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Chapter 5 Rotation Of Rigid Body

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.

> The formulae are listed:

Linear motion

Rotational motion

v = u + at

clip_image002

clip_image004

v2 = u2 + 2as

 

w = w0 + at

a  = clip_image006

a  = w0 t + clip_image008 at2

w2 =w02 + 2aq

> When doing calculation for rotational kinematics, students are not allowed to write the linear motion formulae.

 

Example 1

clip_image010

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 = mr2

I – the moment of inertia

m – mass of a particle

r – perpendicular distance from the axis of rotation.

clip_image012

The moment of inertia of the rigid body is

I = ( m1r12 + m2r22 + m3r32………..miri2)

= clip_image014

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, clip_image016

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.

clip_image018

 clip_image020

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

clip_image022

(b)Axis through one end and perpendicular to rod

clip_image024

Ring

Axis through centre of mass G I = MR 2

clip_image026

Disc

Axis through the centre of mass G

clip_image028

Hollow sphere

Axis through the centre of mass G

clip_image030

Solid sphere

Axis through the centre of the mass G

clip_image032

Solid cylinder

Axis of rotation about centre of cylinder

clip_image034

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 IG 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 = IG +Mh2

IG 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 IX and IY 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 = IX + IY = clip_image036 IX = IY = clip_image038

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

clip_image040

Moment of inertia, I = S mr2

                                   = mr2 + mr2

                                   = 2 mr2

5.4 Angular Momentum

> There are three types of momentum

-Linear momentum

p=mv = mrw

-Angular momentum

p=mvr =mrwr = mr2w

Angular momentum of a rigid body

L = (Smr2)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 = (Smr2)w = Iw

 

5.5 application of Angular Momentum

Conservation of angular momentum, I1w 1 = I2w2,

> Swimmer jumps off from the diving board

clip_image042

-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 = mr2, 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 hard-boiled 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.

clip_image044

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.

> Formula apply – clip_image046

> 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

clip_image048

clip_image050

Acceleration of a hollow sphere

clip_image052

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 m2. 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 clip_image054]

Solution

Example 7

clip_image056

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 clip_image058)

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

G = Fr = la = Iclip_image060 where a = clip_image060[1]

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 m2. 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

clip_image063

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 m2, calculate the acceleration of the mass.

Solution

Translational motion

Rotational motion

Displacement,s

Angular displacement

a  = clip_image065

Velocity, v

Angular velocity w = clip_image067

Acceleration, a

Angular acceleration a = clip_image069

Mass, m

Moment of inertia, I = Smiri2

Linear momentum, p = mv

Angular momentum L = lw

Impulses; Ft = mv – mu

Angular impulse Gt = Iw – Iw0

Work W = Fs

Work W = Gq

Translational kinetic energy = clip_image071mv2

Rotational kinetic energy = clip_image046[1],

Power P = Fv

Power P = Gw

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