, a di-
8. A dielectric with κ = 3 is inserted between the plates of electric material is inserted between the plates, as shown
a capacitor. By what factor will the capacitance increase? in the figure above. What happens to the total charge
(a) No change (b) 1/3 (c) 3 (d) 9 drawn from the battery?
(a) It increases.
9. A dielectric with κ = 3 is inserted between the plates (b) It decreases.
of a capacitor charged with Q . By what factor will the (c) It remains the same.
potential difference V increase? (d) It is impossible to tell.
(a) No change (b) 1/3 (c) 3 (d) 9
10. A dielectric with κ = 3 is inserted between the plates

of a capacitor charged with Q . By what factor will the
electric field increase? 17. What is the equivalent capacitance of the circuit in the
(a) No change (b) 1/3 (c) 3 (d) 9 figure above?
11. By what factor will the energy of a capacitor charged (a) 6 µF (b) 6/11 µF (c) 1/6 µF (d) 4 µF
with Q increase when its capacitance is doubled?
(a) 1/2 (b) 2 (c) 4 (d) 1/4
12. By what factor will the energy of a capacitor charged

with Q increase when the distance between its plates is
doubled?
18. What is the equivalent capacitance of the circuit in the
(a) 1/2 (b) 2 (c) 4 (d) 1/4 figure above?
13. By what factor will the energy of a capacitor increase (a) 6 µF (b) 6/11 µF (c) 1/6 µF (d) 4 µF
when the potential difference V applied to it doubles?
(a) 1/2 (b) 2 (c) 4 (d) 1/4 19. Of two identical capacitors, one has double the charge
and half the potential with respect to the other. By what
14. By what factor will the ratio Q/V increase for a capaci-
factor is the energy of the second capacitor higher than
tor whose capacitance is doubled?
that of the first one?
(a) No change (b) 2 (c) 4 (d) 1/2
(a) Equal (b) 2 (c) 1/2 (d) 4
20. Which of the following can explain the decrease in elec-

tric field when a dielectric is placed between the plates
of a capacitor?
I. An opposite electric field produced by molecular
15. While the capacitors are connected to the circuit, a di- dipoles.
electric material is inserted between the plates, as shown II. The surface charges induced on the dielectric sur-
in the figure above. What happens to the total charge faces.
drawn from the battery? III. Electrons detached from molecules.
(a) It increases. IV. The ionization of molecules.
(b) It decreases. (a) I & II (b) I & III (c) II & IV (d) I & IV
(c) It remains the same.
Problems
17.1 Capacitance
17.1 A capacitor with capacitance 5 pF is to be manufactured
using two conducting plates with surface area 80 cm2 . What
should the distance between the plates be? [A: 14 mm .]
17.2 The distance between the plates of a parallel-plate capac-

itor is 2 mm and each plate has an area of 40 cm2 . Charges Problem 17.8
±5 nC accumulate in its plates when this capacitor is con- 17.8 Each capacitor in the circuit shown in the figure above
nected to a battery. (a) Find the capacitance of the capacitor. has capacitance C=1 µF . What is the equivalent capacitance
(b) Find the potential difference of the plates. (c) Find the between ab ? [A: 15/41 µF .]
electric field between the plates.
[A: (a) 18 pF , (b) 280 V , (c) 140 kV/m .]
17.3 A parallel-plate capacitor with capacitance C = 5 pF

is connected to a 2000 V battery. (a) How much charge is
accumulated? (b) The capacitor is disconnected as charged
and then the distance between the plates is halved. What
will the new capacitance and the new potential difference Problem 17.9
between the plates be? (c) The distance between the plates is 17.9 Each capacitor in the circuit shown in the figure above
halved while the capacitor is connected to the battery. What has capacitance C=1 µF . What is the equivalent capacitance
will the new capacitance and new charge be? between ab ? [A: 6/11 µF .]
[A: (a) 10 nC , (b) 10 pF , 1000 V , (c) 10 pF , 20 nC .]
17.4 The capacitance of a spherical capacitor was found in

Eq. (17.4) as:
4πε0 ab Problem 17.10
C= 17.10 In the circuit shown in the figure we have C1 = 1 ,
b−a
(a) Show that the capacitance is C=4πε0 a when the radius of C2 = 2 , C3 = 3 µF and V = 12 V . (a) Calculate the equiva-
the external conductor goes to infinity ( b → ∞ ). (b) Assume lent capacitance. (b) Find the charge of the equivalent capaci-
that the Earth is a conductor and calculate its capacitance. tor. (c) Find the charge on each capacitor.
(Earth’s radius: 6400 km .) [A: 710 µF .] [A: (a) 1.5 µF , (b) 18 µC , (c) q1 = 6 , q2 = 12 , q3 = 18 µC .]
17.5 A spherical capacitor with 10 pF in capacitance con-

sists of two conducting spheres for which the radius of one is
double that of the other. Calculate the radius of each sphere.
[A: 4.5 and 9 cm .]
17.6 A cylindrical capacitor with 1 nF in capacitance is to Problem 17.11

be manufactured using a coaxial cable for which the radius 17.11 In the circuit shown in the figure we have C1 = 1 ,
of the inner cylinder is 1 mm and the outer is 3 mm . How C2 = 2 , C3 = 3 and C4 = 4 µF . (a) Calculate the equiva-
long should the cable be? [A: 20 m .] lent capacitance between the terminals ab . (a) The terminals
ab are connected to a 12 V potential difference. How much
charge gets accumulated in the equivalent capacitor? (c) How
17.2 Connection of Capacitors
much charge gets accumulated on the capacitor C4 ?
[A: (a) 23/3 µF , (b) 92 µC , (c) 48 µC .]
Problem 17.7
17.7 Each capacitor in the circuit shown in the figure above
has capacitance C=1 µF . What is the equivalent capacitance Problem 17.12
between ab ? [A: 2 µF .] 17.12 The capacitor shown in the figure above with capac-
PROBLEMS 299
itance C1 =4 µC is charged by closing the switch S in the capacitors?

direction of the battery V=12 V . (a) What is the charge of [A: (a) q1 = 20 , q2 = 80 µC , 1 mJ , (b) q01 = 12 , q02 = 48 µC ,
the capacitor C1 ? (b) The switch S is then closed in the 0.36 mJ .]
direction of the capacitors C2 =3 µC and C3 =6 µC . What is
the charge on each capacitor? 17.3 Dielectrics
[A: (a) q1 = 48 µC , (b) q1 = 32 , q2 = q3 = 16 µC .] 17.16 Paraffin is used to fill the volume of a parallel-plate
capacitor whose plates have 5 cm2 in surface area and 2 mm
in distance between them. (a) What is the capacitance of
the capacitor? (b) What is the maximum operating voltage?
(Refer to the table on Page 293 for the dielectric constant and
strength of paraffin.) [A: (a) 4.9 pF , (b) 20 kV .]
Problem 17.13
17.17 There is a distance of 3 mm between the plates of a
17.13 In the circuit shown in the figure above, we have C1 =1 ,
parallel-plate capacitor. Its capacitance is C0 = 5 µF when
C2 =2 , C3 =3 and C4 =4 . (a) In this circuit, the terminals ab
there is vacuum between the plates.
are connected to a 12 V battery when the switch S is open.
(a) The capacitor is connected to a battery with V0 = 24 V .
Calculate the equivalent capacitance and the total charge
What will the charge Q0 and the electric field E0 between
drawn from the battery. (b) The circuit is disconnected from
the plates be?
the battery when charged and then the switch S is closed.
(b) The capacitor is disconnected as charged and a dielectric
Find the equivalent capacitance and the charge on each ca-
layer with dielectric constant κ = 8 is inserted between the
pacitor.
plates. Find the capacitance C , potential V and electric field
[A: (a) 25/12 µF and 25 µC , (b) 2.1 µF and q1 = 8.3 ,
E of the capacitor.
q2 = 16.7 , q3 = 10.7 , q4 = 14.3 µC .]
(c) The same dielectric layer is inserted between the plates
while the capacitor is connected to the battery. Find the
charge Q , potential V and electric field E of the capacitor.
[A: (a) 120 µC and 8 kV/m , (b) 40 µF , 3 V and 1 kV/m , (c)
960 µC , V and E do not change.]
Problem 17.14
17.14 Each of the 3 parallel plates shown in the figure above
have surface area A and the distances in between them are a Problem 17.18
and b . These plates are connected to a potential difference V , 17.18 The parallel-plate capacitor shown in the figure above
as shown in the figure. Prove that the expression for equiva- has capacitance C0 when the space between the plates is
lent capacitance is C = ε0 A(1/a + 1/b) . (Hint: The conductor empty. A dielectric with constant κ is inserted that fills one
plate in the center is the common plate of two capacitors. half of the vacuum between the plates. What will the new
Determine whether these two capacitors are connected in capacitance be? [A: C = (1 + κ)C0 /2 .]
parallel or series.)
17.19 The 5 mm wide space between the plates of a parallel-
plate capacitor is filled with air. (a) What is the maximum
operating voltage? (b) A paper layer is inserted between the
plates of the capacitor. What is the maximum operating volt-
age this time? (Dielectric strengths: Air: Emax = 3×106 V/m ,
paper: 15 × 106 V/m ) [A: (a) 15 kV , (b) 75 kV .]
Problem 17.15
17.15 After being charged under the same potential differ- 17.20 A parallel-plate capacitor is to be manufactured with
ence of 20 V , the capacitors with capacitance C1 =1 µF and a material with a dielectric constant of 12 and a dielectric
C2 =4 µF are connected in reverse, as shown in the figure strength of 5 × 106 V/m such that its capacitance shall be
above, in other words, the negative charged plate of one is 10 pF and it will be able to operate up to 20 kV of potential
connected to the positive charged plate of the other. (a) What difference. What should the surface area and distance be-
are the charges and total energies of the capacitors at the tween the plates of this capacitor be?
start? (b) What are the final charges and total energies of the [A: 3.8 cm2 and 4 mm .]
?
18
CURRENT, RESISTANCE
AND CIRCUITS
The Intel Core i7 chip and the

underlying Sandy Bridge archi-
tecture. Launched in 2011, this
chip contains 2.6 billion transis-
tor circuits.
No other type of power more efficient and practical than electric power has
been discovered in modern technology and industry. Electricity can easily be
transferred and used in all places and under all operating conditions. It can also
be used in sending signals and in operating and controlling machinery.
The electric current lies at the basis of all of these activities. From this chapter
onwards, we shall leave the subject of electrostatics and examine the currents
produced by moving electric charges and their behavior in matter. We will learn
of electrical effects that are much richer and more complex compared to static
charges.
18.1 ELECTRIC CURRENT

The amount of charge flowing through a conductor in a given time is called
the current. More explicitly, if a net charge dq flows through the cross-section
of a conductor in a certain direction during a small time dt , then the current,
indicated with the symbol I , is
Figure 18.1: Charges flowing
dq through a cross-section.
I= (Current) (18.1)
dt

302 18. CURRENT, RESISTANCE AND CIRCUITS
The unit of current is the ampere and is indicated with A :
1 A = 1 C/s
However, as the ampere is the basic unit in the SI system, it is not necessary
to express ampere in terms of coulombs. On the contrary, the coulomb unit is
ampere × seconds.
Note the expression “net charge flowing through in a certain direction” here.
The free electrons in a conductor are continually in motion. When observed in
any cross-section, the average amount of charge flowing through in one direction
is equal to the average amount of charge flowing in the other direction, and
therefore the net charge flowing through is zero. A net charge flow is observed
in a certain direction when this conductor is connected to a potential difference.
Direction of Current
Before the microscopic structure of matter was known, the direction of electric
current was accepted as the direction of motion of positive charges. However, it
was later understood that only the negative electrons were moving. Positively
charged ions do not move; they only vibrate, as they are strongly bonded to each
other and are very heavy.
Figure 18.2: Current is formed Despite this, the direction of current was not changed. According to this
by (−) charged electrons, and + assumption, the direction of current is taken as the direction of motion of positive
charged ions do not move. charges. In actual fact, in a solid medium these are negatively charged electrons
moving in the opposite direction.
This creates no problem macroscopically. Consider this: A +q charge travel-
ing to the right increases the charge on the right by +q . However, a −q charge
traveling to the left also increases the charge on the right by +q . They both lead
to the same result. In this chapter, we will examine many cases as if +q charges
are moving, but the conclusions that we shall reach will not change.
Drift Speed
We can calculate the current in a conducting wire in terms of the speed of
the moving charges inside of the wire. A net electric field E will form inside
of a wire when a potential difference V is applied between the two ends of the
wire. The charges will start to accelerate as a force F=qE will be exerted upon a
charge q in the electric field.
(Let us make one correction here: We had previously stated that E=0 was
always true in a conductor in electrostatics. However, the conditions of elec-
trostatics are not valid any more, as we are examining moving charges and the
electric field may be different from zero.)
Charges under the force F = qE would be expected to accelerate according to
Newton’s law. However, this is not possible in a conducting medium (Figure 18.3).
This is because moving q charges collide with the ions in the medium and lose
energy or change direction. They then accelerate again and repeatedly collide
Figure 18.3: Electrons lose and slow down. You may consider this like an automobile trying to drive along a
energy by colliding with ions road with successive red traffic lights. The automobile accelerates, but halts at a
when accelerating in a conduc- red traffic light and waits; it then starts to accelerate again after a green traffic
tor. light and is then forced to stop again at the next red traffic light.
Electrons thus travel at an average speed inside of a conductor. This is called
the drift speed and is indicated with vd . Now, let us show the relation between
drift speed and current.
18.1. ELECTRIC CURRENT 303
Let ∆q be the amount of charge flowing through the cross-section A of a

conductor during the time interval ∆t (Figure 18.4). The individual charges e
flowing through each with drift speed vd during this ∆t time interval will be
those charges inside of a cylinder with base area A and length L=vd ∆t . Let
the number of free charges inside of a unit volume of the conductor medium be
n=N/V . Accordingly, if the number of e charges inside of the mentioned cylinder
is ∆N , then
∆N = n × volume = n Avd ∆t Figure 18.4: Charges flowing
through the cross-section A dur-
and the amount of charge flowing through the cross-section is ing time interval ∆t , are located
inside of a cylinder with length
∆q = e ∆N = qnAvd ∆t vd ∆t .
If we now use the definition of current, we find the following result:
∆q
I= = enAvd (18.2)
∆t
If current I is known, we can write this expression for speed vd as follows:
I
vd = (drift speed) (18.3)
enA
Let us see the order of magnitude of drift speed using a simple calculation: A
current I = 20 A is flowing through a copper wire with cross-section 1 mm2 .
The copper contains approximately 1029 free electrons per cubic meters. If we
take q = e = 1.6 × 10−19 C as the charge, we find that
I 20
vd = = ≈ 0.001 m/s
enA 1.6 × 10 × 1029 × 10−6
−19
This result may seem surprising. Electrons are moving as slow as 1 mm per
second inside the conductor. However, we see a lamp turn on immediately after
we switch it on. The reason for this is that there are free electrons everywhere in
the conductor. The neighboring charges of the lamp start to move as soon as you
switch it on and do not have to travel a long distance.
Biological Effects of Electric Current
The human body has limited resistance to electric voltage and current. The
effect called electric shock occurs at currents and potential differences above
certain limits and may lead to serious injuries that may result in death. Let us
emphasize the important points to keep in mind for protection from electricity.
• Electric current can flow through the body in two ways: It may flow through
by completing the circuit over the body, as shown in Figure 18.5a. Or by
reaching the ground through the body, as shown in (b).
Figure 18.5: Current can flow
through the body in two ways:
(a) By completing the circuit, (b)
By reaching the ground through
the body.
• The important factor in electric shock is not the voltage but current. You may
come into contact with a very high voltage, but it will not harm the body if a
current does not flow as a result. However, if the high voltage completes the
circuit at another point of the body (for example, if two wires are held with
two hands or a conducting contact is made with the ground), then the circuit
is completed and current starts to flow. (This is why birds are able to land on
electric cables without getting electrocuted.)
• Direct currents higher than 0.3 − 0.5 A and alternating currents higher than
60 milliampere are harmful.
• Electric current can harm the body in two ways:
Burns on the skin and internal organs. The heat energy released
and the ionization that occurs along the current path may damage tissues
and lead to death.
Disruption of the heart and the nervous system. The nervous sys-
tem of our body consists of nerve cells called axons that operate with voltages
as small as 0.1 V . An external electric current disrupts this system and leads
to paralysis in muscles. The heart beat becomes irregular, especially in the
event of a paralysis in the heart ventricles. This is called fibrillation. The
heart may stop as a result of being unable to pump sufficient blood.
80 % of the deaths caused by electric shock are due to burns and the rest due
to heart and nervous system paralysis.
Figure 18.6: The defibrillator • The current flowing through the body depends on the electric resistance of
device used in cardiac arrest.
the body (see Ohm’s law below). This resistance may vary depending on
skin temperature or whether it is moist or dry. The resistance of a dry skin is
around 5 000 − 10 000 Ω but it may drop down to around 1 000 Ω for moist
skin.
• Considering these current and resistance limits, even voltages as low as 10 V
may be harmful.
• Protection. The first rule to be observed in the event of an electric shock
is to cut the contact with the voltage source. You must make sure to isolate
yourself while doing this.
Using three-prong plugs with ground connection will provide the necessary
protection for electric appliances with current leaks. One of the three wires in
these plugs carries the voltage and the other two are neutral. The third wire
is connected both to the ground and to the conductor frame of the electrical
appliance. Any electrical leak thus goes to the ground through this third
path.
Another protection method is to prevent currents flowing to the ground
through the body by wearing shoes with insulating soles (rubber, cork).
18.2 OHM’S LAW AND RESISTANCE

The current flowing through a unit cross-section of a conductor is called
the current density and is indicated with J . If a current I flows through a
cross-sectional area A , the current density will be
I
J= (Current density) (18.4)
A
Its unit is ampere/meter2 = A/m2 .
18.2. OHM’S LAW AND RESISTANCE 305
There is a linear relation between the current density J and the electric field
E in the medium in metals and many other conductors. Discovered for the first
time by the German scientist Ohm, this relation is known as Ohm’s law and is
expressed as follows:
E = ρJ (microscopic Ohm’s law) (18.5)
The ratio ρ is called the resistivity.

Resistivity depends on the type of material and temperature. The following
table lists the resistivities of some materials:
Resistivities of some materials at room temperature (20 ◦ C)

Material Resistivity ρ (Ω·m) Temperature coefficient α (1/C◦ )
Silver 1.5 × 10−8 0.004
Copper 1.7 × 10−8 0.004
Gold 2.4 × 10−8 0.003
Aluminum 2.8 × 10−8 0.004
Tungsten 5.3 × 10−8 0.005
Carbon 3.5 × 10−5 −0.001
Eq. (18.5) is an expression of Ohm’s law at the microscopic level. In order to

turn this into a form that is usable in practical applications, we use the potential
difference V applied on both ends of the conductor (Figure 18.7). The electric
field inside of a conducting wire with cross-section A and length L is
V
E=
L
and, using current density J = I/A , the law (18.5) becomes as follows:
V I L Figure 18.7: Ohm’s law.
=ρ −→ V= ρ I
L A A
From here, we obtain the macroscopic expression of Ohm’s law:
V = RI (macroscopic Ohm’s law) (18.6)
The coefficient R in this law is called the resistance:

L
R=ρ (resistance) (18.7)
A
The resistance of a conductor depends on its type, dimensions and temperature.
The resistance of a conductor increases with its length and decreases with its
cross-sectional area.
The unit of resistance is the ohm and is indicated with the Ω (omega) symbol:
V
1Ω=1
A
When the resistance is used as a circuit element, it is called a resistor and is
indicated with the symbol or .
Variation of Resistivity with Temperature

The resistivity of a conductor increases with temperature and makes current
flow more difficult. Microscopically, the vibration of positive ions usually increases
with temperature; the ions then collide more frequently with the free electrons in
the medium, and thus reduce their drift speed. This, in turn, causes an increase
in the resistance of the medium.
Experimental observations show that the variation of resistivity with temper-
ature can be expressed as follows:
ρ = ρ0 [1 + α (T − T 0 )] (18.8)
The value ρ0 here is the resistivity at a reference temperature T 0 (usually 0 ◦ C

or 20 ◦ C ). The coefficient α is called the temperature coefficient of resistivity.
The values of α for some materials are given in the table on page 305.
The relation of resistance R to temperature is similar:
R = R0 [1 + α (T − T 0 )] (18.9)
Example 18.1
We find the resistivity of copper from the table on Page 305
A 5 Ω resistor is to be made using a copper wire with a 0.1 mm and calculate the required length:
2
cross-section. 5 × 0.1 × 10−6

L= = 29 m
(a) What should the length of the wire be? 1.7 × 10−8
(b) How many amperes of current will flow when this resistor (b) Eq. (18.6) is solved for the current:
is connected to a potential difference of 116 V ? V 116
I= = = 23 A
(c) What will the magnitude of the electric field inside of the R 5
wire be? (c) Assuming that there is uniform electric field inside of the
Answer wire, we write the potential difference in terms of electric
(a) We use Eq. (18.7), which defines the resistance: field:
L RA V 116
R=ρ → L= V =EL → E = = = 4 V/m
A ρ L 29
Example 18.2
E = ρJ
An electric field with a magnitude of 100 V/m is formed when J = I/A is the current flowing through the unit cross-section.
2
a 3 A current flows through a wire with cross-section 2 mm . From here we calculate the resistivity:
I EA
What is the resistivity of the wire? E=ρ → ρ=
A I
Answer We substitute the data and calculate the resistivity:
We write Eq. (18.5), the microscopic expression of Ohm’s law: ρ = 100 × 2 × 10−6 /3 = 6.7 × 10−5 Ω·m
Example 18.3
R = R0 [1 + α (T − T 0 )]
A tungsten wire lamp has 10 Ω in resistance at 20 ◦ C . We find the temperature coefficient of tungsten from the table
(a) What will its resistance be at 600 ◦ C ? on Page 305:
(b) The lamp has 80 Ω in resistance when it is glowing. De- R = 10 [1 + 0.005 × (600 − 20)] = R = 39 Ω
termine the temperature of the lamp. (b) The same formula is used to find temperature T :
Answer 80 = 10 [1 + 0.005 × (T − 20)]
(a) Eq. (18.9) is used: T = 1420 ◦ C
18.3. DIRECT CURRENT CIRCUITS 307
18.3 DIRECT CURRENT CIRCUITS

Electromotive Force (EMF)
There must be a potential difference between the two points of a circuit for
a current to flow through it. A potential difference obtained from other sources
(chemical, mechanical, magnetic, optical, etc.) is called the electromotive force.
It is called emf for short and is indicated with the E symbol.
Emf is not actually a force, but a potential difference, however this name has
stuck for historical reasons. The emf potential difference is actually indicated
with E rather than V , to indicate that it is a source.
Figure 18.8: Various sources of

emf: bicycle dynamos, dry batter-
ies, solar batteries.
As a circuit component, emf is indicated with the symbol .

The most common source of emf is the battery that converts chemical energy
into electrical energy. For example, in a galvanic battery, a potential difference is
produced between two different metal electrodes (zinc and copper) immersed in
an electrolyte liquid. Lead and lead-oxide electrodes are used in car batteries.
In all types of battery, one of the metal electrodes will have higher potential
(Figure 18.9). This positive terminal is called the cathode and the negative terminal
the anode. When this battery is connected to a circuit, the current flows from the
cathode to the anode, in other words, from the positive terminal to the negative Figure 18.9: In a battery, cur-
terminal. The current inside of the battery flows in the reverse direction to rent flows from cathode (+) to
complete the cycle. (As we mentioned earlier, electrons actually flow in the anode (–), but electrons actually
opposite direction.) flow in the reverse direction.
Likewise, solar batteries convert solar energy directly into a potential differ-
ence, just as a generator converts mechanical energy into electrical energy.
The emf value, in other words, the potential difference E of the battery, is
constant and does not vary, even if you increase the dimensions of the battery.
The emf value can only be increased by connecting several batteries in series. For
example, emf is 1.1 V in a galvanic battery, 2.1 V in a lead battery, and 12 V in
an automobile battery with 6 cells.
Internal Resistance and Terminal Voltage
When an emf source is connected to a circuit, while current flows from the
(+) terminal to the (–) terminal on the outside, there must be a current from
the ( − ) terminal to the (+) terminal inside of the electrolyte liquid. An internal
resistance appears, as the mobility of the electrons inside of the liquid will be
limited. Internal resistance is indicated with r .
Now, let us consider a very simple circuit by connecting the emf source E
to a resistor R . In the circuit, it is necessary to add an internal resistance r
for the battery in addition to the resistor R . A current I starting from the (+)
terminal (point b ) with high potential will circulate throughout the whole circuit
(Figure 18.10).
Figure 18.10: A simple circuit
consisting of a resistance R con-
nected to an emf source E with
internal resistance r .
To calculate this circuit, let us start from the (–) terminal of the battery
indicated with a , go along the circuit in the direction of the current and return to
point a . The potential should return to the value Va again at the end of this path.
Let the potential at point a be Va . Since we have to go through the (+)
terminal of the emf source to reach point b , the potential will increase and the
potential of point b will be higher by +E :
Vb = Va + E
When we reach the internal resistance r , according to Ohm’s law, the potential
will drop by rI when going along the resistor in the direction of the current.
Therefore, the potential of point c is
Vc = Vb − r I = Va + E − r I
Likewise, the potential will drop by RI when going from point c through the
resistor R in the direction of the current. Therefore, the potential of point a on
the right will be
Va = Vc − R I = Va + E − r I − R I
We thus return to point a . The graph in Figure 18.11 shows the potential value at
each point along the circuit.
Figure 18.11: The potential dif-

ferences on a circuit as current
flows through the emf and the
resistors.
By simplifying, we get
E−rI −RI = 0
We thus find the current I in a circuit with resistor R by an emf source E with
internal resistance r :
E
I=
R+r
As this circuit operates with the current I flowing, the resistor R will be
subject to the potential difference between the ac terminals of the battery. And
this potential difference will be,
Vac = E − rI (Terminal voltage) (18.10)
This potential difference Vac is called the terminal voltage of the battery. Ter-
minal voltage will always be less than the value of E , because the potential drop
across the internal resistor r should also be taken into account.
18.3. DIRECT CURRENT CIRCUITS 309
Power and Energy in Electric Circuits

Electric circuit components (resistors, capacitors, motors. . . ) convert electrical
energy into other types of energy, such as heat, light and motion, when a current
flows through them. In the most general case, if a current I flows through a
circuit component connected between any two points, there will be a potential
difference V between its terminals. The potential difference shows the change in
the potential energy, i.e., the work performed per unit charge.
Let us suppose that charge dq flows during time interval dt through a circuit
component with potential difference V between its terminals. The change in the
potential energy of this charge will be dU = dq V according to the definition
of potential energy. In other words, the potential energy lost when the charge
dq goes from a high potential to a low potential point on this circuit element,
performs dW = dU = dq V amount of work on this circuit component:
dW = dU = V dq
The work performed per unit time gives us the power:
dW V dq dq
P= = =V
dt dt dt
As the amount of charge flowing during the unit time is I = dq/dt , the expression
for power will be as follows:
P=VI (Power) (18.11)
This formula is the most general expression of power, and it gives us the power
consumed when the potential difference is V and the current flowing through is Figure 18.12: Electric energy is
I on any component of the circuit, including resistors, capacitors, emf, motors, converted to heat in an electric
etc. heater.
To find the energy spent during a certain time interval, we write the definition
of power for dW :
dW = P dt
As a particular case, let us find the expression for the power consumed in a
resistor: As V = R I according to Ohm’s law, we get
P = V I = R I2 (Power converted into heat in a resistor) (18.12)
or, alternately, the amount of energy consumed during time interval dt will be
dW = P dt = R I 2 dt (18.13)
Note that this energy converted into heat is proportional to the square of the
current.
Example 18.4
(c) Calculate the power consumed in the resistors and the
A 15 V battery is connected to a 7 Ω resistor. The internal power provided by the battery to the circuit.
resistance of the battery is 0.5 Ω . Answer
(a) Find the current flowing through the circuit. (a) We add the voltage changes over each circuit component
(b) Find the terminal voltage of the battery. as we make a complete clockwise cycle starting from the
negative terminal of the battery: (c) We use the formula RI 2 to calculate the power consumed
+E − r I − R I = 0 in the resistors:
E 15 PR = RI 2 = 7 × 22 = 28 W and Pr = rI 2 = 0.5 × 22 = 2 W
I= = = 2A
R + r 7 + 0.5 The power provided by the battery is calculated using the
(b) The terminal voltage of the battery is defined in Eq. (18.10): formula (18.11):
Vab = E − rI = 15 − 0.5 × 2 = 14 V P = V I = E I = 15 × 2 = 30 W
Example 18.5
We calculate the power by using the resistance at room tem-
An electric kettle made of tungsten has 50 Ω in resistance at perature: 2
room temperature ( 20 ◦ C ) and operates under 240 V . P = 240 /50 = 1200 W = 1.2 kW
(a) How much power does the kettle consume when it starts (b) As the boiling point of water is 100 C , we have to calcu-
◦
◦
to heat water at 20 C ? late the resistance at that temperature:
(b) How much power does it consume when the water starts R = R0 [1 + α (T − T 0 )]
to boil? We take the temperature coefficient α of tungsten from the
table on page 305:
Answer
R = 50 [1 + 0.005 × (100 − 20)] = 70 Ω
(a) Using the power formula P=V I and Ohm’s law V=RI ,
We calculate the power consumed at this temperature:
the power of the heater can be written as follows:
V 2 2402
P = R I 2 = V 2 /R P= = = 820 W = 0.82 kW
R 70
18.4 COMBINATION OF RESISTORS

A single resistor that performs the function of multiple resistors in a circuit
is called an equivalent resistor. Very complex connections usually require
advance circuit analysis. We shall only discuss circuits consisting of resistances
connected in series and parallel.
Two simple principles are sufficient in circuit solutions:
• The same current flows through resistors on the same branch.
• The same voltage applies to resistors connected to the same two terminals.
Resistors in Series
Suppose that resistors R1 and R2 are connected in series under a potential
difference V , as shown in Figure 18.13. The same current I will flow through
these resistors. Therefore, we write the potential differences and Ohm’s law:
Vac = Vab + Vbc

V = R1 I + R2 I
The equivalent resistor Req should draw the same current I under the same
Figure 18.13: Resistors in se- potential difference V :
ries. V = Req I
Comparing the last two expressions, we find the series connected equivalent
resistor:
Req = R1 + R2
This result is also valid for more than two resistors:
Req = R1 + R2 + · · · + RN (Resistors in series) (18.14)

18.4. COMBINATION OF RESISTORS 311
Resistors in Parallel
Two resistors with resistance R1 and R2 are connected to the same battery
with potential difference V , as shown in Figure 18.14. In this case, each resistor
will be subject to the same V potential difference, but the current flowing through
each will be different. If we use I1 and I2 to indicate these currents, we get
V V
I1 = I2 =
R1 R2
The total current drawn from the battery will thus be I = I1 + I2 .
The equivalent resistor placed between the same terminals should draw the Figure 18.14: Resistors in par-
same total current: allel.
I = I1 + I2
V V V
= +
Req R1 R2
After simplifying, we get the equivalent resistance of resistors in parallel:
1 1 1
= +
Req R1 R2
This result is valid for more than two resistors in parallel:
1 1 1 1
= + + ··· + (Resistors in parallel) (18.15)
Req R1 R2 RN
The equivalent resistance of a parallel connection is less than each resistance.
Example 18.6
calculate the equivalent resistance of these two:
R3 R0 3×6
R00 = = = 2Ω
R3 + R0 3 + 6
This resistor R will be in series connection with R4 :
00
R000 = R4 + R00 = 4 + 2 = 6 Ω
Finally, resistor R000 will be in parallel connection with R2 :
R2 R000 2×6
Req = = = 1.5 Ω
R2 + R000 2 + 6
In the figure, we have R1 =1 , R2 =2 , R3 =3 , R4 =4 and R5 =5 Ω . (b) We first calculate the current drawn by the equivalent
(a) What is the equivalent resistance between ab ? resistor when the terminals ab are connected to the battery:
(b) How much current will flow through resistor R4 when the Vab 12
I= = = 8A
terminals ab are connected to a 12 V source? Req 1.5
Answer Let us assume that this current I enters through the terminal
(a) The resistance of wires is taken to be zero in circuit prob- b . The current will separate here into two branches, which
lems. In other words, we can extend, shorten or even merge we will indicate with I2 and I4 :
these wires at one point. It thus becomes easy to see series I = I2 + I4 = 8 A
or parallel connections. We can immediately calculate the current I2 using Ohm’s
If we examine the circuit in the figure starting from the law, because the resistor R2 is also subject to the potential
inside, we can observe that R1 and R5 are connected in series. difference Vab = 12 V :
Vab 12
Let us use R0 to show their equivalent resistor: I2 = = = 6A
R0 = R1 + R5 = 1 + 5 = 6 Ω R2 2
From here we calculate the current I4 :
If we replace these two resistors with a single resistor R0 , it
I4 = I − I2 = 8 − 6 = 2 A
will be observed to be in parallel connection with R3 . Let us
Example 18.7 R0 R00 5×5

