Circular Motion (Wpe)

CIRCULAR MOTION & WPE Rg.
2019 - 2021
PREFACE
The phrase “centrifugal force” is often heard, and described as a force pulling outward on an
object moving in a circular path. It is likely that you have felt this effect, perhaps on an
amusement park ride, or during a sharp turn in your car. This is not a real force, however. To
convince yourself of this, remember that forces are always interactions between objects. If you
are feeling a “centrifugal force,” with what other object are you interacting? You cannot
identify another object because this is not a real force. This is “fictitious force” that occurs as a
result of your being in a noninertial reference frame.
In this chapter circular motion we deal with the cause which moves a particle in circular
motion. In this booklet we also deal with energy present in Universe in various forms. Every
physical process in the Universe involves energy and energy transfers or transformations. Thus,
energy is an extremely important concept to understand.
This booklet consists of summarized text coupled with sufficient number of solved
examples of varying difficulties, which enables the students to develop problem solving ability
along with emphasis on physical concept. The problems are categorized into five section,
namely Exercise – I (objectives where only one of the option is correct), Exercise – II
(objectives where more than one option may be correct), Exercise – III (matrix matches and
paragraph type questions), Exercise – IV (subjective questions), Exercise- V (old JEE questions)
to help the student assess his understanding of the concept and further improvise on his
problem solving skills. Solutions to all the questions in the booklet are available and will be
provided to the students (at the discretion of the professor). Every possible attempt has been
made to make the booklet flawless. Any suggestions for the improvement of the booklet would
be gratefully accepted and acknowledged.
(Dept. of Physics)
IIT –ian’s PACE
CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI
CIRCULAR MOTION & WPE Rg. 2019 - 2021
IIT–JEE SYLLABUS
Uniform Circular motion; Kinetic and potential energy; Work and power.
CONTENTS
SR. NO. TITLE PAGE NO.


1 THEORY (CIRCULAR MOTION) 1 – 17
2 EXERCISE – 1 18 – 19
3 EXERCISE – 2 20 – 21
4 EXERCISE – 3 22 – 23
5 EXERCISE – 4 23 – 24
6 EXERCISE – 5 25 – 25
7 THEORY (WORK, POWER & ENERGY) 26 – 49
8 EXERCISE –1 50 – 52
9 EXERCISE – 2 53 – 56
10 EXERCISE – 3 57 – 60
11 EXERCISE – 4 60 – 65
12 EXERCISE – 5 65 – 68
13 ANSWER KEY 69 – 71 

INTEGRATION : Tentative Lecture Flow
(Board Syllabus & Booklet Discussion Included)
Lecture no.1 Derivation of centripetal acceleration & tangential acceleration, uniform
circular motion, banking of roads
Lecture no.2 Solved examples on circular motion, conical pendulum, death well.
Lecture no.3 Work energy theorem, energy stored in springs, definition of conservative,
non conservative forces
Lecture no.4 Solved examples on work energy theorem from different frames
Lecture no.5 Motion in a vertical circle

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CIRCULAR MOTION
CIRCULAR MOTION
When a particle moves in a plane such that its distance from a fixed (or moving) point remains constant
then its motion is called as the circular motion with respect to that fixed (or moving) point. That fixed
point is called centre and the distance is called radius.
KINEMATICS OF CIRCULAR MOTION
Y
(a) Angular Position
The angle made by the position vector of a particle undergoing circular
P
motion with a given line (reference line) is called angular position.
Suppose a particle P is moving in a circle of radius r and centre O.  X
O r
The position of the particle P at a given instant may be described by the
angle  between OP and OX. This angle  is called the angular reference
position of the particle. line
(b) Angular Displacement

Definition: Angle rotated by a position vector of the moving particle with some reference line is called
angular displacement.
1. Infinitesimal angular displacement is a vector quantity but finite angular displacement is not.
2. Direction of infinitesimal angular displacement is decided by right hand thumb rule.
y
(c) Angular Velocity 
(i) Average Angular Velocity P

If 1 and 2 are angular position of the particle at time t1 and t2 P
1 2 x
Total Angle of Rotation    
avg  ; av  2 1 
Total time taken t2  t1 t
(ii) Instantaneous Angular Velocity

The rate at which the position vector of a particle w.r.t. the centre rotates at a given instant, is
called its instantaneous angular velocity.
 instantaneous angular velocity

w
t d
 w
lim dt
t 0
(d) Angular Acceleration 

(i) Average Angular Acceleration :
Let 1 and 2 be the instantaneous angular speeds at times t1 and t2 respectively, then the
average angular acceleration avg is defined as
2  1 
avg  
t 2  t1 t
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(ii) Instantaneous Angular Acceleration :
The rate of change of angular velocity at a given instant
 d d 2 d
  lim   2 
t  0 t dt dt d
Important points:
1. If   0 , circular motion is said to be uniform.
2. If the angular acceleration  is constant, we have
1
  0 t   t 2 … (1)
2
f  0  t … (2)
and f2  20  2 … (3)
where  0 and f are the angular velocities at t = 0 and at time t and  is the angular displacement in
time interval t.
y
(e) Relation Between Linear Speed and Angular Speed
P
The linear distance PP travelled by the particle in time Δt is r P

x
Δs  r Δ
s 
or, lim  lim r
t  0  t t  0 t
or, v = r … (4)
Y et B Y'
(f) Unit Vectors Along The Radius and The Tangent A er
Consider a particle moving in a circle. Suppose the particle is at point 

X'
P
P at a given instant. Take the centre of the circle to be the origin,

a line OX as the X-axis and a perpendicular line OY as the O X
Y-axis. The angular position of the particle at this instant is  .
 
Draw a unit vector PA  eˆ r along the outward radius and a unit vector PB  eˆ t along the tangent in
the direction of increasing  . êr is the radial unit vector and êt the tangential unit vector. Let us draw
PX  parallel to the X-axis and PY  parallel to the Y-axis. From the figure,
ê r  iˆ cos   ˆjsin  , … (6)
where i and ˆj are the unit vectors along the X and Y axes respectively. Similarly,,
ê   î sin   ˆjcos 
t
… (7)
VELOCITY AND ACCELERATION IN CIRCULAR MOTION

Consider the situation shown in the above figure.
 
r  OP  OP eˆ r
 r(iˆ cos   ˆjsin ) . … (i)
Differentiating equation (i) with respect to time, the velocity of the particle at time t is

 dr d ˆ
v  [r(i cos   ˆjsin )]
dt dt
  d   d  
 r  î   sin    ˆj  cos   
  dt   dt  
 r [ î sin   jcos ] … (ii)
The term (r  ) is the speed of the particle at time t and the vector in the square bracket is the unit vector
ê t along the tangent. Thus, the velocity of the particle at any instant is along the tangent to the
circle and its magnitude is v = r  .
  
Vectorially, v    r

 dv
The acceleration of the particle at time t is a  .
dt
  d d ˆ 
a  r  [î sin   ˆjcos ]  [ i sin   ˆjcos ]
 dt dt 
 d d  d
 r  î cos   ˆjsin    r eˆ t
 dt dt  dt
d
 2 r[iˆ cos   ˆjsin ]  r eˆ t
dt
dv
 2 reˆ r  eˆ t … (8)
dt
where ê r and ê t are the unit vectors along the radial and tangential directions respectively and v is the
speed of the particle at time t.
UNIFORM CIRCULAR MOTION
If the particle moves in the circle with a uniform speed, we call it a uniform circular motion. In this
dv
case,  0 and equation (8) gives
dt
a  2 r eˆ r
Thus, the acceleration of the particle is in the direction of ê r , that is, towards the centre. The magnitude
of the acceleration is
v2
a  2 r  ... (9)
r
Thus, if a particle moves in a circle of radius r with a constant speed v, its acceleration is v2/r directed
towards the centre. This acceleration is called centripetal acceleration.
Example:
A particle moving on a circular path of radius 2 m covers 15 revolutions in 5 s with uniform
speed. Find magnitude of its acceleration.
Solution:
Particle is moving with uniform speed. Therefore it has only centripetal acceleration.
15 revolution in 5 s  15  2  radians in 5 s.
15  2
So,   rad / s 6 rad / s
5
 Centripetal acceleration = 2 r  (6)2  2  722 m / s 2
NON-UNIFORM CIRCULAR MOTION

If the speed of the particle moving in a circle is not constant, the acceleration has both the radial and the
tangential components:
1. Radial (centripetal) component of acceleration causes the change in the direction of velocity. It
v2
is directed towards the centre and has a magnitude ar  . As the radial component is
r
v2
perpendicular to velocity, it is also called normal component of acceleration denoted as a n  .
r
2. Tangential component of acceleration causes the change in magnitude of velocity. It is directed
along the tangent and its magnitude is decided by the net tangential force acting on the particle.
Its magnitude is given by at as:
dv
at  where v is the speed of the particle.
dt
The tangential acceleration is in the direction of motion if the particle speeds up and
opposite to the direction of motion if the particle slows down.
Note:- 1. In uniform circular motion, tangential component at = 0 because speed does not change.
2. In non-uniform circular motion, the net acceleration is at
a net 
a  a 2r  a 2t
and it makes an angle  with tangent: ar
ar
  tan 1
at
Example: y
A particle moves with constant speed on a circular path shown

in the figure. The instantaneous velocity of the particle is

v  (iˆ  ˆj ) units. Through which quadrant is the particle x
moving if it travels
a) Clockwise
b) Anticlockwise
Solution:
a) Clockwise : Fourth quadrant
45°
b) Anticlockwise: Second quadrant 45°
Example:
A particle travels in a circle of radius 20 cm with uniform angular acceleration. If the speed
changes from 5 m/s to 6m/s in 2 s, find the magnitude of angular acceleration.
Solution:
We know that v  r
When v = 5 m/s,
v 6
When v = 6 m/s,     30 rad / s
r 0.2
As the angular acceleration is uniform, so by applying f  i  t we have 30  25    2
   2.5 rad / s 2
Example:
A particle moves in a circle of radius 2m with time varying speed given by v = 2t2 m/s. Find
a) Magnitude of tangential acceleration of particle at t = 2 s.
b) Magnitude of total acceleration of particle at t = 1 s.
c) Angle between velocity vector of the particle and its total acceleration at t = 2 s.
d) Angle through which it turned from t = 1 s to t = 3 s.
Solution:
dv d
a) Tangential acceleration   (2t 2 )  4t  8 m / s 2
dt dt
b) Total acceleration  a 2t  a c2
2 (2t 2 ) 2
Now a t  4t  4m / s , a c   2t 4  2m / s 2
2
 Total acceleration  42  22  2 5 m / sec2

c) Angle between total acceleration and velocity vector is
4
 ac 
1 1  2t  1
  tan    tan    tan (4)
 at   4t 
d) To find angle swept between t = 1 s to t = 3 s, first we will find distance moved on circular
path during this interval.
ds
 2t 2  ds  2t 2dt
dt
On integrating with appropriate limits we have
s = Distance travelled on circular path during
3
3
2  2t 3  2 3 2 3 52
t :[1,3]   2t dt      3  1  m
1
 3 1 3 3 3
 52 
Arc  3  26
Angle swept      radians
Radius 2 3
DYNAMICS OF CIRCULAR MOTION

In circular motion along any curved path Newton’s law is applied in two perpendicular directions one
along the tangent and other perpendicular to it . i.e. towards centre. The component of net force along
the centre is called centripetal force. The component of net force along the tangent is called tangential
force.
dv
Tangential force (Ft )  ma t  m mr
dt
2 mv 2
Centripetal force (FC )  m r 
r
More about Centripetal Force
Concepts : This is necessary resultant force towards the centre.
mv 2
F  m2 r
r
(i) A body moving with constant speed in a circle is not in equilibrium. There is always a resultant
force directed towards centre of circular path which provides the required centripetal force.
(ii) In the absence of the centripetal force the body will move in a straight line.
(iii) It is not a new kind of force which acts on bodies. Any force which is directed towards the
centre may provide the necessary centripetal force.
(a) Conical Pendulum
A conical pendulum of a string AB (Figure) whose upper end is fixed at A and other end B is tied with
a bob. When the bob is drawn aside and is given a calculated velocity in a perpendicular (horizontal)
direction, it describe a horizontal circle with constant angular speed  in such a way that AB makes a
constant angle  with the vertical. As the string traces the surface of a cone, it is known as conical
pendulum.
Let l be the length of string AB. The forces acting on the bob are (i) Weight mg acting downwards,
(ii) Tension T along the string (horizontal component is Tsin  and vertical components is T cos  ).
The bob is undergoing uniform circular motion. It has only centripetal acceleration towards the centre
of circle.
T cos  = mg (No acceleration along vertical) … (1)
A

The horizontal component provides the centripetal force, l
h
2 T cos 
i.e., T sin  = m  r … (2)
T 
cos  g r T sin 
From eq. (1) and eq. (2)  2 O
B
sin   r
g sin  mg
2  … (3)
r cos 
From figure, r = l sin 
g sin  g
 2   … (4)
l sin  cos  l cos 
2  l cos  
t  2   [Time period] … (5)
  g 
h
But, l cos  = h  t  2  g  … (6)
 
 The time period is independent of the mass of the body and depends only on h, i.e., distance
between point of suspension A and the centre of the circle O.
(b) Death Well
In a death well, a person drives a bicycle or a motor cycle on the vertical surface of a large wooden well.
The walls of death well are at rest. Here, the friction balances the weight of the person while reaction of
the wall provides the centripetal force necessary for circular motion. The forces are shown in the figure.
Therefore,
mv 2
N ; f = mg [No acceleration along vertical]
r f
N r
O
mv 2 Vmin 
rg
   mg ; mg
r 
(c) Circular Turning of roads
When vehicles go through turnings, they travel along a nearly circular arc. There must be some force
which will produce the required centripetal acceleration. If the vehicles travel in a horizontal circular
path, this resultant force is also horizontal. The necessary centripetal force is being provided to the
vehicles in the following three ways.
1. By friction 2. By banking of roads 3. By both friction and banking of roads.
Now let us consider equations of motion in each of the three cases separately.
By friction only
Suppose a car of mass m is moving at a speed v in a horizontal circular arc of radius r. In this case, the
necessary centripetal force to the car will be provided by force of friction f acting towards centre.
mv 2
Thus, f 
r
Further, limiting value of f is N
or fL = N =  mg
mv 2
Therefore, for a safe turn without sliding  fL
r
mv 2 v2
or   mg or  or v  rg
r rg
By Banking of Roads
To avoid dependence on friction, the roads are banked at the turn so that the outer part of the road is
lifted compared to the inner part.
Applying Newton’s second law along the radius and the first law in the vertical direction.
mv 2
N sin   N
r 
mv2
G
r
and N cos  = mg
From these two equations, we get 
2 mg
v
tan   … (i)
rg
or v  rg tan  … (ii)
Note : This is the speed at which car does not slide down even if track is smooth. If track is smooth and
speed is less than rg tan  , vehicle will move down so that r decreases and if speed is more than this,
the vehicle will move up.
By Friction and Banking of Road
If a vehicle is moving on a circular road which is rough and banked, then three forces act on the vehicle,
of these the first force is weight (mg), the second force, is normal reaction N and is perpendicular to
road, the third force, is friction f which can be either inwards or outwards while its magnitude
N N
y
 f 


x f
 mg  mg
can be varied upto a maximum limit (fL = N ).
i) Friction f is outwards if the vehicle is at rest or v = 0. Because in this case the component of
weight mg sin  is balanced by f.
ii) Friction f is inwards if v  rg tan 
iii) Friction f is outwards if v  rg tan  and
iv) Friction f is zero if v  rg tan 

