Mathematical Tools for Physics

7—Operators and Matrices 179

are bothonthe axis, and they have no angular momentum. In the general case as drawn, the vector~Lpoints 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

∫

~r dm

whereM is the total mass. Compare the operatorIusing an origin at the center of mass toI about 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

(17)

The two cross terms vanish, problem 17. What’s left is

I(~ω) =

∫

dm[~r−~rcm]×

(

~ω×[~r−~rcm]

)

+M rcm×

(

~ω×~rcm

)

=Icm(~ω) +M rcm×

(

~ω×~rcm

) (18)

Put this in words and it says that the tensor of inertia about any point is equal to the tensor of inertia about the
center of mass plus the tensor of inertia of a point massMplacedatthe center of mass.