Req = = = 2.5 Ω
R +R
0 00 5+5
In the circuit shown in the figure, R1 =1 , R2 =2 , R3 =3 and We calculate the power consumed in the circuit from the
R4 =4 . The terminals ab are connected to a 12 V source. potential difference:
(a) Calculate the equivalent resistor and the power consumed
V2 122
in the circuit when the switch S is open. P = ab = = 58 W .
(b) Calculate the equivalent resistor and the power consumed Req 2.5
in the circuit when the switch S is closed and compare (b) The pairs (R1 , R3 ) and (R2 , R4 ) will be in parallel when
with item (a). the switch is closed. Let us use R0 and R00 to indicate their
equivalent resistances:
R1 R3 1×3 3
R0 = = = Ω
R1 + R3 1 + 3 4
R2 R4 2×4 4
R00 = = = Ω
R2 + R4 2 + 4 3
Then, these resistors R , R will be in series:
0 00
3 4 25
Req = R0 + R00 = + = = 2.1 Ω
Answer 4 3 12
(a) The pairs (R1 , R4 ) and (R2 , R3 ) are in series connection We calculate the power consumed in the circuit from the
with each other when the switch is open. Let us use R0 and potential difference:
R00 to indicate their equivalent resistances: V2 122
R0 = R1 + R4 = 1 + 4 = 5 Ω P = ab = = 69 W .
Req 2.1
R00 = R2 + R3 = 2 + 3 = 5 Ω The power consumed in the circuit is higher when the switch
Then, these resistors R0 , R00 will be in parallel connection: is closed.
Example 18.8 a potential difference is applied across the points between
which equivalent resistance is requested. The circuit simpli-
fies as shown above if we change the lengths of the connecting
wires and merge them at one point.
It is easily seen from the figure that the resistors R3 and
R6 are connected in parallel. We calculate the equivalent
resistance of these two as follows:
R3 R6 3×6
In the circuit shown in the figure, R1 =1 , R2 =2 , . . . and R0 = = = 2Ω
R1 + R3 3 + 6
R6 =6 Ω . Find the equivalent resistance between points ab . This resistor R will be in series with R4 :
0
R00 = R4 + R0 = 4 + 2 = 6 Ω
This resistor R00 will be in parallel with R2 :
R2 R00 2×6
R000 = = = 1.5 Ω
R2 + R00 2 + 6
This resistor R will be in series with R1 :
000
Riv = R1 + R000 = 1 + 1.5 = 2.5 Ω

Finally, this resistor Riv will be in parallel with R5 :
R5 Riv 5 × 2.5 5
Answer In circuit problems, it is convenient to consider that Req = = = = 1.7 Ω
R5 + Riv 5 + 2.5 3
18.5 ELECTRICAL MEASURING INSTRUMENTS

It is important to know how to connect electrical measuring instruments
that measure resistance, current or emf. An instrument that measures current is
called an ammeter, an instrument that measures potential difference is called a
voltmeter and an instrument that measures emf is called a potentiometer.
The common aspect of all these measuring instruments is that they use the
same instrument, called the galvanometer in different ways. So, let us first
examine the galvanometer.
18.5. ELECTRICAL MEASURING INSTRUMENTS 313
Galvanometer
When a current flows through a wire placed between the poles of a magnet, a
magnetic force acts on the wire (Figure 18.15). As we shall discuss in Chapter 20,
the magnetic force on the wire is proportional to the magnitude of the current.
Also, a torque is exerted if the wire is wound like a coil and the coil gets deflected in
the magnetic field. Likewise, the amount of deflection of the wire is proportional
to the current flowing through it. Thus, the current can be calculated by measuring
the deflection angle on a scale.
However, the wire inside of the galvanometer also has a resistance. This
resistance is part of the circuit, and will thus change the current flowing through
the circuit. The internal resistance of the galvanometer should be very small so that
this effect can be small.
The galvanometer is the basis of all kinds of electrical measuring instruments. Figure 18.15: Galvanometer.
Other quantities can be measured by connecting resistors in series or in parallel
to the galvanometer.
Ammeter
The ammeter used to measure current is actually a galvanometer with a very
small resistor. The ammeter is connected in series by inserting it into the branch
in which the current is to be measured. It is thus ensured that the whole current to
be measured flows through the ammeter. However, the resistance of the ammeter
should be very small, almost zero, such that the current flowing through this
branch is not affected. The current flowing through the ammeter in the circuit
shown in Figure 18.16 will be
Figure 18.16: An ammeter is
E
I= connected in series.
R + RA
The change in the current I can be neglected if the resistance of the ammeter is
RA R .
Voltmeter
The voltmeter used to measure the potential difference between two points is
actually a galvanometer with very large resistance. The voltmeter is connected in
parallel to points a, b of the circuit, and is thus subject to the potential difference
to be measured (Figure 18.17).
This time, some of the current will flow through the voltmeter and the current Figure 18.17: A voltmeter is
flowing through the resistor R will decrease. We calculate the potential difference connected in parallel.
on resistor R :
RI
V=
1 + R/RV
Therefore, the resistance RV of the voltmeter should be very large to ensure that
the potential difference does not deviate too much from its correct value RI .
Potentiometer
A potentiometer is an instrument used to measure the emf voltage of a battery.
As we have seen previously in Eq. (18.10), when a battery connected to the circuit
starts to produce a current, the terminal voltage will be Vab = E − rI and will be
less than E . Since we cannot eliminate the internal resistance r of the battery,
can we take a measurement at zero current?
We can do this with the potentiometer assembly shown in Figure 18.18. The
figure shows a battery with a known value of E and a resistor R . The emf source
whose E x value is to be measured is connected between point b and a variable

point c on the resistor R .
The point c is varied such that a point is found where the current I x flowing
through E x becomes zero. This is possible when E x and E are connected in
reverse and E > E x .
We calculate the same potential difference Vcb at two branches at this zero
condition as follows:
Vcb = Rcb I (Rbc part of resistor R)

Figure 18.18: Potentiometer. Vcb = E x (because I x = 0)
Therefore, the desired E x is equal to the product Rbc I . E x can be calculated by

measuring the current flowing through the circuit and the resistance Rbc .
Today, current, voltage and resistance can all be measured using instruments
called the multimeter or the avometer, which have all these properties in a
digital medium. However, their operating principle is the same.
Figure 18.19: Multimeter.
1. What will the current be if a charge of 6 C flows through
the cross-section of a wire in 2 seconds? (a) I & II (b) I & III (c) I & IV (d) II & III
(a) 6 A (b) 8 A (c) 3 A (d) 1/3 A
5. Which of the following are correct for the biological
effects of electricity?
2. Which of the following are correct for electric current? I. It is dangerous to complete a circuit through the body.
I. Electrons perform accelerated motion in conductors. II. It is dangerous for a current the reach the ground
II. Electrons collide with ions in the conductor and slow through the body.
down. III. There will be no harm if the voltage is high but no
III. Electrons travel at a constant average speed due to current passes through.
collisions. IV. Voltages higher than 10 V can be dangerous.
(a) I & II (b) II & III (c) I & III (d) I, II&III
(a) I & II (b) II & III (c) I & IV (d) All
3. Which of the following is correct for the current in con-

6. Which of the following are correct for the resistance of
ductors?
a wire?
(a) Positive ions move in the direction of the current.
I. Resistance increases with temperature.
(b) Positive ions move in the opposite direction of the
II. Resistance decreases with temperature.
current.
III. Resistance increases with the cross-section area.
(c) Electrons move in the direction of the current.
IV. Resistance decreases with the cross-section area.
(d) Electrons move in the opposite direction of the cur-
rent. (a) I & III (b) I & IV (c) II & III (d) II & IV
4. Which of the following are correct for electromotive 7. Both the length and the cross-section area of a wire are
force (emf)? doubled. By what factor will its resistance increase?
I. It converts another type of power into electrical (a) No change (b) 2 (c) 4 (d) 8
power.
II. It is an electrical force. 8. Which is the expression for Ohm’s law?
III. It provides a constant current to the circuit.
(a) V = RI (b) I = VR (c) V = R/I (d) I = R/V
IV. It provides a constant potential difference to the cir-
cuit.
9. Which of the following are correct for the resistance of 14. How much energy does a light bulb with 5 W of power
a conductor? consume in 1 minute?
I. It is proportional to the resistivity. (a) 5 J (b) 10 J (c) 25 J (d) 300 J
II. It is proportional to the length.
III. It is proportional to the cross-section area.
15. How much power is consumed in a 3 Ω resistor through
IV. It is inversely proportional to the cross-section area.
which 2 A in current flows?
(a) I & II (b) I, II & IV (c) I & III (d) II & IV
(a) 5 W (b) 6 W (c) 12 W (d) 18 W
10. Which of the following will be correct when a 12 V emf
source with an internal resistance of 1 Ω is connected 16. What is the equivalent resistance of the resistors with
to a circuit? resistances of 3 and 6 Ω connected in series?
I. It will provide a constant 12 V potential difference (a) 2 Ω (b) 9 Ω (c) 18 Ω (d) 24 Ω
to the circuit.
II. It will provide a potential difference less than 12 V 17. What is the equivalent resistance of the resistors with
to the circuit. resistances of 3 and 6 Ω connected in parallel?
III. It will provide a constant current to the circuit.
(a) 2 Ω (b) 9 Ω (c) 18 Ω (d) 24 Ω
(a) I & II (b) II & III (c) I & III (d) All
11. When you go along the current through a resistor R in

which a current I is flowing, which of following hap-
pens?
(a) The potential increases by RI. 18. What is the equivalent resistance between ab in the
(b) The potential decreases by RI. figure?
(c) The potential remains constant.
(d) The potential increases by RI 2 . (a) 2 Ω (b) 3 Ω (c) 6 Ω (d) 9 Ω
12. Which two of the following principles are correct for 19. Which of the following are correct for an ammeter?
direct current circuits? I. It is connected to the circuit in series.
I. The same current flows through resistors on the same II. It is connected to the circuit in parallel.
branch. III. It has very low internal resistance.
II. The potential difference is the same in resistors on IV. It has very high internal resistance.
the same branch. (a) I & II (b) I & III (c) II & III (d) II & IV
III. The same current flows through resistors connected
between the same two points.
20. Which of the following are correct for a voltmeter?
IV. The potential difference in resistors connected be-
I. It is connected to the circuit in series.
tween the same two points is the same.
II. It is connected to the circuit in parallel.
(a) I & II (b) II & III (c) I & III (d) I & IV III. It has very low internal resistance.
IV. It has very high internal resistance.
13. A 12 V emf source delivers 3 A when connected to a
circuit. What is the power provided by the source? (a) I & II (b) I & III (c) II & III (d) II & IV
(a) 4 W (b) 15 W (c) 36 W (d) 108 W
Problems
18.2 Ohm’s Law and Resistance consume when its temperature reaches 300 ◦ C ?
(Use the table on Page 305 for the resistivities and temperature [A: (a) 1440 W , (b) 600 W .]
coefficients that may be required in the problems.)
18.4 Combination of Resistors
18.1 A 3 Ω resistor is to be made using an aluminum wire
with a cross-section of 0.2 mm2 . (a) What should the length
of the wire be? (b) How many amperes of current will flow
when this resistor is connected to a potential difference of
60 V ? (c) What will the magnitude of the electric field inside
of the wire be? [A: (a) 21 m , (b) 20 A , (c) 2.8 V/m .]
Problem 18.11
18.2 How much current passes through when a tungsten 18.11 Calculate the equivalent resistance between the termi-
wire with a cross-section area of 1 mm2 and a length of 10 m nals ab shown in the figure. [A: 6 Ω .]
is connected to a 12 V battery? [A: 23 A .]
18.3 A wire with a resistance of 0.02 Ω is to be manufac-

tured out of an aluminum block with a volume of 1 cm3 .
What should the length and cross-section area of the wire
be? [A: 85 cm and 1.2 mm2 .]
18.4 A wire with resistance R is cut into 4 equal parts and

joined side-by-side. What will the resistance of the new wire Problem 18.12
be in terms of R ? [A: R/16 .] 18.12 R1 = 1, R2 = 2, · · · R6 = 6 Ω in the circuit shown in
the figure above. Calculate the equivalent resistance between
18.5 An electric field with a magnitude of 80 V/m exists the terminals ab . [A: 15/13 Ω .]
inside of a wire with a cross-section of 3 mm2 when a 5 A
current flows through. What is the resistivity of the wire?
[A: 4.8 × 10−5 Ω·m .]
18.6 A tungsten wire lamp has 7 Ω in resistance at 20 ◦ C .

(a) What will its resistance be at 500 ◦ C ? (b) The lamp has Problem 18.13
50 Ω in resistance when it is producing light. Determine the 18.13 R1 = 1, R2 = 2, R3 = 3 and R4 = 6 Ω in the circuit
temperature of the lamp. [A: (a) 24 Ω , (b) 1250 ◦ C .] shown in the figure above. (a) Calculate the equivalent resis-
tance. (b) Calculate the current flowing through each resistor.
18.7 At what temperature will the resistivity of tungsten be [A: (a) 2 Ω , (b) I1 = 3 , I2 = 1.5 , I3 = 1 , I4 = 0.5 A .]
four times that of silver? [A: 300 C .]
◦
18.3 Direct Current Circuits

18.8 An 18 V battery is connected to a 8 Ω resistor. The
internal resistance of the battery is 1 Ω . (a) Find the current
flowing through the circuit. (b) Find the terminal voltage of Problem 18.14
the battery. (c) Calculate the power spent on the resistors and 18.14 R1 = 1, R2 = 2, R3 = 3 and R4 = 4 Ω in the circuit
the power provided by the battery to the circuit. shown in the figure. (a) Calculate the equivalent resistance.
[A: (a) 2 A , (b) 16 V , (c) 32 , 4 and 36 W .] (b) Calculate the current flowing through resistor R1 .
[A: (a) 21/16 Ω , (b) I1 = 9 A .]
18.9 When a battery with an emf of 12 V is connected to a
circuit, a 5 A current flows out and the terminal voltage is
11 V . What are the internal resistance of the battery and the
connected resistance? [A: 0.2 Ω and 2.2 Ω .]
18.10 An oven heater made of tungsten has a resistance of

40 Ω at room temperature ( 20 ◦ C ) and operates under 240 V . Problem 18.15
(a) How much power does the oven consume when it starts 18.15 All of the resistors in the circuit above have the same
to cook food at 20 ◦ C ? (b) How much power does the oven value R . (a) Find the equivalent resistance. (b) Calculate the
PROBLEMS 317
potential difference Vab if R=8 Ω and E=26 V .

[A: (a) 13R/8 , (b) 10 V .] Show that
R2 R3
Rx =
R1
when the current on the galvanometer is zero. (Hint: This
is an interesting case in which resistors can be in series and
parallel at the same time. When the galvanometer current is
zero, both the points a and b will have the same potential,
and the current in the upper branch will continue without
Problem 18.16 going to the lower branch. It is sufficient to write potential
18.16 R1 = 1, R2 = 2, R3 = 3 and R4 = 4 Ω in the circuit drops R I equal for each half of the circuit. )
i i
shown above and 6.5 V in voltage is applied between ab . (a)
How much current is drawn from the battery when switch
S is open? (b) How much current is drawn from the battery
when the switch is closed? [A: (a) 8 A , (b) 7.8 A .]
Problem 18.19
18.19 R1 = 1, R2 = 2, R3 = 3, R4 = 4 and R5 = 5 Ω in
Problem 18.17 the circuit shown in the figure. Considering that the current
18.17 58 W power is consumed in the circuit shown in the
flowing through R1 is 1 A , what is the emf value E of the
figure when the resistors, each with resistance R , are con-
battery? [A: 24 V .]
nected to a 45 V -battery. Calculate the value of R .
[A: 25 Ω .]
18.18 Wheatstone bridge. The circuit shown in the figure
below is used to determine an unknown resistance R x . After
a battery is connected to the circuit, the value of the variable
resistor R3 is changed such that the current flowing through
a galvanometer connected between ab is zero.
Problem 18.20
18.20 The ammeter A in the figure above shows the same

value when both of the switches are closed and when both
Problem 18.18 switches are open. Find the resistance R . [A: 60 Ω .]
?
19
MAGNETIC FIELD
The ITER Tokamak reactor, un-

der construction in France as the
world’s largest fusion reactor, is
expected to produce electricity
in 2045. (The man on the bot-
tom right of the figure shows the
scale.)
As no container can contain
the charged particles at 100 mil-
lion ◦ C temperatures in the
medium called plasma, they are
suspended in the air with a very
strong magnetic field (in the yel-
low donut-shaped area in the fig-
ure).
We can better understand mag-
netic field if we first understand
how charged particles move in
magnetic field.
The magnetic needle of a compass will always point north, regardless of the
direction which you turn it. The Chinese and Indian civilizations were already
cognizant of magnetism thousands of years ago. It is said that, in Ancient Greece,
magnetized rocks were found in a region of Anatolia known as Magnesia. Sailors
have used magnetized compasses for navigation for a thousand years.
Understanding magnetism was possible as late as the 17th century. In the
1600s, the English scientist William Gilbert claimed that the Earth was a giant
magnet and that this was how it attracted the poles of a compass. In the 19th cen-
tury, two scientists, the French André Ampere and the Danish Oersted, observed
that a magnet would deflect a current-carrying wire. Later, Michael Faraday
discovered that moving magnets generated emf.
All these discoveries indicated that there was a close relationship between
electricity and magnetism. Ultimately, Scottish scientist James Clerk Maxwell
merged electricity and magnetism under a single theory.
We shall discuss the properties of magnetism and its relation to electrical
forces in this and the next two chapters.

320 19. MAGNETIC FIELD
19.1 MAGNETIC FORCE

We know how a horseshoe or bar magnet attracts nails or pins. This force
cannot have an electrical nature, since both the magnet and the nails are neutral.
From this observation, we can conclude the existence of a new “magnetic force.”
Indeed, we can see the presence of a magnetic field around magnets in the patterns
of iron dust scattered around a bar magnet, as seen in Figure 19.1.
The Coulomb force that we discussed in electrics was caused by electrical
charges. Using the same logic to look for a “magnetic charge” that causes magnetic
Figure 19.1: The magnetic force, we observe that this is not possible. Consider that we take a magnet with
field lines observed in iron dust poles indicated as being North-South (N − S ) and divide it into two pieces. In
around a magnet. such a case, it is observed that each of the new pieces turns into a magnet with
N - S poles (Figure 19.2). We cannot separate one of the poles even if we repeat
this process and go down to the atomic scale. Each atom acts like a small magnet.
Today, in modern physics, it is understood that there is no single magnetic
pole.
So, what is the origin of the magnetic force? As we shall discuss later, mag-
netism has two sources:
Figure 19.2: We cannot isolate • Currents.
magnetic poles even if we keep • Magnetic dipole moments of elementary particles.
splitting the magnet.
We shall learn these concepts in this and subsequent chapters.
In our study of electricity, we had adopted the following modern view: A
charge distribution produces an electric field ~E at every point in space and then
a charge q placed in this field interacts with the electric field as ~F = q~E .
We shall use the same outlook for magnetic field:
• A current generates a magnetic field B ~ at every point in space.
~
• A force F is exerted upon moving charges q and currents I in this magnetic
field.
We shall start with the second item about magnetism, setting aside how a
magnetic field is generated and concentrating first on trying to understand the
force that it exerts.
Magnetic Force Exerted on a Moving Electric Charge

A charge moving between the poles of a magnet is observed to deflect from
its trajectory. The magnetic force exerted upon a charge q traveling with velocity
~v inside of a magnetic field is observed to have the following properties:
• The force is proportional to the charge q and is in the opposite direction for
+q and −q charges.
• The force is proportional to velocity v and perpendicular to it. The force is
zero if the charge is at rest (v = 0) .
• The force is proportional to the magnitude of the magnetic field B and
perpendicular to it.
All these properties indicate that magnetic force can be expressed as a vector
product as follows:
~F = q ~v × B
~

(Magnetic force on a charge) (19.1)
19.1. MAGNETIC FORCE 321
Now, let us get to know the force ~F better by remembering the properties of a
vector product:
• Magnitude of the magnetic force. The magnitude of a vector product
~a × ~b is ab sin θ . Therefore, the force on charge q produced by velocity
vector ~v and magnetic field vector B ~ with angle θ in between, will have the
magnitude
F = qvB sin θ
It can be seen that the force will be zero if the velocity is zero ( v = 0 ) or if it
is parallel to the magnetic field (θ = 0◦ ) .
• Direction of the magnetic force. The orientation of a vector ~c = ~a × ~b is
perpendicular to both ~a and ~b . We use the right-hand rule to determine
which of the two directions in this orientation is to be used:
According to the right-hand rule, when we point our four fingers towards Figure 19.3: Direction of the
the first vector (~a ) and point our palm towards the second vector (~b) , our magnetic force according to the
thumb gives the direction of ~c . right-hand rule.
Likewise, for a positive charge q , the direction of the force is perpendicular
to the plane formed by both ~v and B ~ , and when we point the four fingers of
the right-hand towards the vector ~v and the palm towards the field vector
~ , the thumb gives the direction of the force ~F .
B
When the charge q is negative (−) , the formula (19.1) gives a force in the
direction of −~F . In other words, the magnetic forces exerted upon negative and
positive charges are in opposite directions.
Unit of Magnetic Field
The force formula (19.1) allows us to determine the unit of magnetic field. If we
compare both sides of the equation and substitute (coulomb/second = ampere) ,
we get the unit of magnetic field in the SI system as
N N
1 =1 = 1 tesla = 1 T
C × m/s A·m
The unit of magnetic field was named the tesla (T) in memory of the great Serbian
scientist Nikola Tesla (1857–1943).
The unit of tesla is extremely large for normal magnetic fields. Another unit
of magnetic field is the gauss and
1 gauss = 10−4 T
For example, Earth’s magnetic field is approximately 1 gauss.

Motion of Charged Particles in a Magnetic Field
The fact that magnetic force is perpendicular to velocity ~v causes particles to
move in a specific trajectory. In the simplest case, let us consider a particle with
charge +q entering with velocity ~v perpendicularly into a medium in which the
magnetic field B~ is into the plane of the paper, as shown in Figure 19.4. (In the
following figures in this chapter, the × symbol means into the paper and the Figure 19.4: Magnetic force is
symbol means out of the paper.) always perpendicular to trajec-
According to the formula ~F = q(~v × B~ ) , the force exerted upon the particle tory.
will be perpendicular both to the velocity ~v and the magnetic field B~ . As force is
perpendicular to velocity, it cannot change the magnitude of the velocity, but can
only change its direction. (In other words, magnetic force does not perform work
on the particle.) Even when the particle is slightly deflected, the force will still be
perpendicular to the velocity ~v .
We know already what kind of motion has this property in which the force is
always perpendicular to the velocity vector. This is the circular motion that we
discussed in Section 3.3. In this case, the magnetic force provides the centripetal
force of the circular motion:
v2
Fr = qvB = m
r
From here, we find the radius of the circular motion as follows:
mv
r= (19.2)
qB
Note that the numerator of this formula is p = mv , the momentum of the particle.
In scientific research and technological applications, the radius of the circular
motion of a particle is measured inside of a known magnetic field and either the
velocity v , the charge q or the m of the particle can be determined from this.
Figure 19.5: The most general If the particle does not enter the magnetic field as fully perpendicular, the
spiral motion of a charged parti- component of velocity along the magnetic field does not change. In such a case,
cle. it performs circular motion on one hand and continues to move forward on the
other hand. This is the spiral motion.
Let us briefly introduce the instrument called the mass spectrograph, which
is an important application of this formula.
In this instrument, shown in Figure 19.6, a charge +q generated at a source
of ion is first accelerated to a known velocity v under a potential difference V in
an electric field. The charge then enters perpendicularly into a uniform magnetic
field B and exits by following a semicircular path. The distance 2r of the point at
which it exits is determined using detectors. This data can be used to determine
the mass m of a particle with a known charge q .
The mass spectrograph was used to determine the masses of atoms very pre-
cisely and prove the existence of isotopes. Today, it is frequently used in medicine
Figure 19.6: Mass spectro-
graph. and engineering, in addition to physics, chemistry and biology researches.
Example 19.1
The velocity vector ~v of charge q1 is given in the direction
of the +y axis as the vector B ~ . The vector product of two
parallel vectors is zero:
~F1 = 0 ~)
( ~v k B
Determine only the directions of the forces exerted upon charges

q1 , q2 and q3 thrown with velocity v from the corners of a
cube, in the directions shown in the figure. The magnetic field
vector B~ is in the +y direction.
The velocity of charge q2 is in the +x direction. If we point
Answer We use the formula (19.1) for the magnetic force: the four fingers of our right hand in the +x direction and
~F = q ~v × B
~ ~ ( +y ), our thumb will

point our palm in the direction of B
19.2. MAGNETIC FORCE ON A CURRENT-CARRYING WIRE 323
point in the +z direction. Therefore, the force ~F2 is in the right hand in the ~v ( −z ) direction and point our palm in the
+z direction. ~ ( +y ), our thumb will point in the +x direction.
direction of B
Note that charge q3 is negative. In other words, a force will be The force will be opposite to this direction. Therefore, the
~
exerted in the opposite direction of the one to be found using force F3 is in the −x direction.
the right-hand rule. Again, if we point the four fingers of our The results are shown in the figure above:
Example 19.2 Answer (a) The force exerted upon the proton will
be in the
+x direction according to the formula ~F=q ~v × B
~ , as shown

in the figure. We calculate its magnitude:

F = qvB sin 90◦ = evB
F = 1.6 × 10−19 × 106 × 0.1 = 1.6 × 10−14 N
(b) As force is always perpendicular to velocity, the proton

follows a circular trajectory in the xy -plane. Since magnetic
In an area where the magnetic field has a magnitude of B=0.1 T force provides the centripetal force, we find that
in the direction of the +z -axis, a proton is thrown at a velocity
v2 mv
of 106 m/s in the direction of the +y -axis. (The proton charge m = qvB −→ r=
is e=1.6 × 10−19 C and mass is m p =1.7 × 10−27 kg .) r qB
(a) Find the magnitude and direction of the force exerted upon We substitute the values and calculate the radius as follows:
1.7 × 10−27 × 106
the proton. r= = 0.11 m
(b) Find the radius of the circular trajectory of the proton. 1.6 × 10−19 × 0.1
19.2 MAGNETIC FORCE ON A CURRENT-CARRYING WIRE

A current-carrying wire remains neutral because there are positive charges
in the background of the moving electrons. Therefore, no electric force is exerted
upon a current-carrying wire in an electric field.
However, the case is different when a current-carrying wire is placed in a
magnetic field. Positive ions are not affected by the magnetic field, as they are at
rest; however, a magnetic force is exerted upon the moving electrons, and a net
magnetic force is thus generated on the wire.
We can calculate the force exerted upon a current I starting from the formula
~F=q(~v × B
~ ) exerted upon a charge q . Consider the amount of charge flowing
through the cross-section A during time interval dt (Figure 19.7). According to
the definition I = dq/dt of current, we get
dq = I dt
In section 18.1, we showed that electrons moved with an average drift velocity
vd in conductors. These charges will travel the distance L = vd dt during the Figure 19.7: Force exerted on a
time interval dt . As all these velocities are in the same direction and equal in charge dq in a conductor.
magnitude, we can assume that the total charge dq also moves with velocity
vd . Accordingly, the force exerted upon the total charge dq with velocity ~vd in
magnetic field B~ is
~F = dq ~vd × B
~ = I dt ~vd × B~

If we define ~L = ~vd dt as a displacement vector along the wire and in the direction
of the current, we can find the expression for the force exerted upon a piece of
wire with length L in a magnetic field:
~F = I ~L × B
~

(Magnetic force on a current) (19.3)
Note that the force is perpendicular to both the wire and the magnetic field.
Example 19.3
We calculate its magnitude as follows:
The wires extending along the sides of a cube with side length a , F1 = I LB sin 90◦ = IaB
as shown in the figure, carry the currents with the same magni- The force on I2 : This time, the vector product is zero, as the
~ is in the +y -direction. vector ~L is parallel to vector B
tude: I1 =I2 =I3 =I . The magnetic field B ~:
Calculate the forces exerted upon the currents I1 , I2 and I3 . F2 = 0
The force on I3 : As the vector B ~ in the figure is in the +y
direction, the vector ~F3 perpendicular to it will be parallel to
the xz -plane, in other words, inside that plane of the cube.
Likewise, the vector inside this plane that is perpendicular
to the vector ~L must be along its diagonal. The right-hand
rule gives the direction
√ of the force as downward. We take
the value L = 2a and calculate
√ the magnitude of the force:
Answer F3 = I LB sin 90 = 2IaB
◦
We use the formula (19.3) to calculate the magnetic force All three forces are shown in the figure below.
exerted upon a current-carrying wire:
~F = I ~L × B
~

Here, ~L is a vector with the length of the wire and in the

direction of the current.
The force on I1 : In the figure, the vector ~L is given in the −z
direction and the vector B ~ in the +y direction. The force ~F1
will be in the +x direction according to the right-hand rule.
Example 19.4
current flowing through the rod be such that the force on the
springs is zero?
Answer The magnetic force should be upward to balance
the weight of the rod. As the rod and the magnetic field are
perpendicular, we directly write the magnetic field and set it
equal to the weight W :
F = W → ILB = W
We calculate the current I from here as follows:
A conducting rod with weight W=0.3 N and a length of 50 cm W 0.3
is suspended with conducting springs, as shown in the figure. I= = = 3A
LB 0.5 × 0.2
The region has a magnetic field with a magnitude of 0.2 T into The current must flow from the left to the right so that the
the paper. What should the magnitude and direction of the force is upward according to the right-hand rule.
19.3 MAGNETIC TORQUE ON A CURRENT LOOP – THE ELECTRIC

MOTOR
A conducting loop placed between the poles of a magnet starts to rotate when
a current flows though it. This is the electric motor, which is the instrument that
ensured one of the greatest technological advances in history. Electric motors are
used everywhere, from tiny dentist drills to large cranes, electrical watches to
water pumps, CD drives to submarine propellers.
Figure 19.8 shows a rectangular loop placed inside of a magnetic field B. The
side lengths of the loop are a and b and it can freely rotate about the y -axis.
Figure 19.8: A current loop in When a current I flows through the loop, a force ~F = I (~L × B
~ ) will be exerted
a magnetic field. upon each side. The directions of these forces according to the right-hand rule
19.3. MAGNETIC TORQUE ON A CURRENT LOOP – THE ELECTRIC MOTOR 325
are shown in Figure 19.8. And their magnitudes are
F1 = I b B cos θ (in the −y direction)

F2 = I a B (in the −z direction)
F3 = I b B cos θ (in the +y direction)
F4 = I a B (in the +z direction)
The net force on the loop is zero, as these forces are equal and opposite. Therefore,
the loop performs no translational motion.
However, the total torque of the forces is different from zero, and it may thus
perform a rotational motion. Let us calculate the torques of these forces with
respect to the rotation axis y .
Figure 19.9: The forces on
the sides from different angles:
Views from (a) above, (b) front.
(c) The dipole moment m ~ in the
direction of the normal of the
~.
loop as it makes angle θ with B
Let θ be the angle between the normal of the loop and the magnetic field at
any time. Accordingly, we can write the sum of the moments as follows:
τ = F1 .0 + F2 (b/2) sin θ + F3 .0 + F4 (b/2) sin θ

τ = I ab B sin θ = I A B sin θ (19.4)
In this expression, the moments of the forces F1 and F3 are zero, because they
are parallel to the axis. A = ab is the area of the loop.
The product I A here is defined as the magnetic dipole moment of a current
loop and is defined as a vector perpendicular to the plane of the loop, in the
direction of the normal vector n̂ (Figure 19.9c):
~ = I A n̂
m
More generally, if the loop has N windings, the magnitude of the dipole moment
will be:
m = N IA (Magnetic dipole moment) (19.5)
The aforementioned torque calculation is also valid for geometric shapes
other than a rectangle. As a result, the amount of torque exerted on a current
loop with dipole moment m ~ is found as follows:
~ in a magnetic field B
~τ = m ~
~ ×B (Magnitude: τ = mB sin θ) (19.6)
Electric Motor Figure 19.10: Hybrid automo-