Let us now see how the force of friction and normal reaction change as speed is gradually increased.
N f N
f
 mv 2 
 r 
N sin  f cos
 mg  mg
Figure (a) Figure (b)
In figure (a) : When the car is at rest force of friction is upwards. We can resolve the forces in two
mutually perpendicular directions. Let us resolve them in horizontal and vertical directions.
F H 0  N sin   f cos   0 … (i)
F V 0  N cos   f sin   mg … (ii)
mv 2
In figure (b) : Now the car is given a small speed v, so that a centripetal force is now required in
r
horizontal direction towards centre. So equation (i) will now become,
mv 2
N sin   f cos 
r
mv 2
or we can say, in case (a) N sin  and f cos  were equal while in case (b) their difference is .
r
N will increase and f will decrease. This is because equation number (ii), N cos  + f sin  = mg is still
to be valid, as there is no acceleration along vertical.
So to keep N cos  + f sin  to be constant (= mg) N should increase and f should decrease
(as  = constant)
Now as speed goes on increasing, force of friction first decreases, becomes zero at v  rg tan 
and then reverses its direction.
Maximum velocity with which a vehicle can take a turn on a rough banked road
mv 2max
N sin   f cos   ; N cos   mg  f sin 
R N N cos 

N sin 
f  N (  Velocity is maximum ) 
f cos  
f
2
mv max

N sin   N cos   ... (1) mg
R
N cos   mg  N sin  ... (2)
gR (sin    cos )
Solving for (1) & (2), Vmax 
(cos    sin )
RADIUS OF CURVATURE
Any curved path can be assumed to be made of infinite circular arcs. Radius of curvature at a point is the
radius of the circular arc at a particular point which fits the curve at that point.
mv 2 mv 2 mv 2
Fc = or, R = F = F
R C 
F = Force perpendicular to velocity (centripetal force)

If the equation of trajectory of a particle is given we canfind the radius of curvature of the instantaneous
circle by using the formula ,
3/ 2

 
dy 2 
1  d x 
 
R= .
d2 y
d x2
CENTRIFUGAL FORCE
Centrifugal force is a fictitious force which has to be employed in a
travelling frame of reference attached to a body moving a circular path,
in order to apply Newton’s Laws of motion from that frame.
mv 2
Its magnitude is equal to .
r
When a body is moving in a circular path and the centripetal force vanishes, the body would leave the
circular path. To an observer who is not sharing the motion along the circular path, the body appears to
fly off tangentially at the point of release. To another observer, who is sharing the motion along the
circular path (i.e., the observer is also moving with the body) it appears as if the object has been thrown
off along the radius away from the centre by some force. This force is centrifugal force.
Analysing the dynamics of a body moving in a circle, from a frame of reference that translates with the
2
body, we have to assume a fictitious force directed radially outward, of magnitude m2 r or mv / r ,
apart from any real forces that may be acting on the body.
SOLVED EXAMPLES
Example - 1
A particle describes a horizontal circle on the smooth inner surface of a conical funnel. The
height of the plane of the circle above the vertex is 9.8 cm. Find the speed of the particle.
[Take g = 9.8 m/s2]
Solution:
The forces acting on the particle are shown in the figure. They are
(i) weight mg acting vertically downwards.

(ii) normal reaction N. N

r
Hence, N sin  = m g (acceration along vertical is zero) … (1)
and N cos  = m v2/r … (2)
Dividing equation (1) by equation (2), we get h
mg
gr
tan  = 2 … (3) 
v
tan  = r/h … (4)
Solving (3) and (4), we get

r gr
 or v  gh
h v2
 v = [9.8 (9.8  102)]1/2 = 0.98 m/s.

Example - 2
A man whirls a stone round his head on the end of a string 4.0 m long. Can the string be in a
horizontal plane? If the stone has mass of 0.4 kg and the string will break if the tension in the
string exceeds 8 N, what is the smallest angle the string can make with the horizontal? What
is the speed of the stone? Take g = 10 m/s2. O
Solution:  T T cos 
l

From figure T
T cos  = m g … (1) T sin 
A
mv2 mv2 mg
T sin  =  … (2) mg
r l sin 
From equation (1)
mg
T
cos 
For the string to be horizontal,  must be 90º, cos   0 and the tension infinite - which is impossible. So
the string cannot be in horizontal plane.
The maximum angle  is given by the breaking tension of the string in the equation T cos  = m g.
Here T (maximum) = 8 N, and m = 0.4 kg
 8 cos  = 0.4  g = 0.4  10 = 4;
1
cos  = (4/8) = ,   60o
2
The angle with the horizontal = 90º  60º = 30º
From equation (2)
o0.4  v 2
8sin 60 
4sin 60o
32  sin 2 60o
or v2   80sin 2 60o ;  v  80 sin 60º = 7.7 m/s.
0.4
Example - 3
A 40-kg mass, hanging from a rope of length l, oscillates in a vertical plane with an angular
amplitude  0 . What is the tension in the rope, when it makes an angle  with the vertical? If
the breaking strength of the rope is 80 kg f, what is the maximum angular amplitude with
which the mass can oscillate without breaking the rope ?
Solution:
The situation is shown in the figure
(a) from figure h  l (cos   cos  0 )
0
and v2  2 g h 
T
 2 g l (cos   cos  0 ) ... (1) h
Again T - m g cos  = m v2/l ....(2)
v
2
mg
Substituting the value of v from equation (2) we get
T - m g cos  = m {2 g l (cos  - cos 0 )/l}
or T = m g cos  + 2 m g (cos  - cos 0 ) ; or T = m g (3 cos  - 2 cos 0 )
or T = 40 g (3 cos  - 2 cos 0 ) Newton ; or T = 40 (3 cos  - 2 cos 0 ) kg f.
(b) Let 0 be the maximum amplitude. The maximum tension T will be at mean position where
 = 0.
 Tmax = 40 (3 - 2 cos 0 ) kgf ; But Tmax = 80 kgf
Solving we get 0 = 60º
Example - 4
Figure shows a rod of length 20 cm pivoted near an end and which is made to rotate in a
horizontal plane with a constant angular speed. A ball of mass m is suspended by a string
also of length 20 cm from the other end of the rod. If the angle  made by the string with the
vertical is 30º, find the angular speed of rotation. Take g = 10 m/s2.

l  20 cm
20

cm
mg
Solution:
Let  be the angular speed of rotation. The ball moves in a horizontal circle of radius
(l + l sin  ) and its acceleration towards the centre will be equal to (l + l sin  ) 2 .
Here, we have
T cos  = m g … (1)
and T sin   m 2 (l  l sin  ) … (2)
 2 (l  l sin ) g tan  10  (1 / 3 )
 tan  =
g
or 2 = l (1  sin )  (0.20)[1  1 / 2]
Solving we get,  = 4.4 rad/s.
Example - 5
A large mass M and a small mass m hang at the two ends of a string that passes through a
smooth tube as shown in figure. The mass m moves around in a circular path which lies in the
horizontal plane. The length of the string from the mass m to the top of the tube is l. What
should be the frequency of rotation of the mass m so that mass M remains stationary ?
String
Tube l
Solution:
Various forces acting on mass M and m are as shown in Y
Y
the adjoining figures. From these figures T Tcos 


l

M T m
X X r
T=Mg …(1) Tsin 
Mg
mg
m v2
and T sin  =  m  2 r  m 2 l sin  … (2)
r
from (1) & (2),
Mg sin   m2  sin 
Mg Mg  1 Mg
2  or  ; Frequency = 
ml ml 2 2 ml
Example - 6
A chain of mass m and radius R placed on a smooth table is revolving with a speed v about a vertical axis
coinciding with the symmetry axis of the chain. Find the tension in the chain.
Solution:
To find tension we consider a small elemental length dl of the chain (see figure), subtending an angle d
at the centre. This element (say mass = dm) experiences the centripetal force along radially inward
(dm)v 2
direction, given as Fc 
R
As shown in the figure, tension acts at the edges of this dl tangentially away from the element. If we
d
resolve the two tensions along and perpendicular to the element, the components T cos , will cancel
2
d
each other and the perpendicular components which are radially inward ( 2T sin ) provide the required
2
centripetal force. Thus we have,
d (dm)v 2
2T sin 
2 R
As d is very small, we can use sin d  d , thus
2
 M  v M Mv 2
Td    Rd   [dm   Rd] or T .
 2R  R 2R 2R
Example - 7
A chain of mass m forming a circle of radius R is slipped on a smooth round cone with half
angle  . Find the tension in the chain if it rotates with constant angular velocity  about a
vertical axis coinciding with the symmetry axis of the cone.
Solution:
The situation is shown in the figure below.
Consider a small element of length x of the chain.
Suppose it forms an angle 2  at the centre O.
The mass of this element
m m
m   2 
2 
Different forces acting on this element are shown in the figure.
Resolving these forces horizontally and vertically, we have
2 T sin   N cos  = m .  2 R …(1)
and N sin  = m . g … (2)
m . g
From equation (2), N = sin  . Substituting in (1),
m . g
2 T sin    cos  = m .  2 R
s in 
When  is very small, sin   
m m
 2 T  . g . cot  = .2 R
 
m m mg  2 R 
or T= . g . cot  + . 2 R or T = 2  cot   g 
2 2  
Example - 8
A particle is attached by means of two equally long strings to two points A and B in the same
vertical line and describes a horizontal circle with a uniform angular speed. If the angular
speed of the particle is 2 2 g / h with A B = h, show that the ratio of the tensions of the strings
is 5 : 3.
Solution:
The situation is shown in fig. (a) and various forces on the particle are shown in the fig. (b). From fig. (b),
we have
T1 cos  = T2 cos  + m g (acceleration along vertical is zero)
or (T1  T2) cos  = m g … (1)
2
and T1 sin  + T2 sin  = m  r
and T1 sin  + T2 sin  = m 2 r
or (T1 + T2) sin  = m  2 r … (2)
Figure (a) Figure (b) Figure (c)

From equations (1) and (2), we have
T1  T2  (mg / cos ) … (3)
and T1  T2  (m2 r / sin ) … (4)
h
Taking into account that, r  tan  (fig. (c)), and   2 2 g / h , we have
2
8g h
 2r  . tan   4g tan ... (5)
h 2
Dividing (3) by (4) and using (5), we get
T1  T2 mg tan  1
  
T1  T2 m 2 r 4
T1 2T1 1  4 5
    .
T2 2T2 4  1 3
Example - 9
A hemispherical bowl of radius R is set rotating about its axis of symmetry which is kept
vertical. A small block kept in the bowl rotates with the bowl without slipping on its surface.
If the surface of the bowl is smooth and the angle made by the radius through the block with
the vertical is  , find the angular speed at which the bowl is rotating.
Solution:
The situation is shown in the figure.
Forces acting on the block are
(i) Normal force N
(ii) Weight mg.
The block moves in a horizontal circle with the centre at B.
Its radius is R sin  . Now, applying Newton’s law, we have
N cos  = mg … (1)
N sin  = m  2 r
N sin  = m  2 (R sin  ) … (2)
From (1) and (2),
g

R cos 
Example - 10
Two blocks of masses m1 = 10 kg and m2 = 5 kg, connected to each other by a massless
inextensible string of length 0.3 m, are placed along a diameter of the turn-table. The coefficient
of friction between the table and m1 is 0.5 while there is no friction between m2 and the table.
The table is rotating with an angular velocity of 10 rad/s about a vertical axis passing through
its centre O. The masses are placed along the diameter of the table on either side of the centre
O such that the mass m1 is at a distance of 0.124 m from O. The masses are observed to be at
rest with respect to an observer on the turn table. (g = 9.8 m/s2)
(i) Calculate the frictional force on m1.
(ii) What should be the minimum angular speed of the turn-table so that the masses will slip
from this position ?
(iii) How should the masses be placed with the string remaining taut so that there is no
frictional force acting on the mass m1 ?
Solution:
Givenm1 = 10 kg, m2 = 5 kg,  = 10 rad/s
r = 0.3 m, r1 = 0.124 m
r2 = r  r1 = 0.176 m
(i) Masses m1 and m2 are at rest with respect to rotating table. Let f be the friction between mass m1
and table.
Free-body diagrams of m1 and m2 with respect to table (non-inertial frame of reference) are
shown in the figures below
Equilibrium of m2 gives T = m 2 r2  2 … (1)
Since, m 2 r2  2 < m1r12 (m2r2 < m1r1)

Therefore, m1r12 > T
and friction on m1 will be inwards (towards centre)
Equilibrium of m1 gives f + T = m1r12 … (2)
From equation (1) and (2), we get f = m1r12  m 2 r2  2
= (m1r1  m2r2)  2 … (3)
= (10  0.124  5  0.176)(100) N
= 36 N
Therefore, frictional force on m1 is 36 N, inwards.
Masses will start slipping when f  (m1r1  m 2 r2 ) 2 is greater than fmax
(m1r1  m2r2)  2 > fmax
> m1 g
 Minimum value of  is
m1 g 0.5  10  9.8

 min  
m1 r1  m2 r2 10  0.124  5  0.176
 min  11.67 rad/s
From equation (3), frictional force f = 0 when m1r1 = m2r2
r1 m 2 5 1
or    and r = r1 + r2 = 0.3 m  r1 = 0.1m and r2 = 0.2 m
r2 m1 10 2
i.e., mass m2 should be placed at 0.2 m and m1 at 0.1 m from the centre O.
EXERCISE -1
1. Two cars having masses m1 and m2 move in circles of radii r1 and r2 respectively. If they complete the
circles in equal time, the ratio of their angular speeds 1 / 2 is
(a) m1/m2 (b) r1/r2 (c) m1r1/m2r2 (d) 1.
2. A car moves at a constant speed on a road as shown in the figure. The normal force by the road on the
car is NA and NB at points A and B, respectively. A B
(a) NA = NB (b) NA > NB (c) NA < NB

(d) Insufficient information to decide the relation of NA and NB.
3. A sphere of mass m tied to the end of a string of length l moves along a horizontal circular path with
constant speed v. The work done in a complete circle is :
mv 2 mv 2
(a) zero (b)  2l (c) mg . 2l (d) l
l l
4. A motor car is moving on a circular path of radius 500 meter and has a speed 30 ms–1 at a particualr
instant. Its speed is increasing at the rate of 2 m s-2. What is its acceleration at this instant?
(a) 2.5 m s–2 (b) 2.7 m s–2 (c) 2 m s–2 (d) 4.5 m s–2
5. A person of mass M stands in contact against the wall of a cylindrical drum of radius r rotating with
an angular velocity  . If the coefficient of friction between the wall and the man is  , the minimum
rotational speed of the cylinder which enables the person to remain stuck to the wall when the floor
is removed is :
g r 2g rg
(a) min .  (b) min .  (c) min .  (d) min . 
r g r 
6. A coin is placed on a rotating table. It is observed that the coin just slips when placed at a distance
4r from the centre. On doubling the angular velocity of the table, it will just slip when its distance
from the centre equal to :
r
(a) 4r (b) 2r (c) r (d)
4
7. A mass of 2 kg is whirled in a horizontal circle by means of a string at an initial speed of 5 revolutions
per minute. Keeping the radius constant, the tension in the string is doubled. The new speed is
nearly :
(a) 14 rpm (b) 10 rpm (c) 2.25 rpm (d) 7 rpm
8. A particle is acted upon by a force of constant magnitude which is always perpendicular to the
velocity of the particle. The motion of the particle takes place in a plane. It follows that :
(a) its velocity is constant (b) its acceleration is constant
(c) its kinetic energy is constant (d) it moves in a straight line
9. A car is moving on a circular road of curvature 300 metres. If the coefficient of friction is 0.3 and
acceleration due to gravity is 10 m/s2, the maximum speed the car can have is :
(a) 30 km/hr (b) 81 km/hr (c) 108 km/hr (d) 162 km/hr
10. If a car moves round a circular road of radius r at a constant speed v

