Physics

Session

Rotational Mechanics - 2

Session Objectives

Session Objective

1. Rotational Kinetic energy

2. Moment of Inertia

3. Standard Results of moment of Inertia

4. Radius of gyration

5. Parallel-axis Theorem

6. Perpendicular axis theorem

Rotational Kinetic Energy

Energy possessed due to rotation of a body or a system of particles about an axis of rotation.

Rotational Kinetic Energy –Discrete System

For an ith particle of a rigid body,

For a discrete body

m2

m3

m1

r2 r3

r1

2i i i

1K = mv

2

2 2 2i i i i i

1 1K = mv = mr

2 2

2i iI = mr

21K = I

2

Rotational Kinetic Energy –Coetaneous System

For a continuous body

dmr

21k = I

2

21k = dm.v

2

2 21= dmr

2

21or = dI

2

Moment of Inertia

This property opposes a change in the state of uniform rotation

ParticleI=mr2

m2

m3

m1

r2 r3

r1

Discrete System

I=mr2Continuous body

2I = dm.r

Moment of Inertia

• It has dimensions [ML-2]

• Its SI unit is kg-m2

• It is a tensor quantity

• Depends on mass,shape of the body

• Depends on the distribution of mass for same size and shape

M

L

12

MLI

2

OY O

dr

x

YrL/2

Moment of Inertia of a thin Rod

L / 23L / 22 2

OYL / 2

L / 2

M M rl r dm r dr

L L 3

MAs, dm dr

L

Moment of Inertia of a Disc

2z 2I = dI = dm.r

2z 2

MI = 2 rdr.r

R

2

z

MRI =

2

Limits 0 to R

Standard ResultY

x

ZR

Z

x

Y

ba

Y

x

Z

R

2

x y

2z

MRl l

2I MR

2

x

2

y

2 2z

Mal

12Mb

l12M

l (a b )12

2

x y z

2l l l MR

3

Y

x

ZR

Z

x

Y

ba

Y

x

Z

R

2

x y

2z

MRl l

2I MR

2

x

2

y

2 2z

Mal

12Mb

l12M

l (a b )12

Y

x

Z

R

Z

Y

x

R

2ZYx MR

5

2III

2Z MR

2

1I

X

Y

Z

O

R 2

MRI

4

MRII

2

Z

2

Yx

Standard Results

Radius of Gyration

It is the distance whose square when multiplied by mass gives the moment of inertia of a rigid body.

It is denoted by k

Radius of Gyration

A real rigid body can be replaced by a point mass for rotational motion

k

2I =Mk

Ior k =

M

Rk =

2

For ring

Theorem of Parallel Axis

To calculate moment of inertia about an axis

IcmI

a

2cmI =I +Ma

•The two axes should be parallel to each other.

• One of them should pass through CM.

Theorem of Perpendicular Axis

• Applicable for plane lamina

IZ=IX + IY

• X-Y have to be in a plane perpendicular to Z-axis.

• All the three axis should be perpendicular to each other X Y

Z

Ix Iy + Iz

Iy Ix + Iz

Questions

Class Exercise - 1

Two circular discs A and B of equal masses and thickness are made of metals with densities dA and dB (dA > dB). If their moments of inertia about an axis passing through the centre and normal to the circular faces be IA and IB, then

(a) IA = IB (b) IA > IB

(c) IA < IB (d) IA > IB or IA < IB

Solution

2 2A A A B B BM = ( R )td , M = ( R )td

2

A A A A B2

B B AB B

M R d R d

M R dR d[As MA = MB]

2

A AA

M RI

2

2B B

BM R

I2

2A A B

B B A

I R d

I R d

IB > IA as dA > dB Hence, answer is (b).

Class Exercise - 2

With usual notation, radius of gyration is given by

M I(a) (b)

I M1

(c) I M (d)MI

Solution

I = Mk2 [k radius of gyration]

I

So, kM

Hence, answer is (b).

Class Exercise - 3

The radius of gyration of a disc of mass M and radius R about one of its diameters is

R R(a) (b)

2 2R

(c) (d) R4

SolutionY

X

Z

IZ = IX + IY ZX X Y

Ior I (I I )

2

2

XMR

or I4

2

ZMR

as I2

Solution

2

2MRNow, Mk

4[k Radius of gyration]

R

or, k2

Hence, answer is (a).

Class Exercise - 4

The moment of inertia of a circular disc about one of its diameters is I. Its moment of inertia about an axis perpendicular to the plane of the disc and passing through its centre is

(a) ( 2)l (b) 2I

I I(c) (d)

2 2

Solution

Y

X

Z

IX = IY = I

IZ = IX + IY = 2I

Hence, answer is (b)

Class Exercise - 5

Four spheres of diameter 2a and mass M are placed with their centres on the four corners of a square of side b. The moment of inertia of the system about an axis along one of the sides of the square is

2 2 2 2

2 2 2

4 8(a) Ma 2Mb (b) Ma 2Mb

5 58 4

(c) Ma (d) Ma 4Mb5 5

Solution

A B

CD

Y

Y

b

2B, CM A, CM C, CM D, CM

2I Ma I I I

5

2 2yy A, CM D, CM B, CM C, CMI I I I Mb I Mb

[Parallel axis theorem]

2 2YY

8I Ma 2Mb

5Hence, answer is (b).

Class Exercise - 6

The moment of inertia of a circular ring of radius R and mass M about a tangent in its plane is

22

2 2

MR(a) MR (b)

23

(c) MR (d) 2MR2

Solution

Y

X

Z

X

IZ = MR2

IZ = IX + IY [Perpendicular axis theorem]

ZX X Y

Ior I [I = I ]

2

2

XMR

or I2

Solution

2X XI I MR [Parallel axis theorem]

2X

3or I MR

2

Hence, answer is (c).

Class Exercise - 7

YY

X

Z

O R

a

Z

The adjoining figure shows a disc of mass M and radius R lying in the X-Y plane with its centre on X-axis at a distance ‘a’ from the origin. Then the moment of inertia of the disc about the X-axis is

2 2

2 22 2

MR MR(a) (b)

2 4

R R(c) M a (d) M a

4 2

Solution

ZX

II

2

2

XMR

or I4

Hence, answer is (b).

Class Exercise - 8

In question 7, the moment of inertia of the disc about the Y-axis is

2 2

2 22 2

MR MR(a) (b)

2 4

R R(c) M a (d) M a

4 2

Solution

2Y Y , CMI I Ma

22

YMR

I Ma4

Hence, answer is (c).

Class Exercise - 9

In question 7, the moment of inertia of the disc about the Z-axis is

2 2

2 22 2

MR MR(a) (b)

2 4

R R(c) M a (d) M a

4 2

Solution

2Z Z , CMI I Ma

2

2MRMa

2

Hence, answer is (d).

Class Exercise - 10

The moment of inertia of a cylinder of radius R, length and mass M about an axis passing through its centre of mass and normal to its length is

2 2 2

2 2 2

ML L R(a) (b) M

12 12 4

MR L R(c) (d) M

4 12 2

Solution

dx x

Y Y

CM

Choose an elemental disc at a distance x from the CM of cylinder.

Mass of disc of thickness M

dx dxL

2

YM R

dI dxL 4

Solution

22

YM R M

dI dx dx xL 4 L

[Parallel axis theorem]

L L2 2 232 2

y y L L2 2

MR M MR MLI dI x x

4L L 4 12

Thank you