The torque given with the expression (19.4) above, will try to rotate the loop biles have two motors, one gaso-
line and one electric (right).
counterclockwise with respect to the directions given in the figure, as it is positive.
The electric motor activates at
However, after the loop makes a half rotation, in other words, when the bottom startup and when the speed
side goes to the top, the torque of forces F2 and F4 will be negative this time, as changes.
the currents are in the reverse direction. In this case, the loop will try to rotate in
the reverse direction.
Continuous rotation in the same direction cannot be achieved in this mecha-

nism. It is necessary to reverse the direction of the current at each half rotation.
There are two ways to achieve continuous rotation in electric motors:
1. In alternating current (AC) motors, the current varies as a sinusoidal wave.
After the loop makes a half rotation, the direction of the current reverses
automatically and ensures that the torque remains in the same direction.
2. In direct current (DC) motors, the current that always flows out of the battery
in the same direction passes through a mechanism called the commutator
before reaching the loop (Figure 19.11). A commutator consists of two half
Figure 19.11: The commutator conductor rings, and the side connected to the + charged pole of the battery
reverses the current at each half changes at each half rotation. The direction of the current thus changes at
rotation. each half rotation of the loop and the torque is ensured to remain in the same
direction.
Example 19.5
Among these forces, only the moment of ~F3 is different from

zero with respect to the z -axis. As ~F1 is along the axis, and as
~F2 and ~F4 are parallel to the axis, they do not cause rotation
and have zero torques.
The remaining force ~F3 is in the −x direction and its magni-
tude is calculated using the formula (19.3) and multiplied by
the number of windings N :
The loop shown in the figure consists of 50 windings and can F3 = NIbB sin 90◦ = 50 × 3 × 2 × 2 × 1 = 600 N
rotate around the z -axis. A current of I=3 A is flowing through
The moment arm a cos 37◦ is used when finding the torque
each winding. The dimensions of the loop are a=1 m and
of force F3 with respect to the z -axis:
b=2 m . The magnetic field B = 2 T is towards the y -axis.
(a) Calculate the net torque on the loop by calculating the τ = τ3 = F3 a cos 37◦
torques acting on each side. τ = 600 × 1 × 0.8 = 480 N·m
(b) This time, calculate the net torque by using the magnetic The direction of the torque will be so as to turn from the x -
dipole moment of the loop. to the y -axis.
Answer
(b) The magnetic dipole moment of the loop is calculated with
(a) The forces acting on each side are as shown in the figure the formula (19.5):
according to the right-hand rule:
m = NIA = 50 × 3 × (1 × 2) = 300 A·m2
We then calculate the torque from the formula (19.6). Here,
the angle between the magnetic field and the surface normal
should be taken:
τ = mB sin θ = 300 × 2 × sin 53◦ = 480 N·m
As it can be seen, both calculations give the same result, but
it is easier to work with magnetic moments.
Example 19.6
torque that is trying to rotate it.
Answer
The magnetic dipole moment of the loop is calculated with
the formula (19.5):
m = IA = I πr2 = 25 × 3.14 × 0.402
A current of 25 A is flowing through a circular loop with a m = 12.6 A·m2
radius of 40 cm on the xy -plane. This region has a uniform We then calculate the torque from the formula (19.6).
magnetic field of 0.3 T in a direction with a 45◦ angle with τ = mB sin θ = 12.6 × 0.3 × sin 45◦
the xy -plane. Find the magnetic moment of the loop and the τ = 2.7 N·m
1. Which is the unit of magnetic field in the SI unit system?
(a) tesla (b) gauss (c) ampere (d) einstein
2. What will the direction of the force be exerted upon a

positively charged particle thrown in the +y direction
in a magnetic field that is in the +x direction?
(a) +x (b) +y (c) +z (d) −z 9. Which of the forces exerted upon the charges shown in
the figure above is to the right? (The magnetic field is
into the paper.)
3. Which of the following figures correctly shows the force
acting on charge +q with the velocity ~v in the magnetic (a) q1 (b) q2 (c) q3 (d) q4
~?
field B
10. What will the direction of the force be exerted upon a
current in the +y direction in a magnetic field in the +x
direction?
(a) +x (b) +y (c) +z (d) −z
(a) a & b (b) b & c (c) a & c (d) a & d
11. By what factor does the magnetic force exerted upon
4. If a magnetic force towards the East is exerted upon a a current-carrying wire increase if its length and the
positively charged particle thrown towards the North, magnetic field are both doubled?
in which direction could the magnetic field be? (a) No change (b) 2 (c) 4 (d) 8
(a) Up (b) Down (c) South (d) West
12. By what factor does the magnetic force exerted upon a
5. What will the magnitude and direction of the force be wire increase if the current and the magnetic field are
exerted on a particle with charge 4 C thrown upwards both doubled?
from the ground with a velocity of 3 m/s in a 2 T mag- (a) No change (b) 2 (c) 4 (d) 8
netic field towards the North?
(a) 24 N to the West 13. In which of the following cases will the magnetic force
(b) 24 N to the East exerted upon a current-carrying wire be zero?
(c) 6 N to the West I. If the current is perpendicular to the magnetic field,
(d) 6 N to the East II. If the current is parallel to the magnetic field,
III. If the current is zero,
6. By what factor does the magnetic force acting on an IV. If the magnetic field is constant.
object increase if its velocity and the magnetic field are (a) I & II (b) II & III (c) I & III (d) II & IV
both doubled?
(a) No change (b) 2 (c) 4 (d) 8 14. What will the magnitude and direction of the force be ex-
erted upon a wire with a length of 4 m and carrying 3 A
7. In which of the following cases will the magnetic force in current upward from the ground in a 2 T magnetic
exerted on a charge be zero? field towards the North?
I. If the velocity is perpendicular to the magnetic field, (a) 24 N to the West
II. If the velocity is parallel to the magnetic field, (b) 24 N to the East
III. If the velocity is zero, (c) 6 N to the West
IV. If the magnetic field is constant. (d) 6 N to the East
(a) I & II (b) II & III (c) I & III (d) II & IV
8. If the magnetic force exerted upon a charge +q is 5 N

in the +y direction, what will the force exerted upon a
charge −2q be?
15. Which of the above figures correctly show the force
(a) 10 N in the +y direction ~?
acting on the current I in the magnetic field B
(b) 10 N in the −y direction
(c) 2.5 N in the +y direction (a) a & b (b) b & c (c) a, b & d (d) a & d
(d) 50 N in the −y direction
16. Which section of the wire shown in the figure below has (a) The charges of the particle are equal.
a force exerted upon it towards the right? (b) The masses of the particles are equal.
(c) The charge/mass ratios of the particles are equal.
(d) The product charge×mass of the particles are equal.
19. What is the maximum torque exerted upon a loop with

a cross-section area of 1 m2 and carrying 2 A in current
in a 3 T magnetic field?
(a) I1 (b) I2 (c) I3 (d) I4 (a) 5 N·m (b) 6 N·m (c) 2 N·m (d) 4 N·m
17. A charged particle is in circular motion in a magnetic

20. In what orientation should a current loop be placed in a
field. If its charge and mass are doubled, by what factor
magnetic field for the torque exerted upon it to be zero?
will the radius of the circle change?
(a) Parallel to the magnetic field
(a) No change (b) 2 (c) 4 (d) 1/2
(b) Perpendicular to the magnetic field
(c) 45◦ with the magnetic field.
18. Two charged particles with the same velocity are in cir-
(d) It is never zero.
cular motion with the same radius in a magnetic field.
Which of the following is correct?
Problems
19.1 Magnetic Force on Charges exerted upon it. What is the angle between the velocity and
the magnetic field? [A: 30◦ .]
Problem 19.4
19.4 In a region where the magnetic field has magnitude
B=0.1 T in the direction of the +z -axis, an electron is thrown
Problem 19.1 at a velocity of 2 × 108 m/s along the +x -axis.
19.1 Determine only the directions of the forces exerted
(a) Find the magnitude and direction of the force exerted upon
upon charges q1 , q2 and q3 thrown with velocity v from the
the electron. (b) Calculate the radius of the circular trajectory
corners of a cube, in the directions shown in the figure. The
of the electron. (The electron’s charge is −e= − 1.6 × 10−19 C
magnetic field vector B~ is in the +z direction.
and its mass is me =9.1 × 10−31 kg .)
[A: F1 = 0 , F2 : in the +y direction: F3 in the −x
[A: (a) 3.2 × 10−12 N in +y -direction, (b) 0.01 m .]
direction.]
Problem 19.5
Problem 19.2 19.5 An electron enters the region of a magnetic field shown
19.2 In a region with a magnetic field with a magnitude of in the figure with a velocity of 2 × 108 m/s and exits perpen-
0.4 T in the +x direction, a q=3 µC charge is thrown with dicular to its incoming direction after traveling a quarter of a
a velocity of 5 × 106 m/s and at a 37◦ angle with the mag- circle 1.57 cm long. Calculate the magnitude of the magnetic
netic field. What are the magnitude and direction of the force field in the region. (Hint: Calculate the radius from the length
exerted upon the charge? [A: 3.6 N in the −z direction.] of the quarter circle.) [A: 0.11 T .]
19.3 When a 3 µC charge with a velocity of 106 m/s is 19.6 An electron at rest is first accelerated under an electric
thrown into a B = 0.1 T magnetic field, a 0.15 N force is potential difference of 200 kV , after which it enters a region
PROBLEMS 329
where a perpendicular magnetic field of B=0.1 T is present.

(a) What is velocity of the electron? (b) What will the radius
of the circular path in the magnetic field be?
[A: (a) 2.7 × 108 m/s , (b) 1.5 cm .]
Problem 19.11
19.11 A conducting rod with weight W = 0.5 N and a length
of 40 cm is suspended with conducting springs, as shown in
the figure. The region has a magnetic field with a magnitude
Problem 19.7 of 0.1 T out of the paper. What should the magnitude and
19.7 The proton shown in the figure is thrown into a region direction of the current flowing through the rod be such that
of unknown magnetic field with a velocity of 2 × 106 m/s . the force on the springs is zero. [A: 12.5 A to the left.]
The proton travels a semicircle and exits 20 cm away. Calcu-
late the magnitude and direction of the magnetic field in the
area ( m p = 1.7 × 10−27 kg ). [A: 0.21 T out of the paper.]
19.2 Magnetic Force on Currents
19.8 When a current-carrying wire with a length of 75 cm is Problem 19.12

placed perpendicularly in a magnetic field with a magnitude
19.12 A metal rod with a length of 40 cm , a weight of 0.5 N
of 0.2 T , a 3 N force is exerted upon it. What is the current
and zero resistance is freely placed on two supports, as shown
flowing through the wire? [A: 20 A .]
in the figure. The area has a magnetic field of 0.3 T out of
the paper. What is the minimum value of the resistance R
such that the rod does not jump into the air when the circuit
is connected to a 12 V -battery? [A: 2.9 Ω .]
Problem 19.9
19.9 The wires extending along the sides of a cube with side
length a , as shown in the figure, carry the currents with the
same magnitude: I1 =I2 =I3 =I . The magnetic field B ~ is in the
+z -direction. Calculate the forces exerted upon the currents
Problem 19.13
I1 , I2 and I3 .
[A: F1 = 0 , F2 = IaB : in the −y direction, F3 = IaB : in the 19.13 A constant current of 3 A is given to the circuit in
−x direction] the figure above from a source. A metal rod with a mass of
100 g and a length of 50 cm is freely placed on the friction-
less points ab . There is a magnetic field with a magnitude of
0.2 T into the paper. What will the acceleration of the rod be
when the current is flowing? [A: 3 m/s2 .]
19.3 Magnetic Torque
Problem 19.10
19.10 A 3 A current is flowing through a wire that makes a
53◦ angle with the field lines in an area with a 0.5 T magnetic Problem 19.14
field in the plane shown in the figure. Calculate the magni- 19.14 A 5 A current is flowing through a circular loop with
tude and direction of the force exerted upon a unit length of radius 10 cm in the xy -plane. This region has a magnetic
the wire. [A: 1.2 N into the paper.] field of 0.2 T in a direction at a 37◦ angle with the xy -plane.
Find the magnetic moment of the loop and the torque that is each winding. The dimensions of the loop are a = 1 m and
trying to rotate it. [A: m = 0.16 A·m2 , τ = 0.25 N·m .] b = 2 m . The magnetic field in the region is B=3 T towards
the x -axis. (a) Calculate the magnetic moment of the loop,
(b) Calculate the torque exerted on the loop.
[A: (a) 200 A·m2 , (b) 360 N·m .]
19.16 In the Bohr model of the Helium atom, an electron ro-

tates with a velocity of 6.6×106 m/s around the nucleus in an
orbit with a radius of 1.8 × 10−11 m . (a) In how many seconds
does the electron complete one circle? (b) If we consider the
electron as the charge flowing through a given cross-section,
Problem 19.15 what will the current I be? (c) What is the magnetic moment
19.15 A rectangular loop made of 20 windings can rotate generated by this orbital motion of the electron?
around the z -axis and a current of I=5 A is flowing through [A: 1.7 × 10−17 s , (b) 9.3 mA , (c) 9.5 × 10−24 A·m2 .]
?
20
SOURCES OF MAGNETIC
FIELD
The 2000-ton CMS detector is

being inserted into its socket at
the CERN European Nuclear Re-
search Center. The supercon-
ductor magnet at the center of
the detector (the gray area) can
generate a magnetic field up to
4 T.
We started our study of magnetism in Chapter 19 by examining its effects on

moving charges and currents. But we did not discuss the source of this magnetic
field. Now, we shall examine how a magnetic field is generated.
The source of magnetic field is moving charges and currents. This was first
discovered by the Danish scientist Oersted in 1819 through his observation of the
deflection of a compass needle near a current-carrying wire. Soon afterwards, the
French scientists Biot and Savart were able to establish the form of the magnetic
field produced by any current.
In electrostatics, we had used the Gauss law as a shortcut for calculating
electric field. There is a similar law for magnetic field: We shall examine Ampère’s
law in this chapter. Then, we will look at the magnetic properties of matter on
the atomic scale.

332 20. SOURCES OF MAGNETIC FIELD
20.1 MAGNETIC FIELD OF A CURRENT

As the simplest case, let us consider the magnetic field generated by a straight
wire carrying a current I (Figure 20.1). Observations have revealed the following:
• Magnetic field lines are circles centered around the wire on a plane perpen-
dicular to the wire.
• According to the right-hand rule, the direction of the magnetic field is in
the direction in which the four fingers fold when the thumb points in the
direction of the current.
Figure 20.1: Magnetic field of a • The magnitude of the magnetic field decreases as inversely proportional to
straight wire current. the radius r .
The French scientists Jean-Baptiste Biot and Felix Savart performed meticulous
experiments on the magnetic field of current-carrying wires and discovered the
expression of magnetic field for all kinds of currents. Known as the Biot-Savart
law, this formula gives the contribution of a small current element I d` with
length d` to the magnetic field (Figure 20.2):
The Biot-Savart Law

The contribution of a small current element I d` with length d`
and carrying a current I to the magnetic field at a point at dis-
tance r is
I d` sin θ
dB = k0 (20.1)
r2
The angle θ here is the angle between the position vector ~r and
the direction of d` .
Figure 20.2: The contribution
dB of a piece d` of a current-
carrying wire to the magnetic The constant k0 in this formula has the following value in the SI system:
field at point P located at ~r
is perpendicular to the plane k0 = 10−7 T·m/A (20.2)
formed by these two vectors.
In electrics, we had expressed the constant k in terms of ε0 , the electric permit-
tivity of free space. Likewise, in magnetism, a constant µ0 called the magnetic
permeability of free space is defined as follows:
µ0
k0 = −→ µ0 = 4π × 10−7 T·m/A (20.3)
4π
Let us emphasize the important points of the Biot-Savart law:
• The magnetic field of each small current element I d` is perpendicular to the
plane formed by the element I d` and the position r .
• The right-hand rule gives the direction of the magnetic field among the two
possible perpendicular directions: The thumb is directed towards the current,
the four fingers are pointed towards r and the palm shows the direction of
the vector dB~.
• The magnetic field contribution is directly proportional to the current I and
inversely proportional to r2 , the square of the distance.
• Magnetic field is directly proportional to the sine of the angle between the
element I d` and r .
20.2. MAGNETIC FIELD CALCULATIONS 333
Vector Product Expression of the Biot-Savart Law

In the Biot-Savart formula above, sin θ and the lengths d` and r indicate
that this is actually a vector product. Indeed, if we write the piece d` as a vector
d~` in the direction of the current I and take r̂ as the unit vector in the direction
of r , we get the vector expression of the Biot-Savart law as follows:
~ = k0 I d~` × r̂
dB (Biot-Savart: vector expression) (20.4)
r2
It can be seen that this expression is equivalent to the magnetic field Eq. (20.1) in
terms of both magnitude and direction.
Magnetic Field of a Finite Wire
Every current flows through a finite wire consisting of the sum of small d`
parts. The magnetic field of a whole current will be the limit sum, in other words,
~ vectors above:
the integral of the small dB
I d~` × r̂
Z
~ = k0
B (20.5)
r2
This integral will be taken over the whole wire.
20.2 MAGNETIC FIELD CALCULATIONS

Let us calculate the magnetic fields of the most essential current distributions.
Magnetic Field of an Infinite Straight Wire
Consider a straight wire extending along the x -axis chosen upward, and
carrying current I . Let us calculate the magnetic field at a point P as shown in
Figure 20.3. We choose the origin at the point on the wire that is closest to point
P at a distance a . (This choice will not affect the generalization, as the wire is
infinite.)
Let us choose the length element d` on the wire at distance x and with length
dx . If we write the contribution of this piece to the magnetic field at point P
located at distance r according to Eq. (20.1) of the Biot-Savart law, we get
Figure 20.3: The dB contribu-
I dx sin θ
dB = k0 tion of a length element dx on
r2 an infinite wire to the magnetic
The direction of the magnetic field dB will be into the plane of the paper. The field at point P .
contributions of all other dx elements on the wire to this point P will always
be into the paper, in other words, in the same direction. Therefore, the algebraic
sum of these dB contributions can be directly integrated. Considering that the
wire ranges from (−∞) to (+∞) , we get
dx sin θ
Z ∞
B=k I 0
−∞ r2
Let us write all the variables in the integral in terms of x . If we use the angle
(π − θ) , the complement of angle θ , we get
p
r = x 2 + a2
a a
sin θ = sin(π − θ) = = √
r x + a2
2
When we use these values,

√
dx (a/ x2 + a2 )
Z ∞ Z ∞
dx
B=k I0
= k Ia
0
−∞ x +a
2 2
−∞ x2 + a2
3/2
| {z }
2
2/a
Let us state here that the result is 2/a2 without going into the details of the
Figure 20.4: Magnetic field of a calculation. (It can be calculated by a change of variable x=−a cot θ .) Accordingly,
straight wire. the magnetic field of an infinite straight wire is
2k0 I
B= (Magnetic field of a straight wire) (20.6)
a
The magnetic field lines form concentric circles with the wire as the axis. The
~ will be tangent to these circles (Figure 20.4).
vector B
Magnetic Field of a Current Loop
We wish to find the magnetic field at a point along the axis of a circular loop
with radius a and carrying a current I . Let us choose the small length element
d` along the wire as the arc of circle ds (Figure 20.5). Let us place this arc ds
on the y -axis and in the +z direction. Let point P be located on the x -axis at
distance h . These choices do not affect the generality because of the symmetry
of the loop.
Figure 20.5: The contribution As seen in the figure, the contribution dB of this current element I ds will be
dB of a piece of arc ds to the perpendicular to both ds and the unit vector r̂ , and therefore will be located in
magnetic field. the xy -plane. Also, the angle between ds and r̂ will be θ = 90◦ , because every
vector located on the plane perpendicular to ds will be perpendicular to ds .
If we write this contribution dB according to the Biot-Savart law, we get
I ds sin 90◦ ds
dB = k0 = k0 I 2
r 2 a + h2
Let us separate the vector dB~ into two components, one in the x -direction and
the other perpendicular to it:
dBx = dB cos γ
dB⊥ = dB sin γ
If we move the arc ds around the loop and add each contribution dB~ , the contri-
butions dB⊥ will make a circle around point P and give zero contribution due to
symmetry: Z
dB⊥ = 0 (due to symmetry)
Therefore, the contributions in the direction of the x -axis will give a total magnetic
field in the x -direction:
Z Z
B= dBx = dB cos γ
The γ angles shown in two places in the figure are equal because
√ they are angles
with perpendicular sides. Accordingly, we get cos γ = a/r = a/ h2 + a2 and the
20.2. MAGNETIC FIELD CALCULATIONS 335
integral simplifies as follows:

Z Z
ds a
B = dB cos γ = k0 I √
a + h h2 + a2
2 2
0
k Ia
I
= ds
h2 + a2 3/2

All of the constants are taken outside of the integral here and the sign is used
H
to indicate that the elements ds form a closed curve around the loop.
The sum of the pieces of arc around a complete loop gives the circumference
of the circle. Therefore, if the value of the integral is taken as 2πa , we find the
result as follows:
2πk0 I a2
B= (Magnetic field on the axis of a loop) (20.7)
h2 + a2 3/2

The magnetic field at the center of the loop is found by taking h = 0 in this
formula:
2πk0 I
B= (Magnetic field at the center of a loop) (20.8)
a
Figure 20.6 shows the magnetic field lines of a current loop. The lines are not
parallel to the axis but curve over it at points outside of the axis. Figure 20.6: Magnetic field
Magnetic Field of a Magnetic Dipole lines of the current loop and
magnetic dipole moment.
In section 19.3, we defined the magnetic dipole moment of a current loop as
follows (Eq. 19.5):
m=IA
A is the surface area of the loop. Let us return to Eq. (20.7), which we found
above for the magnetic field on the axis of a current loop:
2πk0 I a2
B=
h2 + a2 3/2

We can easily form a dipole moment in this expression. As the surface area of the
circle is πa2 , we get
2k0 (I πa2 ) 2k0 m

B= =
h2 + a2 3/2 h2 + a2 3/2

At distances very far from the magnetic dipole, in other words, when h a , the
approximate expression of magnetic field becomes as follows:
2k0 m
B≈ (Field of a magnetic dipole at large h ) (20.9)
h3
Considering a current loop as a magnetic dipole has many advantages: An electron
rotating around the nucleus in an atom generates a circular current loop and has
an orbital magnetic dipole moment. As we shall discuss later in this chapter, the
magnetic properties of matter can be explained by taking atoms to be very small
magnetic dipoles.
Example 20.1
find the distance a :
−5 2k0 I 2 × 10−7 × 1
The Earth’s magnetic field is 5 × 10 T . a= =
(a) At what distance from a straight wire carrying a current of B 5 × 10−5
a = 0.004 m = 4 mm
1 A will the magnetic field be equal to this value?
(b) How much current should pass through a circular loop with (b) We use Eq. (20.6), which gives the magnetic field at the
a radius of 1 m for the magnetic field at its center to be equal center of a circular loop:
to this value? 2πk0 I
B=
Answer a
We again set this expression equal to Earth’s magnetic field
(a) We use Eq. (20.6) for the magnetic field of a straight wire: and calculate the current:
2k0 I aB 1 × 5 × 10−5
B= I = = = 80 A
a 2πk0 2 × 3.14 × 10−7
We set this expression equal to the Earth’s magnetic field and
Example 20.2
magnetic fields will appear as viewed from the top of the
paper in the figure above.
Then, we use the formula B = 2k0 I/a to calculate the mag-
netic field of each current:
At point A:
2k0 I1 2 × 10−7 × 100
B1 = = = 0.002 T = 2 mT
a 0.01
In the figure, there is a 2 cm distance between the parallel 2 × 10−7 × 200
currents I1 = 100 A and I2 = 200 A . B2 = = 4 mT
0.01
(a) Calculate the total magnetic field at the midpoint A be- We take the resultant vector of the two opposite vectors at
tween the wires. point A:
(b) Calculate the magnetic field at point B located 1 cm away BA = B1 − B2 = 0.002−0.004 = −2 mT
from the wire I2 .
At point B:
2 × 10−7 × 100
B1 = = 0.00067 T = 0.67 mT
0.03
2 × 10−7 × 200
B2 = = 0, 004 T = 4 mT
0.01
Answer We take the resultant vector of the two vectors in the same
(a) First, let us determine the directions of the magnetic fields direction at point B:
B1 and B2 at both points. Using the right-hand rule, the BB = B1 + B2 = 0.67 + 4 = 4.7 mT
Example 20.3
(b) There is no need to recalculate the value of the magnetic
1 A of current is flowing through a circular conducting loop field here. We only find the ratios of the expressions at the
with a radius of 1 m . center and at distance h . The expression for the magnetic
(a) What is the magnitude of the magnetic field at the center field at point h along the axis of a current loop was given
of the loop? with in Eq. (20.7):
(b) At what point along the axis will the magnetic field drop 2πk0 I a2
down to half of its value at the center? B=
h2 + a2 3/2

Answer Let us use B0 to indicate the expression that we wrote above
(a) We use Eq. (20.8) to find the magnetic field at the center for the field at the center of the loop, and take the ratio of
of the current loop: these two fields:
2πk0 I 2πk0 I a2
B= B
=
a −3/2
3/2 × 2πk0 I = (h/a) + 1
2
a
We substitute the numerical values and calculate the magnetic B0 h +a
2 2
field: We substitute the values B=B0 /2 and a=1 and solve for h :
2 × 3.14 × 10−7 × 1
B= = 6.3 × 10−7 T [h2 + 1]3/2 = 2 → h = 0.77 m
1
20.3. FORCE BETWEEN PARALLEL CURRENTS – AMPERE UNIT 337
Example 20.4 2 × 10−7 × 50

B2 = = 0.002 T = 2 mT
0.005
We then use the right-hand rule to find the directions of these
magnetic field vectors. The direction of both fields will be as
follows:
In the figure, the straight wires perpendicular to the plane of

the paper have currents I1 = 60 A and I2 = 50 A . Calculate
the components of the total magnetic field at point P.
Answer Lastly, we find the sum of these two vectors in terms of their
We first use Eq. (20.6) to calculate the magnitudes of the fields components:
~ =B
B ~1 + B
~2
B1 and B2 at point P:
2k0 I Bx = B1 − B2 cos 53◦ = 4 − 2 × 0.6 = 2.8 mT
B= By = −B2 sin 53◦ = −2 × 0.8 = −1.6 mT
a
2 × 10−7 × 60 The magnitude and direction of the total magnetic field can
B1 = = 0.004 T = 4 mT be calculated from these components if required.
0.003
20.3 FORCE BETWEEN PARALLEL CURRENTS – AMPERE UNIT

In Chapter 19, we calculated the force exerted upon a current placed inside of
a magnetic field. We now know the magnetic field produced by current-carrying
wires. Therefore, we can understand the interaction between two current-carrying
wires: The magnetic field generated by one of the wires will exert a force upon
the second wire.
Let us consider two straight parallel wires with distance d in between (Fig-
ure 20.7). Suppose currents I1 and I2 are flowing through these wires in the same
direction. The magnetic field generated by the current I1 at distance d is given
by the equation (20.6):
2k0 I1
B1 =
d
In Chapter 19, we saw that the force exerted upon the length L of the second
wire in this magnetic field B1 is ~F = I(~L × B
~ ) (equation 19.1):
Figure 20.7: The magnetic field
~F = I2 ~L × B ~1

of two parallel currents exerts a
force upon the other.
As shown in Figure 20.7, the magnetic field B1 is perpendicular to the other wire
with current I2 . In such a case, the force will also be perpendicular to the current
I2 and the field B1 and be towards the wire I1 . In other words, the two wires
will attract each other. Let us calculate the magnitude of the force:
2k0 I1 I1 I2
F = LI2 B1 sin 90◦ = LI2 = 2k0 L
d d
We could have done this calculation for the force exerted upon the current I1 in
the magnetic field generated by the current I2 . In such a case, we would have
found the same force to be equal and in the opposite direction in accordance with
Newton’s third law.
If the currents were taken as opposite to each other (anti-parallel), we would Figure 20.8: Parallel currents
have found that the wires repelled each other. Therefore, parallel currents attract attract and anti-parallel currents
and anti-parallel currents repel each other. repel each other.
The magnitude of the force in both cases is as follows:
2k0 I1 I2
F= L (Force between parallel currents) (20.10)
d
Definition of the Ampere Unit

Let us calculate the expression for force between parallel currents that we
found above, between two equal currents I1 = I2 = 1 A separated by d = 1 m .
The force exerted upon a unit length of wire will be as follows:
F 2k0 I1 I2 2 × 10−7 × 1 × 1
= = = 2 × 10−7 N/m
L d 1m
This expression is the definition of the unit of current ampere (A), which is one
of the basic units in the SI system:
When a force of 2 × 10−7 N/m is exerted per unit length between two parallel
wires separated by 1 m and carrying identical currents, the current flowing through
the wires is 1 ampere (A).
Figure 20.9: The Watt balance This definition allows for the most accurate measuring assembly that can be
used to measure the force be- installed in a laboratory. An instrument called the Watt balance (Figure 20.9) is
tween currents. used to keep a current-carrying weight in balance with the same current, and it
is thus ensured that the current in both wires is identical.
Example 20.5
2 × 10−7 × 1002
F1 = F2 = × 1 = 2N
0.001
We use the practical rule to find the directions of these two
forces: Parallel currents attract and anti-parallel currents re-
pel each other. Accordingly, the forces on current I3 are as
follows:
In the figure, wires perpendicular to the plane of the paper con-

stitute the corners of an equilateral triangle with side lengths
of 1 mm . Since I1 = I2 = I3 = 100 A , calculate the total force
exerted upon 1 m of length on the wire with current I3 .
Answer
The force between parallel currents is given in Eq. (20.10):
2k0 I1 I2 The sum of two equal forces making equal angles with the
F= L horizontal will be towards their angle bisector, in other words,
d
The magnitude of the forces exerted by the currents I1 and horizontal:
I2 are equal, because the magnitude of the currents and dis- F = F1x + F2x = 2F1 cos 60◦
tances are equal: F = 2N
Example 20.6
A current of I2 =4 A flows through the rectangular loop laying
on the same plane as the straight wire carrying a current of
I1 =25 A , as shown in the figure. The distance of the loop to the
near side is a=1 cm , its width is b=3 cm and length is c=5 cm .
Calculate the net force exerted upon the loop.
Answer
We first write the magnetic field of the straight wire:
20.4. AMPERE’S LAW 339
0
2k I1 forces exerted upon four sides are in the directions shown in
B1 =
r the figure:
Its direction is into the plane of the paper inside of the loop.
Eq. (19.1) is used to find the force exerted upon the piece The forces F2 and F4 do not contribute to the total force, as
of current I2 with they are equal and in the opposite direction. Therefore, it is
length L in this magnetic field:
~F = I2 ~L × B
~1 sufficient to calculate the forces F1 and F3 :
c
F1 = I2 LB1 = 2k0 I1 I2
a
5
= 2 × 10−7 × 25 × 4 = 1 × 10−4 N
1
5
F3 = 2 × 10−7 × 25 × 4 = 0.125 × 10−4 N
1+3
The difference of these two forces gives the total force as
being towards the right:
Using the right-hand rule to calculate this vector product, the F = F1 − F3 = 0.875 × 10−4 N
20.4 AMPERE’S LAW