(a) its velocity changes and the acceleration is v/r2 .
(b) there is no force on the car since its speed is constant
(c) the force on the car is towards the centre and its magnitude mv2/r
(d) the force on the car is outward from the centre and its magnitude is mv2/r
11. A small object of mass 0.2 kg is whirled round in horizontal circle in gravity free space. It is tied to a
string of length 0.5 m and rotates with constant angular speed of 4 rad/s. The tension in the string is :
(a) 0.2 N (b) 1.6 N (c) 0.8 N (d) 1.0 N
12. When a girl rides round the corner on her bicycle at a steady speed
(a) there is no force when she takes the turn
(b) when she takes the turn the force on her bicycle is only her weight
(c) the centripetal force is due to the friction between the tyre and the ground
(d) the force on the girl is always outwards.
13. A body moves along an uneven horizontal road surface
with constant speed at all points. The normal reaction B
of the road on the body is :
(a) maximum at A
(b) maximum at B C
A
(c) minimum at C
(d) the same at A , B and C
14. A particle of mass m is tied to a light string and rotated with a speed v along a circular path of radius r.
If T = tension in the string and mg = gravitational force on the particle then the actual forces acting on the
particle are :
(a) mg and T only
(b) mg , T and an additional force of mv2/r directed inwards
(c) mg , T and an additional force of mv2 /r directed outwards
(d) only a force mv2/r directed outwards
15. A long horizontal rod has a bead which can slide along its length, and initially placed at a distance L from
one end A of the rod. The rod is set in angular motion about A with constant angular acceleration . If
the coefficient of friction between the rod and the bead is , and gravity is neglected , then the time after
which the bead starts slipping is :
  1
(a) (b) (c) 
(d) infinitesimal
 
EXERCISE -2
z
1. The upper end of the string of a simple pendulum is fixed to a 2m
vertical z  axis , and set in motion such that the bob moves
along a horizontal circular path of radius 2m , parallel to the
xy plane, 5 m above the origin. The bob has a speed of 3 m/s.
5m
The strings breaks when the bob is vertically above the x-axis,
and it lands on the xy plane at a point ( x , y ) y
O
(a) x = 2m (b) x > 2m (c) y = 3m (d) y = 5 m x
2. A circular race track of radius 300 m is banked at an angle of 15°. If the coefficient of friction between
the wheels of a race-car and the road is 0.2. Choose the correct options (g = 9.8 m/s2)
(a) The optimum speed of the race-car to avoid wear and tear on its tyres is 28.1 m/s
(b) Maximum permissible speed to avoid slipping is 38.1 m/s
(c) The optimum speed of the race-car to avoid wear and tear on its tyres is 38.1 m/s
(d) Maximum permissible speed to avoid slipping is 28.1 m/s
3. Choose the correct statements

(a) The net acceleration of a particle in circular motion is always along the radius of the circle towards
the centre
(b) The velocity vector of a particle at a point is always along the tangent to the path of the particle at
that point
(c) The acceleration vector of a particle in uniform circular motion averaged over one cycle is a null
vector
(d) The acceleration vector of a particle in uniform circular motion averaged over one cycle is a non-
zero vector
4. A vehicle of mass m starts moving along a horizontal circle of radius R such that its speed varies with
distance S covered by the vehicle as v  K S where K is a constant
K2t K2t2
(a) The speed of vehicle is (b) The distance moved by particle is
2 4
(c) The speed of vehicle is K2t2 (d) The distance moved by particle is 2K2t2
5. A string of length  has one end fixed and a particle of mass m is attached to the other end. If the particle
describes a horizontal circle at an angular speed  , in gravity free space.
(a) The tension in the string  m2 
(b) The speed of the particle is 
(c) The resultant force acting on the particle has no vertical component
(d) None of these
y
6. A particle is executing a uniform circular motion on a horizontal surface. Particle v at t = 0
position and velocity at time t = 0 are shown in the figure in the coordinate
system. Which of the indicated variable on the vertical axis is/are correctly x
matched by the graph(s) shown alongside for particle’s motion-
(a) y component of force keeping particle

circle
moving in a circle
(b) Angular velocity of the particle t
t
(c) x component of velocity
t
(d) x coordinate of the particle
7. Particle is placed on a rough disc rotating with uniform angular 

velocity. The friction between block and ground, and tension in m = 0.5 kg
the string in different cases of friction is  = 0.5 rad/sec
O
R=1m
(a) if   0.1 , friction = 1/2 N, Tension = 1/2N
(b) if   0.1 , friction = 1/8 N, Tension = zero
(c) if   1 / 20 , friction = 1/4 N, Tension = 1/4 N
(d) if   1 / 40 , friction = 1/8 N, Tension = zero
EXERCISE -3
1. Match The Column

A particle of mass m moving in a circular path of constant radius r such that its centripetal acceleration
ac is varying with time t as ac = K2rt2, when K is constant
Column - I Column - II
(A) Centripetal force (p) m Kr
(B) Tangential force (q) mK2r2t
(C) Power delivered by centripetal force (r) MK2rt 2
(D) Power delivered by tangential force (s) Zero
COMPREHENSION TYPE
Comprehension - I 
The system shown in the figure is rotated in a horizontal circle about the m1   0.5
vertical axis shown with angular velocity which is gradually increased
m2 0
from zero. Here R = 0.5, m1 = 2kg and m2 = 1 kg.
R
2. What will be the direction of friction acting on upper face of m2 when relative motion between the 2
blocks just starts.
(a) Towards left (b) Towards the right
(c) There will be no friction (d) Cannot be determined
3. Find approximately the minimum value of angular velocity  to start relative slipping between the 2
blocks
(a) 6.3 rad/s (b) 4 rad/s (c) 7.4 rad/s (d) 8.3 rad/s
4. Find the approximate value of tension in the string connecting m1 & m2 when slipping just starts
between the blocks
(a) 20N (b) 30N (c) 40N (d) 50N
Comprehension - II
A road of radius 20m is banked for vehicles going at speed of 10m/s. The road has sufficient friction to
prevent slipping. A vehicle of 200 Kg moves on this road. Take g = 10 m/s2.
5. What will be the magnitude and direction of friction on the vehicle if it negotiates the curve at a constant
speed of 5 m/s.
(a) 200 5 N upwards (b) 300 5 N upwards
(c) 200 5 N downwards (b) 200 5 N upwards
6. What will be the magnitude and direction of friction on the vehicle if it negotiates the curve at a constant
speed of 15 m/s.
(a) 500 5 N downwards (b) 500 5 N upward
(c) 300 5 N downwards (d) 300 5 N upwards
7. What is angle of banking of the road

1 1 1 1
(a) tan (b) tan–12 (c) tan (d) tan–13
2 4
A
Comprehension - III
Bob B of pendulum AB is given an initial velocity 3gL in the
horizontal direction.
B
8. If AB is a massless string, find the angle made by AB with horizontal
when the string is about to slacken 3gL
1 1 1 2 1  1  1  2 
(a) sin (b) sin (c) sin   (d) sin  
4 3 3 5
9. Find maximum height attained by bob from starting point, if AB is massless string
3L 40L 20L
(a) L (b) (c) (d)
2 27 29
10. Find maximum height attained by bob from starting point if AB is a massless rod
3L 40L 20L
(a) L (b) (c) (d)
2 27 29
EXERCISE -4
1. A car moves with constant tangential acceleration a along a horizontal surface circumscribing a circle
of radius R . The coefficient of sliding friction between the wheels of the car and the surface is
 .What distance will the car move without sliding if its initial velocity is zero?

2. An insect crawls up a hemispherical surface very slowly. The coefficient
1
of friction between the insect and the surface is . If the line joining the
3 insect
centre of the hemispherical surface to the insect makes an angle  with
vertical, then find the maximum possible value of  . fixed hemisphere
3. A 70 kg man stands in contact against the inner wall of a hollow cylindrical drum of radius 3m rotating
about its vertical axis with 200 rev/min. The coefficient of friction between the wall and his clothing is
0.15. What is the minimum rotational speed of the cylinder to enable the man to remain struck to the wall
(without falling) when the floor is suddenly removed?
4. Two light inelastic strings AP and BP connect a particle P to fixed points A and B. The point B is vertically
above A and AB = AP =  and BP   3 . The particle P moves in a horizontal circle with constant
speed. The least angular speed of P for both strings to be taut is  . When the angular speed of P is
1 ( ) the tension in the string are equal. Show that 12  2g /  3 .
5. Three point particles P, Q, R move in circle of radius ‘r’ with different

but constant speeds. They start moving at t = 0 from their initial positions
as shown in the figure. The angular velocities (in rad/sec) of P, Q and R are
5, 2 & 3 respectively, in the same sense. Find the time interval after
which they are at same angular position for the first time.
6. A particle is moving along a circular path of radius R in such a way that at any instant magnitude of radial
acceleration & tangential acceleration are equal. If at t = 0 velocity of particle is v0, then find the time
period of first revolution of the particle.
7. Two particle A and B move anticlockwise with the same speed v in a circle of radius R and are diametrically
72v 2
opposite to each other at t = 0. At this instant, A is given a constant acceleration (tangential) a t 
.
25R
Calculate the time in which A collides with B, the angle traced by A, its angular velocity and radial
acceleration at the time of collision.
EXERCISE -5
1. Select the correct alternative (s) : A particle is acted upon by a force of constant magnitude which
is always perpendicular to the velocity of the particle. The motion of the particle takes place in a
plane. It follows that : [IIT- 1987]
(a) Its velocity is constant (b) Its acceleration in constant
(c) Its kinetic energy is constant (d) It moves in a circular path
2. A small block is shot into each of the four tracks as shown below. Each of the tracks rises to the
same height. The speed with which the block enters the track is the same in all cases. At the highest
point of the track, the normal reaction is maximum in - [JEE-2001]
(a) (b) (c) (d)
3. An insect crawls up a hemispherical surface very slowly (see the

figure). The coefficient of friction between the surface and the insect
is 1/3. If the line joining the centre of the hemispherical surface to 
the insect makes an angle  with the vertical, the maximum possible insect
value of  is given by [JEE-2001] fixed hemisphere
(a) cot   3 (b) tan   3 (c) sec   3 (d) cos ec   3
4. A simple pendulum is oscillating without damping. When the displacements of the bob is less than
maximum, its acceleration vector a is correctly shown in [JEE-2002]
(a) (b) (c) (d)
5. A smooth semicircular wire track of radius R is fixed in a vertical plane

(figure).One end of a massless spring of natural length 3R/4 is attached to the
lowest point O of the wire track. A small ring of mass m which can slide on the
track is attached to the other end of the spring. Then ring is held stationary at C
point P such that the spring makes an angle of 60° with the vertical. The spring 60° m
P
constant K = mg/R. Consider the instant when the ring is released O
(i) Draw the free body diagram of the ring

(ii) Determine the tangential acceleration of the ring and the normal reaction. [JEE-1996]
WORK, POWER AND ENERGY

THE CONCEPT OF WORK
The work W done by a constant force F when its point of application undergoes a displacement s is defined
to be
Fig. (1)
W = F s cos  ...(1)
 
Where is the angle between F and s as indicated in figure 1.
Work is a scalar quantity and its SI unit is the joule (J).
1J=1 N–m
Work can also be defined as the dot product of force and its displacement as given by equation (2)
 
W  F. s ...(2)
In terms of rectangular components, the two vectors are
    
F  F i  F j  F k and s  xi  yj  zk
x y z
hence equation (2) may be written as
W  Fx x  Fy y  Fz z ...(3)
The work done by a given force on a body depends only on the magnitude of force and displacement, and the
angle between them. It does not depend on the velocity or the acceleration of the body, or on the presence of
other forces.
  
NOTE : work done = F. s where s is the displacement of the point of application of the force.
Example - 1
A box is moved on a horizontal surface by applying force F = 60 N at an angle = 300 to the
horizontal. What is the work done during the displacement of the box over a distance of 0.5 km.
Solution :
By definition, W = F s cos 
Here F = 60 N; s = 0.5 km = 500 m;  = 300 .
 W = (60)(500)cos 300 = 26 kJ (approx.)
Positive and Negative Work F

<90°
Work done by a force may be positive or negative depending on  s
the angle  between the force and displacement. If the angle is
Fig. (2a) Positive Work done by a factor F
acute ( < 900 ), the work done is positive and the component
of force is parallel to the displacement. (See figure 2 a) F > 90°

0
If the angle  is obtuse ( > 90 ), the component of force in the 
s
direction of displacement is antiparallel to the displacement and the
Fig. (2b) Negative Work
work done by force is negative (see fig. (2 b)). done by a force F
Zero Work Done

It is clear from equation (1) that the work done by a force is zero when
(a) F = 0 (b) s = 0 (c) cos  = 0 i.e.,  = 900
WF = 0 v
WN=0 N
F
v
(a)
(b)
WORK DONE IN VARIOUS SITUATIONS

Case I
The force acting on the particle is constant and the particle moves along a zig-zag path.
Let the particle move from point A to point B along the path as shown. Path followed
 by the particle
A constant force F acts on the particle. The work done over infinitesimal
  
displacementds is given by F . d s and the total work is given by 
  F
W   F.ds 
A F 
     ds
W  F .  d s  F . sAB 
B F

Note:   displacement
ds

 | ds |  pathlength

Here, the work done by the force on the particle depends on the displacement vector A B , not on the path
the particle follows.
Example - 2 O R B
The fig. shows a smooth circular path of radius R in the vertical plane
which subtends an angle of /2 at O. A block of mass m is R
taken from position A to B under the action of a constant force F. F
Find the work done by this force.

A
Solution :
When the block moves from A to B, the displacement of the block in the direction of force is R
Therefore, the work done by the constant force F is
W = FR
 
Alternatively, WR   F . ds   Fds cos 
z
R
or W  Fdx bds cos = dx g
O
or W = FR
Case II
One of the Forces acting on the body is constant in magnitude and directed along
F
the velocity (or opposite to it) all the time.
F
Consider a particle moving along the path as shown
F
in the figure. One of the forces acting on the particle has
A
magnitude F and it acts in the direction of velocity vector 
B

(which is tangent to the path) all the time. (At every point F and ds are in the same direction)
B
  B B
Work done by the force W   F . ds   Fds cos 0o  F  ds  Fd

A A A
Here, d is the distance the length of the path it traces between A and B.
Example - 3 O B
The block in the figure is pulled by a force F which is always tangential
to the cylindrical surface of radius R. Find the work done by this
F
force between A and B.
A
Solution :
In this case, force and displacement are always parallel to each other, i.e.  = 0. The

displacement of the block in the direction of force is R . Thus, the work done by the force
2
is  R  
W  F   FR
 2  2
Case III
The force is always perpendicular to the velocity vector of the particle
Path Path
the
particle
(There are may be other force also)

 
In this case the force vector F is always perpendicular to the d s , therefore
   
F. ds  0 Hence  . ds  0
F
Work done by a force which is always perpendicular to the velocity vector is zero.
Im portant
1. Work is defined for an interval or displacement; there is no term like instantaneous work
similar to instantaneous velocity.
2. For a particular displacement, work done by a force is independent of type of motion i.e.
whether it moves with constant velocity, constant acceleration or retardation etc.
3. For a particular displacement work is independent of time. Work will be same for same
displacement whether the time taken is small or large.
4. When several forces act, work done by a force for a particular displacement is independent of
other forces.
5. A force doesn’t depend on the reference frame, but its displacement depends on the frame.
So, work done by a force is frame-dependent. Work done by a force can be different in
different reference frames.
6. Effect of work is change in K.E. or P. E.
WORK DONE BY VARIOUS FORCES

Work Done by Static Friction
1. Work done by static friction force can be positive negative or zero.
Work done by static friction f during displacement s on

Block A =Wf = - fs ( Negative) ; Block B = Wf = fs ( Positive )
2. Work done by static friction on the whole system (A + B) is zero
i.e., (W f )system  (Wf )A  (Wf )B   fs  fs  zero
Work Done by Kinetic Friction

1. Work done by kinetic friction force can be positive or negative.
sA
Sufficient fk F
Rough A F A F F. B. D.
B B fk
sB
Smooth
Work done by kinetic friction on A =WA = – fKSA ( Negative ) and that on B = WB = fKSB ( Positive )
2. Work done by kinetic friction on a closed system is always negative (WfK)S= –fK . Srel
Srel = Relative displacement between the objects where fK acts
For example : In last case if ( A + B ) is taken as system then work done by kinetic friction on the system.
(Wfk)system =(Wfk)A + (Wfk)B; = –fKSA + fKSB
(Wfk)s = –fK(SA-SB) =-fKSrel
Srel = SA– SB = Relative displacement of A with respect to B
Work Done by Normal Reaction

1. Work done by normal reaction can be positive, negative or zero.
N
F F F
A B A B A N B F.B.D.
Here work done by normal reaction between A & B.

