SP015 CHAPTER 8 : ROTATIONAL MOTION OF RIGID BODY

CHAPTER 8: ROTATIONAL MOTION OF RIGID BODY

8.1 Rotational Kinematics

Parameters in rotational motion

Parameter Definition Equations

Angular An angle through which a point or line
displacement has been rotated in a specified direction
about a specified axis

: Angular displacement
s: arc length
r: radius of the circle

SI Unit: radian

Average angular Rate of change of angular displacement.

velocity,

: the change in angular

displacement
: time interval

SI Unit:
Instantaneous The limit of average angular velocity as
angular the time interval approaches zero.
velocity, (angular velocity at specific time)

SI Unit:

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 SP015 CHAPTER 8 : ROTATIONAL MOTION OF RIGID BODY

Parameter Definition Equation

Average angular The rate of change of angular velocity
acceleration,

: the change in angular

velocity
: time interval

SI Unit:
Instantaneous The limit of the average angular acceleration
angular as time interval approaches zero.
acceleration, (angular acceleration at specific time)

SI Unit:
Sign Convention rotational motion

Positive Anticlockwise
Negative Clockwise
Unit Conversion

Degree Radian No of revolution (rev)

1 revolution
EXAMPLE 1
Express the following angles in radian: a) b) c) d) e) . Give as numerical
values and as a fraction of . [1.05 rad, 1.75 rad, 6.28 rad, 7.77 rad]

EXAMPLE 2
Calculate the angular velocity of: a) the second hand, b) the minute hand, and c) the hour hand,
of a clock. State in rad s-1. What is the angular acceleration in each case? [0.105 rads-1, 1.75 x
10-3 rads-1, 1.45 x 10-4 rads-1]

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 SP015 CHAPTER 8 : ROTATIONAL MOTION OF RIGID BODY

EXAMPLE 3

During a certain period of time, the angular displacement of a swinging door is described by

Where is in radian and t is in seconds. Determine the angular displacement, angular speed and
angular acceleration at time . [53 rad, 22 rads-1, 4 rads-2]

Linear and Rotational Quantities

Quantity Linear Rotational Relation

Displacement
Velocity
Acceleration

Figure 8.1 As a rigid body rotates about fixed Figure 8.2 As a rigid body rotates
axis through point O, the point P has a about a fixed axis through point O,
tangential velocity 𝑣 that is always tangent the point P experiences tangential
Resultant Linear Acceleration (linear) acceleration 𝑎𝑡 , and
to the circular path of radius r.
centripetal (linear) acceleration 𝑎𝑐 .
 Every point on the rotating object has the same angular motion.
 Every point on the rotating object does not have same linear motion.

Resultant Linear Acceleration

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 SP015 CHAPTER 8 : ROTATIONAL MOTION OF RIGID BODY

a) In chapter 6 we found that a point moving in circular path undergoes centripetal (radial)
acceleration,
b) Resultant linear acceleration of point P (figure 8.2) is given by.

√
Analogies between Linear and Rotational motion

a) There are many parallels between the motion equation for rotational motion and those for
linear motion.

b) Every term in linear equation has a corresponding term in the analogous rotational
equations.

Linear motion ( constant) Rotational motion ( constant)

EXAMPLE 4

Centrifuge acceleration. A centrifuge rotor is accelerated from rest to 20 000 rpm in 30s.
a) What is its average angular acceleration? [69.8 rads-1]
b) Through how many revolutions has the centrifuge rotor turned during its acceleration
period, assuming constant angular acceleration? [4999 revolutions]

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 SP015 CHAPTER 8 : ROTATIONAL MOTION OF RIGID BODY

EXAMPLE 5

Rotating Wheel. A wheel rotates with a constant angular acceleration of . If the

angular speed of the wheel is at
a) Through what angular displacement does the wheel rotate in ?[11 rad]
b) Through how may revolutions has the wheel turned during this time? [1.91 rev]
c) What is the angular speed of the wheel at ? [9 rads-1]

