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
.