(i) On A =(WN)A = – NS ( Negative ) ; (ii) On B = (WN)B = NS ( Positive )
2. Work done by normal reaction on the whole system is zero

(WN)system =(WN)A + (WN)B ; (WN)s = –NS + NS = zero
Work Done by Tension Force

1. Work done by tension force can be positive, negative or zero.
T T T T
A B F A B F
S
Here work done by tension force T
(i) On A =(WT)A = TS ; (ii) On B = (WT)B = – TS
2. Work done by tension on the whole system is always zero

(WT)s =(WT)A + (WT)B; (WT)s = TS – TS = zero
Work Done by Gravity
1. Work done by gravity can be postive, negative or zero.
(a) When the object moves vertically downwards, work done by gravity is positive.
(b) When the object moves vertically upwards, work done by gravity is negative.
(c) For horizontal displacement work done by gravity is zero.
2. Work done by gravity is independent of the path, it depends on initial and final vertical coordinates of the object.
 
W  F . S   mg j .(xi  yj  zk )
g g ; Wg   mgy  mg ( yi  y f )
Work Done by Spring Force

1. Work done by spring force can be positive, negative or zero.
2. Spring can do work at its both ends.
3. Spring force is a variable force so work done by spring force is given by Wsf  
Here x is extension or compression in the spring from its natural length.

zxi
xf
(kx )dx 
1 2 1 2
2
kxi  kx f
2
Work Done by Pseudo Force

1. Work done by pseudo force can be positive, negative or zero.
2. Point of application of pseudo force is taken as center of mass of the object for translatory reference
frame.
Work Done by Internal Force

1. Internal force exists for a system of particles. For a single particle, there is no internal force.
2. Internal forces will do work when there is deformation within the system. In the case of rigid body net work
done by internal forces is zero.
3. Work done by internal force can be positive, negative or zero.
E.g. : In the case of compression, work done by internal forces will be negative; in the case of expansion work will
be positive. When internal forces do positive work, energy of system decreases.
Suppose that a man sets himself in motion backward by pushing against the
wall. The forces acting on the man are his weight W, the upward forces
N
n1 and n2 exerted by the ground and the horizontal force N exerted by the
wall. The works of W and of n1 and n2 are zero because they are perpendicular
to the motion. The force N is the unbalanced horizontal force that imparts to
the system a horizontal acceleration. The work of N, however, is zero because n2 n1
there is no motion of its point of application. We are therefore confronted with

a curious situation in which a force is responsible for acceleration, but its work,
being zero, is not equal to the increase in kinetic energy of the system !
The new feature in this situation is that the man is a composite system with several parts that can move in relation
to each other and thus can do work on each other, even in the absence of any interaction with externally applied
forces. Such work is called internal work. Although internal forces play no role in acceleration of the composite
system, their points of application can move so that work is done; thus the man’s kinetic energy can change even
though the external forces do no work.
WORK DONE BY A VARIABLE FORCE

When the magnitude and direction of a force vary in three dimensions, it can be expressed as a function of
 
the position vector F ( r ) , or in terms of the coordinates F ( x, y , z ) . The work done by such a force in an

infinitesimal displacement ds is
 
dW  F . d s ... (5)
The total work done in going from point A to point B as shown in the figure i.e.
z zb
B B
WA  B 

F. d S 

F cos ds g
A A B
ds
In terms of rectangular components, F F
    
F  Fx i + F y j + F z k dx
 A
and d s  dx i + dy j + dz k
z z z
xB yB zB
therefore, WA  B  Fx dx  Fy dy  Fz dz ...(6)
xA yA zA
Graphically, the work done by a variable force F(x) from an initial point xi to final point xf is given by
the area under the force - displacement curve as shown in the figure (9).
Fs F(x)
xi xf
x
x
xi xf
Fig. (8) Fig. (9)
The done by the spring when the displacement of its free end changes from xi to xf is the area of the trapezoid;
Ws  
1
2
e
k x 2f  x i2 j
Area (work) above the x - axis is taken as positive, and vice-versa.
Example - 4
A 5-kg block moves in a straight line on a horizontal frictionless surface under the influence of
a force that varies with position as shown in the figure. Find the work done by this force as the
block moves from the origin to x = 8m.
Solution :
The work from x = 0 to x = 8 m is the area under the curve.
b
W  10  2  g 1
2
b gb
10 4  2  0 
1
2
g
5 8  6  25 J b gb g
Work-Energy Theorem
If a constant force F acts through a displacement x, it does work W = Fx = (ma) x on the particle.
Since the acceleration is constant, we can use the equation of kinematics v 2f  vi2  2ax . Thus,
m  v 2f  vi2  1 1
W  mvf2  mvi2 .....(9)
2 2 2
1
The quantity, K mv 2 is a scalar and is called the kinetic energy of the particle. Kinetic energy is the
2
energy that a particle posses by virtue of its motion.
Thus, equation (9) takes the form

W  K f  K i  K .... (10)
The work done by a force changes the kinetic energy of the particle. This is called the Work-
Energy Theorem.
In general, the net work done by the resultant of all the force acting on the particle is equal to the change
in kinetic energy of a particle.
Wnet  K .... (11)
And taking into account the work done by internal forces, we have
Wint ernal  Wexternal  K
Im portant
1. The kinetic energy of an object is a measure of the amount of work needed to increase its speed from zero
to a given value.
2. The kinetic energy of a particle is the work it can do on its surroundings in coming to rest.
3. Since the velocity and displacement of a particle depend on the frame of reference, the numerical values of
the work and the kinetic energy also depend on the frame.
Example - 5
A block of mass m = 4 kg is dragged 2m along a horizontal surface by a force F = 30 N acting
at 530 to the horizontal. The initial speed is 3 m/s and k = 1/8.
(a) Find the change in kinetic energy of the block
(b) Find its final speed.
Solution :
(a) The force acting on the block are shown in the figure (11).
N
Clearly, WN  0 and Wg  0 , F
whereas WF  FS cos  53°
f
WF  fs   k NS
where N  mg  F sin  mg
The work-energy theorem, Fig. 11: The change in the kinetic
energy of the block is given by the net
work done on it.
K  Wnet  WF  W f
therefore,
b
K  F s cos -  k mg  F sin s g
b gb gb g
 30 2 0.6 
1
8
b
40  24 2  32 J gb g
1 1
(b) Now K = mv 2f  mvi2  32 J
2 2
Since v i =3 m/s.
therefore, v f =5 m/s
Example - 6
A block of mass m = 2 kg is attached to a spring whose spring
constant is k = 8 N/m. The block slides on an incline for

1
which  k  and   37 0 . If the block starts at rest with
8
the spring unextended, what is its speed when it has
moved a distance d = 0.5m down the incline?
Solution :
This problem can be solved by using Newton’s second law. However, the force exerted by the spring varies
with position and, therefore, so does the acceleration. We avoid this difficulty by using the work-energy
theorem. (Unless you are solving a problem in static equilibrium, it is a good idea to think of the energy
approach rather than dynamics when you see a spring in problem.)
The work done by the force due to the spring was found in equation (8).
xi  0 and x f  d
The work done by each of the forces on the block are
Wg   mgd sin
b
W f   k Nd    k mg cos  d g
1
Ws   kd 2
2
1
Of course, WN  0 . The work-energy theorem, with K  mv 2  0 ,
2
b
tells us mgd sin -  k mg cos d  g 1
2
1
kd 2  mv 2
2
3 4 1
putting m = 2 kg; d = 0.5 m; k = 8 N/m; sin  = ; cos = ;  k 
5 5 8
we get, b2gb10gb0.5gb0.6g  18 b20gb0.8gb0.5g  12 b8gb0.5g 2


1
2
bg
2 v2
v = 2 m/s
N
Example - 7
A box of mass m is gently placed on a conveyor fk
belt of a constant speed v. The coefficient of v
mg
kinetic friction is k .
Fig. 13: A box of mass m is gently
placed on conveyor belts moving at
(a) What is the work done by friction? constant velocity v.
(b) How far does the box move before reaching its final speed?
(c) When the box reaches its final speed, how far has the belt moved?
Solution :
(a) When the box is first placed on the belt there will be slipping between the two. But the force of friction on
the box and its displacement are in the same direction. Consequently, the work done by kinetic friction is
positive. Since the final speed of the box is v,
1
W f  K   mv 2 ..... (i)
2
(b) The force of friction is f   k N   k mg and W f   fd

Thus from equation (i),
1
k mgd   mv 2 ..... (ii)
2
v2
Thus, d
2 k g
(c) If the box takes a time t to reach speed v, then v = at where a is the acceleration of box. In this time it will
1 1
move d  at 2  vt .
2 2
Since the belt’s speed is fixed, in time t it moves a distance vt  2d .The belt moves twice as far as the box
while the box is accelerating.
Example - 8
A horizontal force F very slowly lifts the bob of a simple pendulum from a vertical position to a
point at which the string makes an angle  0 to the vertical. The magnitude of the force is varied
so that the bob is essentially in equilibrium at all time.
What is the work done by the force on the bob?
Solution :
Figure (14) is a free body diagram of the system and shows the forces acting on the bob. Since the
acceleration is zero, both the vertical and horizontal components of the forces balance :
F x  F  T sin = 0
F y  T cos   mg  0
Eliminating T we get
F  mg tan .... (i)
This is how the force must vary as a function of
angle in order for the bob to be in equilibrium.

The work done by F in an infinitesimal
displacement

d s along the circular arc is
 
dW = F . d s = Fx dx
 mg tan  dx .... (ii)
dy
From figure (14), we see that  tan  ,
dx
thus dy  tan dx .
Equation (ii) becomes dW  mg dy , therefore,
the total work done from y = 0 to y = y 0 is
z
y0
W= b
mgdy  mgy0  mgL 1  cos   g
0
where the vertical displacement is y0  L 1  cos   . b g
WORK DEPENDS ON THE FRAME OF REFERENCE

Since the magnitude of displacement in a given time interval depends on the velocity of the frame of
reference used to measure the displacement, therefore, the work also depends on the frame of reference.
The Fig.15 shows a trolley moving with a constant velocity V0 along the positive x-axis. On the trolley, a
man is pushing a block with a constant force F. The reference frame x'y' is attached with the trolley and the
frame xy is attached with the ground. The figure 15 (A) shows the initial position of the block with respect
to trolley xi as well as ground. And, the figure 15 (B) shows the final positions.
The work done by the constant force with respect to trolley, that is, in the x'y' frame is
WL' = F (x' f – x' j)
And, the work done with respect to ground, that is, in the xy frame is
W = F (xf – xi )
Since xi = Xi + x' i and xf = Xf + x' f
therefore, W = F[(Xf – Xi ) + (x' f – x' i )]
or W = W' + F(Xf – Xi )
where Xi , Xf : Final position of the trolly w.r.t. observer on ground.
where Xf – Xi is the displacement of the trolley with respect to ground.
y y' F y y' F
x' x'
V0
x´j x´j
x x
Xj xj Xf xf
Fig. 15(A) Fig. 15(B)
Example - 9
A plank of mass M and length L is placed at rest on a smooth horizontal surface. A small block
of mass m projected with a velocity v0 from the left end of it as shown in the fig. The coefficient
of friction between the block and the plank is , and its value is such that the block becomes
stationary with respect to the plank before it reaches the other end.
m
(i) Find the work done by the friction force on the block during
v0 M
the period it slides on the plank. Is the work positive Smooth
or negative.
(ii) Calculate the work done on the plank during the same period. Is the work positive or negative?
(iii) Also, determine the net work done by friction. Is it positive or negative?
Solution :
The free body diagrams of the block and the plank are shown in the fig
N
 N
1 v f =mg
a2
Motion
f=  mg
f
Block : a1 = = –μg Instantaneous velocity, v1 = v0 –  gt
m
f mg  mgt
Plank : a2   Instantaneous velocity  v2 
M M M
Finally both the blocks start moving together,
i.e. v 1 = v 2
 mgt Mv0
then v0   gt  or t
M (M  m)g
mv0
and, the final common velocity is v 
M m
(i) The work done by friction on the block is equal to its change in kinetic energy, i.e.,
2
1 1 1  mv 0  1
W1  K f  K i  mv 2  mv 02 or W1  m   mv 2
2 2 2  m  M  2 0
1 mM(M  2m)v02

2 (M  m) 2
The work done by friction on the block is negative.
(ii) The work done by friction on the plank is given by

2
1 1  mv0  1 m2 M
W2  M    v02
W2  K f  K i  Mv 2  0 or 2  m  M  2  M  m 2
2
The work done by friction on the plank is positive.
1 mM
(iii) The net work done by friction is W  W1  W2   v02
2  M  m
The net work done by friction is negative.
POTENTIAL ENERGY
When we throw a ball upwards with an initial velocity, it rises to a certain height and becomes stationary
for a moment. What happens to the lost kinetic energy? We know with our experience that the ball
returns back in our hands with a speed equal to its initial value. The initial kinetic energy is somehow stored
and is later fully recovered in the form of kinetic energy. The ball must have something at the new height
that it does not have at the previous level. That something by virtue of its position is Potential Energy.
Potential energy is the energy associated with the relative positions of two or more interacting particles.
Potential energy fits well the idea of energy as the capacity to do work. For example, the gravitional
potential energy of an object raised off the ground can be used to compress or expand a spring or to lift
another weight. As a coil spring unwinds, or a straight spring returns to its natural length, the stored
potential energy can be used to do work. For example, if a block is attached to a compressed spring, the
elastic potential energy can be converted into kinetic energy of the block as (Fig.1)
Rest
In the above discussion we have seen that in the case of gravity
and elastic spring, the kinetic energy imparted initially is stored
as potential energy for a short time which is regained, later on.
But this is not true in all cases.
For example, consider block placed at rest on a rough horizontal

surface. If we impart it some initial kinetic energy, it starts sliding
on the surface, the frictional force does negative work on the
block, decreasing its kinetic energy to zero. But it does not Fig. 1: The block gains kinetic
energy when the compressed
come back to our hand no matter how long we wait! spring is released.
The frictional force has used up the kinetic energy in a non-reversible way. The forces, such as gravity
and spring force, which does work in a reversible manner is called a conservative force. In contrast,
the force such a frictional force, which does work in an irreversible manner is called a non-conservative
force. Any constant force (both in magnitude and direction) is always a CONSERVATIVE FORCE.
Important:
1. The work done by a conservative force is independent of path. It depends only on the initial and final positions.
In contrast, the work done by a non-conservative force depends on the path (see figure 3)
2. The work done by a conservative force around any closed path is zero (fig. 2).
The Potential energy is defined only for conservative forces.