EXAMPLE 6

Angular and linear velocities

and accelerations. A carousel
𝑎𝑡
is initially at rest. At it is
given a constant angular
P acceleration ,
O 𝑎𝑐 which increases its angular
velocity for . At ,
determine the magnitude of the
following quantities:
Figure 8.3 Example 6 a) The angular velocity of
the carousel [0.48 rads-1]
b) The linear velocity of a child located 2.5 m from the centre, point P (figure 8.9)[1.2 ms-1]
c) The tangential acceleration of that child [0.15 ms-2]
d) The centripetal acceleration of the child [0.76 ms-2]
e) Total linear acceleration of the child [0.595 ms-2]

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 SP015 CHAPTER 8 : ROTATIONAL MOTION OF RIGID BODY

8.1 Equilibrium of a Uniform Rigid Body

LO 8.1 (a) Define Torque,   rF

LO 8.1 (b) Solve problems related to equilibrium of rigid body

Introduction to torque (moment of force)

a) Consider the common experience of pushing open a door. Figure 8.4 is a top view of a
door that is hinged on the left. Four pushing forces are shown. All of equal strength.
Which of these will be more effective at opening the door?

Top view of door 𝐹

Hinge 𝐹
𝐹 𝐹

Figure 8.4 The four forces are the same strength, but
they have different effects on swinging the door.

b) Force will open the door, but force⃗⃗⃗ , which pushes straight at the hinged, will not.
c) Force will open the door, but not as easily as⃗⃗⃗ .
d) What about force⃗⃗⃗ ? It is perpendicular to the door, it has the same magnitude as , but
you know from experience that pushing close to the hinge (⃗⃗⃗ ) is not as effective as
pushing the outer edge of the door ( ).
e) Therefore, the ability of a force to cause rotation or twisting motion depends on three
factors:
i) The magnitude F of the force.
ii) The distance r from the pivot.
iii) The angle at which the force is applied.
f) The tendency of a force to rotate an object about some axis is measured by a quantity
called torque. Loosely speaking, torque measures the “effectiveness” of the force at
causing an object to rotate about a pivot. Torque is the rotational equivalent of force.

Torque,

a) Definition

Torque is defined as the cross product between the distances of the force from the rotation axis (
) with the acting force ( ).
or 𝜏 : Torque
𝑟: Distance between pivot point and the point of application of force
𝜏 𝑟×𝐹
𝐹 : Force
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b) Equation
Top view of door
𝜏 𝑟𝐹 s n 𝜃
𝑟 𝜃
Pivot
: Angle between r and F
𝐹
c) Torque is a vector quantity.
Figure 8.5 𝑇𝑜𝑟𝑞𝑢𝑒 𝑟𝐹 𝑠𝑖𝑛 𝜃 𝐹𝑟 s n 𝜃
d) The unit of torque is N m.
e) Sign convention of torque:

POSITIVE: turning tendency of the force is anticlockwise.

NEGATIVE: turning tendency of the force is clockwise.

EXAMPLE 7

Determine the net torque on the 2.0-m-long uniform beam shown in figure 8.3. Calculate about:

𝐹 𝑁
𝜃
P

𝜃
C
𝜃

𝐹 𝑁 𝐹 𝑁
Figure 8.6 Example 7
a) Point C, the centre of mass of the beam. ( point C as pivot point) [17.03 Nm]
b) Point P at one end. (Point P as pivot point) [-10.04 Nm]

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 SP015 CHAPTER 8 : ROTATIONAL MOTION OF RIGID BODY

Conditions for equilibrium of rigid body

First condition for equilibrium Second condition for

(translational equilibrium): equilibrium
The net force is zero: (rotational equilibrium):
The net torque is zero:
∑𝑭 𝟎
∑ ,∑
∑ 𝟎
𝑥 𝑦

Problem Solving Strategy: Equilibrium of Rigid Body

(1) Draw free body diagram:

*Show all the forces acting on that object. (Weight, Normal Force, Friction, External Force,
Tension).
*Pick any point you wish as pivot point.
*Label distance r.