In using the Biot-Savart law, we have to take an integral to find the magnetic
field of a current distribution. These integrals are often complicated. However, we
can find the magnetic field without having to integrate if the current distribution
has some symmetry. This is possible with Ampère’s law.
Let us consider a simple case to understand the essence of Ampère’s law.
We had previously calculated the magnetic field of an infinite straight wire. We
had found the magnetic field at distance r from a wire carrying a current I in
Eq. (20.6):
2k0 I Figure 20.10: Closed circular
B=
r curve around a straight wire.
Now, let us take the integral of the magnetic field’s component that is tangent
to the circle with radius r . As shown in Figure 20.10, we multiply the tangent
component of the magnetic field B at each point with the small element of arc
ds at that point and add them over the circle. This is different from the familiar
integral that we take along the x -axis. It is called a line integral, because it
is taken along a curved line (circle). As the magnetic field of an infinite wire is
tangent to the circle with radius r anyway, it is sufficient to take its integral:
I I
B ds = B ds
The magnetic field B was taken outside of the integral, as it has the same value
everywhere on the circle. The integral of small elements of arc ds over the full
circle will be the circumference 2πr . If we also use the expression for magnetic
field, we get
2k0 I
I
B ds = B 2πr = 2πAr = 4πk0 I = µ0 I
rA
In the last expression, we replaced k0 = µ0 /4π with the magnetic permeability of
vacuum µ0 .
It is surprising that the result is independent of radius r . Of course, in this
particular case, this resulted from the fact that the magnetic field of an infinite
wire is inversely proportional to r . However, advanced integration techniques
can be used to prove that the integral of any current distribution over any closed
curve gives the same result in the most general case.
The result would have been zero if this integral had been taken over any
curve that left the current I outside. If we accept this without a proof, Ampère’s
law is expressed as follows:
Ampère’s Law
The line integral of the tangential component of the magnetic
field over any closed curve is proportional to the net current
crossing any surface bounded by the closed curve:
I
~ · d~s = µ0 Iencl
B (20.11)
In this formula, Iencl indicates the net current enclosed by the closed loop, in
other words, if the current in one direction is positive, the current in the negative
direction is included as negative. Currents outside of the loop are not taken into
Figure 20.11: The currents consideration. The scalar product B ~ · d~s shows that the projection of the magnetic
I1 , I2 are taken into considera- field along the path is to be taken.
tion according to Ampère’s law. It is not necessary to carry out this integral when applying Ampère’s law. We
The current I3 outside of the take the magnetic field B outside of the integral by examining the symmetry of
loop is not taken into considera-
the problem, and this makes it easier to calculate the left-hand side. It is only
tion.
required to add the currents on the right-hand side.
Example 20.7
perpendicular to both d` and r .)
The current I flowing through an infinite cylindrical conductor According to Ampère’s law Eq. (20.11), the integral of
with radius R is evenly distributed throughout its cross-section. this tangential magnetic field along the circle with radius r
Find the magnetic field outside and inside of the cylinder. should be equal
I to the enclosed net current:
B
As B is parallel to the path, we have B ~ · d~s = B ds and take
the constant value B outside of the integral. The remaining
integral is equal to the circumference of the circle 2πr . The
current on the right-hand side is the total current I :
µ0 I 2k0 I
B (2πr) = µ0 I → B = =
2πr r
This result is the same as the straight wire expression.
Answer
Inside of the wire: Let us again consider an imaginary circle
It is always necessary to find a symmetry to apply Ampère’s
with radius r < R . Again, according to symmetry, the mag-
law. The most basic symmetry in magnetism is the magnetic
netic field on the circle should have equal value at every point
field of a straight wire current. The magnetic field of a straight
on the circle and be tangent to the circle. We write Ampère’s
wire is generated as tangent to circles centered around the
law as follows:I
wire. This should be the starting point if nothing disrupts
this symmetry. B ds = B (2πr) = µ0 Iencl
Outside of the wire: Let us consider an imaginary circle with This time, the current inside of the circle is less. As the current
radius r > R . Looking from a point on this circle, both halves is uniformly distributed over the cross-section of the cylin-
of the wire will give the same contribution because of the der with radius R , we use proportions to find the amount
symmetry. Therefore, the magnetic field should be in a plane enclosed inside of the radius r :
perpendicular to the wire. Again, looking from a point on this Iencl =
I
· πr2
circle, the half sections along the wire would give the same πR2
contributions that would result in a magnetic field along the We substitute this expression and find B :
µ0 Ir 2k0 I
tangent of the circle. (We could also deduce that dB is in B= = r
the tangential direction from the Biot-Savart law, dB being 2πR2 R2
20.4. AMPERE’S LAW 341
Example 20.8
Solenoid. Calculate the magnetic field inside and outside of

a solenoid with infinite length and carrying a current I . The
number of windings per unit length is n .
Now, let us apply Ampère’s law along the rectangular

path abcd shown in the figure above:
Answer I hZ b Z c Z d Z a i
~ · d~s =
B + + + ~ · d~s = µ0 Iencl
B
We obtain a solenoid by taking a conducting wire and wind- a b c d
ing it spirally around a long cylinder. Recall from the topic Only the first of these integrals will be non-zero. This is be-
of electrics that a plane capacitor generated uniform elec- cause, in the 2nd and 4th integrals (paths bc and da ), the
tric field. Likewise, we use a solenoid to generate a uniform vector B~ will be perpendicular to the path d~s and the scalar
magnetic field. product will be zero. The third integral is completely outside
of the solenoid where it was shown that B = 0 . As the value
In Section 20.2, we found the magnetic field of a circular
of B is the same along ab , we take it outside of the integral.
current loop and showed that it had the following distribu-
If we use L to indicate the distance ab , the left-hand side
tion:
becomes as follows:
B L = µ0 Iencl
Now let us find the value of Iencl on the right-hand side.
Since n is the number of windings of the solenoid per unit
length, the number of windings inside of the rectangle will
be nL . As current I is passing through each one, the amount
of current inside will be nLI :
Now, let us consider that we place a second identical current B L = µ0 nLI
loop next to it. Performing vector summation of the magnetic From here, we find the expression for the magnetic field inside
field at every point, we observe that lines in the interior re- of the solenoid:
gion get more and more parallel to the axis and exit the loop
much later.
B = µ0 nI (Magnetic field of a solenoid) (20.12)
When infinitely many such loops are placed side by side,
the magnetic field inside of the solenoid becomes parallel to Note that the result is not dependent on radius. The mag-
the axis and approaches zero outside. As a result, we obtain netic field has this uniform value at every point inside of the
the following structure: volume of the solenoid.
Example 20.9
around the toroid in the direction of the axes of loops.
Now let us consider an imaginary loop with radius r inside
of the toroid and
I apply Ampère’s law:
B
As B~ is parallel to the path along the circle, the scalar product
becomes B ds and we take B outside of the integral, because
it does not vary along the path. Also, Iencl on the right-hand
side is the sum of N currents with value I :
Toroid. Calculate the magnetic field at a distance r inside of a
B (2πr) = µ0 NI
solenoid consisting of N total windings wound in the shape of
From here, we find the magnetic field of the toroid:
a toroid (donut) carrying a current I .
µ0 NI 2k0 NI
Answer Let us consider the toroid as consisting of N cur- B= =
2πr r
rent loops. As the magnetic field of each loop on its axis is If we had performed the same operation for a circle outside
along that axis, the structure of the toroid will not change of the toroid, we would have found the magnetic field there
this symmetry and the magnetic field inside will circulate to be zero.
20.5 MAGNETIC PROPERTIES OF MATTER

It is possible to turn a non-magnetic iron nail into a magnet. This can be
performed in two ways: first, by keeping the nail in contact with the poles of a
horseshoe magnet for a while, as shown in Figure 20.12-a. When we detach it
later, we observe that the nail also starts to attract needles, in other words, that it
has gained permanent magnetism.
Figure 20.12: We magnetize a
nail through two methods: (a)
Contact with a permanent mag-
net, (b) Placing it inside of the
magnetic field of a solenoid.
The second way is to place a non-magnetic iron nail in the gap between
the windings of a solenoid and send a current through the solenoid for a while
(Figure 20.12-b). This nail is again observed to gain magnetic property later.
These observations show that magnetism is intrinsic to matter. What is the
source of this magnetism? Why can magnetism be permanent in some materials,
such as iron and nickel? We need to examine the atomic structure of matter in
order to answer these questions.
Magnetic Dipole Moment of Atoms
Negatively charged electrons (e− ) in an atom rotate around a positive nucleus
(Figure 20.13). In classic physics, they are assumed to rotate in a circular path with
radius r at speed v . Accordingly, we can consider that the electron generates a
current around the nucleus. Since the current is the amount of charge passing
through a cross-section per unit time, one pass of the charge e taking place in
one period T =2πr/v will give a current of:
Figure 20.13: The current and ∆q e e ev
magnetic moment m ~ generated I= = = =
by the orbital motion of an elec- ∆t T 2πr/v 2πr
tron. (In modern physics, this model is not exactly correct, because neither the speed
nor the orbit radius of the electron are well-defined quantities. Nevertheless, the
classical model is useful to have a rough idea.)
We had studied this current loop previously (in Section 20.2) and calculated
its magnetic field and its magnetic dipole moment. If we write the definition of
the magnetic dipole moment here for the electron, we get
ev
m=IA= πr2 = 12 evr
2πr
Therefore, atoms have a magnetic dipole moment caused by the rotation of
electrons in their orbit.
In Chapter 7, we defined the angular momentum of a point mass as L=mvr .
If we use me here to indicate the mass of the electron, we get
e
mL = L
2me
We thus see that there is a relation between angular momentum and magnetic
Figure 20.14: The magnetic
dipole moment: Each rotating particle generates an orbital magnetic moment
moment of the atom consists of that depends on its angular momentum.
the contributions of spin and or- Electrons also have another angular moment called the spin, which is intrinsic
bit. to them (Figure 20.14). The spin angular moment has no classic explanation, but
20.5. MAGNETIC PROPERTIES OF MATTER 343
you may consider the electron as a spinning top rotating about its own axis.
Therefore, in addition to the orbital magnetic moment, it is necessary to add a
spin magnetic moment, indicated with m s :
~ =m
m ~L+m
~s
Consequently, every atom generates a magnetic field caused by its total

magnetic dipole moment m ~ . When placed in an external magnetic field, it
contributes to and changes the net surrounding magnetic field.
~)
Magnetization ( M
In section 19.3, we found the torque exerted upon the magnetic dipole moment
of a current-carrying loop in an external magnetic field (Eq. 19.6):
τ = mB sin θ
The atoms of a material at the macroscopic scale usually have magnetic moments
in random directions, and therefore do not produce a net macroscopic magnetic
moment. However, when this object is placed in an external magnetic field B ~0,
the torque exerted upon the magnetic moments of the atoms try to rotate them.
The object thus gains a macroscopic magnetic moment.
The net magnetic moment per unit volume is called magnetization :
~i
P
M~ = im (20.13)
V
The magnetic field generated by this magnetization is found to have the value
~ M =µ0 M
B ~ . Here, µ0 is the magnetic permeability of free space that we defined
earlier.
Therefore, the net magnetic field for a material medium is as follows:
~ =B
B ~ 0 + µ0 M
~
Let us write this expression as follows to bring the contributions to magnetic field
in the same dimension:
~
 
~B = µ0  B0 + M
~ 

µ0
The first term in the brackets is the contribution of external currents to the
magnetic field inside of the matter, but it is written in the magnetization dimension.
This term is called the magnetic field strength vector and is indicated with H ~:
~
~ = B0
H (20.14)
µ0
Accordingly, the magnetic field inside the material can be written as follows:
~ = µ0 H ~ +M
~

B (20.15)
~ = 0 , this
If there is no magnetization in the medium, in other words, if M
~
expression gives the magnetic field B0 in free space.
Paramagnetism and Diamagnetism

The magnetic moments of the electrons in a given atom usually cancel each
other out and do not generate a net magnetic moment. However, some materials
(aluminum, platinum, calcium, sodium, etc.) have a non-zero atomic magnetic
moment. The torque generated when these types of materials are placed in an
external magnetic field will try to rotate the magnetic moment in the direction of
the field, and a contribution that increases the external magnetic field will thus be
generated. This is called paramagnetism.
On the other hand, in certain other materials (gold, silver, copper, lead, etc.),
although the magnetic moment is zero, an opposing reaction is given when they
are placed in an external magnetic field. The orbit of the electrons changes and a
magnetic moment is generated again. However, this magnetic moment gives a
Figure 20.15: The change of
magnetic field in paramagnetic contribution that reduces the external magnetic field. This is called diamagnetism.
and diamagnetic media.
Figure 20.15 shows the change in magnetic field in these two types of medium.
Magnetic field increases in a paramagnetic medium (dense field lines) and de-
creases in a diamagnetic medium (sparse field lines).
In both types of magnetization, the magnetization vector M ~ of the medium
~
is proportional to the magnetic field strength vector H :
~ = χm H
M ~
The dimensionless constant χm here is called magnetic susceptibility. χm is

positive for paramagnetic media and negative for diamagnetic media.
Accordingly, if we write Eq. (20.15), which we found above for magnetic field
~,
B
~ = µ0 H ~ +M~ = µ0 (1 + χm ) H
~

B
Just as we defined ε to replace ε0 for the electric field, here we define the
magnetic permeability of the medium, indicated with µ , as follows:
µ = (1 + χm ) µ0 (20.16)
Depending on the type of medium,
in a paramagnetic medium : µ > µ0

in a diamagnetic medium : µ < µ0
~ that takes into account the magnetic properties
We thus find the expression for B
of the medium:
~ = µH
B ~ (20.17)
Ferromagnetism
The magnetism of paramagnetic and diamagnetic materials occurs only when
an external magnetic field is applied. The magnetization disappears when the
external magnetic field is removed.
However, in five metals (iron, nickel, cobalt, gadolinium and dysprosium) and
certain oxide alloys, magnetism does not disappear when the external magnetic
field is removed. This permanent magnetization is called ferromagnetism. Fer-
romagnetic materials are used in the structure of credit cards, computer memories,
speakers, motor cores, compasses, etc.
20.5. MAGNETIC PROPERTIES OF MATTER 345
Ferromagnetic materials have very strong magnetism. Their magnetization

values M can be more than 1000 times compared to paramagnetic materials.
However, this magnetization continues until a certain critical temperature,
called the Curie temperature. (Curie temperature for iron is 770 ◦ C .) When
this temperature is exceeded, the material again returns to the paramagnetic
phase with a sudden phase transition.
There is no satisfactory classic explanation for ferromagnetism; one needs
quantum theory for that. Let us briefly provide some essential concepts here as
information.
When the microscopic structure of ferromagnetic materials is examined, it is
observed that it consists of small regions where magnetic moments of the atoms
are aligned together. These regions are called domains (Figure 20.16). Although
these domains have a net magnetism, they do not lead to a net magnetism, because
each one is in a random direction. When placed in an external magnetic field,
the domains in the direction of the field are observed to enlarge and those in the Figure 20.16: Ferromagnetic
other directions are observed to shrink. In other words, the domain walls move domain structure in the NdFeB
so as to increase the net magnetization. crystal.
A very interesting change occurs in magnetism when we place a ferromagnetic
material in an external magnetic field B0 (Figure 20.17). As the external field B0
is increased, the magnetization M also increases (curve ab in the figure). This
increase continues up until a value called the saturation (point b ). At this point,
the magnetic moments of all of the atoms are in parallel with the external field
and no further increase takes place.
Then, when B0 is decreased, this time the value of M decreases along a
different path (curve bc ). As shown in the figure, although B0 = 0 at point c , a
permanent magnetization value Mr , called remanence, remains. If the external
magnetic field is reversed, (curve cd ), magnetization also decreases and forms in Figure 20.17: Hysteresis curve
the reverse direction, slightly delayed. When we start to reduce the magnitude of ferromagnetic magnetization
of the external field in the negative direction (curve de ), this time, a permanent as a function of applied mag-
magnetization occurs in the reverse direction (point e ). We thus observe a closed netic field B0 .
curve with different forward and backward paths. This is called a hysteresis
curve.
The hysteresis curve also explains why ferromagnetic materials are used in
memory chips. As seen in the curve, the magnetization value of the ferromagnetic
material has the value Mr if the magnetic field is reduced in one direction and
the value −Mr if reduced in the other direction. In other words, it has a memory
and remembers how it reached zero magnetic field.
Earth’s Magnetic Field
Earth’s magnetic field is what directs the needle of a compass towards the
North Pole. Measurements show that the Earth has an average magnetic field of
10−4 T . Examining the distribution of this field from space, we can see that it acts
as if a giant magnet was placed at the center of the earth. However, note that the
North pole of this magnet (N) is at the geographic South pole (Figure 20.18). In
other words, the magnetic field lines come out of Earth’s South pole and merge
at the North pole. The North pole of a compass can only be attracted to the
geographical North in such a case. The magnetic field of the Earth is not exactly Figure 20.18: The Earth is a gi-
parallel to the surface and also has a component that is perpendicular to the ant magnet, with the N-pole lo-
surface. cated at the south.
The magnetic poles where the field lines converge do not coincide with the
geographical North and South Poles. The location of the magnetic pole varies
with time. Research on rocks show that they vary over 1000-year periods. For
example, according to measurements conducted in 2005, the magnetic North Pole
was located around Ellesmere Island in Northern Canada, with latitude 82.7°N
(north) and longitude 114.4°W (west), has been shifting towards Russia since then.
Magnetic field also shows local variations on Earth. The magnetic field of every
Figure 20.19: Homing pigeons region has been mapped for use in maritime and air transportation.
can find their home from thou- The source of Earth’s magnetic field is one of the greatest mysteries in science.
sands of miles away. It has been At first, it was considered that layers of iron and nickel at the center of the Earth
established that they use small were the source; however, we know that ferromagnetism disappears at very
magnetic crystals in their beaks high temperatures. Today, it is considered to be caused by convection currents
to detect Earth’s magnetic field.
generated by electrically charged underground molten lava.
1. Which of the following are the source of magnetic field?
I. Magnetic charges.
II. Currents. (a) I & II (b) I & III (c) II & III (d) II & IV
III. Magnetic moments of atoms.
8. What is the force exerted upon 1 m in length of one of
(a) All (b) I & II (c) I & III (d) II & III two parallel wires separated by 1 m and each carrying
1 A of current?
I. Parallel currents repel each other. (a) 1 × 10−7 N (b) 2 × 10−7 N (c) 3 × 10−7 N
II. Anti-parallel currents attract each other.
9. Which of the following is Ampère’s law?
III. Parallel currents attract each other.
IV. Anti-parallel currents repel each other. (a) The integral of the magnitude of the magnetic field
over any closed curve is proportional to the net current
(a) I & II (b) III & IV (c) I & IV (d) II & III
enclosed by the curve.
3. By what factor does the magnetic field increase when (b) The integral of the tangential component of the
the current flowing through an infinite wire is doubled? magnetic field over a straight line is proportional to
the net current in that wire.
(a) No change (b) 2 (c) 4 (d) 1/2
(c) The integral of the magnetic field over any volume is
proportional to the net current enclosed by the volume.
4. The radius of a current-carrying loop is doubled. By
(d) The integral of the tangential component of the
what factor will the magnetic field at the center of the
magnetic field over any closed curve is proportional to
loop increase?
the net current enclosed by the curve.
(a) No change (b) 2 (c) 4 (d) 1/2
10. By what factor does the magnetic field increase when
5. What is the magnetic field at a 1 m distance from a the current of a solenoid is doubled?
straight wire carrying 1 A in current?
(a) No change (b) 2 (c) 4 (d) 1/2
(a) 1×10−7 T (b) 2×10−7 T (c) 3 × 10−7 T (d) 4×10−7 T
11. By what factor does the magnetic field increase when
6. What is the expression for the magnetic field at the cen- the number of windings per unit length of a solenoid is
ter of a circular current loop? doubled?
(a) k0 I/a (b) k0 I/a2 (c) 2k0 I/a (d) 2πk0 I/a (a) No change (b) 2 (c) 4 (d) 1/2
7. In which of the following figures are the forces between 12. Which of the following is correct?
parallel currents shown correctly?
(a) Magnetic field is zero inside of a solenoid.
(b) Magnetic field is zero outside of a solenoid.
(c) Magnetic field is constant outside of a solenoid.
(d) Magnetic field is infinite outside of a solenoid.
PROBLEMS 347
13. What is the expression for the magnetic field of a (a) I (b) I & II (c) I & III (d) I & IV
solenoid?
(a) 12 µ0 nI 2 (b) µ0 nI 2 (c) µ0 nI (d) µ0 n2 I 18. Which of the following are correct for ferromagnetism?
I. Magnetism increases with temperature.
14. Which of the following produces a uniform magnetic II. Magnetism decreases with temperature.
field? III. Magnetism disappears above a certain temperature.
(a) Straight wire current. (a) I (b) II (c) I & III (d) II & III
(b) Circular current loop.
(c) Solenoid. 19. Which of the following is correct when a ferromagnetic
(d) Toroid. material is placed inside of an external magnetic field?
15. Which of the following is correct in terms of the magne- (a) It gets magnetized in the direction of the external
tization properties of matter? magnetic field.
(b) Magnetization increases with the external magnetic
(a) For paramagnetic materials: µ > µ0 field.
(b) For diamagnetic materials: µ < µ0 (c) Magnetization does not increase after a certain
(c) For ferro magnetic materials: µ µ0 saturation value.
(d) All of the above. (d) All of the above.
16. Which of the following is the source of the magnetiza-
20. Which of the following are correct for Earth’s mag-
tion property of materials?
netism?
I. The orbital motion of atomic electrons.
I. The geographic North Pole is the magnetic North
II. The spin property of electrons.
Pole.
III. The merging of magnetized domains.
II. The geographic South Pole is the magnetic North
(a) I (b) I & II (c) I & III (d) I, II & III Pole.
III. The locations of the magnetic poles vary over time.
17. Which of the following are correct? IV. The locations of magnetic poles do not vary over
I. Magnetic field increases inside paramagnets. time.
II. Magnetic field increases inside diamagnets.
III. Magnetic field decreases inside paramagnets. (a) I & III (b) II & III (c) I & IV (d) III & IV
IV. Magnetic field decreases inside diamagnets.
Problems
20.2 Magnetic Field Calculations at point P located at the center of the circle.
[A: 4 mT , out of the paper .]
20.1 (a) What is the magnetic field at a distance of 1 cm
from a straight wire carrying a current of 10 A ? (b) How
much current should be flowing through a loop current with
a radius of 1 cm such that a 1 mT magnetic field is generated
at its center? [A: (a) 0.2 mT , (b) 16 A .]
Problem 20.3
20.3 In the figure, there is a 2 cm distance between the anti-
parallel currents I1 =300 A and I2 =50 A . (a) Calculate the
total magnetic field at the midpoint A between the wires.
(b) Calculate the magnetic field at point B located 1 cm away
Problem 20.2
from the wire I2 . [A: (a) 7 mT , (b) 1 mT .]
20.2 A part of an infinite wire carrying a current I = 50 A
is bent into a circle with radius a = 1 cm , as shown in the 20.4 A current of 1 A is flowing through a conducting cir-
figure. Find the magnitude and direction of the magnetic field cular loop with a radius of 1 m . (a) What is the magnitude
of the magnetic field located at the center of the loop? (b) At

what point along the axis will the magnetic field drop down
to a quarter of its value at the center?
[A: (a) 6.3 × 10−7 T , (b) h = 1.2 m .]
Problem 20.9
20.9 The straight wire currents perpendicular to the plane
of the paper and located on the corners of a square with side
a = 1 mm are equal and have I = 10 A magnitude. Calculate
Problem 20.5 the magnitude and direction of the total magnetic field at
20.5 Helmholtz coils. Two identical circular loops, each point P located at the center of the square. (Hint: The prob-
with radius a=1 cm and carrying the same I=100 A -currents lem will not be as lengthy as it looks if you first determine
in the same direction, are placed on an axis with 2a distance the direction of each magnetic field.)
in between, as shown in the figure. Calculate the magnetic [A: 8 mT to the right.]
field at the midpoint between the loops. (This assembly is
used to obtain uniform magnetic field in a small region.)
[A: 4.4 mT .]
Problem 20.6
20.6 The currents on two straight wires perpendicular to the
plane of the paper, as shown in the figure, are I1 = 1 A and Problem 20.10
20.10 The four straight wire currents on the plane of the
I2 = 5 A , respectively, and are separated by 1 m . At what
paper are equal and have magnitude I = 50 A . Calculate the
point on the x -axis (left, center, right) will the total magnetic
magnitude and direction of the total magnetic field at point
field be zero? [A: 0.25 m on the left.]
P located at the center of the square with a side length of
20 cm . [A: 0.2 mT into the paper.]
20.3 Force Between Parallel Currents
Problem 20.7
20.7 The two straight wires perpendicular to the plane of the
paper, as shown in the figure above, are separated by 6 mm Problem 20.11
and have equal currents of I = 100 A in opposite directions. 20.11 In the figure, wires perpendicular to the plane of the
Calculate the magnitude and direction of the total magnetic paper constitute the corners of an equilateral triangle with
field at point P at 4 mm in distance on the perpendicular a side length of 1 mm . Since I1 = I2 = I3 = 60 A , calculate
bisector of the currents. [A: 6.4 mT to the right.] the magnitude and direction of the total force exerted upon
1 m of length on the wire with current I3 .
[A: 1.2 N upward.]
Problem 20.8
20.8 The two straight wires perpendicular to the plane of
the paper, as shown in the figure above, are separated by Problem 20.12
5 mm and have equal currents of I = 60 A . Calculate the 20.12 In the figure, wires perpendicular to the plane of the
magnitude and direction of the total magnetic field at point paper constitute the corners of a square with a side length of
P , which forms a right triangle with the currents. 1 mm . Since the magnitude of the current passing through
[A: 5 mT .] each wire is I = 100 A , calculate the magnitude and direction
PROBLEMS 349
of the total force exerted upon 1 m of length on the 4th wire.

[A: 1.4 N towards the inside from the diagonal.]
Problem 20.15
20.15 Of the two coaxial infinite cylindrical shells shown
in the figure, the one with radius a carries a current 3I and
the one with radius b carries a current I in the opposite
direction. Calculate the magnetic field in all three regions.
[A: B = 0 for r < a , B = 6k0 I/r for a > r > b , B = 4k0 I/r
Problem 20.13 for r > b .]
20.13 A current I2 =10 A flows through the rectangular loop
laying on the same plane as the straight wire carrying a cur-
rent of I1 =50 A as shown in the figure. The distance of the
loop to the near side is a=1 cm , its width is b=2 cm and
length is c=60 cm . Calculate the net force exerted upon the
loop. [A: 0.004 N to the left.]
Problem 20.16
20.16 The total current I is evenly distributed inside of the
20.4 Ampère’s Law infinite cylindrical region with inner radius a and outer ra-
dius b as shown in the figure. Calculate the magnetic field in
20.14 A current I flows through an infinite cylindrical shell all three regions.
with radius R . Calculate the magnetic field at a distance r r2 − a2
inside and outside of the cylinder. [A: 0 for r < a , B = 2k0 I 2 for a < r b .]
0 0
?
21
FARADAY’S LAW –
INDUCTION
Solar-powered aircraft Helios

cruising the Hawaiian skies.
Developed by NASA, the so-
lar batteries located on the
wings of this plane supply
each of its 14 electric motors
with 1.5-kilowatt (2 HP) of
power.
What physics laws govern
the operation of electric mo-
tors and generators? We will
find the answer to these ques-
tions in Faraday’s law.
Towards the mid-19th century, it was established that electric currents would
produce magnetic fields. Therefore, many people naturally asked the question,
“Could magnetic field also lead to electric field?”. Finally, in 1831, the British
scientist Michael Faraday and the American scientist Joseph Henry succeeded in
producing electric current through a changing magnetic field. No revolution has,
to date, been greater than this one. Motors, generators, transformers, wireless
energy, signal transmission, etc. were all developed as a result of this discovery.
The discovery of magnetic induction completed another great circle in science.
The British scientist James Clerk Maxwell was able unify electricity and mag-
netism in a single electromagnetic theory in which electrics and magnetism
induce and affect each other.

352 21. FARADAY’S LAW – INDUCTION
21.1 FARADAY’S LAW

A current is observed to be induced in a conducting loop when we move a bar
magnet near it (Figure 21.1a). No current is induced if the magnet is stationary,
regardless of its strength. This experiment shows us that current flows when
there is a change in the magnetic field lines passing through the loop.
Figure 21.1: Two methods of in-
ducing current: (a) A current is
induced when a magnet moves
near a solenoid. (b) A current
is induced in one of two adja-
cent circuits when the current
changes in one.
Moving a bar magnet is not the only way to change the magnetic field lines.
Two circuits are shown in Figure 21.1b. The first circuit has a loop connected to a
battery and, in the other circuit, there is a loop connected solely to a galvanometer.
When the switch in the first circuit is closed, a brief current is observed in the
second circuit. Here, a change in the magnetic field of the first loop also produces
an electric current in the second loop.
Induction of current means that an emf is induced as if a battery were con-
nected to the loop. In order to understand this effect, let us consider that we are
moving a conducting rod with velocity ~v towards the right on a plane perpen-
dicular to the magnetic field (Figure 21.2). Recall the expression for magnetic
force:
~F = q (~v × B
~)
As the force ~F exerted upon a free charge +q inside of the conductor will be
Figure 21.2: Conducting rod perpendicular to both the magnetic field and the velocity, it will be towards the
moving in a magnetic field. top of the conductor. The +q charges will thus gather at the top of the conductor
and leave a negative end at the bottom. A potential difference will thus be induced
between the two ends of the conductor.
Faraday’s Law
We had discussed the concept of flux in examining Gauss’ law in electrics.
Magnetic flux is also defined here as the number of magnetic field lines crossing
an area A enclosed by a conducting wire, as follows:
Z
ΦB = B dA cos θ (Magnetic flux) (21.1)
A
The angle θ here is the angle between the magnetic field vector and the surface
normal n̂ (Figure 21.3).
Faraday’s law states that a variation in magnetic flux induces an emf:
Figure 21.3: Magnetic flux Faraday’s Law

crossing a surface enclosed by
a conductor. The potential difference induced in a circuit is directly propor-
tional to the time rate of change of the number of magnetic field
lines passing through the circuit:
dΦB
E=− (21.2)
dt
21.1. FARADAY’S LAW 353
We shall explain the meaning of the negative sign in this expression using Lenz’s
law.
Faraday’s law includes several ways in which the magnetic flux may vary:
• The magnetic field B may vary.
• The surface area A may vary.
• The angle θ between the surface and the magnetic field may vary.
• All or some of the above may vary simultaneously.
Lenz’s Law
The negative sign (−) in Faraday’s law is placed symbolically in the equation.
This sign tells us the direction of the emf or that of the current to flow through
the circuit. This direction is determined by Lenz’s Law:
Lenz’s Law
The current produced in a loop by the induced emf flows in a
direction such that its magnetic field opposes the change in orig-
inal magnetic flux.
In order to apply Lenz’s law, let us first recall the magnetic field at the center
of a circular loop carrying a current. According to the right-hand rule, when
the four fingers are curved in the direction of current I , the thumb points in the
direction of the magnetic field B.
Now let us consider the bar magnet in Figure 21.4a. The magnetic flux crossing
the surface of the loop increases as the magnet approaches. The current opposing
this must be induced in such a direction that its magnetic field B0 is generated
in the opposite direction, and thus decreases the increasing flux. According to
the right-hand rule, this current I 0 should be in the direction shown in the figure.
Likewise, the current produced will be in the opposite direction as the bar magnet
moves away.
Figure 21.4: (a) Magnetic flux
increases as the bar magnet ap-
proaches, therefore the induced
current should be in the direc-
tion that generates an opposing
magnetic field. (b) The current
I 0 in the bottom circuit will be
induced in the opposite direction
as the current I increases in the
top circuit, and in the same direc-
tion as I decreases.
Let us again consider two loops affecting each other, as shown in Figure 21.4b.
If the current I increases in the loop connected to the battery, its magnetic field B
will increase. The induced current to oppose this in the second loop will generate
a current I 0 with a magnetic field B0 in the opposite direction. Likewise, if the
current I and field B decrease, the magnetic field in the second loop should be
in the same direction so as to compensate for this decrease. Therefore, a current
I 0 is induced in the same direction.
Let us now look at some examples to see how Faraday’s law and Lenz’s law
are applied.
Example 21.1
ΦB = BA cos 0◦ = πa2 B
We write the magnetic field as B = 0.1t , as a function of time,
and take its derivative (we ignore the negative sign):
d[πa2 (0.1t)]
E= = 0.1πa2
dt
We use Ohm’s law to find the current generated in the circuit
by this electromotor force:
I = E/R = 0.1πa2 /R
The conducting loop with resistance R = 1 Ω , as shown in the I = 0.1 × 3.14 × 32 /1 = 2.8 A
figure, is located in a magnetic field B directed into the plane The direction of the current should be such as to oppose the
of the paper. increase in the magnetic field, in other words, its own mag-
(a) The magnetic field starts to increase as B = 0.1 t (tesla) , netic field B0 should be out of the paper. According to the
while the radius of the loop is constant at a=3 m . Deter- right-hand rule, the current on the loop should be counter-
mine the value and direction of the current induced in the clockwise.
loop. (b) This time, the variation in magnetic flux is caused by the
(b) When the magnetic value is constant at B=0.1 , the radius increase in the cross-section area A :
of the loop starts to increase as a = 3t (meters). Find the ΦB = BA = B(πa2 ) = πB(3t)2
value and the direction of the current at t=1 s . The induction emf is calculated using Faraday’s formula:
Answer dΦB d(9πBt2 )
E= = = 18πBt
We write Eq. (21.2) for Faraday’s law: dt dt
dΦB We use Ohm’s law to find the induced current at time t=1 s :
E=− I = E/R = 18πBt/R = 18 × 3.14 × 0.1 × 1/1 = 5.7 A
dt
The magnetic flux ΦB here is calculated using Eq. (21.1). As Again, according to Lenz’s law, the current will be counter-
B is constant and parallel to the surface normal, we get clockwise.
Example 21.2
ing A :
The rectangular loop with the dimensions given in the figure ΦB = BA cos θ
is located in a region where a uniform magnetic field B=1 THere, θ is the angle between the magnetic field and the sur-
face normal. As seen in the figure, the angle between B ~ and
in the +y direction is present. The loop plane has a 37◦ angle
the loop surface is 53 , and therefore the angle with the nor-
with the x -axis. The magnitude of the magnetic field is reduced
◦
mal perpendicular to the surface is θ = 37◦ . We therefore

down to zero in 0.3 s . Calculate the value and direction of the
calculate the magnetic flux at the start:
current induced in the loop. The resistance of the loop is R=4 Ω .
ΦB = 1 × (2 × 3) × cos 37◦ = 5.2
We apply Faraday’s law:
E = ∆ΦB /∆t = (5.2 − 0)/0.3 = 17.4 V
We use Ohm’s law to find the induced current:
I = E/R = 17.4/4 = 4.4 A
As magnetic flux is decreasing, according to Lenz’s law, the
direction of the magnetic field induced on the loop should
be in a direction that prevents the decrease in B ~ , in other
words, in the same direction. Therefore, the current will be
Answer We write the expression of the magnetic flux cross- downward in the front side.
Example 21.3
into the plane of the paper is present. The side of the loop with
length L=50 cm is pulled to the right at velocity v=4 m/s . Find
the value and direction of the induced current.
Answer
In this problem, the variation in the magnetic flux crossing
the loop is caused by the change in the surface area of the
loop. The side with length L of the rectangular loop is con-
The rectangular loop in the figure has a resistance R=2 Ω and stant and the length of the other side varies as x = vt . Let us
is located in a region where B=0.2 T in magnetic field directed write the magnetic flux:
21.2. GENERATORS AND TRANSFORMERS 355
ΦB = BA = BL(vt) The magnetic flux crossing the loop is increasing. Therefore,

We take the derivative of this expression with respect to time the magnetic field generated by the induction current shall
t and calculate the current from the emf using Ohm’s law: be in the direction that decreases this, thus out of the plane
E 1 d(BLvt) BLv of the paper. The current that gives such a magnetic field
I= = =
R R dt R should be counterclockwise.
I = 0.2 × 0.5 × 4/2 = 0.2 A
Example 21.4
ΦB = BA = µ0 n I (πa2 ) = µ0 nπa2 I
We use Faraday’s law to find the induction emf:
dΦB dI
E= = µ0 nπa2
dt dt
The current I decreases exponentially. Its derivative will be
equal to itself:
dI d −t
= 5 e = −5 e−t
dt dt
The number of windings per unit length of the solenoid shown The value of this derivative at t = 0 s will be 5 (we ignore
in the figure is n=500 turns/meter and the current it carries the negative sign). We substitute this value and calculate the
varies with time as I=5 e−t (ampere). A conducting loop with current I 0 induced in the loop:
E µ0 nπa2 dI

radius a=50 cm placed inside of the volume of the solenoid has I0 = =
resistance R=1 Ω . Determine the value and direction of the R R dt t=0
current flowing through the loop at t=0 s . 4π × 10−7 × 500 × π × 0.52 × 5
I0 =
1
Answer I 0 = 0.0025 A = 2.5 mA
As we discussed in Chapter 20, the magnetic field inside of a The direction of this current I 0 should be in the direction
solenoid is uniformly distributed and given in Eq. (20.12): that compensates the decrease in the magnetic field of the
B = µ0 n I solenoid, in other words, in the same direction as the current
The magnetic flux crossing the loop with radius a is: I.
Example 21.5
ΦB = BA cos θ
B and A are constant here, but the angle θ varies as θ = ωt
depending on the angular velocity ω . We find the expression
for the induction emf using Faraday’s law:
dΦB d(cos ωt)
E= = BA = −ωBA sin ωt
dt dt
In the figure, a rectangular loop with dimensions a=1 m and We use Ohm’s law to find the current without considering
b=2 m is placed in a uniform magnetic field B=0.1 T directed the negative sign:
into the paper. The loop is rotating around its axis with an E ωBA
I= = sin ωt
angular velocity of ω=3 rad/s . Calculate the current induced R R
on the loop with resistance R=1 Ω as a function of time t . We substitute the numerical values:
3 × 0.1 × (1 × 2)
Answer I= sin 3t
1
We use the formula for the magnetic flux crossing a loop: I = 0.6 sin 3t
21.2 GENERATORS AND TRANSFORMERS

Generators
A device that converts mechanical energy into electric energy is called a
generator or a dynamo. The energy of water falling from a height in hydroelectric
power plants and the energy of steam in thermoelectric and nuclear power plants
are converted into electric power.
When examining the electric motor in Chapter 16, we discussed how a torque
was exerted on a current-carrying loop inside of a magnetic field. Generators
perform the opposite of that which motors do, in other words, electricity is Figure 21.5: Diagram of a sim-
generated by a loop rotating in a magnetic field. ple generator.
Figure 21.5 shows the diagram of a simplified alternating current generator.