The change in potential energy as a particle moves from point A to point B is equal to the negative of
the work done by the associated conservative force
U  U B  U A   WC
Using definition of work
z
B
U B  U A   FC . ds ... (1)
A
From equation (1), we see that starting with potential energy U A , we end up with potential energy. UB at
point B, because WC has the same value for all paths. When a block slides along a rough floor, the work
done by the force of friction on the block depends on the length of the path taken from point A to point
B. There is no unique value for the work done. So one cannot assign unique values for potential energy
at each point. Hence, non-conservative force can not have potential energy. When the forces within a
system are conservative, external work done on the system is stored as potential energy and is fully
recoverable.
Note : The potential energy is always defined with respect to a reference point.
Spring Potential Energy

The work done by the spring force when the displacement of the free end changes from xi to xf is given by
equation.
1
e
Ws   k x 2f  x i2
2
j
By definition d
Ws   U s   U f  U i , i
therefore, U f  Ui 
1
2
e
k x 2f  xi2 j US 
1 2
2
kx ...(2)
If we assume the potential energy stored in the spring at equilibrium
is zero and all the displacements are measured from equilibrium,
then xi = 0 and U i = 0.
Thus, final energy stored in the spring is U f 

1 2
2
kx d x f  x i
The potential energy function for an ideal spring is a parabolic
function as in (Fig. 4).
Example - 10
A uniform rod of mass M and length L is held vertically upright on
a horizontal surface as shown in the figure (5). Find the potential
energy of the rod if the zero potential energy level is assumed at
the horizontal surface.
Solution :
Since the parts of the rod are at different levels with respect to the horizontal
surface, therefore, we have to use the integration to find its potential
energy. Consider a small element of length dy at a height y from the
horizontal.
M
Mass of element is dm  dy
L
Its potential energy is given by
b g
dU  dm gy or dU 
M
L
gydy
On integrating, we get
z
L L
Mg Mg y 2 1
U ydy or U or U MgL
L L 2 2
0 0
FG L IJ
Note: The potential energy of the rod is equal to the product of Mg and height of the center of mass
H 2K
from the surface.
CONSERVATION OF MECHANICAL ENERGY

From the previous section we know that the work done by a conservative force in terms of the change in
potential energy is given by
U   WC ...(1)
Where U is the potential energy and WC is the work done by a conservative force. From the work-energy
theorem, we know that
Wnet  K
Where Wnet represents the sum of work done by all the forces acting on the mass.
If a particle is subjected to only conservative forces, then
WC  Wnet  K ...(2)
Thus, the equation (1) becomes,
U    or U  K  0 ... (3)
Important Concept: - The equation (3) tells us that the total change in potential energy plus the total
change in kinetic energy is zero if only conservative forces are acting on the system.
 (K+U) = 0 ... (4)
or E = 0
where E = K+U
The quantity E = K + U is called the total mechanical energy.
Example - 11
Two blocks with masses m1 = 3 kg and m2 = 5 kg are connected by a light
string that slides over a frictionless pulley as shown in figure. Initially,
m2 is held 5 m off the floor while m1 is on the floor. The system is then
m2
released. At what speed does m2 hit the floor?
Solution : m1 h
The initial and final configurations are shown in the figure (b & c). (a)
It is convenient to set U g = 0 at the floor. Initially, only m2 has potential energy. As it falls, it loses potential
energy and gains kinetic energy. At the same time, m1 gains potential energy and kinetic energy. Just before
m2 lands, it has only kinetic energy. Let v the final speed of each mass. Then, using the law of conservation
of mechanical energy.
K f  U f  Ki  U i
1
2
b g
m1  m2 v 2  m1gh  0  m2 gh
v2 
b g
2 m2  m1 gh
m1  m2
Putting m1 = 3 kg; m2 = 5 kg; h = 5 m and g = 10 m/s2
2
We get v 
b gb gb g
2 5  3 10 5
or v = 5 m/s
5 3
Example - 12
A chain of length l = 80 cm and mass m = 2 kg hanging from the
I0
end of a plane so that the length l0 of the vertical segment is
50 cm as shown in the figure. The other end of the chain is fixed
by a nail. At a certain instant, the nail is pushed out, what is the velocity of the chain at the
moment it completely slides off the plane? Neglect the friction.
Solution : y
We assume the zero potential energy level at the horizontal plane.
O x
The initial and final configuration of the chain are shown
I0
in the figure.
Initially, Ki = 0 Initial
FG m l IJ g FG  l IJ ml02
H l K H 2K
0
Ui  0  0 or Ui   g
2l
Note:- that the part of chain lying over the table has zero potential energy.
1
Finally, K f  mv 2
2 y
Where v is the final velocity of the chain. O x

l
and U f = -mg
2 I
Using the law of energy conservation
v
K f  U f  Ki  U i Final
ml02 g
1
2
l
mv 2  mg  0 
2 2l
or v
g 2
l
e
l  l02 j
putting l = 0.8 m; l 0 = 0.5 m; g = 10 m/s2 ,
we get v 5m / s or v = 2.23 m/s
Example - 13
Two blocks with masses m1 = 2 kg and m2 = 3 kg hang on either
side of a pulley as shown in figure. Block m1 is on an incline
m1
( = 300) and is attached to a spring whose stiffness constant k
m2
is 40 N/m. The system is released from rest with the spring in

its natural length. Find
(a) the maximum extension of the spring
(b) the speed of m 1 when the extension is 0.5m. Ignore friction and mass of the pulley.
[Take g = 9.8 m/s2]
Solution :
To use E f = E i we would need to assign the initial heights of the blocks arbitrary values h 1 and h 2 . The
corresponding potential energies, m 1 gh 1 and m 2 gh 2 would appear in both E i and E f and hence would
cancel. We avoid this process by using the form K + U = 0 instead, since it does not require
U = 0 reference level.
(a) At the maximum extension xmax , the blocks come to rest, and thus K = 0. Next, we must find the
changes in U g and U S. When m2 falls by xmax , the spring extends by xmax and m1 rises by xmax sin.
Therefore, K  U g   U S  0
b
0   m2 gxmax  m1gx max sin  g 1 2
2
kx max  0
Thus, xmax 
2g
k
b g
m2  m1 sin  0.98 m
(b) In this case the change in kinetic energy is K =

1
2
b
m1  m2 v 2 . g
The change in potential energy has the same form as in part (a), but with xmax replaced by x = 0.5 m.
1
2
b g b 1
m1  m2 v 2   m2 gx  m1gx sin  kx 2  0
2
g
Putting the given values we find v = 1.39 m/s.
A
Example - 14
A body slides down an inclined surface which ends into a
vertical loop of radius R = 40 cm. What must be the height
H
H of the inclined surface for the body not to fall at the R
uppermost point of the loop? Assume friction to be absent.
Solution :
Let v be the velocity of the particle at the highest point. According to Newton’s Second Law, the net force
Fnet  N  mg provides the centripetal force.
mv 2
Therefore, N  mg 
R
The body is not detached from the loop if N  0 . In the limiting case, N = 0.
mv 2
That is mg  or v 2  gR
R
Applying energy conservation at the initial and highest point of the loop, we get
b g 21 mv
mgH  mg 2 R  2
Using v 2 = gR,
mgH  mg b 2 Rg  mb gRg
1 5
we obtain, or H R  2.5 R
2 2
putting R = 40 cm = 0.4 m, we get H = (2.5)(0.4) = 1m.
Example - 15
The bob of a simple pendulum of length L = 2 m has a mass m = 2 kg and a speed v = 1 m/s
when the string is at 350 to the vertical. Find the tension in the string at
(a) the lowest point in its swing
(b) the highest point

Solution :
The problem requires to use dynamics and the conservation of mechanical
L
energy. The forces on the bob are shown in the figure. The acceleration
T
has both radial and tangential components. The equation for y L(1–cos)
tangential component, F x  mg sin = ma x

mg Ug = 0
Since the bob is moving in a circular path of radius L, the equation for the radial component is
mv 2
F y  T  mg cos =
L
.... (i)
To find the tension we need the speed, which can be found from the conservation law. We set U g = 0 at the
lowest point.
The mechanical energy is

1
E  mv 2  mgL 1  cos 
2
b g .... (ii)
1
2
b gb
2kg 1 m / s g  b 2kg gb9.8 N / kg gb 2mgb1  0.8g  9 J
2
1 2
(a) At the lowest point = 0; hence (ii) becomes E  mvmax 0
2
Since E = 9 J we find v max = 3 m/s.
2
mvmax
Now we have the speed we can find the tension at  = 0. From (i), T  mg  , from which we
L
find T = 20 + 9 = 29 N
(b) At the highest point v = 0; hence, (ii) becomes E = 0 + mgL(1 - cosmax )
31
Using E = 9 J lead to cosmax =
40
Since v = 0, therefore, T = mgcosmax = 15.5 N
Example - 16
A block of mass m = 0.2 kg is held against, but not attached to a
d
spring (k = 50 N/m) which is compressed by 20 cm, as shown x x=0
in figure, when the block is released, the block slides 50 cm
Us = 0
up the rough incline before coming to rest. Find Ug = 0
(a) the force of friction
(b) the speed of the block as it leaves the spring. 37°
Solution :
We could set U g = 0 at x = 0, but if we use the lowest point instead, all subsequent values are positive.
Again the total energy has three terms E = K + U g + U S.
1 2
(a) We s e t x 0 = 0 . 2 m a n d d = 0 . 5 m . Bo th K i a n d K f a re ze r o, s o Ei  kx0 and
2
E f = mgdsin. From work-energy theorem, E f  Ei  Wnc
1 2
mgd sin  kx 0   fd
2
Thus f = 0.82 N
1
(b) The initial energy Ei is the same as above, but the final value at x = 0 is E f  mv 2  mgx0 sin
2
From work-energy theorem, E f  Ei  Wnc
1 1
mv 2  mgx 0 sin   kx02   fx0
2 2
After solving we get, v = 2.45 m/s.
RELATION BETWEEN POTENTIAL ENERGY & FORCE

Potential energy U of a particle basically depends on its position, rather we can say its co-ordinates x,
y and z. Thus, potential energy of a particle is a function of its co-ordinates x, y and z i.e,
U = U (x, y, z)
Now suppose potential energy function is known to us, then the force (of course conservative) acting on it
is given by :
  U ˆ U ˆ U 
F   i j kˆ 
 x y z 
U
Here is the partial derivation of U with respect to x. On the other hand, if the force acting on the
x
particle is given to us and we want to find potential energy function, then we will use
 
dU   F . d r or z z
i
f
dU  
i
f 
F. d r
But while using this equation, at least the potential enegy of the particle at some reference point should be
known to us. For example, if we are given that U = 0 at r = 
Then z 
b x ,y ,z g
dU  
z 
b x ,y , z g 
F.d r

or b g bg
U x , y, z  U   
z
b x ,y , z g 
F.d r

or b
U x , y, z   g z 
b x ,y ,z g 
F. d r

as U () = 0
Similarly, if we are given that U = U 0 at origin. Then
z b x ,y ,z g
b 0,0,0g
dU  
z
b x ,y , z g  
b 0,0,0g
F.d r or b g b
U x, y , z  U 0,0,0   g z b x ,y ,z g  
b 0,0,0g
F. d r
or b g
U x , y, z  U 0  
z b x, y ,z g  
b 0,0,0g
F. d r or b g
U x , y, z  U 0 
z b x, y ,z g  
b0,0,0g
F. d r
CONSERVATIVE FORCES AND POTENTIAL ENERGY FUNCTIONS

We now consider how we can find a conservative force if we are given the associated potential energy function.
According to equation, an infinitesimal change in potential energy dU is related to work done by the conservative
force Fc in an infinitesimal displacement ds as follows
 
dU  FC . d s ...(6)
In one dimension, the above equation reduces to dU   Fx . dx
dU
Thus, Fx   ...(7)
dx
Let us see how equation (7) can be used for the common known cases:
For gravitational potential energy,
dU
U g  mgy Fy     mg
dy
For spring potential energy,
1 2 dU
US  kx Fx     kx
2 dx
A conservative force can be derived from a scalar potential energy function
V(r)
r1 r2
r0 r
O
F
r1
r0 r2 r
O
Fig. From the given potential energy function U(r) one can find the radial component of the force from Fr = -dU/dr,
which is the negative of the slope of the U(r) curve. A positive force means repulsion, and a negative force means
attraction.
Consider an arbitrary potential energy function U(r) as shown in figure (a).

The radial component of the associated conservative force is negative of the slope of the potential energy
dU
function, that is Fr  
dr
The force function may be plotted qualitatively as shown in the figure (b). If F r>0, the force is
directed toward positive r, which means repulsion, whereas F r<0 means attraction.
The following important points can be easily noticed by looking at the potential energy and force
diagrams.
(r > r2 ): Fr > 0 The particle is weakly repelled.
b g
r  r2 : Fr  0 . At the maximum point of potential energy function, the particle would be in unstable
equilibrium. If the particle were slightly displaced either to the left or to the right, it would tend to move
away from this point.
(r0 < r <r2 ): Fr < 0. The force is attractive, being strongest at r1 where the slope is greatest.
(r = r0): Fr = 0. At the minimum point of the potential energy function, the particle would be in stable equilibrium.
If slightly displaced in either direction, it would tend to return to this point.
(r < r0 ): Fr > 0. The particles repel each other. The repulsive force becomes stronger as r is reduced (since
the slope of U(r) gets steeper).
We have following conditions for equilibrium
(i) For stable equilibrium
dU d 2U
0 & is positive
dx dx 2
potential energy is minimum in this case
(ii) For unstable equilibrium

dU d 2U
0 & is negative
dx dx2
potential energy is maximum in this case
(iii) For Neutral equilibrium
dU d 2U
0 & 0
dx dx 2
potential energy is constant
Example - 17
The potential energy function for the force between two atoms in a diatomic molecule can be
a b
expressed approximately as U( r )  12

r r6
Where a and b are constant and r is the separation between the atoms.
(a) Determine the force function F(r).
(b) Find the value of r for which the molecule will be in the stable equilibrium.
Solution :
(a) The force between the two atoms is given by
dU 12a 6b
F( r )   or F( r )  
dr r 13 r7
d 2U
(b) For stable equilibrium F(r) = 0 and 0
dr 2
12a 6b FG 2a IJ 1/ 6
Thus, 
r 13

r7
 0 or r
HbK
Motion in a vertical circle
Suppose a particle of mass m is attached to an inextensible light string of length R. The particle is moving
in a vertical circle of radius R about a fixed point O. It is imparted a velocity u in horizontal direction at
lowest point A. Let v be its velocity at point B of the circle as shown in figure. Here
h = R(1 – cos ) ... (i)
From conservation of mechanical energy
1
2
e
m u 2  v 2  mghj
or v 2 = u2 – 2gh ... (ii)
The necessary centripetal force is provided by the resultant

of tension T and mg cos 
O
mv 2
T  mg cos   ... (iii)  v
R R T
h
B
e j
mg cos 
Condition of Looping the Loop u  5 gR A u mg cos 
The particle will complete the circle if the string does not slack even at the highest point ( = ). Thus,
tension in the string should be greater than or equal to zero (T  0) at  = . In critical case substituting
T = 0 and  =  in Eq. (iii), we get

2
mvmin 2
mg  or vmin  gR or vmin  gR (at highest point)
R
Substituting  =  in Eq. (i), h=2R
2 2
Therefore, from Eq. (ii) umin  vmin  2 gh
or, 2
umin b g
 gR  2 g 2 R  5 gR or umin  5 gR
Thus, if u  5 gR , the particle will complete the circle. At u  5gR , velocity at highest point is v  gR
and tension in the string is zero.
Substituting  = 00 and v  u  5 gR in Eq. (i) and (ii), we get T = 6 mg or in the critical condition
tension in the string at lowest position is 6 mg. This is shown in figure.
If u  5 gR , following two cases are possible
Condition of Leaving the Circle e 2 gR  u  5 gR j

If u  5 gR , the tension in the string will become zero before reaching the highest point. From Eq. (iii),
tension in the string becomes zero (T = 0)
 v2 2gh  u 2
where, cos  = or cos  =
Rg Rg
Substituting, this value of cos  in Eq. (i), we get
2 gh  u 2 u 2  Rg
Rg
 1
h
R
or h=
3g
 h1 say b g ... (iv)
or we can say that at height h1 tension in the string becomes zero. Further, if u  5 gR , velocity of the
particle becomes zero when
u2
2
0 = u – 2gh or h h2 (say) ... (v)
2g
i.e., at height h2 velocity of particle becomes zero.
Now, the particle will leave the circle if tension in the string becomes zero but velocity is not zero, or
T= 0 but v  0. This is possible only when h 1 < h 2
u 2  Rg u2
or  or 2u 2  2 Rg  3u 2
3g 2g
or u 2  2 Rg or u 2 Rg
Therefore, if 2 gR  u  5gR , the particle leaves the circle.
From Eq. (iv), we can see that h > R if u2 > 2gR. Thus, the particle will leave the circle when h > R or 900
<< 1800 . This situation is shown in the figure.
2 Rg  u  5gR or 900 <<1800
Note: That after leaving the circle, the particle will follow a parabolic path.
Condition of Oscillation e0  u  2 gR j
The particle will oscillate if velocity of the particle becomes zero but tension in the string is not zero or
v = 0, but T  0. This is possible when h2 < h1
u2 u 2  Rg
or  or 3u2 <2u2 +2Rg
2g 3g
or u2 <2Rg or u 2 Rg
Further, if h1 = h2 , u  2 Rg and tension and velocity both becomes zero simultaneously..
Further, from Eq. (iv), we can see that h  R if u  2 Rg . Thus, for 0  u  2 gR , particle oscillates in
0 0
lower half of the circle (0 < 90 ). This situation is shown in the figure.
0u 2 gR or 00 <900
Note: The above three conditions have been derived for a particle moving in a vertical circle attached to a
string. The same conditions apply if a particle moves inside a smooth spherical shell of radius R. The
only difference is that the tension is replaced by the normal reaction N.
Condition of Looping the Loop is
u 5 gR ;
v gR ;
N=0
Condition of Leaving the Circle
2 gR  u  5 gR
Condition of Oscillation is
V=0
N0
0u 2 gR
h R
u
Ver tical Circular Motion Connected by Rods and Tubes:
If a particle of mass m is connected to a light rod and whirled in a vertical circle of radius R, then to
complete the circle, the minimum velocity of the particle at the bottommost point is not 5gR . Because in
this case, Velocity of the particle at the topmost point can be zero also. Using conservation of mechanical
energy between point A and B as shown in Fig (i), We get
1
2
e j
m u 2  v 2  mgh or
1
2
b g
mu 2  mg 2 R
 u  2 gR
Therefore, the minimum value of u in this case is 2 gR .