(2) Choose Coordinate Axis (x and y components):

*Construct a table to resolve the forces into x and y components.

(3) Calculate torque by each force in the same table.

(4) Apply the condition for equilibrium of rigid

body
∗∑ ,∑ 𝑥 ,∑ 𝑦
∗ ∑𝜏

(5) Solve these equations for

the unknowns.

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 SP015 CHAPTER 8 : ROTATIONAL MOTION OF RIGID BODY

EXAMPLE 8

Balancing a seesaw. A board of mass M = 2.0 kg serves as a seesaw for two students (Aiman
and Boniface). Aiman has a mass of 50 kg and sits 2.5 m from pivot point P.

Figure 8.6 Aiman and Boniface on a seesaw.

a) At what distance x from the pivot point must Boniface of mass 45 kg, place himself to
balance the seesaw? Assume the board is uniform and centred over the pivot.[2.78 m]
b) Calculate force exerted by pivot P. [951.57 N]

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 SP015 CHAPTER 8 : ROTATIONAL MOTION OF RIGID BODY

EXAMPLE 9

Ladder. A 5.0-m-long ladder leans against a smooth wall

at a point 4.0 m above a cement floor as shown in figure
8.6. The ladder is uniform and has mass .
Assuming the wall is frictionless (but the floor is not), 4m
determine the forces exerted on the ladder by the floor
and by the wall. [125.73 N]

Figure 8.6 A ladder leaning

against a wall.

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 SP015 CHAPTER 8 : ROTATIONAL MOTION OF RIGID BODY

EXAMPLE 10

Hinged beam and cable. A uniform beam 2.20 m long with mass , is mounted by a
small hinge on a wall as shown in figure 8.7. The beam is held in a horizontal position by a cable
that makes an angle . The beam support a sign of mass suspended from
its end.

a) Sketch free body diagram of the beam.

b) Determine
i. The tension in the cable, T [794.61 N]
ii. Force exerted by the hinged on the 𝜃
beam, S [10.100]

Kafe
Merinding
Figure 8.7

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 SP015 CHAPTER 8 : ROTATIONAL MOTION OF RIGID BODY

8.3 Rotational Dynamics

LO 8.3(a) Define & use moment of inertia of a uniform rigid body (Sphere, cylinder, ring disc and
rod)
LO 8.3(b) State and use torque τ = Iα
Moment of Inertia, I

Definition

Moment of Inertia, I is defined as the sum of the products

of the mass, m of each particle and the square of its
respective distance, r from the rotation axis.

a) Moment of inertia is a scalar quantity

b) SI Unit:
c) The factor which effects the moment of inertia, I of a rigid body:
i. The mass of the body.
ii. The shape of the body.
iii. The position of the rotation axis.
Moment of inertia of various bodies

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 SP015 CHAPTER 8 : ROTATIONAL MOTION OF RIGID BODY

EXAMPLE 11 y
1.50 m

0.50 m

m m

0.50 m x

M M

Figure 8.8
Calculate the moment of inertia of the array of point objects shown in figure 8.11 about:
a) The vertical axis (y axis) [6.625 kgm2]
b) The horizontal axis (x axis) [0.6625 kgm2]
Assume , and the objects are wired together by a very light rigid piece
of wire. The array is rectangular and is split through the middle by the horizontal axis (x axis).

Torque and angular acceleration

a) When a rigid object is subject to a net torque (∑𝜏 ), it undergoes an angular
acceleration, .
b) The angular acceleration is directly proportional to the net torque:
∑𝜏
The relationship is analogous to ∑ (Newton’s Second Law)
c) The angular acceleration is inversely proportional to the analogy of the mass in a rotating
system. This mass analog is call the moment of inertia of the object, I.
d) Therefore:
∑𝜏 𝐼𝛼
i. The angular acceleration is directly proportional to net torque
ii. The angular acceleration is inversely proportional to the moment of inertia of the
object.