A rectangular loop with cross-sectional area A is rotated between the poles of a
magnet. The magnetic field B of the magnet is constant in the indicated direction.
The variation in the magnetic flux crossing the loop is caused only by the change
in angle θ due to the rotational motion. Let us write the magnetic flux for N
windings:
ΦB = NBA cos θ
If the loop rotates with angular momentum ω , the angle will increase over time
as θ = ωt . The magnetic flux thus varies sinusoidally:
ΦB = NBA cos ω t
Applying Faraday’s law, the derivative of flux ΦB gives the induction emf:
dΦB
E=− = NBAω sin ωt (21.3)
dt
The sinusoidal variation of this current in time is given in Figure 21.7, with the
Figure 21.6: The angle θ be- various directions of the loop plane indicated. This is an alternating current, in
tween the loop plan and the other words, it varies between the values ±NBAω during a full 360◦ rotation of
magnetic field. the loop, positive in the first half and negative in the second half.
Figure 21.7: The values of the

sinusoidal current depending on
the direction of thee loop in the
generator.
Alternating current is the most practical type of current in energy transmis-

sion, as it can be supplied to circuits directly as it is generated. A commutator
must be used as it is in motors, if direct current needs to be generated from this
mechanism.
Transformers
The electricity generated in all kinds of power plants must be transmitted
to the point of use. Heat losses must be minimized during transmission. If the
resistance on the transmission lines is R , the heat loss will be RI 2 , proportional
to the square of the current. Therefore, it is necessary to transmit the power with
lower currents, or higher voltages, according to the power expression P=V I .
Hence, very high values of voltage such as 100 000–500 000 Volts are carried by
high voltage lines coming from power plants. This voltage is then further reduced
in cities and, in the end, brought down to the 220 or 120 volts used in homes
and the industry. Another advantage of alternating current is that it is easy to
reduce or increase the voltage. The device that performs this function is called a
transformer.
21.2. GENERATORS AND TRANSFORMERS 357
Figure 21.8 shows the structure of a transformer. Two coils of wire are wound
around a core of iron ring. The first coil, called the primary, consists of N1
windings and the second coil, called the secondary, consists of N2 windings.
Suppose that a current I1 is flowing through the first circuit. A very high
percentage of the magnetic field to be generated by this current will be concen-
trated inside of the core and almost all of it will pass through the secondary coil
on the other side. Accordingly, the flux crossing each winding will be the same
Figure 21.8: The primary and
everywhere:
secondary windings of a trans-
ΦB1 = ΦB2 = ΦB former.
Now, if we write the total emf induced by the varying current for both circuits
according to Faraday’s law, we get:
dΦB1 dΦB2
E1 = N1 E 2 = N2
dt dt
As ΦB1 = ΦB2 at each time t , their derivatives will also be equal. From here, we
obtain the following formula between the emf of both circuits:
Figure 21.9: A transformer
N2 densely wound to prevent mag-
E2 = E1 (21.4)
N1 netic flux losses.
This formula shows how the emf varies in the secondary circuit depending on
the number of windings. For example, if we wish to double the voltage, we need
to have N2 = 2N1 , in other words, the number of windings in the second circuit
should be double. Likewise, the number of windings in the second circuit should
be lower to reduce the voltage.
Energy losses are inevitable in a real transformer. Transformers therefore
heat up when operating and must be cooled. There are several causes for these
losses. First, the flux will be less in the secondary circuit if the magnetic flux is
not fully inside of the ferromagnetic core. Transformers are wound very densely
and without any gaps to prevent this.
Another cause of energy losses is eddy currents. Eddy currents in random Figure 21.10: Eddy currents in-
directions are induced in the ferromagnetic iron core when the transformer duced inside the transformer.
operates. These moving charges also take away some of the energy. To prevent
the forming of eddy currents, the iron core is laminated, in other words, made of
many insulated iron sheets.
Example 21.6
E = NBAω sin ωt
The rotating coil of a generator is made of N windings, each We calculate the number of windings N from the maximum
with cross-section area of 20 cm2 , and is rotating with an angu- value ( sin ωt=1 ) of this sinusoidal voltage:
lar speed of 6000 rpm inside of a 0.5 T magnetic field. What E 12
N= =
is the minimum number of windings required such that the BAω 0.5 × 0.0020 × (6000 × 2π/60)
maximum voltage generated by the generator is 12 V ? N = 19.1
Answer We use Eq. (21.3) to find the emf produced by the If we convert this into a whole number, we get the number
generator: of required windings as N = 20 .
Example 21.7
ampere current entering the transformer be at the output?
The primary coil of a transformer has 200 windings and its Answer The relation between the input and output voltages
secondary coil has 50 windings. What will a 240-volt and 20- of the transformer is found using Eq. (21.4):
N2 Considering that an ideal transformer has no power loss, the
E2 = E1
N1 power at the input and output are equal:
We calculate the output voltage from this formula: E1 I1 = E2 I2
50
E2 = × 240 = 60 V From here, we find the output current to be I2 = 80 A .
200
Example 21.8
Accordingly, the current I1 when the power P is transmitted
Power loss on transmission lines. 1 MW (megawatt) in at the voltage V1 is:
power generated at a dam is transmitted to a city using two P 106
I1 = = = 200 A
different methods: V1 5000
(a) What will the power converted into heat be on the 1 Ω We calculate the power converted into heat with this current
resistance of the line when this power is transmitted at aon the resistor R :
voltage of 5000 V = 5 kV ? P01 = R I12 = 1 × 2002 = 40 000 W = 40 kW
(b) What will the power converted into heat be on the 1 Ω re-
(b) We repeat the same calculation when the same power is
sistance of the line when this same power is first increased
transmitted at voltage V2 =500 kV :
to a voltage of 500 000 V = 500 kV using a transformer?
P 106
Answer I2 = = = 2A
V2 5 × 105
(a) The general power formula was found in Section 18.4: We calculate the power converted into heat with this current
P=VI on the resistor R :
Also, the power converted into heat on a resistor R was found
P02 = R I22 = 1 × 22 = 4 W
to be
The power loss is less with transmission at higher voltage.
P0 = R I 2
21.3 INDUCTANCE – MAGNETIC ENERGY

Let us consider two coils that are very close to each other. Let us focus on
the magnetic flux ΦB2 passing through the second coil and due to the current I1
in the first coil. When the current I1 is varied, the flux ΦB2 will also vary and,
according to Faraday’s law, an emf E2 will be induced in the second coil:
dΦB2
E2 = −
dt
If the geometric shapes and positions of the coils are constant, the induced
Figure 21.11: Magnetic flux be- magnetic flux will be proportional only to the current I1 :
tween two coils.
ΦB2 = M21 I1 (21.5)
We can see this feature from the expressions for the magnetic field of a coil and
solenoid that we found earlier. In this case, the value of the emf induced in the
second circuit can be written as follows, in terms of I1 instead of ΦB1 :
dI1
E2 = −M21
dt
The coefficient M21 here, into which we gathered all of the constants is called
the mutual induction coefficient or mutual inductance. Its unit is 1 T·m2 /A =
1 Henry (H) . This coefficient is dependent on the geometric structure of the given
current loops, the number of windings and the µ permeability of the medium.
We can use the same reasoning for the flux of the second coil on the first
one. The emf value E1 induced on the first coil by the variation in the current I2
flowing through the second coil, will be,
dI2
E1 = −M12
dt
21.3. INDUCTANCE – MAGNETIC ENERGY 359
It can be proven that these two coefficients M21 and M12 are equal even if the
coils are not identical. We accept this result here without proof and define a single
mutual inductance for both coils:
M21 = M12 = M (Mutual inductance)
Thus, instead of the induction flux ΦB , we express the induction emf values in
terms of the directly measurable current:
dI2 dI1
E1 = −M and E2 = −M (21.6)
dt dt
Self-inductance
When you shut down the current of a building by turning off the main lever
switch, you will notice that a spark jumps between the two poles of the switch.
How can an emf of about 1000 Volts be induced in a circuit that is switched off,
when it is not possible for a spark to jump at 220 Volts? Faraday’s law holds the
answer to this. As the circuit is switched off, during that small duration, an emf
opposed to the decrease in magnetic flux is induced on itself. We describe this
with the concept of self-induction.
Consider the simple circuit in Figure 21.12 consisting of a coil with only
two windings. When the switch is closed, the current and its magnetic field are
constant. You will note that the magnetic flux of each winding passes through
the other winding. Now, suppose that we open the switch and cut the current.
The magnetic field of the second winding will decrease, and thus its magnetic
flux crossing the first winding will also decrease. According to Faraday’s law, an Figure 21.12: A circuit with
opposite emf will be induced in the first winding against this decrease. As both two windings to understand self-
windings are part of the same circuit, we can consider this induction effect as the induction.
self-induction of the circuit.
Every current loop opposes the variation in the current that it carries. It does
this by producing an opposing induction emf. This emf value is proportional to
the variation in the current I flowing through the coil. Likewise, if we gather all
of the geometric factors together and write the magnetic flux flowing through
the coil as proportional to current I , we get
ΦB = L I (21.7)
Thus, the self-induced emf generated in the circuit will be proportional to

the time rate of change of the current:
dI
E = −L ( L : Self-inductance) (21.8)
dt
L is called the self-induction coefficient or simply the inductance. Its unit is,
again, the Henry (H), as in mutual inductance.
In alternating current circuits, windings with a fixed inductance value are
used as a circuit component called the inductor and indicated with the sym-
bol . The potential difference between the terminals of an inductor
carrying a current I is
dI
Vab = −L (21.9)
dt
The potential difference will be opposite to the current if the current is increasing
and in the same direction if it is decreasing.
Magnetic Energy
In Chapter 18, the most general expression for energy in electric circuits was
found to be (equation 18.11):
P=VI
Using the expression that we found above for the potential difference between
the terminals of a coil carrying current I , we get
dI
P = LI
dt
The work performed on this coil during the time interval dt will be stored as an
increase in the potential energy of the coil:
dI
dU = dW = P dt = L I d@t
@
d@t
@
dU = L I dI
When the current goes from an initial value of zero to a final value of I , the
magnetic energy stored in the coil will be the integral of this expression:
Z I
U=L I dI = 1
2 L I2 (Magnetic energy of a coil) (21.10)
0
This magnetic energy is stored in the coil and may later be returned to the circuit.
Example 21.9
µ0 N 2 A
ΦB = NBA = N (µ0 NI/`) A = I
`
(a) Find the expression for the inductance of a solenoid. The inductance was defined in Eq. (21.8) as:
(b) Calculate the inductance of a solenoid with 500 windings, ΦB = L I
20 cm in length and a 10 cm2 cross-section area. We find the expression for inductance by comparing the last
(c) Calculate the energy stored when a 40 A current flows two expressions:
through this solenoid. µ0 N 2 A
L=
Answer `
(b) We substitute the numerical values in the expression L
(a) We had found the expression for the magnetic field of a
as follows:
solenoid in Chapter 20: 4π × 10−7 × 5002 × 0.0010
B = µ0 nI L=
0.20
n is the number of windings per unit length and, for a L = 0.0016 H = 1.6 mH
solenoid with length ` and N windings, it is n = N/` . (c) The energy stored in a coil was given with the formula
As the magnetic field is constant inside of the cross-section of (21.10):
the solenoid, the magnetic flux is calculated for N windings: U = 12 L I 2 = 12 × 0.0016 × 402 = 1.3 J
Example 21.10
(a) Find the mutual inductance of the solenoid and the coil.
(b) Find the coefficient M for N1 =500 , N2 =20 , `=10 cm
and A=40 cm2 .
Answer
(a) We write the magnetic field of a solenoid carrying a current
I1 :
B = µ0 nI1 = µ0 (N1 /`) I1
A wire is wound N2 times around a solenoid with length ` , The flux of this magnetic field flowing through a coil with
cross-section area A and N1 windings. N2 windings is
21.4. RLC CIRCUITS 361
µ0 N1 N2 A µ0 N1 N2 A
ΦB2 = N2 BA = I1 M=
` `
The mutual inductance was defined in Eq. (21.5) as follows: (b) M is calculated using the given numerical values:
4π × 10−7 × 500 × 20 × 0.0040
ΦB2 = M I1 M=
0.10
We find the coefficient M by comparing the last two formulas: M = 0.0005 H = 0.5 mH
21.4 RLC CIRCUITS

Let us recall the circuit components that we have examined thus far:
Resistor : R Vab = R I
q
Capacitor : C Vab =
C
dI
Coil : L Vab = L
dt
L and C components are not used in a direct current circuit, because no direct
current can flow through a capacitor and the coil gives no reaction to a direct
current. However, these components are activated in alternating current circuits
and generate a potential difference when a current flows through them.
The topic of alternating currents is very broad and simply cannot be examined
thoroughly in this course. We shall merely briefly summarize some results here
using what we have learned.
Figure 21.13 shows L, R, C circuit components connected in series to an
emf source. In order to find the current flowing through this circuit, we add the
potential differences on each component by going from the positive pole of the
generator:
Figure 21.13: An RLC circuit.
Vad = Vab + Vbc + Vcd
dI q
E = L +RI +
dt C
We cannot use the equation in this form, as it includes both the unknown current
I and charge q . We take the derivate of the equation, and recalling that dq/dt = I ,
we get
dE d2 I dI 1 dq
= L 2 +R +
dt dt dt C |{z} dt
I
Rearranging the terms, we get:
1 dE
L I 00 + R I 0 + I= (21.11)
C dt
This expression is a differential equation that gives the unknown current I(t) .
We can solve this equation and find the current if the emf function E(t) on the
right-hand side is given.
Before solving this equation, note a very important similarity: Recall the
equation x(t) giving the position of the mass when we examined damped and
driven oscillatory motion in mechanics (Eq. 9.30):
d2 x dx
m 2
+b + k x = F(t)
dt dt
Comparing the two equations, we notice that the mass m in mechanics is replaced
by the coil L , the friction b is replaced by the resistance R and the spring k is
replaced by the capacitor 1/C . Let us remember the solutions that we found in
mechanics: The system makes a damped oscillation motion if there is no external
force ( F = 0 ). Likewise, resonance was observed when the frequency ω of the
external force was near the natural frequency.
The RLC circuit also has damped oscillation motion (Figure 21.14) and reso-
Figure 21.14: Damped motion nance solutions, depending on the applied potential difference E(t) . This topic is
in an RLC circuit. examined in more detail in courses on circuit analysis.
1. According to Faraday’s law, the emf induced in a coil is:
IV. All of the above.
(a) Proportional to the magnetic field crossing the coil.
(a) I & II (b) II & III (c) I & III (d) IV
(b) Proportional to the magnetic flux crossing the coil.
(c) Proportional to the time rate of change of the mag-
6. The magnetic flux crossing a coil is varying as ΦB =t3 +
netic flux crossing the coil.
2t2 (T·m2 ) . What is the emf induced at time t=1 s ?
(d) Proportional to the cross-sectional area of the coil.
(a) 0 (b) 3 (c) 5 (d) 7
2. According to Lenz’s law, the direction of the current
7. What will happen to the generated emf if the coil of a
produced by an induction emf is:
generator rotates faster?
(a) Opposite to the current generating it.
(a) It will increase.
(b) In the same direction as the current generating it.
(c) In opposite direction to the change in the current
(c) It will remain the same.
generating it.
(d) In the same direction as the change in the current
generating it. 8. When will an induction emf be induced in a coil?
(a) When a constant current flows.
3. The current I flows through the linear wire in the fig- (b) When the current is maximum.
ure below. In which of the two conductor coils will an (c) When the current varies over time.
induction emf be produced? (d) When the current is zero.
9. Which of the following are correct for the inductance of

a solenoid?
I. It is directly proportional to the cross-section area.
II. It is directly proportional to the number of windings.
III. It is directly proportional to the square of the number
(a) A (b) B (c) A and B (d) None
of windings.
IV. It is inversely proportional to the length.
4. In which of the following situations will an induction
(a) I, III & IV (b) I & IV (c) II & IV (d) I & II
emf be produced in a coil?
I. If the magnetic field crossing the coil varies. 10. In which direction will a current be induced on the coil
II. If the surface area of the coil varies. when the current in the straight wire shown in the figure
III. If the magnetic field is changing direction. below is decreased?
(a) I & II (b) II &III (c) I & III (d) IV
5. The maximum value of the induction emf inducted on

a rotating coil in a magnetic field depends on which of
the following? (a) Clockwise.
I. The magnitude of the magnetic field. (b) Counterclockwise.
II. The surface area of the coil. (c) No current is induced.
III. Angular momentum. (d) It is impossible to tell.
PROBLEMS 363
III. It is directly proportional to the inductance.

(a) I & III (b) II & III (c) I & II (d) IV
11. In which direction will the induction current flow on 15. The magnetic flux crossing a coil decreases from 6 T·m2
the branch ab of the circuit shown in the figure above, to zero in 2 seconds. What will the induction emf be?
when it moves to the right?
(a) 2 (b) 3 (c) 8 (d) 12
(a) Upward.
(b) Downward.
16. By what factor will the inductance increase when the
(c) The current will be zero.
number of windings of a solenoid is doubled?
(a) 2 (b) 4 (c) 1/4 (d) 1/2
12. Which of the following are correct for the number of
windings in a transformer? 17. By what factor will the magnetic energy increase if the
I. The secondary emf will be larger if N2 > N1 . inductance of a coil is doubled and the current flowing
II. The primary emf will be larger if N2 > N1 . is tripled?
III. The secondary current will be larger if N2 > N1 . (a) 6 (b) 16 (c) 18 (d) 36
IV. The primary current will be larger if N2 > N1 .
(a) I & III (b) II & IV (c) I & IV (d) II & III 18. The number of primary windings of a transformer is dou-
bled and the number of secondary windings is increased
by a factor of 4. By what factor will the secondary volt-
13. Energy losses are caused by which of the following in a
age increase?
transformer?
I. Resistance of the wire. (a) 2 (b) 4 (c) 8 (d) 1/2
II. Loss of magnetic flux.
III. Eddy currents. 19. How much energy will be stored when a 10 A current
IV. All of the above. flows through a coil with inductance L = 0.5 H ?
(a) I & III (b) II & III (c) I & II (d) IV (a) 2 J (b) 10 J (c) 25 J (d) 50 J
14. Which of the following are correct for the magnetic 20. Which of the following is the expression for mutual
energy stored in a coil? inductance?
I. It is directly proportional to the flowing current. µ0 N1 N2 A µ0 N1 A µ0 A µ0 A2
(a) (b) (c) (d)
II. It is directly proportional to the square of the flowing ` N2 ` N1 N2 ` N1 N2 `
current.
Problems
21.1 Faraday’s law
21.1 The magnetic flux crossing a coil varies as

ΦB = t3 + 2t (tesla·m2 )
Calculate the induction emf in the coil at time t=1 s .
[A: 5 V .] Problem 21.3
21.3 The conducting coil with resistance R=2 Ω , as shown

21.2 The magnetic field crossing a coil with a cross-section in the figure above, is located in a magnetic field B=0.2 T
area of 50 cm2 and 200 windings, is raised from zero to 0.4 T directed into the plane of the paper. The radius of the coil
in 2 seconds. What is the emf inducted on the coil? starts to decrease as a = 3/t (meters). Find the value and the
[A: 0.2 V .] direction of the current at t=2 s . [A: 0.24 A , clockwise.]
Problem 21.4
21.4 The coil with radius a=50 cm and resistance R=10 Ω
in the figure above is placed inside of a magnetic field that Problem 21.8 and Problem 21.9
varies as B=0.2t5 T . The magnetic field makes a 53◦ angle
21.9 How much force is required to pull the mobile side
with the plane of the coil. Find the value and the direction of
of the rectangular coil shown in the figure above with the
the current induced in the coil at time t=1 s .
same constant velocity of v=6 m/s ? (Hint: It is sufficient to
[A: 0.063 A , clockwise.]
overcome the force exerted upon the current-carrying wire
in a magnetic field.) [A: F = 0.016 N .]
Problem 21.5
21.5 The conducting coil with resistance R=10 Ω in the fig-
ure is placed in a magnetic field directed out of the plane of the
paper. The radius of the coil decreases as a(t)=2e−3t (meters) , Problem 21.10
while the magnetic field increases as B(t)=0.1 et (tesla) . Cal- 21.10 The mobile side with a length of 20 cm of a coil lo-
culate the value and direction of the induction current in the cated vertically in a B=0.5 T magnetic field directed out of
coil at time t=0 . [A: 0.63 A , counterclockwise.] the plane of the paper, as shown in the figure, has weight
W=0.1 N . The total resistance of the coil is R=10 Ω . What
is the limit velocity of the mobile side when it is released
for from rest? (Hint: The mobile side travels with constant
velocity when the magnetic force exerted upon its current is
equal to its weight.) [A: 100 m/s .]
Problem 21.6
21.6 The rectangular loop with the dimensions given in the
figure above is located in a region where a uniform magnetic
field B=0.5 T in the −y direction is present. The loop plane
makes a 30◦ angle with the x -axis. The magnitude of the
Problem 21.11
magnetic field is reduced down to zero in 0.1 s . Calculate the 21.11 The number of windings per unit length of the solenoid
value and direction of the current induced in the loop. The shown in the figure is n=800 turns/meter and the current
resistance of the loop is R=1 Ω . it carries decreases as I1 =5/t2 (ampere). A coil with radius
[A: 8.7 mA , upward on the front side.] a=20 cm and 25 windings placed inside of the volume of the
21.7 A coil with 100 windings has a cross-section area of solenoid has resistance R=1 Ω . Determine the value and di-
40 cm2 and is placed perpendicularly to a B=0.4 T magnetic rection of the current flowing through the inner coil at t=1 s .
field. The direction of the coil is suddenly reversed by 180◦ . [A: 0.032 A , in the same direction as I1 .]
Considering that this rotation takes place in 0.1 s , what will
the average induction emf inducted on the coil be?
[A: 3.2 V .]
21.8 The rectangular coil in the figure below has resistance

R=10 Ω and is located inside of a B=0.4 T magnetic field Problem 21.12
directed out of the plane of the paper. The side of the coil 21.12 In the figure above, a rectangular coil with resistance
with length L=40 cm is pulled to the left with a velocity R=10 Ω and dimensions a=1 m and b=3 m is placed inside
of v=6 m/s . Find the value and direction of the induction of a uniform magnetic field B=0.5 T directed out of the paper.
current in the coil. [A: 0.1 A , clockwise.] The coil is rotating about its axis with an angular velocity
PROBLEMS 365
of ω=5 rad/s . Calculate the current induced in the coil as a converted into heat be on the 1 Ω of resistance of the line
function of time t . [A: I = 0.75 sin 5t .] when this same power is first increased to a voltage of 500V
using a transformer? [A: (A) 25 W , (b) 4 W .]
21.2 Generators and Transformers
21.13 The rotating coil of a generator is made of N windings, 21.3 Inductance – Magnetic Energy
each with cross-sectional area 40 cm2 and is rotating with 21.17 Of two coaxial solenoids, the current on the first one
an angular velocity of 100 rad/s inside of a 0.2 T magnetic decreases from 5 A to zero in 0.01 s and a 2000 V emf is
field. What is the minimum number of windings required generated on the second solenoid. Determine the mutual
such that the maximum voltage generated by the generator inductance of this system. [A: M = 4 H .]
is 24 V ? [A: N = 300 windings.]
21.18 A 3 A current is flowing through a coil with 2 H in
21.14 The primary circuit of a transformer has 100 windings inductance. (a) How much energy is stored in the coil? (b)
and its secondary circuit has 40 windings. What will a 120- The current increases by 40 A/s each second. What is the
volt and 10-ampere current entering the transformer be at emf induced in the coil? [A: (a) 9 J , (b) 80 V .]
the output? [A: 48 V and 25 A .]
21.19 (a) Calculate the inductance of a solenoid with 600
21.15 A 24 V and 20 A current supplied to the primary cir- windings, 50 cm in length and a 20 cm2 cross-sectional area.
cuit of a transformer is required to generate a 1 A current. (b) Calculate the energy stored when a 10 A current flows
What will the ratio N2 /N1 and the output voltage be? through this solenoid. [A: (a) 1.8 mH , (b) 0.09 J .]
[A: N2 /N1 = 20 and E2 = 480 V .]
21.16 The 1000W in power generated in a generator is 21.20 A conducting wire is wound N2 =50 times around a
transmitted using two methods. (a) What will the power con- solenoid with length `=20 cm , cross-section area A=100 cm2
verted into heat be on the 1 Ω of resistance of the line when and N1 =800 windings. Calculate the mutual inductance of
transmitted at a voltage of 200V ? (b) What will the power the system. [A: M = 2.5 mH .]
?
22
GEOMETRIC OPTICS
The Hubble Space Telescope,

with the Earth in the back-
ground. This 11-ton spacecraft
contains a telescope 2.4 m in
diameter. It is able to clearly
see the furthest celestial objects
in space, as its orbit at 560 km
in altitude is located outside of
Earth’s atmosphere. The Hubble
telescope will remain in service
until 2030.
In this chapter, we shall introduce Optics, one of the most important branches
of physics. All human activity in nature starts with the sense of sight. The sense
of sight uses a property of materials called light. The light emitted from the
objects around us forms an image in our eyes. We thus know that such an object
is there even if we are unable to touch it.
We will first learn the structure of light. Understanding that light is an
electromagnetic wave was one of the most important developments in the history
of science. Many technical and optic instruments were able to be developed after
this wave structure was understood.
We will also discuss the laws governing the way in which light behaves when
passing from one medium into another or when being reflected. We will work
out to calculate the formation of images by mirrors and lenses, which facilitates
daily life and technology.