Same is the case when a particle is compelled to move inside a smooth vertical tube as shown in Fig. (ii).
B
v= 0 T=0
O h  2R
R
R
u
(as v = 0) A
u  2 gR u  2 gR
d <<R
Fig. (i) Fig. (ii)
POWER
Power is defined as the rate at which work is done. If an amount of work W is done in a time interval t,
then the average power is defined to be
W
Pav 
t
The SI unit of power is J/s which is given the name watt (W) in the honour of James Watt.
Thus, 1 W = 1 J/s.
dW
The instantaneous power is the limiting value of Pav as t  0 ; that is P 
dt
  
The work done by force F on an object that has an infinitesimal displacement d s is dW = F . d s . Since the

ds 
velocity of the object is v  , the instantaneous power may be written as
dt

dW  d s  
P  F. or P  F. v
dt dt
Since the work and energy are closely related, a more general definition of power is the rate of energy
transfer from one body to another, or the rate at which energy is transformed from one form to another.
dE
P
dt
Example - 18
A particle of mass m is moving in a circular path of constant radius r such that its centripetal
acceleration ar is varying with time t as ar = k2rt2, where k is a constant. What is the power
delivered to the particle by the forces acting on it.
Solution :
v2
Let v be the instantaneous speed of the particle, then centripetal acceleration is given by ar 
r
Since a r = k 2 rt2 is given,
v2
therefore,  k 2rt 2 or v = krt v=0
r T0
dv O
The tangential acceleration is given by at   kr
dt
The tangential force is Ft = ma t = mkr
R P
Work done by centripetal force is always zero. hR
Hence, it doesn’t contribute to the power. A u
b gb g
Hence, power delivered is P  Ft v  mkr krt or P  mk 2r 2t
Example - 19
A small body of mass m is located on a horizontal plane at the point O. The body acquires a horizontal
velocity u. Find :
(a) the mean power developed by the friction force during the whole time of motion, if the friction coefficient
is k.
(b) The maximum instantaneous power developed by the friction force, if the friction coefficient varies as
k =  x, where  is a constant and x is the distance from the point O.
Solution :
(a) The frictional force acting on the body f r   mg
Retardation provided by this force is a   g
u u
Total time taken by the body to come to rest is t ; or 
a g
Total change in kinetic energy due to friction is
1
Mean power can be given as E  0  mu 2
2
1
 mu 2
Net gain in kinetic energy E 2 1
Mean Power   ; or   mu  g
Total time of motion t u 2
g
(b) When friction coefficient is k =  x, the friction force on the body when it is at a distance x from the
point O is
f r  xmg
Retardation due to this force is a  gx
dv
or v  gx ; or v dv  gx dx
dx
Integrating the above expression for velocity at a distance x from point O, gives
v x
 v dv    gx dx
u 0
; or v 2  u 2  gx 2
Instantaneous power due to friction force at a distance x from point O is
P=F.v ; or  mgx (u 2  gx 2 ) ... (i)
dp
This power is maximum when  0 , thus
dx
dp mgx
 2 2 1/ 2
 gx  mg[u 2  gx 2 ]1/ 2  0 ;
dx [u  gx ]
u
or x ... (ii)
2g
Equation (ii) gives the value of x at which instantaneous power is maximum. Using above value of x in
equation (i) gives the maximum instantaneous power as
u 1
u2  u 2
2
Pmax  mg ; Pmax   mu 2 g
2g 2
EXERCISE -1
1. A particle moves along the x-axis from x = 0 to x = 5m under the influence of a force given by
F = 7 - 2x + 3x2 N. The work done in the process is
(a) 107 J (b) 270 J (c) 100 J (d) 135 J
2. The total work done on a particle is equal to the change in its kinetic energy
(a) always
(b) only if the forces acting on the body are conservative
(c) only if the forces acting on the body are gravitational
(d) only if the forces acting on the body are elastic
3. A body of mass m is moving in a circle of radius r with a constant speed v. The force on the body is
mv2/r and is directed towards the centre. What is the work done by the force in moving the body half the
circumference of the circle :
mv 2 mv 2 r 2
(a)  r (b) zero (c) (d)
r r2 mv 2
4. A particle moves with a velocity v  (5iˆ  3jˆ  6k)

ˆ m/s under the influence of a constant force

F  (10iˆ  10ˆj  20k)N
ˆ . The instantaneous power applied to the particle is
(a) 200 W (b) 140 W (c) 40 W (d) 170 W
5. A chain is held on a frictionless table with one third of its length hanging over the edge. If the chain has a
length L and mass M, how much work is required to pull the hanging part back on the table
MgL MgL MgL
(a) MgL (b) (c) (d)
3 9 18
6. There is a smooth hemispherical bowl of radius R. A block of mass m slides from the rim of the bowl to
the bottom. The velocity of the block at the bottom will be
(a) Rg (b) 2Rg (c) 2Rg (d) Rg
7. A spring of force-constant k is cut into two pieces such that one piece is double the length of the other.
Then the longer piece will have a force-constant of
(a) (2/3)k (b) (3/2)k (c) 3k (d) 6k
8. An object of mass m is tied to a string of length L and a variable

horizontal force is applied on it which starts at zero and gradually
increases until the string makes an angle  with the vertical. 
L
This process is done slowly. Work done by the force F is
(a) mgL(1 - sin  ) (b) mgL
F
(c) mgL(1 - cos  ) (d) mgL(1 + cos  ) m
9. A body of mass m is projected at an angle with an initial velocity u. The mean power of gravity over the
whole time of journey is
1 1
(a) mg cos  (b) mg u cos  (c) mgu sin  (d) zero
2 2
10. A particle moves in a straight line with retardation proportional to its displacement. Its loss of kinetic
energy for any displacement ‘x’ is proportional to
(a) x (b) x2 (c) ln x (d) ex
11. Under the action of a force, a 2kg body moves such that its position x as a function of time is given by
t3
x , where x is in meter and t is in second. The work done by the force in the first two second is
3
(a) 1600 J (b) 160 J (c) 16 J (d) 1.6 J
12. If a simple pendulum of length l has maximum angular displacement  then the maximum kinetic energy
of its bob of mass m is
1 l 1 g 1  g 
(a) m  (b) m  (c) mgl(1 - cos  ) (d)   m 
2  g  2 l 2  l 
13. A person pulls a bucket of water from a well of depth h. If the mass of uniform rope is m and that of the
bucket full of water M, the work done by the person is
 m 1  m M 
(a)  M   gh (b) ( M  m) gh (c)  M   gh (d)   m  gh
 2 2  2  2 
14. A car starts from rest and moves on a surface in which the coefficient of friction between the road and the
tyres increases linearly with distance (x). The car moves with the maximum possible acceleration. The
kinetic energy (E) of the car will depends on x as
1 1
(a) E  (b) E  (c) E  x (d) E  x 2
x2 x
15. A cord is used to lower vertically a block of mass M a distance d at a constant downward acceleration
of g/4. Then the work done by the cord on the block is
Mgd Mgd 3Mgd  3Mgd
(a) (b)  (c) (d)
4 4 4 4
16. If the block in the figure is pulled by a constant force F which is always tangential O F B
to the surface then, the work done by this force between A and B is
F
F
(a) FR (b) 2 FR
A

(c) FR (d) FR
2
17. Power supplied to a particle of mass 2 kg varies with time as p = 3 t2/2 watt. Here t is in seconds.
If velocity of particle at t = 0 is v = 0, the velocity of particle at time t = 2s will be
(a) 1 m/s (b) 4 m/s (c) 2 m/s (d) 2 2 m/s
18. A ball is released from the top of a tower. The ratio of work done by force of gravity in first, second and
third second of the motion of ball is
(a) 1 : 2 : 3 (b) 1 : 4 : 16 (c) 1 : 3 : 5 (d) 1 : 9 : 25
19. The potential energy function for the force between two atoms of a diatomic molecule is given by
U(x) = a/x12 - b/x6, where a and b are positive constant and x is the distance between the atoms. The
minimum energy needed to break up the molecule is
(a) 2a2 / b (b) b / 2a2 (c) b2 / 4a (d) - b2 / 4a
20. A block of mass m, placed on a rough horizontal plane is pulled by a constant power P. The friction
coefficient between the block and the surface is  . Maximum velocity of the block will be
(a) P / mg (b) mg / P (c) mgP (d) P / mg
21. A body pulls a 5 kg block 20m along a horizontal surface at a constant speed with a force directed 45º
above the horizontal. If the coefficient of kinetic friction is 0.2, how much work does the boy do on the
block ? Take g = 9.8m/s2.
(a) 32.5 J (b) 113.5 J (c) 163.3 J (d) 25.00 J
22. The masses of 1g and 2g are moving with equal kinetic energies. The ratio of magnitude of their momenta
is
(a) 4 : 1 (b) 2 :1 (c) 1: 2 (d) 1 : 16
23. A simple pendulum has a string of length l and bob of mass m. When the bob is at its lowest position, it
is given the minimum horizontal speed necessary for it to move in a circular path about the point of
suspension. The tension in the string at the lowest position of the bob is
(a) 3 mg (b) 4 mg (c) 5 mg (d) 6 mg
24. A mass of 2 kg falls from a height of 40 cm on a spring of force constant 1960 N/m. The spring is
compressed by (g = 9.8m/s2)
(a) 10 cm (b) 0.4 cm (c) 0.01 cm (d) 0.04 cm

25. A force F  k ( yiˆ  xˆj ) , where k is a positive constant, acts on a particle moving in the xy plane.
Starting from the origin, the particle is taken along the positive x-axis to the point (a, 0), and then parallel
to the y-axis to the point (a, a). The total work done by the force on the particle is
(a) –2ka2 (b) 2ka2 (c) –ka2 (d) ka2
EXERCISE -2
Multiple Choice Questions with One or More Choices Correct
1. A particle is acted upon by a force of constant magnitude which is always perpendicular to the velocity
of the particle. The motion of the particle take place in a plane. It follows that
(a) the velocity of the particle is constant
(b) the acceleration of the particle is constant
(c) the kinetic energy of the particle is constant
(d) the particle moves in a circular path.
2. Two inclined frictionless tracks of different inclinations P A
Q
meet at A from where two block P and Q of different
masses are allowed to slide down from rest at the same
time, one on each track, as shown in figure.
(a) Both blocks will reach the bottom at the same time
(b) Block Q will reach the bottom earlier than block P θ1 θ2
(c) Both blocks reach the bottom with the same speed B C
(d) Block Q will reach the bottom with a higher speed than block P
3. Choose the correct statements from the following:
(a) When a conservative force does positive work on a body, its potential energy increases.
(b) When a body does work against friction, its kinetic energy decreases.
(c) The rate of change of total momentum of a many-particle system is proportional to the net external
force acting on the system.
(d) The rate of change of total momentum of many-particle system is proportional to the net internal
force acting on the system.
4. A body is subjected to a constant force F in newton given by
F  î  2 ĵ  3k̂
where î , ĵ and k̂ are unit vectors along x, y and z axes respectively. The work done by this force in
moving the body through a distance of
(a) 4m along the y-axis is 12 J. (b) 3m along the y-axis is 6 J.
(c) 4m along the z-axis and then 3m along the y-axis is 18 J.
(d) 4m along the z-axis and then 3m along the y-axis is (12) 2  (6) 2 J.
F(N)
5. Figure shows the force F acting on a body as a function of x. The work
done in moving the body
5
(a) from x = 0 to x = 1m is 2.5 J.

(b) from x = 1 m to x = 3 m is 10 J.
(c) from x = 0 to x = 4m is 15 J. 1 2 3 4 x(m)
(d) from x = 0 to x = 4 m is 12.5 J.
6. A block of mass 2kg, initially at rest on a horizontal floor, moves under the action of a horizontal force of
10N. The coefficient of friction between the block and the floor is 0.2. If g = 10 ms-2
(a) the work done by the applied force in 4 s is 240 J.
(b) the work done by the frictional force in 4 s is 96 J.
(c) the work done by the net force in 4 s is 336 J.
(d) the change in kinetic energy of the block in 4 s is 144 J.
7. Which of the following can be negative

(a) Kinetic energy (b) Potential energy (c) Mechanical energy (d) All of these
8. Which of the following may not be conserved for an isolated system
(a) Energy (b) Potential energy (c) Mechanical energy (d) Kinetic energy
9. A block of mass 2 kg is hanging over a smooth and light pulley through a light
string. The other end of the string is pulled by a constant force F = 40N. The
kinetic energy of the particle increases 40 J in a given interval of time. Then
(g = 10 m/s2)
(a) tension in the string is 40 N F

(b) displacement of the block in the given interval of time is 2m
(c) work done by gravity is - 20 J
2 kg
(b) work done by tension is 80 J
10. A particle moves in a straight line with constant acceleration under a constant force F .
Select the correct alternative (s)
(a) Power developed by this force varies linearly with time
(b) Power developed by this force varies parabolically with time
(c) Power developed by this force varies linearly with displacement
(d) Power developed by this force varies parabolically with displacement
11. One end of a light spring of force constant k is fixed to a wall and other end is tied to a block placed on
1 2
a smooth horizontal surface. For displacement x, the work done by the spring is kx .
2
The possible case (s) may be
(a) the spring was initially stretched by a distance x and finally was in its natural length
(b) the spring was initially in its natural length and finally it was compressed by a distance x
(c) the spring was initially compressed by a distance x and finally was in its natural length
(d) the spring was initially in its natural length and finally stretched by a distance x
12. In projectile motion, power of gravitational force
(a) is constant throughout
(b) is negative for first half, zero at topmost point and positive for rest half
(c) varies linearly with time (d) is positive for complete path
13. The potential energy U of a particle of mass 1 kg moving in x-y plane obeys the law U = 3x + 4y, where
(x, y) are the co-ordinates of the particles in metre. If the particle is at rest at (6, 4) at time t = 0 then
(a) the particle has constant acceleration
(b) the particle has zero acceleration
(c) the speed of particle when its crosses the y-axis 10 m/s
(d) co-ordinate of particle at t = 1s are (4.5, 2)
14. A block is suspended by an ideal spring of force constant k. If the block is pulled down by applying a
constant force F and if the maximum displacement of the block from its initial mean position of rest is x0
then :
3F
(a) increase in energy stored in spring is k x 02 (b) x 0 
2K
2F
(c) x 0  (d) work done by applied force F is Fx0
k
15. A smooth track in the form of a quarter circle of radius 6 m lies in a
vertical plane. A particle moves from P1 to P2 under the action of force
    6m P2
F1 , F2 and F3 . Force F1 is always towards P2 and is always 20N in
 
magnitude. Force F2 always acts tangentially and is always 15N in F2
 6m 
magnitude. Force F3 always acts horizontally and is of magnitude 30N. F1 
F3
Select the correct alternative (s)
P1
 