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 SP015 CHAPTER 8 : ROTATIONAL MOTION OF RIGID BODY

Problem solving Strategy: Rotational Motion

1) As always, draw a clear and complete diagram

2) Choose the object or objects that will be the system to be studied.
3) Draw a free body diagram for the object under consideration (or for each object, of
more than one), showing all (and only) the forces acting on that object and exactly
where they act, so you can determine the torque due to each.
4) Identify the axis of rotation and determine the torques about it. Assign the correct
sign for each torque.
5) Apply ∑𝝉 𝑰𝜶
6) Also apply ∑𝑭 𝒎𝒂, and other laws or principle as needed.
7) Solve the resulting equation(s) for the unknown(s).

EXAMPLE 12

Forces, and are applied tangentially to a

disk with radius 36 cm and mass 5.0 kg as shown in figure 8.9.
Calculate:
a) The net torque on the disk [-1.692 Nm]
𝑟 b) The magnitude of angular acceleration of the disk [5.22 rads-1]
(Use moment of inertia about the centre mass, )

𝐹
𝐹
Figure 8.9

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 SP015 CHAPTER 8 : ROTATIONAL MOTION OF RIGID BODY

EXAMPLE 13

An object of mass is suspended from a frictionless

pulley of radius by a light string as shown in
figure 8.10. The pulley has a moment of inertia, I of 0.020 kg m2
about the axis of the pulley. The object is released from rest.
Assume that the string does not slip on the pulley. After ,
determine:
a) The linear acceleration of the object [7.36 ms-2]
b) The angular acceleration of the pulley [36.8 rads-1]
c) The tension in the string [3.675 N]
d) The linear velocity of the object.[22.08 ms-1]

Figure 8.10 Example 13

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 SP015 CHAPTER 8 : ROTATIONAL MOTION OF RIGID BODY

8.4 Conservation of Angular Momentum

LO 8.4(a) Define and use angular momentum, L = Iω

LO 8.4(b) State and use principle of conservation of angular momentum
Definition

a) Angular momentum is defined as the product of angular velocity of a body and its
moment of inertia about the rotation axis.

𝐿 𝐼𝜔
L : angular momentum
I : moment of inertia of a body
ω : angular velocity
b) Angular momentum is a vector quantity
c) SI Unit: kg m2 s-1

Principle of conservation of angular momentum

Principle of conservation of angular momentum states the total angular momentum of a system
remains constant if the net external torque acting on it is zero.

∑𝜏
∑𝐿𝑖 ∑𝐿𝑓

Example of conservation of angular momentum:

Spinning Ice skater:

Figure 8.11 A skater doing spin on ice, illustrating

conservation of angular momentum (a) 𝐼 is large so
𝜔 is small; (b) 𝐼 is small so 𝜔 is larger.

(a) (b)

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EXAMPLE 14

A horizontal platform in the shape of a circular disk

rotates freely in a horizontal plan about a frictionless,
vertical axle (figure 8.12). The platform has mass
and a radius . A student whose
mass is walks slowly from the rim of the
disk toward its centre. If the angular speed of the system
is 2.0 rad s-1 when the student is at the rim, what is the
angular speed when she reaches a point
from the centre? [4.1 rads-1]

(Moment of inertia of the platform )

Figure 8.12 As the students walks towards
the centre of the rotating platform, the
angular velocity of the system increases
because the angular momentum of the
system remain constant

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EXAMPLE 15

A student sits on a freely rotating stool holding two

dumbbells, each of mass 3.0 kg (figure 8.13). When
his arms are extended horizontally (figure 8.19 a), the
dumbbells are 1.00 m from the axis of rotation and the
student rotates with angular speed of 0.750 rad s-1.
The moment of inertia of the student plus stool is 3.00
kg m2 and is assumed to be constant. The student pulls
the dumbbells inward horizontally to a position 0.300
m from the rotation axis (figure 8.13b).

a) Find the new angular speed of the student

(a) (b)
[1.91 rads-1]
Figure 8.13 Example 15

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