7Rigid bodies

IA Dynamics and Relativity

7.3 Calculating the moment of inertia

For a solid body, we usually want to think of it as a continuous substance with

a mass density, instead of individual point particles. So we replace the sum of

particles by a volume integral weighted by the mass density ρ(r).

Definition (Mass, center of mass and moment of inertia). The mass is

M =

Z

ρ dV.

The center of mass is

R =

1

M

Z

ρr dV

The moment of inertia is

I =

Z

ρs

2

dV =

Z

ρ|

ˆ

n × r|

2

dV.

In theory, we can study inhomogeneous bodies with varying

ρ

, but usually

we mainly consider homogeneous ones with constant ρ throughout.

Example

(Thin circular ring)

.

Suppose the ring has mass

M

and radius

a

, and

a rotation axis through the center, perpendicular to the plane of the ring.

a

Then the moment of inertia is

I = M a

2

.

Example

(Thin rod)

.

Suppose a rod has mass

M

and length

`

. It rotates

through one end, perpendicular to the rod.

`

The mass per unit length is M/`. So the moment of inertia is

I =

Z

`

0

M

`

x

2

dx =

1

3

M`

2

.

Example

(Thin disc)

.

Consider a disc of mass

M

and radius

a

, with a rotation

axis through the center, perpendicular to the plane of the disc.

a

Then

I =

Z

2π

0

Z

a

0

M

πa

2

|{z}

mass per unit length

r

2

|{z}

s

2

r dr dθ

| {z }

area element

=

M

πa

2

Z

a

0

r

3

dr

Z

2π

0

dθ

=

M

πa

2

1

4

a

4

(2π)

=

1

2

Ma

2

.

Now suppose that the rotation axis is in the plane of the disc instead (also

rotating through the center). Then

I =

Z

2π

0

Z

a

0

M

πa

2

|{z}

mass per unit length

(r sin θ)

2

| {z }

s

2

r dr dθ

| {z }

area element

=

M

πa

2

Z

a

0

r

3

dr

Z

2π

0

sin

2

θ dθ

=

M

πa

2

1

4

a

4

π

=

1

4

Ma

2

.

Example.

Consider a solid sphere with mass

M

, radius

a

, with a rotation axis

though the center.

a

Using spherical polar coordinates (r, θ, φ) based on the rotation axis,

I =

Z

2π

0

Z

π

0

Z

a

0

M

4

3

πa

3

|{z}

ρ

(r sin θ)

2

| {z }

s

2

r

2

sin θ dr dθ dφ

| {z }

volume element

=

M

4

3

πa

3

Z

a

0

r

4

dr

Z

π

0

(1 − cos

2

) sin θ dθ

Z

2π

0

dφ

=

M

4

3

πa

3

·

1

5

a

5

·

4

3

· 2π

=

2

5

Ma

2

.

Usually, finding the moment of inertia involves doing complicated integrals.

We will now come up with two theorems that help us find moments of inertia.

Theorem (Perpendicular axis theorem). For a two-dimensional object (a lam-

ina), and three perpendicular axes

x, y, z

through the same spot, with

z

normal

to the plane,

I

z

= I

x

+ I

y

,

where I

z

is the moment of inertia about the z axis.

x

y

z

Note that this does not apply to 3D objects! For example, in a sphere,

I

x

= I

y

= I

z

.

Proof. Let ρ be the mass per unit volume. Then

I

x

=

Z

ρy

2

dA

I

y

=

Z

ρx

2

dA

I

z

=

Z

ρ(x

2

+ y

2

) dA = I

x

+ I

y

.

Example. For a disc, I

x

= I

y

by symmetry. So I

z

= 2I

x

.

Theorem

(Parallel axis theorem)

.

If a rigid body of mass

M

has moment of

inertia

I

C

about an axis passing through the center of mass, then its moment of

inertia about a parallel axis a distance d away is

I = I

C

+ Md

2

.

CM

d

Proof.

With a convenient choice of Cartesian coordinates such that the center of

mass is at the origin and the two rotation axes are x = y = 0 and x = d, y = 0,

I

C

=

Z

ρ(x

2

+ y

2

) dV,

and

Z

ρr dV = 0.

So

I =

Z

ρ((x − d)

2

+ y

2

) dV

=

Z

ρ(x

2

+ y

2

) dV − 2d

Z

ρx dV +

Z

d

2

ρ dV

= I

c

+ 0 + Md

2

= I

c

+ Md

2

.

Example.

Take a disc of mass

M

and radius

a

, and rotation axis through a

point on the circumference, perpendicular to the plane of the disc. Then

a

I = I

c

+ Ma

2

=

1

2

Ma

2

+ Ma

2

=

3

2

Ma

2

.