368 22. GEOMETRIC OPTICS
22.1 THE WAVE NATURE OF LIGHT

Various answers have been given throughout history to the question of “What
is light?” The road of the theories of light is very interesting. The particle model
set forth by Newton was accepted until the 19th century. Newton asserted that
sources of light would emit particles too small to be seen, which would then cause
images to be formed when they reached the eye, and he could explain reflection
and refraction of light accordingly. Despite the fact that, during the same years,
Huygens and Fermat had claimed that light could be a wave, they did not receive
much recognition.
However, the wave-model of light began to stand out in the early 19th century,
when Young and Fresnel revealed light’s interference and diffraction properties.
This was because interference and diffraction properties cannot be explained with
the particle model.
Later, in 1873, James Clerk Maxwell published his electromagnetic theory,
merging electricity and magnetism into a single theory. He suggested the concept
of the electromagnetic wave for the first time. According to Maxwell, the electric
field E and the magnetic field B are both jointly included in an electromagnetic
wave (Figure 22.1), as transverse sinusoidal waves (perpendicular to the direction
of propagation).
Figure 22.1: The fields E and B

constituting an electromagnetic
wave are in transverse sinusoidal
wave motions perpendicular to
the direction of propagation.
Maxwell found the following expression for the propagation speed of electro-
magnetic waves in free space in terms of the electric permittivity and magnetic
permeability of free space:
1
v= √
ε0 µ0
When calculated, this expression was found to be equal to the speed of light:
1
c= √ = 2.9979 × 108 m/s ≈ 3 × 108 m/s
ε0 µ0
After coming up with this result, Maxwell went one step further and asserted
that light was also an electromagnetic field. His ideas were soon proven experi-
mentally. In 1887, Heinrich Hertz managed to generate an electromagnetic wave
in laboratory.
Today, we know of all kinds of electromagnetic waves, from radio waves to
microwaves, infrared rays to X- and gamma rays. Let us show these waves on a
graph called a spectrum, sorted according to their frequencies (Figure 22.2):
• Radio waves are waves whose wavelength ranges from 10 cm to 10 km
and that can easily cross the atmosphere. They are used in radio and TV
broadcasts.
22.1. THE WAVE NATURE OF LIGHT 369
Figure 22.2: Electromagnetic

wave spectrum.
• Microwaves have shorter wavelengths and are used in radars, wireless

internet, kitchen ovens, etc.
• Infrared rays are used in short-range communication and fiber-optic com-
munication. They are quickly absorbed by the atmosphere.
• Visible light is the range of wavelength to which the human eye is sensitive.
It ranges from red (700 nm) up to violet (400 nm).
• Ultraviolet rays have higher energy than visible light and may cause an Figure 22.3: The range of visi-
ionizing impact on tissues and lead to sunburns. ble light.
• X-rays are also dangerous for tissues and are used in medicine and the
industry because they can easily pass through matter.
• Gamma rays are the electromagnetic radiation with the highest energy.
They come from outer space or are emitted in nuclear reactions. They are
used in medicine for diagnosis and cancer treatment.
This model of electromagnetic waves was one of the great successes of classic
physics. However, some observations at the start of the 20th century revived
the particle model. New effects such as the photoelectric effect and Compton
scattering indicated that light behaved not like a wave, but like a billiard ball.
These uncertainties were resolved in 1926 with quantum theory. According
to quantum theory, not only light, but all particles have a double nature, called
the wave-particle structure. They exhibited their particle nature in certain cases
and their wave nature in certain other cases.
However, the classic electromagnetic wave model could explain all aspects of
light, except for the phenomena of scattering, emission and absorption of light.
The Ray Model of Light
Light emitted from any source propagates into space as an electromagnetic
wave. The locus of the points of this wave with the same phase is called a wave
front (Figure 22.4). A wave front expands in space as a spherical surface centered
at the source of light, if there is no obstruction. We can thus visualize light as a
successions of wave fronts.
Virtual lines assumed to be perpendicular to the wave front and in the direction
of propagation are called rays. In a homogeneous medium, rays are straight lines
drawn from the source. Rays may be curved in non-homogeneous media, for Figure 22.4: Wave fronts, and
example, when passing through the atmosphere. However, they shall always rays propagating perpendicular
propagate as perpendicular to the wave front. The ray model is easier to work to them.
with than wave fronts.
Wave fronts can approximately be taken as planes at very far distances from
the source of light (Figure 22.5) where the radius of the sphere is very large. Such
a wave with the source at infinity is called a plane wave. Accordingly, the rays
become parallel to each other.
Huygens–Fresnel Principle
What will happen when a propagating light wave meets an obstacle? Huygens
and Fresnel explained the propagation of light in a medium with the following
principle:
Every point of a medium on which a light is incident constitutes a new source of
Figure 22.5: Wave fronts be- waves, and the wave front emitted therefrom spreads evenly in every direction.
come planes and rays become Consider two adjacent rooms. Let there be a hole in the wall between the two
parallel at very great distances rooms. When someone speaks in the first room, the sound will be heard in the
from the source.
second room as if emitted from this hole. Even if the speaker changes position in
the room, in the second room, the sound wave will always be heard as if coming
from the same point. This complies with the Huygens–Fresnel principle.
Figure 22.6 shows what happens when a plane wave meets two slits in a
screen. The light spreads as new spherical wave fronts are emitted from these
slits.
The Speed of Light and Index of Refraction
The speed of propagation of light is a universal physical constant:
c = 2.9979 × 108 m/s ≈ 3 × 108 m/s
Figure 22.6: The Huygens–

Special techniques are required to measure such a high speed. Many techniques
Fresnel principle: Each slit pro- have been developed throughout history to measure the speed of light. The most
duces new wave fronts. commonly known method is the apparatus developed by the French scientists
Fizeau and Foucault.
In the Fizeau–Foucault apparatus, the light emitted from a very distant source
passes between the teeth of a rotating wheel and is converted into light pulses
(Figure 22.7). A pulse is observed after it is reflected back from a mirror and again
passes between the teeth of the wheel. When a pulse is reflected from the mirror,
it gets stopped if the rotation speed of the wheel is not right. By varying the
rotation speed, it is ensured that a pulse passes through the teeth while both
going and returning. When this is achieved, the speed of light can be calculated
in terms of the distance between the mirror and the wheel, the rotation speed of
the wheel and the number of teeth.
Figure 22.7: The Fizeau– (Note: The expression above, “measuring the speed of light,” became meaning-
Foucault apparatus for less after 1983. If we remember the modern definition of the basic unit of length
measuring the speed of light. (meter) as “the distance traveled by light in vacuum in 1/299 792 458 seconds,"
we understand that the speed of light is a given universal constant; there is no
question of measuring it. What these experiments really measure is the length of
the meter unit. This outlook is only correct historically: Before 1983, the meter
unit was defined differently, and then the speed of light was measured.)
The speed of light decreases when the light passes through a gas or liquid
environment. It is easy to understand the reason for this: Recalling the definition
√
for the speed of light c = 1/ ε0 µ0 , as ε > ε0 and µ > µ0 is true for every
medium, the speed of light will always be less than its value in vacuum.
This degree of slowing is expressed with a dimensionless coefficient. If we
use v to indicate the speed of light in the medium and c to indicate its speed in
vacuum, the ratio
c
n= (index of refraction) (22.1)
v
22.2. REFLECTION AND REFRACTION 371
is called the index of refraction of the medium. The index of refraction depends
on the type of medium and the wavelength of the incident light. The index of
refraction of some media are as follows:
Index of refraction of some media.

n n
Vacuum 1 Acrylic glass 1.49
Air 1.0003 Flint glass 1.60
Water 1.3330 Crown glass 1.52
Ice 1.309 Silicon 3.96
Recall the relation between the velocity v , wavelength λ and frequency f

(or period T ) of a wave:
v
λ = vT = (22.2)
f
The frequency of a light wave does not change when passing from one medium
into another. Frequency is a quantity determined by the source generating the
wave. (When you generate a wave by vibrating a rope, the up and down vibration
frequency of the rope does not vary with the propagation of the wave.) Therefore,
if we write the above formula for two media and divide both sides, we get
λ1 v1
= (22.3)
λ2 v2
22.2 REFLECTION AND REFRACTION

A light ray that reaches the interface surface between a medium with index
of refraction n1 and another medium with index of refraction n2 will be partially
reflected and be partially refracted (bent in the second medium). Let us first give
the laws of reflection and refraction:
Laws of Reflection and Refraction

1. Reflected and refracted rays are in the plane of incidence
formed by the the incident ray and the normal to the surface.
2. The angle of incidence and the angle of reflection measured
from the normal to the surface are equal:
θ1 = θ10 (22.4)
3. The relation between the angle of incidence and the angle of

refraction is determined with Snell’s law:
n1 sin θ1 = n2 sin θ2 (Snell’s law) (22.5)

Figure 22.8: Reflection and re-
Each of these laws can be proven using wave fronts. Let us prove the refraction fraction.
law here as an example.
Consider two incident light rays with the same angle of incidence θ1 (Fig-
ure 22.9). When the first ray reaches point A on the interface, the point B, which
is on the same wave front still has to cover distance BC . While this second ray
travels distance BC at velocity v1 , the ray at point A will start to travel at velocity
v2 in the second medium. The distances traveled in the same time interval ∆t
will be as follows:
BC = v1 ∆t AD = v2 ∆t
The new wave front thus becomes DC . Let us write the distances BC and AD
in terms of the common side AC and the angles:
BC = v1 ∆t = AC sin θ1 AD = v2 ∆t = AC sin θ2
Figure 22.9: Successive wave
We now divide both sides:
fronts travel slower in the sec- v1 sin θ1
ond medium. =
v2 sin θ2
Lastly, we replace velocities v using the formula n = c/v :
v1 n2 sin θ1
= =
v2 n1 sin θ2
This is the Snell’s law in Eq. (22.5).
Let us emphasize the important points of reflection and refraction:
• If the incident ray is perpendicular (θ1 =0) , then θ2 =0 and the ray goes into
the second medium without refraction.
• If the rays fall on an interface where n1 < n2 (e.g., air-water), there is always
refraction in the second medium. In this case, we have θ2 < θ1 and the
Figure 22.10: Light coming out refracted ray approaches the surface normal. This is what causes a rod
of water shows a pencil as if bro- submerged in water to seem as if it is broken (Figure 22.10).
ken.
• Total internal reflection. An interesting thing happens when rays try to
exit a denser medium, e.g., from water into the air (n1 > n2 ) . In this situation,
the exiting rays bend away from the normal. As the angle of incidence θ1
increases, the angle of refraction θ2 approaches 90◦ (Figure 22.11). When θ1
is greater than this value, no refraction occurs and all of the light is reflected
internally. Let us write Eq. (22.5) for θ2 :
n1
sin θ1 = sin θ2
n2
As n1 /n2 > 1 when leaving a denser medium, this equation has no solutions
Figure 22.11: Total internal re- for angles that make the right-hand side greater than 1. Therefore, for a
flection and critical angle θc . solution to exist, we must have
n1
sin θ1 6 1
n2
In particular, the maximum incident angle that makes the right-hand side
equal to 1 is called the critical angle and is indicated with θc :
n2
sin θc = (Critical angle) (22.6)
n1
In conclusion, it is possible to exit from a denser medium only with angles
θ1 < θc . The light cannot exit the medium and is fully reflected back at angles
greater than this. This interesting phenomenon is called total internal
Figure 22.12: Light entering reflection.
one end of a fiber optic cable is The most important application of total internal reflection is fiber optic
transmitted to the other side by cables. Both visible light and other electromagnetic waves can be transmitted
total internal reflection. through fiber optic cables with very slight losses.
22.2. REFLECTION AND REFRACTION 373
Dispersion of Light
White light incident on a glass prism is dispersed into all of the colors of
visible light after passing through the glass medium. As seen in the adjacent
figure, all of the components of visible light from red to violet are refracted at
different degrees. Red light, which has a greater wavelength, is refracted less and
violet light, which has a lesser wavelength, is refracted more.
This property, which shows that the index of refraction is different at each
wavelength, is called the dispersion of light:
n = n (λ) (Dispersion) (22.7)
As a rule, index of refraction decreases with wavelength.

Recall that the index of refraction depends on the speed of light with the Figure 22.13: Dispersion of
formula n = c/v . The speed of light c in vacuum is the same in all wavelengths, light.
however, its velocity v is different at each wavelength. According to the figure
above, the speed of light in a medium with a larger wavelength (red) is greater
than that of violet light.
Rainbow
The dispersion property of light combined with the total internal reflection
effect that we discussed above produces one of the most spectacular scenes in
nature. A rainbow that contains all of the colors and that seems to be suspended
in the air is observed when the sun comes out after rain.
Let us explain in a figure how a rainbow is formed. Consider the paths traveled
by the red and violet components inside of a ray of sunlight incident on a spherical
droplet of water, as shown in Figure 22.15a. According to dispersion property,
violet light will be refracted more than red light in the droplet. When these two Figure 22.14: Rainbow.
rays reach the back surface of the sphere, in certain droplets, the two rays incident
at angles higher than the critical angle cannot exit the droplet, and thus reflect
inside due to total internal reflection. After a few reflections, these two rays will
finally exit the droplet and continue on their paths.
Figure 22.15: (a) Total internal
reflection inside of a spherical
water droplet. (b) The image of a
rainbow formed in the eye by re-
flections from droplets at various
angles.
Let us now examine the effect generated when the red and violet rays from
two such droplets reach our eyes. As shown in Figure 22.15b, from two droplets
located at different locations, we observe the light from the top one as red and
the light from the bottom one as violet. The droplets in the area between these
two ends are lined up so as to reflect all of the other colors of visible light.
These particular reflections take place only within a certain range of degree.
Other reflections do not reach our eyes. Also, we observe the same phenomenon
within the same degree range from different directions. As a result, a colorful
image appears in an arch-shaped region.
Example 22.1
(b) Frequency remains constant when light passes into an-
The yellow light of sodium, which has a 598 nm wavelength other medium, however, its speed and wavelength changes.
in vacuum, enters a glass medium with an index of refraction We use Eq. (22.1), in which we defined the index of refraction:
c c
of 1.6 . (a) What is the frequency of this light in vacuum? n= → v=
v n
(b) What is its speed in a glass medium? (c) What is its wave- From here, we calculate the speed of light in a glass medium:
length in a glass medium? c 3 × 108
v= = = 1.9 × 108 m/s
Answer n 1.6
(c) The wavelength in a glass medium can either be calcu-
(a) The relation between the speed, frequency and wavelength lated using λ = v/ f in terms of speed and frequency or by
of a wave is found using Eq. (22.2): taking the ratio of these values in the two media. There is
v
f = less calculational error in taking the ratio:
λ λ2 v2
We calculate frequency by taking the speed of light in vacuum =
λ1 v1
as c : v2 1.9
c 3 × 108 λ2 = λ1 = × 598 = 379 nm .
f = = = 5 × 1014 Hz v1 3
λ 598 × 10−9
Example 22.2
We use tables or a calculator to find the angle with a known
A light ray is sent from air to glass ( n = 1.60 ) with a 30 angle sinus value:
◦
of incidence. (a) What is the angle of refraction in the glass θ2 = 18◦ .

medium? (b) The light is sent at such an angle that its angle of (b) In Snell’s law, we take θ2 = θ1 /2 and solve for θ1 :
refraction is half its angle of incidence: θ2 = θ1 /2 . What is the sin θ1 = n sin θ1 /2
angle of incidence? Using the half angle formula sin θ1 =2 sin θ1 /2 cos θ1 /2 from
trigonometry and simplifying, we get
Answer θ1 n
(a) We use Snell’s law ( n1 = 1, n2 = n ): 2 cos θ1 /2 = n → cos =
2 2
sin θ1 = n sin θ2 From here, we calculate the cosine of the half angle:
We substitute the values and solve for θ2 : θ1 1.6 θ1
cos = = 0.8 → = 37◦ .
1 1 2 2 2
sin θ2 = sin θ1 = × sin 30 = 0.31 From here, we find that θ1 = 74◦ .
n 1.6
Example 22.3
θc among those sent from inside of the water can pass into
The light emitted from a lamp located at the bottom of a 10 m - the air, while the rest undergo total internal reflection. We
deep-pool produces an illuminated circle at the surface of the therefore first find the critical angle.
water. What is the radius of this circle? (The index of refraction Here, the first medium is water ( n1 =n ) and the second
of water is 1.33 .) medium is air ( n2 =1 ). Let us write Snell’s law:
n sin θ1 = sin θ2
We find the critical angle by taking the maximum value of
the angle of refraction in air as θ2 = 90◦ :
n sin θc = 1 × sin 90◦ = 1
1 1
sin θc = = = 0.75 → θc = 49◦
n 1.33
We write the expression for the radius R by examining the
Answer right triangle in the figure:
As shown in the figure, only the rays up to the critical angle R = h tan θc = 10 × tan 49◦ = 10 × 1.15 = 11.5 m .
Example 22.4
The apex angle of a prism is 90◦ . A ray that enters the prism
with an angle of incidence of 53◦ from point A on one side
exits from the other side at point B as parallel to the surface.
Determine the index of refraction of the glass.
Answer
We write Snell’s law for the first surface:
22.3. IMAGE BY REFLECTION – MIRRORS 375
sin 53◦ = n sin θ2 (1) 1

sin θc = → n sin(90◦ − θ2 ) = n cos θ2 = 1 (2)
n
We need a second equation, as both n and θ2 are unknown. We divide equations (1) and (2) on both sides:
For this purpose, let us look at the second surface. As the n sin θ2 sin 53◦
refracted rays go parallel to the surface at point B, there is = → tan θ2 = sin 53◦ = 0.8
n cos θ2 1
total internal reflection, in other words, the rays are incident
From here, we find θ2 = 39◦ and calculate n by substituting
on the surface at the critical angle θc . As seen in the drawing
in the second equation:
in the figure, we have θc = 90◦ − θ2 . Let us write the critical 1
angle formula: n= = 1.28
cos 39◦
22.3 IMAGE BY REFLECTION – MIRRORS

Formation of an Image by Reflection
In the optics sense, an object is a body from which light rays originate. It
can be a lamp, in other words, generate its own light. Other objects may reflect
back the rays that they receive from the environment. When light rays diverging
from an object reach the intersection of two media, they are partially reflected
and partially transmitted to the second medium through refraction. An image is
formed at the position where either reflected or refracted rays converge. It is a
real image if the rays really intersect and a virtual image if only the extensions
of the rays intersect.
(A practical method for determining whether the image is real or virtual is to
place a screen at the location of the image. A real image will form on the screen,
but a virtual image will not.)
In this section, we shall examine the properties of images formed through
reflection. Mirrors are used to obtain images through reflection. Mirrors are
surfaces that are polished or that have their reflection properties increased through
coating by another material and reflect almost all of the incident light. The most
commonly used types are plane and spherical mirrors.
Plane Mirror
When we look at a plane mirror, what we see makes it look as if an exact
copy of whatever object is standing in front of the mirror is behind the mirror
(Figure 22.16). According to our definition above, this is a virtual image, because
the two rays are not really intersecting; rather it is as if their extensions are
intersecting at a point behind the mirror. In the case of a plane mirror, the image
is always virtual. However, the image has the same length, direction and distance
from mirror as the object.
In Figure 22.16, the distance of the object from the mirror is indicated with Figure 22.16: The image is vir-
p and the distance of the image to the mirror is indicated with q . The absolute tual in a plane mirror because
value of these two distances is equal. However, the object and image positions the extensions of the rays inter-
are taken as algebraic numbers in the calculations, as we shall see later. sect behind the mirror.
Sign rule 1: According to the assumption made in optics, the positions of real
objects and images are taken as positive and the positions of virtual images and
objects are taken as negative.
Accordingly, the relation between the positions of the object and the image
in a plane mirror should be expressed as follows:
q = −p
Spherical Concave Mirror

In a concave mirror, the rays are reflected from the inner surface of a sphere.
If we use C to indicate the center of the sphere with radius R , the line extending
from the middle point O on the mirror to C and beyond is called the principal
axis: CO = R .
At least two rays from the object are sufficient to obtain an image in a concave
mirror. We know two such rays from our laws of reflection:
1. The ray drawn from the object to the middle point O will be reflected back
at an equal angle with the principal axis (Figure 22.17, ray 1).
2. The ray passing through the center C of the sphere will be reflected back on
itself, since it falls perpendicularly on the mirror (Figure 22.17, ray 2).
Figure 22.17: Coordinates in a A real image forms at the intersection of these two rays. If they do not
concave mirror. intersect, their extension beyond the mirror will form a virtual image, to be
examined later.
Let us use the coordinates shown in Figure 22.17 to find the position and
height of the image. Let p be the position of the object at point P as measured
from the mirror and q be the position of the image at point Q .
Also, let us use h to indicate the height of the object and h0 to indicate the
height of the image. We make a new assumption here for the heights:
Sign rule 2: The height of an inverted image is expressed with a negative sign.
This means that the h0 shown in the figure has a negative sign.
As the right triangles OPP0 and OQQ0 facing the same angle θ are similar
triangles, the ratios of their sides are equal:
h p
0
= (22.8)
−h q
Note that, as the height of the inverted image h0 is negative, an extra negative
sign was placed before it to make it positive.
This time, let us use the similarity of the triangles CPP0 and CQQ0 facing
the same angle α . Remember to take the height of the inverted image as (−h0 ) .
When writing the ratios of the sides, we get
h p−R
0
=
−h R−q
Eliminating h/h0 from the last two formulas, we get
p p−R
=
q R−q
From here, we obtain the formula that gives us the position of the image q :
1 1 2
+ = (22.9)
p q R
Focal point: When the object is very far away, in particular, when p→∞ in the
equation above, we have 1/p = 0 and the image position is:
R
q=
2
22.3. IMAGE BY REFLECTION – MIRRORS 377
This special point is called the focal point F and its distance to the mirror is
called the focal length f =R/2 . The focal point is the midpoint between the
mirror and the center of the mirror.
Likewise, when the object is placed at the focal point, p= f =R/2 in the equa-
tion above, we have 1/q=0 and q→∞ , in other words, the image is at infinity.
In conclusion, the concave mirror equation in terms of the focal length f is:
1 1 1
+ = (Concave mirror equation) (22.10)
p q f
Magnification: The ratio of the image height h0 to the object height h is called
the magnification and is indicated with M = h0 /h . Eq. (22.8) gives us the magnifi-
cation in terms of positions:
h0 p
M= =− (Magnification) (22.11)
h q
Also, solving the equation (22.10) for q , we get
pf
q=
p− f
Figure 22.18: Images in an con-

cave mirror: (a) The image is real
and smaller if the object is be-
yond the center, (b) The image
is real and larger if the object is
between the center and the focal
point, (c) The image is virtual if
the the object is closer than the
focal point.
These two equations show us the possibilities for the image position and the
magnification:
• If p > f , we have q > 0 and M < 0 . In other words, beyond the focal point,
we have a real, inverted image (Figure 22.18a and b).
• If p < f , we have q < 0 and M > 0 . In other words, inside of the focal
point, we have a virtual, upright image (Figure 22.18c).
Concave mirrors are used to collect light or to send light to faraway points.
They are used to focus light in telescopes, lasers and solar panels. On the other
hand, light emitted from the focal point in automobile headlights and flashlights Figure 22.19: The concave mir-
can be sent as parallel beams over very great distances. Also, they provide a ror used in the Hubble space
bigger virtual image in dentist’s and shaving mirrors. telescope.
Example 22.5
Answer A diagram should always be drawn in mirror cal-
culations to get an idea about the image. Certain particular
The focal length of a concave mirror is f =30 cm . (a) Where will
rays are useful for obtaining the image. Three particular rays
the image of an object with 1 m in size placed 120 cm away
are shown in the following figure:
from the mirror form and what will its size be? (b) Answer
the same question for an object with 1 m in size placed 10 cm
away from the mirror.
We use the magnification M that we defined with for-

mula (22.11) to find the size of the image:
h0 q 40 1
M= =− =− =−
h p 120 3
The negative sign indicates that the image is inverted. There-
fore, a triply diminished inverted image is formed.
(b) This time, the object is placed between the focal point and
the mirror. The image is formed as follows, if we take two
particular rays, one from the center and one from the focal
point:
1st ray: Each ray coming from infinity will be reflected from
the focal point.
2nd ray: The ray incident at the central point (O) of the
mirror will be reflected at an equal angle.
3rd ray: The ray passing through the focal point (F) will be
reflected as parallel.
Only two of these three rays are sufficient, because the image
The image position is calculated using the same formula:
will form at the intersection of two rays. 10 × 30
q= = −15 cm
(a) We write the concave mirror equation (22.10): 10 − 30
1 1 1 pf The image is virtual, because q is negative.
+ = → q= We then calculate the magnification:
p q f p− f
h0 q (−15)
The focal length is taken as positive if the mirror is concave: M= =− =− = +1.5
f = 30 cm . Accordingly, we calculate the image of an object h p 10
The positive sign means that the image is upright, in other
p = 120 cm away:
words, in the same direction as the object. Therefore, an
120 × 30
q= = 40 cm image enlarged by 1.5 times is formed in the same direction.
120 − 30
The image is real, because q is positive.
Convex Mirror
A convex mirror reflects rays from the outer surface of a sphere. We again take
two rays to obtain the image of a convex mirror. As shown in Figure 22.20, since
one of the rays is directed towards the center C of the sphere, it is perpendicularly
incident to the surface of the mirror and is reflected back in the same direction.
When the other ray is drawn towards the point O on the principal axis, it will be
reflected at an equal angle with the principal axis.
Figure 22.20: The image is al-

ways virtual in a convex mirror.
Reflected rays will never intersect at any point, regardless of which other
rays are taken, but their extensions intersect behind the mirror. Therefore, the
image is always virtual in a convex mirror.
There is no need to derive a new equation for the convex mirror. In a concave
mirror, the distance p of the virtual image behind the mirror was taken with a
negative sign. Likewise, the same formula will be valid again if we take the radius
R , and therefore the focal point f , as negative, since the center of a concave
22.4. IMAGE BY REFRACTION– LENSES 379
mirror is behind the mirror:

1 1 1
+ = (Convex mirror equation, f negative) (22.12)
p q f
It can easily be proven that the image is always virtual, in the same direction as
the object and smaller in convex mirrors. Solving the same equation for q and
writing the magnification as well, we get
pf h0 q −f
q= M= =− =
p− f h p p− f
Considering that f is always negative in the formula on the left, the numerator
will always be negative and the denominator will always be positive. Therefore,
q will always be negative, in other words, the image will always be virtual. Also,
since the denominator is always larger, the magnification will be less than 1, in
other words, the image will always be smaller than the object.
Convex mirrors can contain the image of a broader area, because the image is Figure 22.21: A wing mirror on
smaller than the object. Due to this property, they are used in the wing mirrors a car.
and rear-view mirrors of automobiles.
Example 22.6
(a) We write the formula (22.10) that we found for the convex
The focal length of a convex mirror is 20 cm . Where will the mirror:
image of an object with 1 m in size placed 60 cm away from 1 1 1 pf
the mirror form and what will its size be? + = → q=
p q f p− f
The focal length should be taken as negative, because the mir-
Answer
ror is convex: f = −20 cm . We calculate the image position
The image will form as follows if we take two particular rays, accordingly:
one reflecting from the midpoint at an equal angle, and the 60 × (−20) 120
other directed towards the center and reflected back on itself: q= =− = −15 cm
60 − (−20) 80
The image is virtual, because q is negative.
We then calculate the magnification:
h0 q (−15) 3
M= =− =− =+
h p 20 4
The positive sign means that the image is upright, in other
words, in the same direction as the object. Therefore, an
image diminished by 3/4 is formed in the same direction.
22.4 IMAGE BY REFRACTION– LENSES

Light rays exiting from an object may again intersect to form an image after
refracting into another medium. In glass lenses, rays first enter the glass medium
and then get refracted from there to continue on their paths, subsequently going
back into the first medium. Therefore, they get refracted twice.
Refraction from a Spherical Surface
In order to understand the image formed by a lens, let us first solve half of
the problem and consider the image in a semi-infinite glass medium.
In Figure 22.22, rays coming from a medium with index of refraction n1 are
incident on a spherical surface with radius R and then pass into the second
medium with index of refraction n2 .
Figure 22.22: Coordinates in a

spherical surface.
This time, let us take the object as a point P on the principal axis. Let one
of the rays emitted from that point be the principal axis itself. This ray directly
passes into the second medium when it is incident on the surface at a right angle.
Let the other ray start at angle α with the principle axis and be incident to the
surface of the glass at point A at angle θ1 with the normal.
According to Snell’s law, this ray will get refracted by a certain angle θ2 . The
image will form at the point Q where this ray intersects the principal axis.
If a ray with a very small angle α is chosen, the ray will be incident to the
surface almost perpendicularly, in other words, θ1 , and therefore θ2 , will be very
small. Accordingly, Snell’s law is simplified at very small angles where we can
take sin θ ≈ θ :
n1 sin θ1 = n2 sin θ2 =⇒ n1 θ1 ≈ n2 θ2
In a triangle, an exterior angle is equal to the sum of the two opposite interior
angles. Applying this property to the triangles PAC and QAC , we get
θ1 = α + β and β = θ2 + γ
Eliminating θ1 and θ2 from these three formulas, we get
n1 α + n2 γ = (n2 − n1 )β (22.13)
As angles α, β, γ are very small, using the approximate expression tan θ ≈ θ , we

can calculate the tangents of these angles from Figure 22.22. Using the perpendic-
ular d facing all three angles, we get
d d d
α ≈ tan α ≈ β ≈ tan β ≈ γ ≈ tan γ ≈
p R q
We substitute the values of these three angles in the equation (22.13) and simplify:
n1 n2 n2 − n1
+ = (22.14)
p q R
This formula is valid for all n1 , n2 values. Also, this proof was made for a convex
spherical surface. But it can also be shown to be valid for a concave spherical
surface. In order to do this, we agree on the sign of the radius R of the sphere as
follows: R is taken as positive if the center C of the sphere surface is inside of the
medium n2 and R is taken as negative if it is outside. Then, this formula will also
be valid for refraction from a concave surface.
This expression shall be used later to obtain the lens formula.
Refraction at a Plane Surface

Let us apply the above formula to a special case in which the interface is
a plane. If we consider a plane as a sphere with infinite radius, when we take
R → ∞ in the formula (22.14), we get
n1 n2
+ =0
p q
From here, we find the position of the image:
!
n2
q=− p (22.15)
n1
According to this result, the image will always be virtual on a plane surface. For
example, a pool seems to be shallower when looked at from the outside (n1 > n2 ) . Figure 22.23: A fish under wa-
Likewise, a fish looking at us from under the water (n1 < n2 ) sees us as further ter seems nearer when looked at
away. from above.
Thin Lenses
Glass lenses with two spherical surfaces are used in all kinds of optical instru-
ments. The image may be real or virtual, depending on whether these surfaces are
convex or concave. Lenses with convex surfaces are called converging lenses
and those with concave surfaces are called diverging lenses. This designation
will be understood later.
We have to use the formula (22.14) twice in order to obtain a formula that is
valid for all lenses. To simplify the calculations, let us take the first medium as air
and the index of refraction as n ≈ 1 .
Figure 22.24: Coordinates for a

lens with thickness t .
Let us consider a lens with radius of the first surface R1 , radius of the second
surface R2 and thickness t (Figure 22.24). Let p1 be the distance between an
object at point P outside of this lens. Let q1 be the position of the intermediate
virtual image ( Q0 ) formed on the same side as P ( q1 is negative). Accordingly,
if we apply Eq. (22.14) to a surface with radius R1 at the interface between air
( n1 = 1 ) and glass ( n2 = n ), we get
1 n n−1
+ =
p1 q1 R1
The rays intersecting at this virtual image q1 then continue on their paths and
arrive at the second surface located at distance t . If the thickness of the glass is
t , this intermediate image is considered as an object at position p2 = t − q1 for
the second surface. We apply the equation (22.14), this time to the surface with
radius R2 at the interface of glass ( n1 = n ) and air ( n2 = 1 ):
n 1 1−n
+ =
t − q1 q2 R2
Note that R2 is negative, because it is inside of the glass.
Now, we take t ≈ 0 , assuming that the lens is thin and we add these last two
equations on both sides and eliminate q1 . Also, if we remove the subscripts and
rename them as the first object ( p1 = p ) and the final image ( q2 = q ), we get the
following result:
!
1 1 1 1
+ = (n − 1) − (Lens makers’ equation) (22.16)
p q R1 R2
This result is known as the lens makers’ equation. We can write it in terms of
the focal length f to make it more practical. As q = f when p → ∞ , substituting
these values, we get !
1 1 1
0 + = (n − 1) −
f R1 R2
and f is positive for a converging (convex) lens because we take R1 > 0 and
Figure 22.25: Converging lens. R2 < 0 . f is negative for a diverging (concave) because R1 < 0 and R2 > 0 .
Every lens has two focal points, because the rays can go in both directions.
In conclusion, the thin lens equation for both converging and diverging lenses
is as follows:
1 1 1
+ = (Thin lens equation) (22.17)
p q f
f is positive for a converging lens and f is negative for a diverging lens. The
Figure 22.26: Diverging lens. image is real if q is positive and virtual if it is negative.
Magnification ( M )
As in mirrors, the ratio of the height h0 of the image to the height h of the
object is the magnification, and its relation with the positions is as follows:
h0 q
M= =− (22.18)
h p
If M is positive, the image is in the same direction as the object and virtual; if M
is negative, the image is inverted and real.
Power of a Lens ( D ).
Opticians use the optical power of a lens, rather than its focal length. The
power of a lens is just the inverse of the focal length f , which is expressed in
meter units:
1
D=
f (meters)
The unit of lens power is the diopter. For example, a lens with D = 3 diopter
will focus parallel incident rays to f = 1/D = 1/3 m = 33 cm .
Example 22.7 q 120

M=− =− = −2
p 60
The focal length of a thin converging lens is f = 40 cm .
The negative sign indicates that the image is inverted. There-
(a) Where will the image of an object with a size of 1 m and
fore, a doubly magnified inverted image is formed.
placed 60 cm away from the lens form? And what will its size
be? (b) This time, the object is placed between the focal point
(b) Answer the same question for the same object placed 30 cm and the lens. The image is formed as follows if we take two
away from the lens. particular rays, one of which passes through the center of the
Answer lens without being refracted and one coming from infinity as
(a) To form the image, let us consider two special rays, one parallel to the axis and passing through the focal point.
coming from infinity in parallel to the axis and going through
the focal point and the other passing through the center of
the lens without being refracted. Accordingly, the image is
formed as follows:
The image position is again calculated using the same for-

We write Eq. (22.17), which we found for lenses: mula:
1 1 1 pf 30 × 40
+ = → q= q= = −120 cm
p q f p− f 30 − 40
The focal length is taken as positive, as the lens is converging: The image is virtual, because q is negative.
f = 40 cm . Accordingly, we calculate the image of an object We then calculate the magnification:
p = 60 cm away: q (−120)
60 × 40 M=− =− = +4
q= = 120 cm p 30
60 − 40
The image is real, because q is positive. The positive sign means that the image is upright, in other
We use the magnification M that we defined with for- words, in the same direction as the object. Therefore, a
mula (22.11) to find the size of the image: quadruply enlarged image is formed in the same direction.
Example 22.8
We write the thin lens equation:
1 1 1 pf
+ = → q=
p q f p− f
The focal length should be taken as negative, because the lens
is diverging: f = −50 cm . We calculate the image position
accordingly:
75 × (−50) 3750
q= =− = −30 cm
75 − (−50) 125
The image is virtual, because q is negative.
The focal length of a diverging lens is 50 cm . Where will the We then calculate the magnification:
image of an object with a size of 1 m placed 75 cm away from q (−30) 2
the lens form and what will its size be? M=− =− = + = 0.4
p 75 5
Answer The positive sign means that the image is upright, in other
The image will form as follows if we take two particular rays, words, in the same direction as the object. Therefore, an
one coming from infinity and reflecting from the focal point image diminished by 2/5 is formed in the same direction.
and the other directed towards the center and passing without
being refracted:
Example 22.9
unknowns p and q will be as follows:
An object is placed 16 cm away from a screen. It is required p + q = 16
that we obtain the image of the object on the screen using a Second, these p, q values fulfill the lens formula:
converging lens with a focal length of 3 cm . How far away 1 1 1 1
+ = =
from the object should the lens be placed? p q f 3
If we get q = 16 − p from the first formula and substitute it
in this equation, we obtain an equation for p :
1 1 1
+ = → p2 − 16p + 48 = 0
p 16 − p 3
We find the roots of this 2nd degree equation and calculate q
Answer starting from p . There are two solutions:
( (
The image should be real ( q positive), because it must ap- 4 cm 12 cm
p= → q=
pear on the screen. Therefore, the first relation between the 12 cm 4 cm
Example 22.10 q
M=−
p
An object placed in front of a converging lens with a focal length
The magnification should be taken as negative, because the
of 24 cm has an image that is inverted and 3 times magnified.
image is inverted:
Find the position of the object. q
Answer −3 = − → q = 3p
p
We establish and solve two equations for the two unknowns We substitute this expression for q in the lens equation and
(p, q) in this problem. The first one of these is the lens equa- solve for p :
tion: 1 1 1 4 × 24
1 1 1 1 + = → p= = 32 cm
+ = = p 3p 24 3
p q f 24
The other equation is obtained from the magnification: And the position of the image is found as q = 3p = 96 cm .
22.5 OPTICAL INSTRUMENTS