(a) Work done by F1 is 120 J (b) Work done by F2 is 45 
 
(c) Work done by F3 is 180 J (d) F1 is conservative in nature
16. A small spherical ball is suspended through a string of length l. The whole arrangement is placed in a
vehicle which is moving with velocity v. Now suddenly the vehicle stops and balls starts moving along
circular path. If tension in the string at the highest point is twice the weight of the ball then
(a) v  5 g l (b) v  7 g l
(c) velocity of the ball at highest point is g l (d) velocity of the ball at the highest point is 3 g l

17. A particle is acted upon by a conservative force F  (6î - 6ˆj)N (no other force is acting on the particle).
Under the influence of this force particle moves from (0, 0) to (–3m, 4m) then :
(a) work done by the force is 3 J (b) work done by the force is –42 J
(c) at (0, 0) speed of the particle must be zero (d) at (0, 0) speed of the particle must not be zero
18. Power of a non-zero force acting on block varies with P(watts)

time t as shown in figure. Then angle between force
acting on the block and its velocity is 10
(a) acute at t = 1s
(b) 90º at t = 3s 6 8 10
t(s)
(c) obtuse at t = 7s 2 4
(d) change in kinetic energy from t = 0 to t = 10s is 20 J
-10
A B
19. U (r )   : Where r is the distance from the centre of the force and A and B are positive constants.
r2 r
2A
(a) The equlibrium distance is given by
B
B2
(b) The work required to move the particle from equilibrium distance to infinity is
4A
B
(c) The work required to move the particle from equilibrium distance to infinity is
4A
(d) Only (A)
20. In the figure shown upper block is given a velocity of 6m/s and Rough
lower block 3 m/s. When relative motion between them is 1kg 6m/s
stopped. 2kg 3m/s
Smooth
(a) Work done by friction on upper block is negative
(b) Work done by friction on both blocks is positive
(c) Work done by friction on upper block is –10J
(d) Net work done by friction is zero.
s C
21. Displacement time graph of a particle moving in a straight line is as B
shown in figure. select the current alternative(s):
(a) Work done by all the forces in region OA and BC is positive A
(b) Work done by all the forces in region AB is zero
(c) Work done by all the forces in region BC is negative t
O
(d) Work done by all the forces in region OA is negative.
22. A small block of mass m is released from rest from position A inside a smooth A R
hemispherical bowl of radius R as shown in figure. Choose the wrong option :
(a) acceleration of block is constant throughout (b) acceleration of block is g at A
(c) acceleration of block is 3g at B (d) acceleration of block is 2g at B. B
23. A force F = – kx3 is acting on a block moving along x-axis. Here, k is a positive constant. Work done by
this force is :
(a) positive in displacing block from x = 3 to x = 1
(b) positive in displacing block from x = –1 to x =3
(c) negative in displacing block from x = 3 to x = 1
(d) negative in displacing block from x = –1 to x =3.
EXERCISE -3
C
1. Match the following:
Table 1 Table 2 B
(A) Work done by all the forces (p) Change in potential energy
(B) Negative of work done by (q) Change in kinetic energy
internal conservative forces (r) Change in mechanical energy A
(C) Work done by external forces (s) None
+non conservative internal forces
2. A particle is suspended from a string of length R. It is given a velocity u  3 gR at the bottom.

Match the following:
Table 1 Table 2
(A) Velocity at B (p) 7mg
(B) Velocity at C (q) 5gR
(C) Tension in string at B (r) 7gR

(D) Tension in string at C (s) 5mg
(t) None
3. A force F = kx (where k is a positive constant) is acting on a particle. Work done:
Table 1 Table 2
(A) In displacing the body from (P) Negative
x = 2 to x = 4
(B) In displacing the body from (Q) Positive
x = -4 to x = -2
(C) In displacing the body from (R) Zero
x = -2 to x = +2
4. A block of mass m is stationary with respect to a rough wedge as shown in figure. Starting from
rest in time t, m  1kg, θ  30º , a  2m/s 2 , t  4s  work done on block:
Table 1 Table 2
(A) By gravity (p) 144 J
(B) By normal reaction (q) 32 J
(C) By friction (r) 56 J a
(D) By all forces (s) 48 J
(t) None

Comprehension - 1
Two physicists, both of mass 50kg, climb up identical ropes suspended form the ceiling of a gymnasium.
The ropes are 15m long. Physicist 1 reaches the top twice as quickly as physicist 2 does. After
physicist 2 also reaches the top, they argue about who did more work against gravity:
Physicist 1:-
“I did more work fighting gravity, because I was overcoming gravity more quickly. Your climb was
lazier, and therefore, you did less work.”
Physicist 2:-
“No way. I did more work fighting gravity, because I spend more time climbing the rope. Since we
both ended up at the same height, but I spent more time getting there, I had to work harder.”
5. Which physicist, if either, did more work against gravity while climbing from the floor to the ceiling?
(a) physicist 1 did more work (b) physicist 2 did more work
(c) Both physicist did the same work (d) neither physicist did any work
6. Physicist 1 climbed her rope in 30s. What average power did she exert fighting gravity?
(a) 25W (b) 250W (c) 1,500W (d) 15,000W
7. Physicist 2 started at rest from the floor and ended at rest near the ceiling. Which of the following best
expresses the net energy transfer during this process?
(a) chemical to kinetic (b) chemical to potential
(c) kinetic to chemical (d) kinetic to potential
8. Physicist 2 now lets go of the rope and falls onto a heavily-padded cushion, safely coming to rest. During
this process, the energy transfer is best described as:
(a) potential to kinetic to chemical (b) potential to kinetic to heat
(c) kinetic to potential to chemical (d) kinetic to potential to heat
9. When physicist 2 has fallen one third of the way from the ceiling to the floor, her kinetic energy is
approximately:
(a) 10 J (b) 250 J (c) 1000 J (d) 2500 J
Comprehension - 2
B
A ball tied to a string of length R is given a speed V, when its
hanging vertically. At some point B during its circular motion, R

the string gets slacked & after that the body passes through
its initial point.
10. Find angle  .
A V
(a) 30o (b) 60o (c) 45o (d) 270o
11. Find v.
43 3 2 2 3 7 gR
(a) gR (b) gR (c) (d) 5gR
2 2 2
12. Find speed at point B.
gR 3 gR gR
(a) (b) (c) (d) gR
2 2 2
Comprehension - 3
B
A
A chain of length  and mass m is in position as shown in the figure.
All surfaces are frictionless. h
13. Find speed of end A when it is at point B.
     2h h 
(a) 2 g  h    (h  ) ln    (b) gh  1  
  h     
  
(c) 2 gh ln   (d) 3gh 1  
h  h
14. Find max kinetic energy of chain before end A reaches point B.
mgh 2mgh mgh
(a) (b) (c) (d) e mgh
e e 2e
15. Find energy lost as heat before end A reaches B.
mgh     mgh    
(a)    h  h ln    (b)    h  h ln   
   h     h 
mgh    
(c) 0 (d)    h  h ln   
   h 
Comprehension - 4
A single conservative force acts on a 1kg particle that moves along x  axis. Potential energy U(x) is
given by U(x) = 20 + (x – 3)2, where x is in m. At x = 0, particle has kinetic energy of 20J.
16. What is the mechanical energy of system
(a) 20J (b) 49 J (c) 100 J (d) 75 J
17. Find the greatest & least value of x between which particle can move  29  5.4 
(a) 8.4m & –2.4m (b) 8.4 m & 4.4 m (c) 7m & 3m (d) 8m & 4m
18. Find maximum kinetic energy of particle & value of x at which it occurs.
(a) 20J & x = 0m (b) 29J & x = 3m (c) 49J & x = 1m (d) 29J & x = 0m
19. Find value of x at which body is in equilibrium

(a) 0m (b) –2m (c) 3m (d) 1m
Comprehension - 5
The work done by all forces on a body is equal to change of kinetic energy of the body. This is true for
both constant and variable force (variable in both magnitude and direction). For a particle W =  K. For
a system, Wint + Wext =  Kcm or Wext + Wnonconservative =  K+  U
In the absence of external and nonconservative forces, total mechanical energy of system remains
conserved.
20. I - Work done in raising a box on a platform depends on how fast it is raised
II - Work done by force depends on the frame of reference
(a) I - False II - True (b) I - False II - False
(c) I - True II - False (d) I - True II - True.
21. The total energy of a system is
(a) always conserved in presence of external forces.
(b) always conserved in absence of internal forces.
(c) always conserved in presence of no external force and no internal non conservative force
(d) none of the above.
22. Work done in motion of a body over a closed loop is
(a) always zero for any force (b) always zero for conservative forces
(c) only (a) (d) None of the above.
Comprehension - 6
    
Work done by a constant force acting on a particle is defined as W  F.r where  r  rf  ri , rf is the final

position of particle and ri is the initial position of particle. If the force is variable work done is defined as


rf
W   F.dr , where  is the force at any general position  between r and r , and  is representing
 F r i f dr
ri

infinitesimal small displacement of particle. dr  dxiˆ  dyjˆ  dzkˆ .
23. A particle is thrown from a tower of height h at an angle  above horizontal. Work done by gravity during
its time of flight is (mass of particle is m)
(a) 0 (b) mgh cos  (c) mgh sin  (d) mgh.

24. Given: F  2iˆ  4ˆj, ri  2iˆ  3jˆ and rf  3iˆ  4ˆj Work done is (use SI units)
(a) 4 J (b) 6 J (c) 10 J (d) zero.
25. A particle thrown from ground with a velocity v at an angle  above horizontal. Work done by gravity in
2vsin 
t equals
g
mgv 2 sin 2  mgv 2 sin 2  2mgv 2 sin 2 

(a) 2g
(b) g
(c) zero (d) g
EXERCISE -4
SUBJECTIVE QUESTIONS F
1. A block of mass m=2 kg is pulled by a force F = 40 N upwards through a height m

h = 2 m. Find the work done on the block by the applied force F and its weight
mg. (g = 10 m/s2).
2. Two unequal masses of 1 kg and 2 kg are attached at the two ends of a light inextensible
string passing over a smooth pulley as shown in figure. If the system is released from rest,
find the work done by string on the left block in 1s. Take g = 10 m/s 2. 1kg
2kg
3. A force F = (2+x) acts on a particle in x-direction where F is in newton and x in metre. Find the
work done by this force during a displacement from x = 1.0 m to x=2.0 m.
F
4. A 1.8 kg block is moved at constant speed over a surface for which coefficient 45°
1
of friction   . It is pulled by a force F acting at 45 0 with horizontal as
4
shown in figure. The block is displaced by 2 m. Find the work done on
the block by
(a) the force F (b) friction (c) gravity. F(N)
10
–4 –2
5. Force acting on a particle varies with displacement as shown in figure.
2 4 x(m)
Find the work done by this force on the particle from –10
x = – 4 m to x = +4 m.
6. An object of mass m is tied to a string of length I and a variable horizontal force q
is applied on which moves the bob slowly until the string makes an angle  with
the vertical. Find the work done by the force F. F
m
7. A body of mass ‘m’ was slowly hauled up the hill as shown in the figure m F h
by a force F which at each point was directed along a tangent to the

trajectory. Find the work performed by this force, if the height of the
l
hill is h, the length of its base is l and the coefficient of friction is .
8. Velocity-time graph of a particle of mass 2 kg moving in a straight line is as v(m/s)
20
shown in figure. Find the work done by all the forces acting on the particle.
t(s)
2
9. A particle of mass m moves on a straight line with its velocity varying with the distance travelled according to
the equation v   x , where  is a constant. Find the total work done by all the forces during a displacement
from x = 0 to x = b.
10. Consider the situation shown in figure. Mass of block A is m and that of block B is 2m. The force constant of
spring is K. Friction is absent everywhere. System is released from rest with the spring unstretched. Find
(a) the maximum extension of the spring xm

A
xm
(b) the speed of block A when the extension in the spring is x 
2
xm
(c) net acceleration of block B when extension in the spring is x 
4 B
A
11. In the figure block A is released from rest when the spring is in
its natural length. For the block B of mass m to leave a contact with
the ground at some stage what should be the minimum mass of block A? B m
12. Two blocks A and B each having mass of 0.32 kg are connected
by a light string passing over a smooth
S
pulley as shown in Fig.: The horizontal
surface on which the block A slides is smooth. The block A is
A m
attached to a spring of force constant 40 N/m whose other end
is fixed to a support 0.40 m above the horizontal surface. Initially,
when the system is released to move, the spring is vertical and m B
unstretched. Find the velocity of the block A at the instant it breaks
off the surface below it. [Take g=10 m/s 2.]
13. A block of mass m slides on a frictionless table. It is constrained to move

inside a ring of radius R which is fixed to the table. At t = 0, the block is moving 0
along the inner circumference of the ring with velocity v0. The coefficient of friction
R
between the block and the ring is . Find the speed and position of the block as a
function of t.
14. The two small spheres, each of mass m, are connected by a cord of length 2b (measured from the centres of
the spheres ) and are intially at rest on a smooth horizontal surface in the position shown. If a vertical force of
constant magnitude F is applied to the centre A of the cord, determine the velocity v of each sphere when
they collide as  approaches 900. What is the maximum value of F for which the spheres do not lose contact
with the surface ?
15. A string with one end fixed on a rigid wall, passing over a fixed
1m C 1m B
frictionless pulley at a distance of 2 m from the wall, has a point
M
mass M of 2 kg attached to it at a distance of 1 m from the wall.
A
A mass m of 0.5 kg is attached to the free end. The system is
initially held at rest so that the string is horizontal between the wall and
pulley and vertical beyond the pulley as shown in figure. m D
What will the be the speed with which the point mass M will hit the wall
when the system is released ?
(g = 10 m/s 2 )
e j e j
16. An ob je ct i s di spla c e d from positi on v e c tor r1  2 i  3 j m to r2  4i  6 j m unde r a forc e
e j
F  3 x 2 i  2 yj N . Find the work done by this force.
17. One of the forces acting on a certain particle depends on the particle’s
position in the xy-plane. This force F expressed in newtons, is given by
e je j
the expression F  xyi  xyj 1 N / m where x and y are in metres.
2
Find the work done by this force when the particle is moved from
O to C through three different paths:
(i) OC (ii) OAC (iii) OBC
Is F a conservative force.
18. A uniform chain of length l and mass m is placed in three positions as shown in figure. Find the
gravitational potential energy of the chain with reference to the ground in all the three cases.
19. A block A of mass m is held at rest on a smooth horizontal floor.

