7Rigid bodies
IA Dynamics and Relativity
7.2 Moment of inertia
In general, consider a rigid body in which all
N
particles rotate about the same
axis with the same angular velocity:
˙
r
i
= ω × r
i
.
This ensures that
d
dt
|r
i
− r
j
|
2
= 2(
˙
r
i
−
˙
r
j
) · (r
i
− r
j
) = 2
ω × (r
i
− r
j
)
· (r
i
− r
j
) = 0,
as required for a rigid body.
Similar to what we had above, the rotational kinetic energy is
T =
1
2
N
X
i=1
m
i
|
˙
r
i
|
2
=
1
2
Iω
2
,
where
Definition
(Moment of inertia)
.
The moment of inertia of a rigid body about
the rotation axis
ˆ
n is
I =
N
X
i=1
m
i
s
2
i
=
N
X
i=1
m
i
|
ˆ
n × r
i
|
2
.
Again, we define the angular momentum of the system:
Definition. The angular momentum is
L =
X
i
m
i
r
i
×
˙
r
i
=
X
i
m
i
r
i
× (ω × r
i
).
Note that our definitions for angular motion are analogous to those for linear
motion. The moment of inertia
I
is defined such that
T
=
1
2
Iω
2
. Ideally, we
would want the momentum to be
L
=
Iω
. However, this not true. In fact,
L
need not be parallel to ω.
What is true, is that the component of
L
parallel to
ω
is equal to
Iω
. Write
ω = ω
ˆ
n. Then we have
L ·
ˆ
n = ω
X
i
m
i
ˆ
n · (r
i
× (
ˆ
n × r
i
))
= ω
X
i
m(
ˆ
n × r
i
) · (
ˆ
n × r
i
)
= Iω.
What does L itself look like? Using vector identities, we have
L =
X
i
m
i
(r
i
· r
i
)ω − (r
i
· ω)r
i
Note that this is a linear function of ω. So we can write
L = Iω,
where we abuse notation to use
I
for the inertia tensor. This is represented by a
symmetric matrix with components
I
jk
=
X
i
m
i
(|r
i
|
2
δ
jk
− (r
i
)
j
(r
i
)
k
),
where i refers to the index of the particle, and j, k are dummy suffixes.
If the body rotates about a principal axis, i.e. one of the three orthogonal
eigenvectors of
I
, then
L
will be parallel to
ω
. Usually, the principal axes lie on
the axes of rotational symmetry of the body.