Eye
The human eye (Figure 22.27), which is the most perfect optical instrument,
can see objects and distinguish colors within a very broad range. The eye is
shaped like a sphere with a 2.5 cm radius, and the cornea is located at the front of
its external surface, called the sclera, and acts as a window. A liquid-filled anterior
chamber is located behind the cornea. Then, there is the transparent pupil located
at the center of a diaphragm with an open center, called the iris.
The pupil contracts and expands to adjust the amount of light entering the
eye. Behind it is located the eye lens with an index of refraction of 1.396. Ciliary
muscles around it hold the lens. Then, there is the liquid-filled vitreous chamber.
Figure 22.27: Human eye. The rear of the eye is the retina, which is covered with many nerve ends
(neurons). The fovea, where the nerve ends are densest, allows us to see clearly.
The optic disk region, located at the back, where the nerve ends are gathered, is a
blind spot, and rays incident here are not sent to the brain. The optic nerve, to
which the nerve ends are connected, transmits the signals that it receives to the
brain.
The eye allows us to see by forming the image of an object on the retina. For
this purpose, the muscles that hold the lens vary the focal length of the lens by
relaxing and contracting, and ensure that the rays intersect on the retina. The
state in which the muscles are relaxed is when the eye is focused on infinity. The
muscles contract more as the object approaches the eye and it ultimately cannot
22.5. OPTICAL INSTRUMENTS 385
see clearly after a certain point, called the near point. The near point is around 25
cm in adults. The muscles lose their flexibility with age and the near point may
recede up to 2-3 meters.
Figure 22.28: In a nearsighted

eye, the image forms before the
retina and is corrected with a
diverging lens. In a farsighted
eye, the image forms behind the
retina and is corrected with a con-
verging lens.
A net image is not formed on the retina in defective eyes (Figure 22.28).
The eye has nearsightedness (myopia), if the image forms before the retina
and farsightedness (hyperopia) if it forms behind the retina. The defect is
corrected by using a diverging lens for nearsightedness and a converging lens for
farsightedness.
Astigmatism, another eye defect, is caused by the difference in curvature
in horizontal and vertical planes. An astigmatic eye does not see horizontal and Figure 22.29: You have astig-
vertical lines with the same clarity (Figure 22.29). matism if you do not see these
lines with equal thickness when
Microscope
you look with one eye.
We bring very small objects close to our eyes in order to see them. However,
we cannot bring them closer than the near point of the eye, because the eye
cannot focus on the object after the near point. In such a case, we use a simple
magnifying glass or a more complex microscope.
Microscopes consist of at least two converging lenses. The one called the
objective lens is located near the object and has a focal length of about 2−40 mm .
The one on the side of the eye is called the eyepiece lens and has a larger focal
length. The operating principle of the microscope is roughly as follows: If we
can obtain a larger and real image of the small object near the focal point of the
eyepiece, we can produce the final image as much larger and at the near point of
the eye.
Figure 22.30: Microscope. The

object is placed just beyond
the focal point of the objective
lens. The intermediate image is
formed near the focal point of the
eyepiece. The final image is at
the near point.
In the microscope diagram in Figure 22.30, the object is placed just in front
of the focal point of the objective lens. A larger and real intermediate image is
thus formed. The diverging rays from this image are considered as a new object
for the eyepiece. Then, the eyepiece’s distance is varied to ensure that this real
image is closer to the focal point of the eyepiece.
Recall the formula for magnification: If we use h to indicate the height and p
to indicate the position of the object, and h0 for the height and q for the position
of the image, we get
h0 q
M= =−
h p
First, let us calculate Mo , the magnification at the objective lens: We can take
p ≈ fo , because the object is placed near the focal point of the objective lens. If
we ignore the negative size, the magnification of the objective is as follows:
q1
Mo =
fo
We take the following into consideration when calculating the magnification for
the eyepiece: The angular magnification is what is important for the eye. Even if
the size of the object is small, the angle at which the eye sees it becomes large.
For example, a needle close to the near point of the eye is perceived as larger
than a distant pole. For this reason, it is required that the final image given by the
eyepiece be close to the near point of the eye, in other words, at approximately
25 cm in distance. Accordingly, the magnification of the eyepiece is (for p2 = fe
and q2 = 25 cm ),
25 cm
Me =
fe
Combining these results, the total magnification of the microscope is the product
of these two magnifications:
(25 cm) q1
M = Mo Me = (22.19)
fo fe
Microscopes operating with visible light have limited magnification. Microscopes
Figure 22.31: A microscope using electron beams are much more powerful.
made in 1879. Telescope
Telescopes are used to see celestial bodies and other distant objects much
larger and in detail. There are two types of telescopes: Refracting telescopes,
which only use lenses and reflecting telescopes, which use a combination of lenses
and mirrors.
Figure 22.32: A telescope forms

a larger image of an object lo-
cated again at infinity.
The operating principle of the telescope is to form an image of an object

located at infinity that is again located at infinity. The real image given by the
objective lens is taken by the eyepiece and a final image is formed at infinity
which is magnified further. In the diagram shown in Figure 22.32, incident rays
from infinity form an inverted and real image at the focal point of the objective.
If the focal point of the eyepiece is located right at this point, it will form a virtual
image at infinity with the rays that it receives from this intermediate image.
The angular magnification of the telescope is defined as the angular ratio of
the angle θ covered by the image to the angle θ0 of the object observed with a
naked eye:
θ
m=
θ0
As seen in the figure, the angle θ0 covered by the object in the objective lens is
equal to the angle covered by the intermediate image in the objective lens. The
tangent of this angle is taken approximately as tan θ ≈ θ for small angles. If we
use fo to indicate the focal length of the objective lens, we get
h0
θ0 ≈ tan θ0 =
fo
Likewise, the angle θ covered by the final image in our eye is equal to the angle
covered in the eyepiece by the ray drawn as parallel from the intermediate image
to final image. Its tangent is also taken as an approximation for small angles. If
we use fe to show the focal length of the eyepiece, we get
h0
θ ≈ tan θ =
fe
We use these two formulas to find the angular magnification of the telescope as
follows:
θ fo
m= = (22.20)
θ0 fe
It is necessary to use very large lenses, as the magnification of the telescope is Figure 22.33: The Mount Gra-
proportional to the focal length of the objective lens. However, reflecting tele- ham telescope in Arizona is the
scopes that utilize mirrors are preferred, because large lenses are both expensive largest reflecting telescope in
to produce and difficult to install. Likewise, radio wave, X-ray or gamma ray the world with a diameter of
telescopes are used in astronomy, rather than visible light. 2 × 8.4 meter.
1. Which of the following is correct for electromagnetic (a) Microwave-X-ray-infrared-ultraviolet
waves? (b) Infrared-ultraviolet-X-ray-microwave
(a) The electric field is perpendicular and the magnetic (c) Microwave-infrared-ultraviolet-X-ray
field is parallel to the direction of propagation. (d) Infrared-visible light-gamma rays-microwave
(b) The magnetic field is perpendicular and the electric
field is parallel to the direction of propagation.
(c) The electric and magnetic fields are perpendicular 4. Which of the following are correct?
to the direction of propagation. I. Wavelength decreases as the speed of light decreases.
(d) The electric and magnetic fields are parallel to the II. The speed of light decreases as the index of refraction
direction of propagation. of the medium increases.
III. Wavelength decreases as the index of refraction of
2. What is the speed of light in a medium with index of the medium increases.
refraction n=1.5 ?
(a) I & II (b) I & III (c) II & III (d) All
(a) c/2 (b) 2c/3 (c) 3c/2 (d) 3c/4
3. Which of the following orderings of electromagnetic 5. Which of the following is the Huygens-Fresnel princi-
waves is done according to increasing frequency? ple?
(a) Every point of a medium on which light is incident 12. What is the focal length of a concave mirror with a 40 cm
reflects the wave. radius of curvature?
(b) Every point of a medium on which light is incident (a) 10 cm (b) 20 cm (c) 40 cm (d) 80 cm
refracts the wave.
(c) Every point of a medium on which light is incident 13. What is the focal length of a convex mirror with a 20 cm
constitutes a new source of waves. radius of curvature?
(d) Two waves, one passing and one reflected, are (a) -20 cm (b) -10 cm (c) 10 cm (d) 40 cm
formed at every point of a medium on which light
is incident. 14. When will the image of a real object on a converging
lens be virtual?
6. Which of the following is Snell’s law? (a) Object is between the center and the focal point,
(a) n1 cos θ1 = n2 cos θ2 (b) Object is between the focal point and the lens,
(b) n1 sin θ1 = n2 sin θ2 (c) Object is between infinity and the focal point.
(c) n1 sin θ2 = n2 sin θ1 (d) Object is between infinity and the center.
(d) n1 / sin θ1 = n2 / sin θ2
15. When will the image of a real object on a diverging lens
7. Which of the following is correct for a light ray sent be real?
from the water medium into the air? (a) Object is between the center and the focal point,
(a) It never passes into the air. (b) Object is between infinity and the center,
(b) It always passes into the air. (c) Object is between the focal point and the lens,
(c) It passes at angles larger than a critical angle. (d) Never.
(d) It passes at angles smaller than a critical angle.
8. Which of the following is correct for a ray incident on I. The image is always virtual on a convex mirror.
the interface between water ( n=1.3 ) and glass ( n=1.5 )? II. The image is always upright on a convex mirror.
(a) The rays always pass from glass to water. III. The image is always virtual on a concave mirror.
(b) The rays always pass from water to glass. IV. The image is always inverted on a concave mirror.
(c) The rays always pass in both ways. (a) I & II (b) I & III (c) II & III (d) II & IV
17. Which of the following is correct for the magnification
9. In which of the following cases will the image of a real M?
object on a mirror be real? (a) The image is inverted and real if M > 0.
I. If it is observed on a screen, (b) The image is upright and real if M > 0.
II. If the reflected rays intersect at a point, (c) The image is inverted and virtual if M > 0.
III. If the image is inverted. (d) The image is upright and virtual if M > 0.
(a) I & II (b) I & III (c) II & III (d) All 18. Which of the following is correct for the magnification
10. When will the image of a real object on a concave mirror M?
(a) The image is inverted and real if M < 0.
be virtual?
(a) When the object is between the center and the focal (b) The image is upright and real if M < 0.
point, (c) The image is inverted and virtual if M < 0.
(b) When the object is between the focal point and the (d) The image is upright and virtual if M < 0.
mirror, 19. Which of the following is correct?
(c) When the object is between infinity and the focal (a) The image is formed before the retina in a near-
point. sighted eye.
(d) When the object is between infinity and the center. (b) The image is formed behind the retina in a near-
11. When will the image of a real object on a convex mirror sighted eye.
be real? (c) The image is formed before the retina in a farsighted
(a) When the object is between the center and the focal eye.
point, (d) The image is formed behind the retina in an astig-
(b) When the object is between the focal point and the matic eye.
mirror, 20. By what factor will the magnification of a microscope
(c) When the object is between infinity and the focal increase if its objective lens magnifies by 5 times and its
point, eyepiece magnifies by 20 times?
(d) Never.
(a) 25 (b) 30 (c) 50 (d) 100
PROBLEMS 389
Problems
22.2 Reflection and refraction
point B on the lateral surface? (Hint: What is the relation be-
tween the critical angle at point B and the angle of incidence
22.1 A light with a 600 nm wavelength in vacuum enters a
at point A ?) [A: 78◦ .]
glass medium with an index of refraction 1.5 . (a) What is the
frequency of this light in vacuum? (b) What is its speed in a 22.8 A fish at the bottom of a 4 m deep lake looks upward
glass medium? (c) What is its wavelength in a glass medium? toward the surface of the water. What will be the radius of
[A: (a) 5 × 1014 Hz , (b) 2 × 108 m/s , (d) 400 nm .] the surface area from which this fish can see incident rays?
(The index of refraction of water is 1.33 .) [A: 4.6 m .]
22.2 What is the wavelength in glass of a light with a 500 nm
wavelength in water? ( nwater = 1.33 , nglass = 1.5 .) 22.3 Image by Reflection - Mirrors
[A: 440 nm .]
22.9 The focal length of a concave mirror is f =30 cm .
22.3 A light ray has a angle of refraction of 37◦ when sent (a) Where will the image of an object 1 m in size placed 40 cm
from air towards glass at a 60◦ angle of incidence. Determine away from the mirror form and what will its size and type
the index of refraction of the glass. [A: 1.45 .] be? (b) Answer the same question for an object 1 m in size
placed 15 cm away.
[A: (a) q = 120 cm , real, inverted and 3 times larger. (b)
q = −30 cm , virtual, upright and 2 times larger.]
Problem 22.4 22.10 The focal length of a convex mirror is 10 cm . Where

will the image of an object 1 m in size placed 40 cm away
22.4 A prism with an apex angle of 90◦ is made of a trans-
from the mirror form and what will its size and type be?
parent plastic material with n=1.25 . A ray enters perpen-
[A: q = −8 cm , virtual, upright and 5 times smaller.]
dicularly from one face of the prism, as shown in the figure
above. What is the minimum value of the angle α such that 22.11 When an object is placed 25 cm in front of a concave
the ray cannot exit from the hypotenuse? [A: 37◦ .] mirror, its image is inverted and 4 times enlarged. What is
the focal length of the mirror? [A: f = 20 cm .]
22.5 Water and glass layers with parallel surfaces are super-
imposed in the figure below. What will the angle of refraction 22.12 When an image is placed before a convex mirror with a
at the glass layer be for a ray that enters the water from the focal length of 40 cm , its image is upright and 4 times smaller.
air with an angle of incidence of 37◦ ? [A: 24◦ .] What is the position of the object? [A: p = 120 cm .]
22.13 There is 90 cm in distance between an object and a

screen. At what distance from the object should a concave
mirror with a 24 cm focal length be placed such that a clear
image is obtained on the screen? [A: p = 30 cm .]
22.14 The image of an object placed before a convex mirror is

Problem 22.5 and Problem 22.6 15 cm behind the mirror, upright and 4 times smaller. What
22.6 At what maximum angle can rays sent upward from is the focal length of the mirror? [A: f = −20 cm .]
the glass layer in the figure above pass on to the water layer?
22.4 Image by Refraction - Lenses
(Hint: At the limit, the ray emerges into the air parallel to the
surface.) [A: 42 .] 22.15 The focal length of a thin converging lens is f = 30 cm .
◦
(a) Where will the image of an object 1 m in size placed 90 cm

away from the lens form and what will its size and type be?
(b) Answer the same question for an object placed 10 cm
away.
[A: (a) q = 45 cm , real, inverted and triply diminished, (b)
q = −15 cm , virtual, upright and magnified 1.5 times.]
Problem 22.7 22.16 The focal length of a diverging lens is 20 cm . Where

22.7 What should the angle of incidence θ1 be such that the will the image of an object 1 m in size placed 60 cm away
light entering from the top face of the glass block shown in from the lens form and what will its size and type be?
the figure with index of refraction n = 1.4 does not exit from [A: q = −15 cm , virtual, upright and 4 times diminished.]
22.17 An object is placed 125 cm away from a screen. We 22.19 The image of an object placed before a converging
want to obtain the image of the object on the screen using a lens is 200 cm away, inverted and 4 times larger. What is the
converging lens with a focal length of 20 cm . How far away focal length of the lens? [A: 40 cm .]
from the object should the lens be placed?
[A: p = 25 cm and 100 cm .]
22.20 An object is placed 45 cm away from a screen. The
22.18 An object placed in front of a converging lens with a image obtained on the screen by a converging lens placed in
focal length of 30 cm has an image that is inverted and twice between is inverted and twice larger. What is the focal length
as large. Find the position of the object. [A: p = 45 cm .] of the lens? [A: f = 10 cm .]
?
23
WAVE OPTICS
The beautiful colors of this pea-

cock are actually not the colors
of its feathers. They result from
an effect called interference of
the light that results from the
structure of the feathers.
In geometric optics, light was considered as a ray propagating along a line.

This approach can be sufficient to obtain images with mirrors and lenses, but it
fails to explain many other properties of light. We can see all of the colors of the
rainbow on a soap bubble, on the surface of a compact disk or on the feathers of
a peacock. This is caused by the interference effects of the light wave.
Therefore, we need to dwell further on the wave structure of light. The two
most notable properties that distinguish a wave from a particle are its ability to
display interference and diffraction effects. These effects constitute the basis of
many techniques used in medicine and industry: X-rays, thin films, holography,
etc. In this chapter, we will discuss the interference and diffraction effects that
result from the wave properties of light.

392 23. WAVE OPTICS
23.1 YOUNG’S DOUBLE-SLIT EXPERIMENT

In chapter 11, we discussed the fact that waves display interference effects.
When two waves superpose at a point at the same phase, they undergo constructive
interference and their amplitudes are added. If they superpose at opposite phases,
they undergo destructive interference, in other words, their amplitudes cancel each
other out.
Light, which is an electromagnetic wave, should also display interference
effects. However, it is very difficult to observe the interference effect, because the
wavelength of light is very small ( ∼ 10−7 m ) and very precise experiments must
be conducted.
In 1801, the English scientist Thomas Young (1773-1829) was able to demon-
strate the interference of light experimentally. This setup, called Young’s double-
slit experiment, needs two sources emitting identical waves at the same phase
(Figure 23.1). For this purpose, an obstacle with two slits is placed in front of a
Figure 23.1: Young’s double- monochromatic (i.e., single wavelength) light source S. According to Huygens’
slit experiment setup. principle, after the light reaches these slits, two waves with the same phase will
be emitted from sources S 1 and S 2 .
When these two spherical waves arrive at a point on the screen, they will
produce constructive interference if they arrive with the same phase and cause
maximum illumination at that point. If they reach the screen at opposite phases,
they will produce destructive interference and that point will remain dark. There-
fore, dark and light interference fringes are observed on the screen.
Figure 23.2: Young’s experi-

ment. (a) The wave from a single
source generates two waves with
the same phase after the slits S 1
and S 2 . (b) Coordinates.
Now let us calculate the positions of the interference fringes in Young’s

experiment. In the diagram in Figure 23.2, the distance between the slits is
indicated with d and the distance of the screen is indicated with D . Let the
distances traveled by two waves to a point P on the screen at distance y from
the center O of the screen be r1 and r2 .
Let us draw a perpendicular line from the source S 1 to the path r2 and form
the right triangle S 1 S 2 B. If the screen is very far, in other words, if the paths
r1 and r2 are very large with respect to d , the side S 2 B will be the difference
between the two paths. Considering the right triangle S 1 S 2 B, we get
S 2 B = d sin θ
If this path difference is equal to integer multiples of the wavelength, in other
words, equal to any one of the values λ, 2λ, 3λ . . . , there is constructive interfer-
ence and a bright fringe appears at point P . Therefore, the angle θm of the mth
bright fringe will satisfy the following relation:
d sin θm = mλ (m = 0, ±1, ±2, ±3 . . .) (Bright fringes) (23.1)
23.1. YOUNG’S DOUBLE-SLIT EXPERIMENT 393
However, if the path difference is equal to odd multiples of a half wavelength, in

other words, equal to any one of the values λ/2, 3λ/2, 5λ/2 . . . , there is destructive
interference and P is in a dark fringe:

d sin θm = m + 12 λ (m = 0, ±1, ±2 . . .) (Dark fringes) (23.2)
In order to simplify this formula, let us consider the right triangle AOP formed
by drawing a line from the midpoint A of the slits in the figure to P . As the two
sides of this triangle are perpendicular to two sides of the triangle S 1 S 2 B, they
are similar triangles and their angle θ is equal. For small angles, both the sine
and the tangent can be taken as equal to the radian value of the angle:
ym
θm ≈ sin θm ≈ tan θm = (23.3)
D
This angle is used in the above expressions for dark and light fringes. We can
combine both cases into a single formula:
 λD
 m d (Bright)



ym =  (m = 0, ±1, ±2 . . .) (23.4)

λD
 m + 12 (Dark)



d
Therefore, if the distance D of the screen, the distance d between the slits and
the wavelength λ of the used light are known, the type and number of fringes
at distance ym can be calculated. This experiment is actually mostly used to
measure the wavelength of light.
Example 23.1
Accordingly, we find the distance between two successive
In a Young’s experiment conducted with an Argon laser using fringes:
green light with a wavelength of 514 nm , the distance between λD
y2 − y1 =
the slits is 0.1 mm and the screen is placed 3 m away. d
(a) What is the distance between two bright fringes? 514 × 10−9 × 3
(b) What is the distance between the 3rd bright fringe and the y2 − y1 = = 0.015 m = 1.5 cm
0.1 × 10−3
7th dark fringe?
(b) We write Eq. (23.4) for the 7th dark and 3rd bright fringes
Answer
and calculate the difference:
(a) The positions of the bright fringes are given with Eq. (23.4): λD
λD y7 − y3 = (7.5 − 3) × = 4.5 × 1.5 = 6.8 cm
ym = m d
d
Example 23.2
We use the value sin θm ≈θm for small angles and calculate the
In a Young’s experiment conducted with a ruby laser using red angular difference ∆θ between two successive maximums:
λ
light with a wavelength of 695 nm , the distance between the ∆θ = θ2 − θ1 =
slits is 0.1 mm . d
695 × 10−9
(a) What is the angular difference between two bright fringes? ∆θ = = 0.007 radians
(b) How many bright fringes are there in an angular area of 0.1 × 10−3
We convert this value into degrees by multiplying by 180/π :
±10◦ from the center? ∆θ = 0.007 × 180/π = 0.4◦
Answer (b) After finding the angle covered between two maximums,
(a) We use Eq. (23.1), which we developed for angular posi- we calculate the number of maximums within the given range:
tions: 10
N =2× = 50
d sin θm = mλ 0.4
394 23. WAVE OPTICS
Example 23.3 bright fringes:

λD
In a Young’s experiment, two sources generating two different ym = m
lights with wavelengths λ1 = 440 nm and λ2 = 695 nm are We write this expressiond
for both wavelengths at the same
used. The distance between the slits is 0.1 mm and the distance maximum m=6 and calculate the difference in distance:
from the screen is 5 m . What is the distance between the 6th D
∆y6 = 6 (λ2 − λ1 )
bright fringes of both types of light? d
5
Answer ∆y6 = 6 (695 − 440) × 10−9 ×
0.1 × 10−3
We use the formula (23.4) that gives us the positions of the ∆y6 = 0.077 m = 7.7 cm
23.2 INTERFERENCE IN THIN FILMS

You can see all of the colors of the rainbow on soap bubbles and in oil slicks
on puddles of rainwater. This effect is caused by the interference of the light
reflected from the front and rear surfaces of a thin film layer.
Let us consider a thin film layer with thickness t . As shown in Figure 23.3,
some of the light incident on this surface will get reflected from the front surface
and the rest will get refracted and pass on to the second medium. When the ray
inside of the film reaches the back surface of the film, some of it will get reflected
again and return to the front surface. This part of the returning ray that shoots
out into the air generates interference with the first reflected ray.
Let us calculate this interference for a normal, in other words, perpendicular,
incident light on a thin film with refractive index n .
Figure 23.3: Two lights re- Let us emphasize a very important point before starting the calculation: We
flected from a thin film layer. had discussed the following property when examining the reflection and transfer
of waves in Chapter 10: When a wave is incident from a less dense medium to a
denser medium ( n1 <n2 ), the wave is reflected with a phase difference of 180◦ , in
other words, as an inverted wave. This phase difference is observed only in the
reflection of a wave going from a less dense medium into a denser medium, and
not in the reverse case.
According to this property, the first ray coming from the air and reflected
from the first surface will get reflected with a phase difference of 180◦ . However,
two different cases can occur when the ray that passes inside of the film reaches
the back surface:
• If the film layer is denser than the rear medium (Figure 23.4):
In this case, no phase difference occurs in the second ray reflected from this
surface and it gets reflected as it is. Therefore, the phase difference between
the first and second rays is caused by two factors:
I The 180◦ phase difference from the first reflection,
I The distance 2t taken when going back and forth in the film medium.
Figure 23.4: The phase differ- The interference would have been destructive if there was only the 180◦
ence of the second light will phase difference from the reflection. However, the path difference from going
vary depending on whether the back and forth in the film medium shall be added to this. If the wavelength
film layer is more or less dens of the light in air is λ , its wavelength in a medium with refractive index n is
than the rear medium. λn = λ/n . Therefore, we should compare the path difference inside of the
film with this wavelength.
We reach the following conclusions considering that the phase difference is
caused by two effects:
23.2. INTERFERENCE IN THIN FILMS 395
I The interference is constructive if the path difference 2t is equal to

half multiples of λn and the light at that wavelength is observed at
maximum magnitude:
2t = (m + 12 ) λn (m = 0, 1, 2 . . .) (Maximum) (23.5)
I The interference is destructive if the path difference 2t is equal to full

multiples of λn and the light disappears at that wavelength and is not
observed:
2t = m λn (m = 1, 2, 3 . . .) (Minimum) (23.6)
Note that these formulas are opposite to those in Young’s experiment. The
reason for this is that a phase difference of 180◦ from reflection is added
here to the phase difference with the second wave.
• If the film layer is less dense than the rear medium (Figure 23.4):
In this case, a 180◦ phase difference will arise in the rays reflected from both
the front and the rear surfaces. Therefore, the phase differences caused by
the surface effects will cancel each outer out and only the phase difference
from the path difference in the film medium will be taken into consideration.
In this case, we reverse Eqs. (23.5) and (23.6) that we found above for con-
structive and destructive interference. In other words, we use the maximum
(constructive interference) formula for the minimum (destructive interference).
This is the reason why various colors are observed on soap bubbles and on
oil slicks. The fact that, for example, the color blue is reflected from a point of a Figure 23.5: The colors re-
thin film reflecting white light, is caused by the fact that all of the other colors flected from a thin oil slick are
are canceled due to destructive interference. a result of interference.
Example 23.4
without reflection means destructive interference. Therefore,
An oil slick with a refractive index of 1.5 is spread over a water the formula (23.6) should be used:
puddle (n = 1.33) . What is the minimum value of the oil 2t = m λn
slick thickness such that the yellow light with a wavelength of If the path difference is equal to one wavelength for m=1 , it
600 nm within the white light coming from air passes through shall be multiples of the wavelength for the other m values.
without getting reflected? Also, in the oil medium, we take the wavelength as λm = λ/n :
λ λ 600
Answer Only a phase difference of 180◦ will occur in the 2t = 1 × → t= = = 200 nm
n 2n 2 × 1.5
first surface, because the refractive index of the oil slick is Yellow light will not get reflected at multiples of this thick-
greater than the water underneath. Also, passing through ness.
Example 23.5
the refractive index of the film layer is less than the refractive
A thin transparent film with refractive index n=1.3 is spread index of the rear glass.
on a window glass with a refractive index of 1.5 . The thickness In this case, we must invert Eqs. (23.5-23.6) for destructive
of the film should be multiples of which value such that the interference of the red light:
reflected light does not contain red)? (Take the wavelength of 2t = (m + 12 ) λn (Minimum)
red light as 700 nm .) We take m = 0 and λn = λ/n and calculate the thickness:
λ λ 700
Answer Two 180◦ phase differences will occur this time, as 2t = 12 × → t= = = 135 nm
n 4n 4 × 1.3
396 23. WAVE OPTICS
23.3 DIFFRACTION FROM A SINGLE SLIT

In geometric optics, light was considered as rays propagating in straight lines.
If this was correct, an obstacle placed in front of a source of light would have a
shadow with sharp edges. But this is not the case. If you closely examine the
shadow of a razor blade in Figure 23.6, you will notice that there are second, third,
etc., shadows on the sides. These are called diffraction fringes.
Diffraction is the general name given to the effects caused by the deviation
of light from its linear path. In addition to the double-slit experiment that we
discussed earlier, the wave nature of light can be observed even in the shadow of
Figure 23.6: Diffraction in the a single slit.
shadow of a razor blade (Lecture Let us consider the single-slit experiment so as to understand diffraction at
Demonstration Services, Har- the simplest level. In the setup shown in Figure 23.7, a monochromatic light
vard Science Center). emitted from a source passes through an obstacle with only a single slit and is
then projected onto a screen. On the screen, in addition to an image of the slit at
the center, we also observe weaker second, third, etc., slit images on both sides of
the center.
To understand the diffraction effect in this experiment, the width of the slit
should be taken into consideration. (The widths of the slits were ignored in
Young’s experiment.) Let us use Figure 23.8 to explain the diffraction fringes
formed by rays from a single slit with width a on a very distant screen.
Figure 23.7: Diffraction in a sin- According to Huygens’s principle, secondary waves with the same phase are
gle slit. generated at each point of a slit on which light is incident. It is clear that the
center of the screen will be bright, as all of these waves will reach the center in
the same phase.
Let us consider the first dark fringe right next to the center. As all light rays
reaching this dark fringe from various parts of the slit are very distant from the
screen, they will all have started with the same angle θ . Let us match every point
at the top half of the slit with a point at the bottom half.
As shown in Figure 23.8, the wave from the top point and the wave from
just below the central point (the two black circles in the figure) should have
a phase difference of λ/2 , because they generate destructive interference on
the dark fringe on the screen. After these two points, the two points following
them in the top and bottom halves of the slit should also have a phase difference
of λ/2 . Therefore, all matching pairs of points will have a phase difference of
Figure 23.8: Two matching half a wavelength. As shown in Figure 23.8, as the path difference between two
(black) points in the top and bot- matching waves is (a/2) sin θ , the condition for the first dark fringe is written as
tom half. follows:
a λ λ
sin θ = −→ sin θ =
2 2 a
This method can be repeated by considering that the slit is divided into 4 parts,
6 parts, 8 parts, etc. It can be shown that the dark fringes will have the value
sin θ=2λ/a, 3λ/a , etc., each time. In conclusion, the position of dark fringes in
single-slit diffraction will fulfill the following condition:
Figure 23.9: Diffraction fringes λ
in a single slit. sin θ = m (m = 1, 2, 3 . . .) (Condition for dark fringes) (23.7)
a
Likewise, we use the small angle approach to find the distances ym of the fringes
from the center on a screen located at distance D :
y
sin θ ≈ tan θ =
D
23.3. DIFFRACTION FROM A SINGLE SLIT 397
λD
ym = m (m = 1, 2, 3 . . .) (Positions of dark fringes) (23.8)
a
We can determine the positions of the bright fringes using a similar method:
λD
ym = m + 1
2 (m = 1, 2, 3 . . .) (Positions of bright fringes) (23.9)
a
Intensity of Light in Diffraction
The intensity of light varies gradually between dark and light areas in diffrac-
tion fringes. The calculation for the distribution of the resultant intensity of two
waves at a certain point is long and complex, and we shall therefore present it
here without proof. If we use I0 to indicate the intensity of light at the central
point θ = 0 , the expression for the light intensity at any angle is as follows:
#2
sin(πa sin θ/λ)
"
I = I0 (23.10)
πa sin θ/λ
Figure 23.10: Distribution of
Notice that this function also includes the condition for dark fringes. This curve the intensity of light in diffrac-
is shown in Figure 23.10. tion fringes.
Example 23.6
ym = m λD/a (m = 1, 2, 3 . . .)
A red light with a wavelength of 700 nm is sent through a We calculate for m = 1 :
λD 700 × 10−9 × 5
0.1 mm wide slit. A screen is located 5 m away. y1 = = = 0.035 m = 35 mm .
a 0.1 × 10−3
(a) What is the position of the first dark fringe?
(b) The bright fringe at the center is limited by the first dark
(b) What is the width of the bright area at the center?
fringes at both sides. Therefore, its width is 2 times the value
Answer that we found in item (a):
(a) We use Eq. (23.8), which we found for dark fringes: ∆y = 2 × y1 = 70 mm .
Example 23.7 λ1 D
y1 (m = 5) = 5
a
A red light with wavelength λ1 = 700 nm and a green light Eq. (23.9) for bright fringes gives us, for m=3 of the green
with wavelength λ2 = 650 nm are simultaneously sent through light:
a slit. On a screen that is 2 m away, the distance between the λ2 D
y2 (m = 3) = 3 + 12
5th dark fringe of the red light and the 3rd bright fringe of the a
green light is observed to be 2 mm . Determine the width of the We equate the difference between these two positions to
slit. 2 mm and solve for the unknown a :
y1 (m = 5) − y2 (m =
3) = 0.002
Answer 5λ1 − 3.5λ2 D
Eq. (23.8) for dark fringes gives us, for m=5 of the red light: a= = 0.0012 m = 1.2 mm
0.002
Resolution
The image formed formed in our eye or on photographic film by optical
instruments such as microscopes, telescopes or spectroscopes must be very sharp.
However, there is a limit to the ability of an instrument to distinguish between
objects that are close together, regardless of how perfectly it is manufactured.
This ability, limited by the diffraction of light, is called the resolution or the
resolving power of the instrument.
The limit of the resolution is caused by diffraction. Let us consider two light
sources S 1 and S 2 located near each other (e.g., two neighboring stars in the
sky). Let the rays from these two sources pass through a slit with width a and be
398 23. WAVE OPTICS
Figure 23.11: (a) Two resolvable
images. (b) Two images on the
limit of resolution, (c) Two unre-
solvable images. Note that the
central maximum of one coin-
cides with the first minimum of
the other.
projected onto a screen. As we discussed in the topic of single-slit diffraction, the
image of each on the screen will show a distribution around a central maximum
(Figure 23.11). The centers of these two images become unresolvable if their
central maximums overlap.
The limit of resolution is accepted as the limit at which the central maximum of
one of the images coincides with the first minimum of the other image (Figure 23.11b).
We obtain the minimum angle θ at this limit using Eq. (23.8), which we obtained
for single-slit diffraction. The expression giving the first minimum was found to
be the following:
λ
sin θ =
a
We can take the angle (in radian) instead of the sine for small angles. Accordingly,
the smallest angle θmin that can be resolved by the optical instrument is:
λ
θmin = (for rectangular slit) (23.11)
a
This formula is valid for a rectangular slit. The formula changes slightly when
calculated for the circular objective lenses of optical instruments:
λ
θmin = 1.22 (For circular slit) (23.12)
a
a is the diameter of the circle here.
The smaller the value of θm , the higher resolution of the instrument. Accord-
ingly, to increase resolution, we should either use light with a smaller wavelength
or increase the diameter of the objective lens.
Example 23.8
The angle θmin here should be equal to the angle covered by
A camera is required to resolve two points separated by 1 cm at the width of 1 cm located 100 meters away:
a distance of 100 m in the photo that it shoots. As the average θmin = 0.01/100 = 0.0001 radians
wavelength is 500 nm within the visible light range, what is We use these values to calculate the diameter a of the circular
the minimum value of the diameter of the objective lens? objective lens:
Answer λ 500 × 10−9
a = 1.22 = 1.22 ×
We use Eq. (23.12) given for circular slits: θmin 0.0001
θmin = 1.22λ/a a = 0, 006 m = 6 mm
Problems
23.1 Young’s Double-Slit Experiment
(a) What is the distance between two bright fringes? (b) What
23.1 In a Young’s experiment conducted with a ruby laser is the distance between the 3rd bright fringe and the 5th dark
using red light with a wavelength of 700 nm , the distance be- fringe? [A: (a) 3.5 mm , (b) 8.8 mm .]
tween the slits is 0.1 mm and the screen is placed 5 m away.
PROBLEMS 399
23.2 In a Young’s experiment conducted with a He-Cd laser thickness of the oil slick such that red light with a wave-
using red light with a wavelength of 450 nm , the distance be- length of 700 nm within the white light coming from air is
tween the slits is 0.1 mm . (a) What is the angular difference not reflected? [A: 250 nm .]
between the fringes? (b) How many bright fringes are there
in an angular area of ±1◦ from the center? 23.7 A coat of thin film with refractive index n=1.3 is ap-
plied to an objective lens with a refractive index of 1.6 . The
[A: (a) 0.27◦ , (b) 7.8 ≈ 8 fringes.]
thickness of the film should be multiples of which value such
23.3 In a Young’s experiment, sources generating two lights that the reflected light does not contain blue)? (Take the
with wavelengths λ1 = 600 nm and λ2 = 650 nm are used. wavelength of blue light as 400 nm .) [A: 77 nm .]
The distance between the slits is 0.1 mm and the distance
from the screen is 3 m . What is the distance between the 5th
23.3 Diffraction from a Single Slit
bright fringes of both types of light? [A: 7.5 mm .] 23.8 A light with a wavelength of 600 nm is sent through a
0.1 mm wide slit. (a) On a screen 4 m away, what is the width
23.4 A Young’s experiment is designed with a light with a of the central area at the center? (b) What is the position of
700 nm wavelength such that the angle between the bright the 7th dark fringe? [A:(a) 4.8 cm , (b) 17 cm .]
fringes is 0.01◦ . What should the distance between the two
slits be? [A: 4 mm .] 23.9 Two lights with wavelengths λ1 =500 nm and
λ2 =600 nm are simultaneously sent through a slit. On a
23.5 The angle between the bright fringes is 2◦ in a Young’s screen that is 3 m away, the distance between the 7th bright
experiment. What will the angle be when this experimental fringe of the first light and the 5th dark fringe of the second
setup is placed under water with refractive index n = 1.33 ? light is observed to be 3 mm . Determine the width of the slit.
(Hint: Take the ratios and work with the wavelength in water.) [A: 2 mm .]
[A: 1.5◦ .]
23.10 A telescope is required to resolve two points at a dis-
tance of 100 km and separated by 1 cm . As the average
23.2 Interference in Thin Films
wavelength is 500 nm within the visible light range, what is
23.6 An oil slick with a refractive index of 1.4 is located the minimum value of the diameter of the telescope?
on a puddle of water (n = 1.33) . What is the minimum [A: 6.1 m .]
APPENDIX A. PHYSICAL CONSTANTS
Quantity Symbol Value
Speed of light in vacuum c 2.997 92 × 108 m/s
Charge of electron e 1.602 19 × 10−19 C
Gravitational constant G 6.6726 × 10−11 N m2 /kg2
Avogadro’s number NA 6.022 05 × 1023

