14Tensors of rank 2
IA Vector Calculus
14.2 The inertia tensor
Consider masses
m
α
with positions
r
α
, all rotating with angular velocity
ω
about
0. So the velocities are v
α
= ω × r
α
. The total angular momentum is
L =
X
α
r
α
× m
α
v
α
=
X
α
m
α
r
α
× (ω × r
α
)
=
X
α
m
α
(|r
α
|
2
ω − (r
α
· ω)r
α
).
by vector identities. In components, we have
L
i
= I
ij
ω
j
,
where
Definition (Inertia tensor). The inertia tensor is
I
ij
=
X
α
m
α
[|r
α
|
2
δ
ij
− (r
α
)
i
(r
α
)
j
].
For a rigid body occupying volume
V
with mass density
ρ
(
r
), we replace the
sum with an integral to obtain
I
ij
=
Z
V
ρ(r)(x
k
x
k
δ
ij
− x
i
x
j
) dV.
By inspection, I is a symmetric tensor.
Example.
Consider a rotating cylinder with uniform density
ρ
0
. The total
mass is 2`πa
2
ρ
0
.
x
1
x
3
x
2
2`
a
Use cylindrical polar coordinates:
x
1
= r cos θ
x
2
= r sin θ
x
3
= x
3
dV = r dr dθ dx
3
We have
I
33
=
Z
V
ρ
0
(x
2
1
+ x
2
2
) dV
= ρ
0
Z
a
0
Z
2π
0
Z
`
−`
r
2
(r dr dθ dx
2
)
= ρ
0
· 2π · 2`
r
4
4
a
0
= ε
0
π`a
4
.
Similarly, we have
I
11
=
Z
V
ρ
0
(x
2
2
+ x
2
3
) dV
= ρ
0
Z
a
0
Z
2π
0
Z
`
−`
(r
2
sin
2
θ + x
2
3
)r dr dθ dx
3
= ρ
0
Z
a
0
Z
2π
0
r
r
2
sin
2
θ [x
3
]
`
−`
+
x
3
3
3
`
−`
!
dθ dr
= ρ
0
Z
a
0
Z
2π
0
r
r
2
sin
2
θ2` +
2
3
`
3
dθ dr
= ρ
0
2πa ·
2
3
`
3
+ 2`
Z
a
0
r
2
dr
Z
2π
0
sin
2
θ
= ρ
0
πa
2
`
a
2
2
+
2
3
`
2
By symmetry, the result for I
22
is the same.
How about the off-diagonal elements?
I
13
= −
Z
V
ρ
0
x
1
x
3
dV
= −ρ
0
Z
a
0
Z
`
−`
Z
2π
0
r
2
cos θx
3
dr dx
3
dθ
= 0
Since
R
2π
0
d
θ cos θ
= 0. Similarly, the other off-diagonal elements are all 0. So
the non-zero components are
I
33
=
1
2
Ma
2
I
11
= I
22
= M
a
2
4
+
`
2
3
In the particular case where ` =
a
√
3
2
, we have I
ij
=
1
2
ma
2
δ
ij
. So in this case,
L =
1
2
Ma
2
ω
for rotation about any axis.