Introduction
Particles make a rigid body. The relative positions of a particles remain static when the body moves or acted upon by a force. In this chapter, no new law is introduced in rotational motion. Newton’s laws of motion are adapted to this topic.
5.1 Rotational Kinematics
> It involves the movement of an object during rotating.
Linear motion 
Rotational motion 
v = u + at^{} v^{2} = u^{2} + 2as 
w = w_{0 }+ at w^{2} =w_{0}^{2} + 2aq 
> When doing calculation for rotational kinematics, students are not allowed to write the linear motion formulae.
Example 1
A disc is rotated from rest at vertical axis at the centre O with the angular acceleration of 4.0 rads^{1} . The total angular displacement is 18 rad. Calculate:
(a) The angular velocity.
(b) The duration for the movement.
Solution:
Example 2
A ball with diameter 4 cm rolls down from rest. After 3.0 s the angular velocity of the ball is 1.5 rads^{1} 1. Calculate:
(a) The angular displacement.
(b) The linear displacement along the plane.
(c) The linear velocity.
Solution
5.2 Rotational Dynamics
> Definition
Moment of inertia is define as the resistance of a body to change to its rotational motion.
> Formulae
I = mr^{2}
I – the moment of inertia
m – mass of a particle
r – perpendicular distance from the axis of rotation.
The moment of inertia of the rigid body is
I = ( m_{1}r_{1}^{2} + m_{2}r_{2}^{2} + m_{3}r_{3}^{2}………..m_{i}r_{i}^{2})
The I, the moment of inertia depends on:
The mass.
The distribution of the mass from the axis of rotation (shape of the body).
The position of the axis of rotation.
> the centre of mass of a rigid body,
The point where the mass of the body is assumed to be concentrated
If a force is applied through [lie Centre of mass
– the motion is a linear motion with no rotational motion.
– the body accelerates in the direction of the force.
Example 3
Particles with mass 0.30 kg and 0.60 kg are at 30.0 cm and 40.0 cm mark respectively on a uniform metre rule of mass 0.2 kg. Find the position of the centre of mass of the system.
Solution
Moment of inertia of different shape of body
Shape 
Illustration 
Formulae 
Uniform rod 
(a) Axis through the centre of the mass G and perpendicular to rod (b)Axis through one end and perpendicular to rod 