 8.31451 J/(mol K)


Universal gas constant

R 

 0.0820578 litre atm/(mol K)

Permittivity of vacuum ε0 8.854 187 × 10−12 C2 /(N m2 )

Coulomb force constant k = 1/(4πε0 ) 8.987 551 78 × 109 N m2 /C2
Permeability of vacuum µ0 4π × 10−7 N/A2
Electron mass me 9.109 390 × 10−31 kg
Proton mass mp 1.672 623 × 10−27 kg
Neutron mass mn 1.674 929 × 10−27 kg
Planck constant h 6.626 18 × 10−34 J s
Bohr radius aB = ~2 /(ke2 me ) 5.291 773 × 10−11 m
APPENDIX B. USEFUL MATHEMATICAL RELATIONS

Derivative Integral Trigonometry
Z Z Z
d d f dg
( f + g) = + ( f + g) dx = f dx + g dx sin2 α + cos2 α = 1
dx dx dx
Z Z
d dg df
( f g) = f +g f g0 dx = f g − g f 0 dx sin(−α) = − sin α
dx dx dx
Z
dx
=1 dx = 1 cos(−α) = cos α
dx
xn+1
Z
d n
x = nxn−1 xn dx = sin 2α = 2 sin α cos α
dx n+1
Z
d 1 dx
ln x = = ln x cos 2α = cos2 α − sin2 α = 2 cos2 α − 1
dx x x
2 tan α
Z
d x
e = ex e x dx = e x tan 2α =
dx 1 − tan2 α
Z
d
sin x = cos x sin x dx = − cos x sin(α ± β) = sin α cos β ± cos α sin β
dx
Z
d
cos x = − sin x cos x dx = sin x cos(α ± β) = cos α cos β ∓ sin α sin β
dx
Z
d 1
tan x = tan x dx = − ln |cos x| Law of Cosines: a2 = b2 + c2 − 2bc cos A
dx cos2 x
Z
d 1 a b c
cot x = − 2 cot x dx = ln |sin x| Law of Sines: = =
dx sin x sin A sin B sin C

B. Karaoglu, Classical Physics, https://doi.org/10.1007/978-3-030-38456-2
402
C. TRIGONOMETRIC TABLE
For angles greater than 45°, use the right-hand column together with the bottom labels.
Degree Radian Sin Cos Tan Cot ⇐=

00 0.000 0.000 1.000 0.000 ∞ 1.571 90
01 0.018 0.018 1.000 0.018 57.290 1.553 89
02 0.035 0.035 0.999 0.035 28.636 1.536 88
03 0.052 0.052 0.999 0.052 19.081 1.518 87
04 0.070 0.070 0.998 0.070 14.301 1.501 86
05 0.087 0.087 0.996 0.088 11.430 1.484 85
06 0.105 0.105 0.995 0.105 9.514 1.466 84
07 0.122 0.122 0.993 0.123 8.144 1.449 83
08 0.140 0.139 0.990 0.141 7.115 1.431 82
09 0.157 0.156 0.988 0.158 6.314 1.414 81
10 0.175 0.174 0.985 0.176 5.671 1.395 80
11 0.192 0.191 0.982 0.194 5.145 1.379 79
12 0.209 0.208 0.978 0.213 4.705 1.361 78
13 0.227 0.225 0.974 0.231 4.332 1.344 77
14 0.244 0.242 0.970 0.249 4.011 1.327 76
15 0.262 0.259 0.966 0.268 3.732 1.309 75
16 0.279 0.276 0.961 0.287 3.487 1.292 74
17 0.297 0.292 0.956 0.306 3.271 1.274 73
18 0.314 0.309 0.951 0.325 3.078 1.257 72
19 0.332 0.326 0.946 0.344 2.904 1.239 71
20 0.349 0.342 0.940 0.364 2.748 1.222 70
21 0.367 0.358 0.934 0.384 2.605 1.204 69
22 0.384 0.375 0.927 0.404 2.475 1.187 68
23 0.401 0.391 0.921 0.425 2.356 1.169 67
24 0.419 0.407 0.914 0.445 2.246 1.152 66
25 0.436 0.423 0.906 0.466 2.145 1.135 65
26 0.454 0.438 0.899 0.488 2.050 1.117 64
27 0.471 0.454 0.891 0.510 1.963 1.100 63
28 0.489 0.470 0.883 0.532 1.881 1.082 62
29 0.506 0.485 0.875 0.554 1.804 1.065 61
30 0.524 0.500 0.866 0.577 1.732 1.047 60
31 0.541 0.515 0.857 0.601 1.664 1.030 59
32 0.559 0.530 0.848 0.625 1.600 1.012 58
33 0.576 0.545 0.839 0.649 1.540 0.995 57
34 0.593 0.559 0.829 0.675 1.483 0.977 56
35 0.611 0.574 0.819 0.700 1.428 0.960 55
36 0.628 0.588 0.809 0.727 1.376 0.943 54
37 0.646 0.602 0.799 0.754 1.327 0.925 53
38 0.663 0.616 0.788 0.781 1.280 0.908 52
39 0.681 0.629 0.777 0.810 1.235 0.890 51
40 0.698 0.643 0.766 0.839 1.192 0.873 50
41 0.716 0.656 0.755 0.869 1.150 0.855 49
42 0.733 0.669 0.743 0.900 1.111 0.838 48
43 0.751 0.682 0.731 0.933 1.072 0.820 47
44 0.768 0.695 0.719 0.966 1.036 0.803 46
45 0.785 0.707 0.707 1.000 1.000 0.785 45
Cos Sin Cot Tan Radian Degree
1 H D. PERIODIC TABLE 2 He
1.01 4.00
Lanthanides that start at 58 and Actinides that start at 90 are shown separately.
0.00009 0.00018
3 Li 4 Be 5 B 6 C 7 N 8 O 9 F 10 Ne
6.94 9.01 10.81 12.01 14.01 16.00 19.00 20.18
0.534 1.85 2.34 2.267 0.0013 0.0014 0.0017 0.0009
11 Na 12 Mg Atomic number and symbol - 13 Al 14 Si 15 P 16 S 17 Cl 18 Ar

22.99 24.31 Atomic mass (g) - 26.98 28.09 30.97 32.06 35.45 39.95
0.971 1.738 Density (g/cm3 ) - 2.698 2.330 1.82 2.067 0.0032 0.0018
19 K 20 Ca 21 Sc 22 Ti 23 V 24 Cr 25 Mn 26 Fe 27 Co 28 Ni 29 Cu 30 Zn 31 Ga 32 Ge 33 As 34 Se 35 Br 36 Kr
39.10 40.08 44.96 47.90 50.94 52.00 54.94 55.85 58.93 58.70 63.55 65.38 69.72 72.59 74.92 78.96 79.90 83.80
0.862 1.54 2.989 4.54 6.11 7.15 7.44 7.874 8.86 8.912 8.96 7.134 5.907 5.323 5.776 4.809 3.122 0.0037
37 Rb 38 Sr 39 Y 40 Zr 41 Nb 42 Mo 43 Tc 44 Ru 45 Rh 46 Pd 47 Ag 48 Cd 49 In 50 Sn 51 Sb 52 Te 52 I 54 Xe
85.47 87.62 88.91 91.22 92.91 95.94 98 101.07 102.91 106.04 107.87 112.41 114.82 118.69 121.75 127.60 126.90 131.30
1.532 2.64 4.469 6.506 8.57 10.22 11.5 12.37 12.41 12.02 10.501 8.69 7.31 7.287 6.685 6.232 4.93 0.0059
55 Cs 56 Ba 57 La 72 Hf 73 Ta 74 W 75 Re 76 Os 77 Ir 78 Pt 79 Au 80 Hg 81 Tl 82 Pb 83 Bi 84 Po 85 At 86 Rn
132.91 137.33 138.91 178.49 180.95 183.85 186.21 190.2 192.22 195.09 196.97 200.59 204.37 207.2 208.98 209 210 222
1.873 3.594 6.145 13.31 16.654 19.25 21.02 22.61 22.56 21.46 19.282 13.534 11.85 11.342 9.807 9.32 7 0.0097
87 Fr 88 Ra 89 Ac
58 Ce 59 Pr 60 Nd 61 Pm 62 Sm 63 Eu 64 Gd 65 Tb 66 Dy 67 Ho 68 Er 69 Tm 70 Yb 71 Lu
403
223 226.03 227.03
1.87 5.5 10.07 140.12 140.91 144.24 145 150.4 151.96 157.25 158.93 162.50 164.93 167.26 168.93 173.04 174.97
6.77 6.773 7.007 7.26 7.52 5.243 7.895 8.229 8.55 8.795 9.066 9.321 6.965 9.84
90 Th 91 Pa 92 U 93 Np 94 Pu 95 Am 96 Cm 97 Bk 98 Cf 99 Es 100 Fm 101 Md 102 No 103 Lr

232.04 231.04 238.03 237.05 244 243 247 247 251 252 257 258 259 260
11.72 15.37 18.95 20.45 19.84 13.69 13.51 14.79 15.1 13.5 – – – –
404
PICTURE CREDITS
Chapter and picture numbers are indicated below in boldface.
Chapter 1: Header picture: NASA, 1.1: US Navy Imagery and Bureau International des
Poids et des Mésures.
Chapter 2: Header picture: Potiyama, 2.5: Michael Maggs.
Chapter 3: Header picture: Geraint Otis Warlow.
Chapter 4: Header picture: European Space Agency (ESA), 4.1: jon oropeza.
Chapter 5: Header picture: Coasterman1234, 5.13: Olliver Mallich.
Chapter 6: Header picture: Dave Jackson, 6.4: Alsterdrache, 6.5: Brady Holt, 6.12:
Erik Forsberg.
Chapter 7: Header picture: NOAA.
Chapter 8: Header picture: Fer.filol.
Chapter 9: Header picture: Dan Pancamo, 9.12: Leonard G., 9.15: BrokenSphere.
Chapter 10: Header picture: Shalom Jacobovitz, 10.17: TaitaFkm.
Chapter 11: Header picture: Saffron Blaze, 11.1: Pia von Lützau and Stefan-XP, 11.7:
fleecetraveler, pGordon and Aranya Sen, 11.10: Jose Olgon, 11.11: Orhan Kamer, 11.13:
kurtsik, 11.16: Rob Bulmahn, 11.19: WikipedianProlific, Mvapdevila and neepster,
11.22: Joao Miranda, 11.23: Michael Belisle.
Chapter 12: Header picture: Andries Oudshoorn, 12.2: Andres Rueda and makerbot,
12.4: Markus Schweiss, 12.6: CrazyD, 12.8: IMPRESS Education, 12.10: NASA, 12.12:
Jens Buurgard Nielsen, 12.13: Linda Hermans Killam (IPAC).
Chapter 13: Header picture: Christoph Ehlen, 13.15: Timitrius.
Chapter 14: Header picture: Angela Yuriko Smith, 14.1: Didier Desouens, 14.4:
Lawrence Berkeley National Lab, 14.8: Peo, 14.9: ElBibliomata.
Chapter 15: Header picture: Beverly and Pack.
Chapter 16: Header picture: Jon ’ShakataGaNai’ Davis.
Chapter 17: Header picture: Courtesy-Sandia National Laboratories, 17.1: Eric Schrader,
17.2: Steve Jurvetson, Pilaf, and Arnold Reinhold, 17.11: CommScope, Jom and Hellisp,
17.12: Bert Hickman, 17.14: 2Tales, 17.15: Anynobody.
Chapter 18: Header picture: Intel Corp. 18.6: Yury Masloboev, 18.8: Ralf Roletschek,
18.12: timlewisnm, 18.19: André Karwath.
Chapter 19: Header picture: ITER, 19.10: Hatsukari715.
Chapter 20: Header picture: CERN, 20.9: Richard Steiner (NIST), 20.16: Gorchy.
Chapter 21: Header picture: NASA, 21.9: wdwd.
Chapter 22: Header picture: NASA, 22.2: Inductiveload, NASA 22.7: Brews ohare,
22.10: Velual, 22.12: BigRiz, 22.19: NASA, 22.21: Johan, 22.27: Holly Fischer, 22.33:
NASA.
Chapter 23: Header picture: Doug Hodel, 23.5: John, 23.6: Lecture Demonstration
Services-Harvard Science Center, 23.9: Gisling, 23.10: Magnus Manske.
Index
absolute error, 5 cylindrical, 285 contact angle, 188

acceleration, 27 energy, 290 convection, 209
centripetal, 47 in parallel, 287 converging lens, 381, 382
relative, 49 in series, 288 convex mirror, 378
tangential, 47 parallel-plate, 284, 285 coordinate system, 10
acceleration vector, 41 spherical, 286 coulomb (unit of charge), 241
adhesion force, 188 capillarity, 188 Coulomb’s constant k , 243
adiabatic process, 223 Carnot cycle, 233 Coulomb’s law, 242
ammeter, 313 center of mass, 104–105 critical angle, 372
Ampère’s law, 340 centripetal acceleration, 47, 69, 117 Critically damped, 154
ampere (unit of current), 302, 338 centripetal force, 69 Curie temperature, 345
amplitude, 145 charge density, 250 current, 301–304
angular acceleration, 115 circular motion, 45, 69 cycle, 229
angular frequency, 145 centripetal acceleration, 47 Carnot, 233
angular kinematics, 114–117 tangential acceleration, 47 Diesel, 232
angular momentum, 128 coefficient of linear expansion, 205 Otto, 229
angular position, 114 cohesion force, 188 cylindrical capacitor, 285
angular velocity, 115 collisions, 99–101
anomalous behavior of water, 206 dam wall, 183
elastic, 99 damped harmonic motion, 153
Archimedes’ buoyant force, 186 inelastic, 100
atom damping coefficient, 154
one dimension, 99 density, 180
closed shell, 242
totally inelastic, 101 diamagnetism, 344
free electron, 242
two dimensions, 102 dielectric, 291–293
magnetic dipole moment, 342
commutator, 326 dielectric constant ( κ ), 292
structure, 242
compression ratio, 230 dielectric strength, 293
average acceleration, 27
concave mirror, 376 Diesel cycle, 232
average acceleration vector, 41
concave mirror equation, 377 diffraction, 396–398
average velocity, 25
conduction, 208 fringes, 396
average velocity vector, 40
Avogadro’s Number, 210 conductor, 241 intensity of light, 397
conductors, 264 diffraction fringes, 396
Bernoulli’s equation, 192 equipotential surface, 277 dimension, 1
Biot-Savart law, 332 conservation of angular momentum, diopter, 382
129 direct current circuit, 307–309
calorie (cal), 202 conservation of charge, 240 energy, 309
capacitance, 284 conservative force, 85 power, 309
capacitor, 284–290 constructive interference, 167 dispersion, 373

B. Karaoglu, Classical Physics, https://doi.org/10.1007/978-3-030-38456-2
406 INDEX
dispersion of light, 373 remanence, 345 index of refraction, 371

displacement, 24 saturation, 345 induced surface charge, 292
displacement vector, 40 first condition of equilibrium, 136 inductance, 359
Doppler effect, 172–174 first law of thermodynamics, 220 inertia, 57
drift speed, 303 flow rate, 190 inertial reference frames, 58
driven harmonic motion, 155 force, 57 infrared rays, 369
resonance, 155 free electron, 242 instantaneous axis of rotation, 127
free fall, 32 insulator, 241
Earth’s magnetic field, 345 free-body diagram, 64 interference, 392–395
eddy currents, 357 frequency, 146 thin film, 394
efficiency, 229 friction force, 62–63 Young’s experiment, 392
elastic potential energy, 86 fundamental frequency, 170 interference fringes, 392
electric charge, 240–241 internal combustion engine, 229
electron, 241 galvanometer, 313 internal energy, 220
electric current, 301–304 gamma rays, 369 ideal gas, 222, 227
biological impacts, 303 gauge pressure, 183 internal resistance, 307
electric dipole, 248, 275, 291 gauss (unit of magnetic field), 321 International System of Units (SI), 2
electric dipole moment, 291 Gauss’s law, 258–260 kilogram, 3
electric dipole potential, 275 generator, 355 meter, 3
electric field, 244–250 gravitational acceleration, 31 second, 3
conductors, 264 gravitational potential energy, 86, 87
dipole, 248 joule (unit of work), 78
harmonic motion
field lines, 246
damped, 153 kilogram, 3
point charge, 245
driven, 155 kinetic energy, 83
electric flux, 258 physical pendulum, 152 rotational motion, 126
electric motor, 325 simple, 143–150
electric permittivity ε0 , 243, 292 simple pendulum, 151 laminar flow, 190
electric potential harmonics, 169 latent heat, 203
conductors, 277 heat, 202 fusion, 203
constant electric field, 272 heat engine, 229 vaporization, 203
dipole, 275 efficiency, 229 law of conservation of energy, 88
equipotential surface, 277 heat transfer, 207 law of conservation of momentum, 98
gradient, 278 conduction, 208 law of gravitation, 60
point charge, 273 convection, 209 laws of reflection and refraction, 371
electric potential energy, 270 radiation, 209 lens
electromagnetic wave, 162 heat transfer coefficient, 208 converging, 381
electromotive force (emf), 307 Helmholtz coils, 348 diverging, 382
electron charge, 241 Hooke’s law, 81 magnification, 382
electrostatic force, 242 horsepower (HP), 82 lens makers’ equation, 382
emf (electromotive force), 307 Huygens-Fresnel principle, 370 lens power D , 382
equation of continuity, 191 hydrostatic pressure, 181, 192 Lenz’s law, 353
equation of state, 211 hysteresis curve, 345 light
equation of state for ideal gas, 211 dispersion, 373
equipotential surface, 277 ideal gas, 210–211 electromagnetic wave, 368
escape velocity, 90 assumption, 211 laws of reflection and refraction,
eye, 384 equation of state, 211 371
astigmatism, 385 internal energy, 227 plane wave, 370
farsightedness, 385 monatomic, 227 ray model, 369
near point, 385 root-mean-square speed, 228 wave front, 369
nearsightedness, 385 specific heat, 222 wave nature, 368
image lightning, 294
farad (capacitance unit), 284 real, 375 longitudinal wave, 162
Faraday’s law, 352 virtual, 375
ferromagnetism, 344 Impulse, 96 magnetic dipole, 335
hysteresis curve, 345 impulse-momentum theorem, 96 magnetic dipole moment, 325
INDEX 407
magnetic domains, 345 Newton’s first law, 56 rays, 369

magnetic field Newton’s laws, 55–59 reflection and transmission of waves,
circular loop, 335 The first law, 56 170
Earth, 345 The second law, 57 relative acceleration, 49
infinite wire, 334 The third law, 58 relative error, 5
magnetic dipole, 335 Newton’s second law, 57 relative motion, 48–50
solenoid, 341 Newton’s third law, 58 relative velocity, 49
toroid, 341 nonconservative force, 85 addition rule, 49
~ , 343
magnetic field strength vector H normal reaction force, 61 resistance, 305
magnetic force, 320, 323 resistivity, 305
parallel currents, 338 Ohm’s law, 305 change with temperature, 306
magnetic permeability µ , 344 Otto cycle, 229 temperature coefficient, 306
magnetic permeability of free space resistor
parallel axis theorem, 124
µ0 , 332 in parallel, 311
parallel-plate capacitor, 284, 285
magnetic susceptibility χm , 344 in series, 310
paramagnetism, 344
magnetic torque, 324 resolution, 397
Pascal’s principle, 182
magnetization, 343 resonance, 155, 362
period, 47, 145
magnification, 382 natural frequency, 155
periodic wave, 162
manometer, 182 right-hand rule, 16
phase angle, 150
margin of error, 5 rigid body, 113
phase change, 203
mass spectrograph, 322 physical pendulum, 152 rigid body motion, 121
mechanical wave, 162 period formula, 153 RLC circuit, 361
meter, 3 Pitot tube, 193 rocket motion, 107, 108
microscope, 385 plane mirror, 375 rolling motion, 126
microwave, 369 plane wave, 370 root-mean-square speed, 228
mirror point charge, 245 rotational kinetic energy , 126
concave, 376 polar molecules, 291 round off rule, 6
convex, 378 polarization, 292
plane, 375 position, 24 scalar product, 14
molar mass M , 210 position vector, 40 second, 3
mole, 210 potential energy, 84–87 second condition of equilibrium, 136
moment (torque), 118 elastic, 86 second law of thermodynamics, 232
moment of inertia, 121, 123 electric, 270 shock wave, 174
momentum, 96 gravity, 86, 87 significant figures, 6
motion potential gradient, 278 simple harmonic motion, 143–150
circular, 45 potential of point charge, 273 acceleration, 147
constant acceleration, 28 potentiometer, 313 differential equation, 144
constant angular acceleration, power, 82 energy, 149
115 pressure, 180 equation, 145
free fall, 32 gauge, 183 phase angle, 150
projectile motion, 42 hydrostatic, 181 velocity, 147
relative, 48 kinetic calculation, 225 simple pendulum, 151–152
rolling, 126 pressure gauge, 182 sinusoidal wave, 164
wave, 162 pressure units Snell’s law, 371
motion of center of mass, 107 atmosphere, 181 solenoid, 341
motion with constant acceleration, 28 bar, 181 specific heat, 202
velocity formula without time, height of mercury, 181 ideal gas, 222
29 millibar, 181 speed of light, 370
motion with constant angular accel- projectile motion, 42 speed of sound, 163
eration, 115 trajectory equation, 43 spherical capacitor, 286
mutual inductance, 358 spin, 342
radian (unit of angle), 115 spin magnetic moment, 343
natural frequency, 155, 170 radiation, 209 spring constant, 81
near point, 385 radio waves, 368 standing wave, 168
newton (N) unit of force, 57 rainbow, 373 antinode, 169
408 INDEX
fundamental frequency, 170 transverse wave, 162 wave motion, 162–175

harmonics, 169 triple point of water, 201 amplitude, 164
natural frequency, 170 turbulent flow, 190 constructive interference, 167
node, 169 destructive interference, 167
state variables, 200 ultraviolet rays, 369 Doppler effect, 172–174
static equilibrium, 135–137 uniform circular motion, 45 function, 163
first condition, 136 uniform linear motion, 29 interference, 166
second condition, 136 unit of electric charge, 241 longitudinal wave, 162
steady flow, 190 unit vectors, 12 periodic, 162
Stefan-Boltzmann law, 209 units, 2 pulse, 162
Stokes’ law, 190 universal gas constant R , 211 reflection and transmission, 170
superposition principle, 166 universal gravitational constant, 60 shock wave, 174
surface tension, 187 sinusoidal, 164
van de Graaff generator, 293
surface tension coefficient, 187 standing, 168
van der Waals force, 187
transverse wave, 162
vector, 8–18
tangential acceleration, 47, 117 wave pulse, 162
addition, 8
telescope, 386 wave speed, 162
components, 10
temperature, 200 electromagnetic, 163
scalar product, 14
temperature gradient, 208 liquid, 163
unit vectors, 12
temperature scale vector product, 16 sound, 163
celsius ( ◦ C ), 201 vector addition string, 163
fahrenheit ( ◦ F ), 201 parallelogram rule, 8 wavelength, 164
kelvin ( K ), 201 triangle rule, 8 wavenumber, 165
tension force, 64 using components, 13 weight, 59
terminal voltage, 308 vector product, 16 work, 78–81
tesla (unit of magnetic field), 321 velocity, 25 adiabatic, 224
thermal expansion, 205 velocity formula without time, 29 performed by gas, 218
thermodynamic equilibrium, 200 velocity vector, 40 scalar product expression, 78
thermometer Venturi tube, 192 spring force, 81
gas, 200 virtual image, 375 variable force, 79
mercury, 200 viscosity, 189 work performed by gas, 218
thermocouple, 201 viscosity coefficient, 189 at constant pressure (isobaric),
thin lens equation, 382 visible light, 369 218
timbre, 170 volt (unit of potential), 271 at constant temperature (isother-
toroid, 341 voltmeter, 313 mal), 219
torque (moment), 118 at constant volume, 219
Torricelli’s formula, 193 watt (power unit), 82 work-energy theorem, 83
total internal reflection, 372 wave amplitude, 164
total mechanical energy, 88 wave front, 369 X-rays, 369
trajectory equation, 43 wave function, 163
transformer, 356 wave interference, 166 Young’s double-slit experiment, 392

Fisica Classica

Uploaded by

Copyright:

Available Formats

Fisica Classica

Uploaded by

Document Information

Original Title

Copyright

Available Formats

Share this document

Share or Embed Document

Sharing Options

Did you find this document useful?

Is this content inappropriate?

Copyright:

Available Formats

Fisica Classica

Uploaded by

Copyright:

Available Formats

Bekir Karaoglu

Translated from Turkish by Mehmet Karaoglu

ISBN 978-3-030-38455-5 ISBN 978-3-030-38456-2 (eBook )

© Springer Nature Switzerland AG 2020

7 ROTATIONAL MOTION 113

12.2 HEAT . . . . . . . . . . . . . . . . . . . 202 18.4 COMBINATION OF RESISTORS . . . . 310

The Russian Soyuz spaceship

1.1 DIMENSIONS AND UNITS

© Springer Nature Switzerland AG 2020 1

Physical quantity Dimension

day, month, year,

The dimension of certain quantities can be expressed in terms of more basic

Surface area = width×length = (length)2

Then, a dimension standard or unit is established for each dimension. For

The basic units of the SI system

Figure 1.1: The laser assembly

• Kilogram: The mass of a cylindrical prototype of the platinum-iridium alloy

Some derived units

1 foot (ft) = 12 inch (in) = 0.3048 meter

1.2 PRECISION AND SIGNIFICANT FIGURES

• In multiplications and divisions, relative errors are added:

• In multiplications and divisions, the lowest number of significant

3.4567 × 2.7 = 9.33309 = 9.3

Of course, the natural numbers ( 1, 2, 3, 4 . . . ) that are used for counting do

Example 1.1 ∆A 0.1 0.2

Figure 1.5: The parallelogram

Figure 1.6: The triangle rule.

In the more useful triangle rule, one of the vectors ( A ~ or B

Figure 1.7: (a) Adding many vec-

shows us that D x = 3, Dy = −5, and Dz = 6 .

Addition Using Vector Components

Working on components through the use of conventional addition and subtraction

A good application of a scalar product is in finding the angle between two

and its magnitude and direction are:

Right-Hand Rule: The right-hand rule needs to be well understood, as it

• If the two vectors are parallel ( θ = 0 ) or anti-parallel ( θ = 180◦ ), then the

Expression of Vector Product in Terms of Components

The circular permutation technique is used to memorize this formula. Consider

Accordingly, when writing the C x component, follow it with the y -component of

Another way of expressing vector product is to write it as a determinant:

The figure shows vectors drawn at the corners of a cube with

1. How many significant figures does the number 0.003804

3. The mass of a metal coin is (8.2 ± 0.1) grams. What is ~ ·A

16. Which of the following vector algebra operations does

1.1 Dimensions and Units 1.3 Vectors

1.2 The unit light-year, a unit of length used in astronomy,

1.4 Express the following data using prefixes of basic units:

1.2 Precision and Significant Figures

1.14 The vectors ~F = −2ı̂ + 3 ̂ + 6 k̂ and G~ = 4ı̂ − 7 ̂ − 4 k̂

1.15 (a) Calculate the magnitudes of the vectors A ~ = 2ı̂ −

The masterpiece of Japanese

© Springer Nature Switzerland AG 2020 23

2.1 POSITION, DISPLACEMENT, VELOCITY AND ACCELERATION

x = x(t), y = y(t), z = z(t)

Therefore, the position of an object at point P is its x coordinate at time t :

x = x(t) (Position) (2.1)

The property that determines motion is the change in position. Displacement

Figure 2.2: Displacement ∆x .

Definition: If the position of an object is x1 at time t1 and its position is x2