7—Operators and Matrices 150Don’t count on all such results factoring so nicely.
In this basis, the angular velocity~ωhas just one component, so what isL~?
(
m 1 r 12 +m 2 r^22
)
cos^2 α −cosαsinα 0
−cosαsinα sin^2 α 0
0 0 1
0
ω
0
=
(
m 1 r^21 +m 2 r^22
)
−ωcosαsinα
ωsin^2 α
0
Translate this into vector form:
L~=(m 1 r^21 +m 2 r^22 )ωsinα(−~e 1 cosα+~e 2 sinα) (7.19)
Whenα = 90◦, then cosα = 0and the angular momentum points along they-axis. This is the
symmetric special case where everything lines up along one axis. Notice that ifα= 0then everything
vanishes, but then the masses are bothonthe axis, and they have no angular momentum. In the general
case as drawn, the vectorL~points to the upper left, perpendicular to the line between the masses.
Parallel Axis Theorem
When you know the tensor of inertia about one origin, you can relate the result to the tensor about a
different origin.
The center of mass of an object is
~rcm=
1
M
∫
~rdm (7.20)
whereMis the total mass. Compare the operatorIusing an origin at the center of mass toIabout
another origin.
~r
~rcm
~r−~rcm
I(~ω) =
∫
dm~r×(~ω×~r) =
∫
dm[~r−~rcm+~rcm]×
(
~ω×[~r−~rcm+~rcm]
)
=
∫
dm[~r−~rcm]×
(
~ω×[~r−~rcm]
)
+
∫
dm~rcm×
(
~ω×~rcm
)
+two cross terms(7.21)
The two cross terms vanish, problem7.17. What’s left is