Ring 
Axis through centre of mass G I = MR 2 

Disc 
Axis through the centre of mass G


Hollow sphere 
Axis through the centre of mass G 

Solid sphere 
Axis through the centre of the mass G 

Solid cylinder 
Axis of rotation about centre of cylinder 

5.3 Theorems to Find Moment of Inertia, I of A Rigid Body
> Parallel axis theorem
If the moment of inertia of a rigid body I_{G} about an axis through its centre of mass G is known, the moment of inertia of the body about an axis which is parallel to the first axis can be obtained by using the parallel axis theorem.
I = I_{G} +Mh^{2}
I_{G} refers to the moment of inertia about the axis through the centre of mass G, and h is the distance between the two axes.
> Perpendicular axis theorem
Suppose I_{X} and I_{Y} are the moment of inertia of a body about axis OX and OY respectively, the OX is perpendicular to OY. The the moment of inertia I about an axis OZ which is perpendicular to both OX and OY is given by
I = I_{X} + I_{Y} = I_{X} = I_{Y} =
Example 4
A chlorine molecule consists of two atoms each of mass 3.15 x 10^{32} kg. The atoms are separated by a distance of 1.78 x 10^{11} m. What is the moment of inertia of a chlorine molecule about an axis through its centre of mass and perpendicular to the line joining the two atoms`?
Solution
Moment of inertia, I = S mr^{2}
= mr^{2} + mr^{2}
= 2 mr^{2}
5.4 Angular Momentum
> There are three types of momentum
Linear momentum
p=mv = mrw
Angular momentum
p=mvr =mrwr = mr^{2}w
Angular momentum of a rigid body
L = (Smr^{2})w
=Iw
Example 5
Calculate the angular momentum of a bowling ball of mass 5.5 kg which moves in a circle of radius 0.15 m with angular velocity 3.0 rads^{1}.
Solution
Angular momentum, L = (Smr^{2})w = Iw
5.5 application of Angular Momentum
Conservation of angular momentum, I_{1}w _{1} = I_{2}w_{2},
> Swimmer jumps off from the diving board
The swimmer starts to rotate because a torque is produced by his weight.
– During rotation, he pulls himself closer to his axis of rotation. This mechanism reduces his moment of inertia, I. Then, his angular velocity, w increases and the swimmer is able to make more rotations.
Before hitting water, the swimmer needs to straighten his body, the moment of inertia, I, increases, and the angular velocity w decreases.
– When the swimmer falls, the path taken by his centre of mass is a parabolic path (the path of projectile).
> Determining if an egg is hard boiled or half boiled
When a raw egg is spun, the yolk which is denser moves away from the axis of rotation. Since moment of inertia I = mr^{2}, when the distance r from the axis of rotation increases, I increases.
– When I increases, w decreases. – It is hard to spin a raw egg.
– When a hardboiled egg is spun, its yolk does not move away from the axis of rotation.
The moment of inertia, I remains constant and the egg spins easily.
> A ballet dancer
– When she starts to spin and stretches her arm, the moment of inertia I is large and the angular velocity w is small.
– When she lifts her arms above her head, the moment of inertia decreases and the angular of velocity m increases.
5.6 Rotational Kinetic Energy
> Definition – Rotational kinetic energy is the total energy of all the particle in the body when a rigid body rotates about a fixed axis.
> Application kinetic energy of a rolling body
Velocity v of the centre of mass where R = radius of rolling body w= angular velocity 
v = Rw 
Total kinetic energy = rotational kinetic energy + translational kinetic energy Kinetic energy of a hollow sphere 

Acceleration of a hollow sphere 
Example 5
the moment of inertia of a ball of mass 30 kg and radius 0.50 m about its axis is 3.2 kg m^{2}. What is its total kinetic energy when it rolls along a horizontal surface with a speed 3.0 ms^{1}.
Solution
Example 6
A uniform rod, length 2.40 m is hinged freely at one terminal. The rod wised so that it is 1 horizontal and then released from rest so that it rotates about an axis in a vertical plan. Calculate by using l(length) and g the angular velocity when the rod is vertical.
[moment inertia of the rod about its axis ]
Solution
Example 7
A cylinder of radius 10 cm and mass 600 g rolls on a floor without slipping with a speed v. The cylinder then rolls up an inclined plane.
(a) Calculate the speed of v if the maximum height h reached by the cylinder is 3 m.
(b) After 5 second the cylinder rolls down. Calculate the angle of inclination of the inclined plane, q and the deceleration.
(Given that the moment of inertia of a cylinder about its axis is )
Solution
Torque
•Definition – Torque or the moment of the force is the turning effect of force.
• Formula Torque,G
Force x Perpendicular distance of the force from the axis of rotation = F x r
Torque is measured in the unit of Nm
Example 8
CD player rotates about a vertical axis with uniform angular velocity of 30 revolutions per minute. A CD is placed onto the recorder. The friction between the CD and the recorder causes the CD to accelerate uniformly until angular velocity of 30 rev per minute in 5.0 s. The moment (d f inertia of the CD about its axis of rotation is 2.0 x 10^{2} kg m^{2}. Calculate:
(a) The torque on the CD.
(b) Rotational kinetic energy gained by the CD.
(c) The total work done by the torque in 5.0 s.
From the findings of b and c, what inference can be concluded?
Solution
Example 9
The diagram shows a disc that is attached to an object with the mass of 1.0 kg. When the mass is released from rest to fall vertically, the disc starts to rotate. The radius of the disc is 12.0 cm and its moment of inertia about the axis of rotation is 0.030 kg m^{2}, calculate the acceleration of the mass.
Solution
Translational motion 
Rotational motion 
Displacement,s 
Angular displacement 
Velocity, v 

Acceleration, a 

Mass, m 
Moment of inertia, I = Sm_{i}r_{i}^{2} 
Linear momentum, p = mv 
Angular momentum L = lw 
Impulses; Ft = mv – mu 
Angular impulse Gt = Iw – Iw_{0} 
Work W = Fs 
Work W = Gq 
Power P = Fv 
Power P = Gw 
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