A light frictionless, small pulley is fixed at a height of 6 m from the
floor. A light inextensible string of length 16m connected with
A passes over the pulley and another identical block B is hung
from the string. Initial height of B is 5m from the floor as shown
in Fig. When the system is released from rest, B starts to move
vertically downwards and A slides on the floor towards right.
(a) If at an instant string makes an angle  with horizontal, calculate relation

between velocity u of A and v of B
(b) Calculate v when B strikes the floor.
(g = 10 m/s 2)
20. One end of a light spring of natural length d and spring constant k is fixed
on a rigid wall and the other is fixed to a smooth ring of mass m which can
slide without friction on a vertical rod fixed at a distance d from the wall.
Initially the spring makes an angle of 370 with the horizontal as shown
in Fig. When the system is released from rest, find the speed of the ring
when the spring becomes horizontal.
[sin 370 = 3/5]
21. A small body of mass m is allowed to slide on an inclined

A
frictionless track from rest position as shown in the figure. What
must be the minimum height, so tha t the b ody ma y H
successfully complete the loop of radius ‘r’. If h is double
h
of that minimum height, find the resultant force on the block at r
position H.
22. A heavy particle hanging from a fixed point by a light inextensible string of length l is projected horizontally
with speed gl . Find the speed of the particle and the inclination of the string with the vertical at the
instant when the tension in the string is equal to the weight of the particle.
23. A bob is suspended from a crane by a cable of lengthl = 5m. The crane
and load are moving at a constant speed v0. The crane is stopped by a
bumper and the bob on the cable swings out an angle of 600.
Find the initial speed V0.(g = 9.8 m/s 2)
24. A small body slides down an inclined surface passing into a loop from the
minimum height ensuring that the body does not leave the surface of the
loop as shown in the figure. What symmetrical segment with an angle  A B
< 900 can be cut out of the loop for the body to reach point B after  
travelling a certain distance in the air ?
25. A person rolls a small ball with speed u along the floor from point
A. If x = 3R, determine the required speed u so that the ball
returns to A after rolling on the circular surface in the vertical plane
from B to C and becoming a projectile after C. What is the minimum
value of x for which the game could be played if contact must be
maintained at point C. Neglect friction.
26. A smooth semicircular thin tube AB of radius r is fixed in a vertical plane

and contains a heavy flexible uniform chain as shown. Assuming a slight
disturbance to start the chain in motion, find the velocity v with which it
will emerge from the open end B
27. A skier plans to ski a smooth fixed hemisphere of radius R. He starts from
rest from a curved smooth surface of height (R/4). The angle  at which
he leaves the hemisphere is?
28. A loop of light inextensible string passes over smooth small pulleys
A and B. Two masses m and M are attached to the points O and
C respectively. Then the condition between m, M and  so that m and
M will cross each other will be?
[Take AB = 2L and AC = AB = L]
29. A small particle of mass m is lying in the

bottom of a smooth trench of radius R as shown. F
It is being pulled from above with the help of a

massless string passing over a smooth pulley and a
constant force is being applied at the other end.
If the particle does not leave contact with the curved
surface, what will be its velocity when it hits the pulley.
30. An object of mass m is thrown thrown with a velocity V0 toward ax

V0 k
a spring. The coefficient of friction varies with displacement as
shown. Find the value of V0 such that it will compress the
spring by a distance L x=0
L
31. a horizontal plane supports a plank with a bar of mass 1kg
placed on it and attached by a light elastic non deformed o
cord of length 40 cm to a point O as shown in fig.

The coefficient of friction between the bar and plank is 0.2.
The plank is slowly shifted to right until the bar starts sliding
over it. It occurs at the moment when the cord deviates from F
o
the vertical by 30 . Find the work that has been done till that
moment by the friction acting on the bar in the reference
frame fixed to the plane.
32. In the figure shown a smooth pipe is bent to form a circle

and is kept on floor with its axis horizontal. Two identical
 
particles are released from the top and they start sliding down
a) At what angular displacement of each particle does the normal
reaction between the pipe and the floor becomes minimum?
b) what is the least possible ratio of each particle to the mass
of pipe so that the pipe loses contact with the floor at some
point of time during the motion of the particles
EXERCISE -5
Previous Years’ JEE Problems
1. A particle of mass m is moving in a circular path of constant radius r such that its centripetal
acceleration ac is varying with time t as ac = k2rt2, where k is a constant. The power delivered
to the particle by the force acting on it is (IIT-JEE 1994)
(mk 4r 2 t 5 )
(a) 2mk 2 r 2 (b) mk 2 r 2 t (c) (d) Zero
3
2. A spring of force constant k is cut into two pieces such that one piece is double the length of
the other. Then the long piece will have a force constant of (IIT-JEE 1999)
(a) (2/3)k (b) (3/2)k (c) 3k (d) 6k
3. A particle, which is constrained to move along x-axis, is subjected to a force in the same
direction which varies with the distance x of the particle from the origin as F(x) = –kx + ax3.
Here, k and a are positive constants. For x  0, the functional form of the potential energy U(x)
of the particle is (IIT-JEE 2002)
(a) (b)
(c) (d)
4. An ideal spring with spring constant k is hung from the ceiling and a block of mass M is
attached to its lower end. The mass is released with the spring intially unstretched. Then the
maximum extension in the spring is (IIT-JEE 2002)
(a) 4Mg/k (b) 2Mg/k (c) Mg/k (d) Mg/2k
5. If W1, W2 and W3 represent the work done in moving a particle

from A to B along three different paths 1, 2 and 3 respectively
(as shown) in the gravitational field of a point mass m. Find the
correct relation between W1, W2 and W3.(IIT-JEE 2003)
(a) W1 > W2 > W3 (b) W1 = W2 = W3
(c) W1 < W2 < W3 (d) W2 > W1 > W3
6. A particle is placed at the origin and a force F = kx is acting on it (where k is a positive

constant). If U(0) = 0, the graph of U(x) versus x will be (where U is the potential energy
function) (IIT-JEE 2004)
(a) (b)
(c) (d)
7. A bob of mass M is suspended by a massless string of

length L. The horizontal velocity v at position A is just
sufficient to make it reach the point B.
The angle  at which the speed of the bob is half of that
at A, satisfies (IIT-JEE 2008)
  
(a)   (b) 
4 4 2
 3 3
(c)  (d) 
2 4 4
8. A block (B) is attached to two unstretched springs S1 and S2 with

spring constants k and 4k, respectively. The other ends
are attached to two supports M1 and M2 not attached to the
walls. The springs and supports have negligible mass. There is no
friction anywhere. The block B is displaced towards wall 1 by a
small distance x and released. The block returns and moves a
maximum distance y towards wall 2.
Displacements x and y are measured with respect to the
equilibrium position of the block B. The ration y/x is
(IIT-JEE 2008)
(a) 4 (b) 2 (c) 1/2 (d) 1/4
9. A piece of wire is bent in the shape of a parabola y = kx2 (y-axis vertical) with a bead of mass m on it. The bead
can slide on the wire without friction. It stays at the lowest point of the parabola when the wire is at rest. The
wire is now accelerated parallel to the x-axis with a constant acceleration a. The distance of the new equilibrium
position of the bead, where the bead can stay at rest with respect to the wire, from the y-axis is (IIT-JEE 2009)
a a 2a a
(a) (b) (c) (d)
gk 2gk gk 4gk
10. A string, with end fixed on a rigid wall, passing over a fixed frictionless
pulley at a distance of 2m from the wall, has a point mass M = 2 kg
attached to it at a distance of 1 m from the wall. A mass m = 0.5
kg attached at the free end is held at rest so that the string is
horizontal between the wall and the pulley and the vertical beyond
the pulley. What will be the speed with which the mass M will hit the
wall when the mass m is released ? (Take g = 9.8 m/s2)
(IIT-JEE 1985)
11. A particle is suspended vertically from a point O by an inextensible massless
string of length L. A vertical line AB is at a distance L/8 from O as shown in
figure. The object is given a horizontal velocity u. At some point, its motion
ceases to be circular and eventually the object passes through the line AB.
At the instant of crossing AB, its velocity is horizontal. Find u.
(IIT-JEE 1999)
12. A spherical ball of mass m is kept at the highest point in the space
between two fixed, concentric spheres A and B (as shown in figure). The
smaller sphere A has a radius R and the space between the two spheres
has a width d. The ball has a diameter very slightly less than d.
All surfaces are frictionless. The ball is given a gentle push (towards
the right in the figure). The angle made by the radius vector of the
ball with the upward vertical is denoted by .
(a) Express the total normal reaction force exerted by the spheres on the ball as a function of angle .
(b) Let NA and NB denote the magnitudes of the normal reaction forces on the ball exerted by the spheres A and
B, respectively. Sketch the variations of NA and NB as function of cos in the range 0     by drawing two
separate graphs in your answer book, taking cos on the horizontal axis. (IIT-JEE 2002)
13. A light inextensible string that goes over a smooth fixed pulley as
shown in the figure connects two blocks of masses 0.36 kg and 0.72
kg. Taking g = 10 ms–2, find the work done (in Joule) by string on the
block of mass 0.36 kg during the first second after the system is
released from rest. (IIT-JEE 2009)
14. A block of mass 2 kg is free to move along the x - axis. It is at rest and from t = 0 onwards it is subjected to a time
- dependent force F (t) in the x - direction. The force F (t) varies with t as shown in the figure. The kinetic energy
of the block after 4.5 second is: (IIT - JEE 2010)
F (t)
4N
4.5 s
t
O 3s
(a) 4.5 J (B) 7.5 J (c) 5.06 J (d) 14.06 J
15. A block of mass 0.18 kg is attached to a spring of force constant 2 N/m. The coefficient of friction between the
block and the floor is 0.1. Initially the block is at rest and the spring is unstreched. An impulse is given to the
block as shown in the figure. The block slides a distance of 0.06 m and comes to rest for the first time. The initial
N
velocity of the block in m/s is   . Then N is (IIT - JEE 2011)
10
ANSWER KEY
CIRCULAR MOTION
EXERCISE 1
1. (d) 2. (c) 3. (a) 4. (b) 5. (a) 6. (c) 7. (d)
8. (c) 9. (c) 10. (c) 11. (b) 12. (c) 13. (a) 14. (a)
15. (a)
EXERCISE 2
1. (a, c) 2. (a, b) 3. (b, c) 4. (a, b)
5. (a, b, c) 6. (a, b, d) 7. (b, d)
EXERCISE 3
1. A-r, B-p; C-s; D-q 2. (b) 3. (a) 4. (b)
5. (b) 6. (a) 7. (a) 8. (c)
9. (c) 10. (b)
EXERCISE 4
R 2g 2  a 2
1. 2. cot  = 3 3. 4.7s–1
2a
R 5R 11 17v 289v
5. 1.5 sec 6. (1  e 2 ) 7. , , ,
V0 6v 6 5R 25R
EXERCISE 5
1. (c, d) 2. (a) 3. (a) 4. (c)
5 3 3mg
5. (i) (ii) aT = g, N =
8 8
WORK, POWER ENERGY

EXERCISE 1
1. (d) 2. (a) 3. (b) 4. (a) 5. (d)
6. (b) 7. (b) 8. (c) 9. (d) 10. (b)
11. (c) 12. (c) 13. (a) 14. (d) 15. (d)
16. (c) 17. (c) 18. (c) 19. (c) 20. (d)
21. (c) 22. (c) 23. (d) 24. (a) 25. (d)
EXERCISE 2
1. (c, d) 2. (b, c) 3. (b, c) 4. (b, c)
5. (a, b, c) 6. (a, d) 7. (b, c) 8. (b, c, d)
9. (a, b, d) 10. (a, d) 11. (a, c) 12. (b, c)
13. (a, c, d) 14. (c, d) 15. (b, c, d) 16. (b, d)
17. (b, d) 18. (a, c, d) 19. (a, b) 20. (a, c)
21. (b, c) 22. (a, c) 23. (a, d)
EXERCISE 3
1. (A - q), (B - s), (C - r) 2. (A - r), (B - q), (C - p), (D - t
3. (A - q), (B - p), (C - r) 4. (A - t), (B - p), (C - s), (D - q)
5. (c) 6. (b) 7. (b) 8. (b) 9. (d)
10. (a) 11. (c) 12. (a) 13. (c) 14. (a)
15. (b) 16. (b) 17. (a) 18. (b) 19. (c)
20. (a) 21. (c) 22. (b) 23. (d) 24. (b)
25. (c)
EXERCISE 4
200
1. 80J, –40J 2. J 3. 3.5 J
9
4. (a) 7.2 J (b) -7.2 J (c) zero
5. 30 J
1
6. mgl (1  cos ) 7. mgh   mgl 8. – 400 J 9. m 2b
2
4mg m g
10. (a) xm  ; (b) v  2 g 3 K ; (c) a  (downwards)
K 3
m 1 1  R
11. 12. 1.54 m/s 13.   t ; s  In(1  v0t / R )
2 v v0 R 
Fb
14. v (1  sin ) , F  2mg 15. 3.29 ms–1
m
2 1 1
16. 83 J 17. J , J , J , No
3 2 2
mgl m 2 l 
18. (a) mgh (b) (c) R g sin  
2 l R
3g k 5
19. u  v sec , ( 40 / 41) m / s 20. vd  21. hmin  r ; F  6mg
2d 16m 2
gl
22. cos 1 (2 / 3) 23. 7 m/s 24. 0° and 60°
3
5 2 
25. gr , 2 R 26. 2 gr (  ) 27. cos 1 (5 / 6)
2  2
m  1 F k
28. 2 1 29. 2 R( 2  g) 30. L  4ag
M  3 m m
31. .09 J 32. cos 1 (1/ 3) , 3/2
EXERCISE 5
1. (b) 2. (b) 3. (d) 4. (b) 5. (b)

6. (a) 7. (d) 8. (c) 9. (b)
 3 3
10. v = 3.29 m/s 11. u  gL  2  
 2 
13. 8J 14. (c) 15. (4)

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CIRCULAR MOTION & WPE Rg.

IIT –ian’s PACE

SR. NO. TITLE PAGE NO.

1 THEORY (CIRCULAR MOTION) 1 – 17

7 THEORY (WORK, POWER & ENERGY) 26 – 49

(b) Angular Displacement

(i) Average Angular Velocity P

(ii) Instantaneous Angular Velocity

(d) Angular Acceleration 

Consider a particle moving in a circle. Suppose the particle is at point 

VELOCITY AND ACCELERATION IN CIRCULAR MOTION

 r [ ˆi sin   jcos ] … (ii)

 Centripetal acceleration = 2 r  (6)2  2  722 m / s 2

NON-UNIFORM CIRCULAR MOTION

A particle moves with constant speed on a circular path shown

a) Clockwise : Fourth quadrant

b) Anticlockwise: Second quadrant 45°

 Total acceleration  42  22  2 5 m / sec2

DYNAMICS OF CIRCULAR MOTION

can be varied upto a maximum limit (fL = N ).

iii) Friction f is outwards if v  rg tan  and

iv) Friction f is zero if v  rg tan 

Figure (a) Figure (b)

F V 0  N cos   f sin   mg … (ii)

F = Force perpendicular to velocity (centripetal force)

(i) weight mg acting vertically downwards.

Solving (3) and (4), we get

 v = [9.8 (9.8  102)]1/2 = 0.98 m/s.

the adjoining figures. From these figures T Tcos 

Figure (a) Figure (b) Figure (c)

Equilibrium of m2 gives T = m 2 r2  2 … (1)

Since, m 2 r2  2 < m1r12 (m2r2 < m1r1)

(m1r1  m2r2)  2 > fmax

m1 g 0.5  10  9.8

 min  11.67 rad/s

From equation (3), frictional force f = 0 when m1r1 = m2r2

(a) NA = NB (b) NA > NB (c) NA < NB

10. If a car moves round a circular road of radius r at a constant speed v

(a) x = 2m (b) x > 2m (c) y = 3m (d) y = 5 m x

3. Choose the correct statements

(a) y component of force keeping particle

(b) Angular velocity of the particle t

7. Particle is placed on a rough disc rotating with uniform angular 

1. Match The Column

7. What is angle of banking of the road

5. Three point particles P, Q, R move in circle of radius ‘r’ with different

(a) (b) (c) (d)

3. An insect crawls up a hemispherical surface very slowly (see the

(a) cot   3 (b) tan   3 (c) sec   3 (d) cos ec   3

(a) (b) (c) (d)

5. A smooth semicircular wire track of radius R is fixed in a vertical plane

constant K = mg/R. Consider the instant when the ring is released O

(i) Draw the free body diagram of the ring

WORK, POWER AND ENERGY

Positive and Negative Work F

of force is parallel to the displacement. (See figure 2 a) F > 90°

Zero Work Done

WORK DONE IN VARIOUS SITUATIONS

taken from position A to B under the action of a constant force F. F

Find the work done by this force.