Mathematical Tools for Physics

7—Operators and Matrices 171

supporting rod is attached to the axis, then~r×~pfor the mass on the right is pointing up and to the left. For the
other mass both~rand~pare reversed, so the cross product is in exactly the same direction as for the first mass.
The total angular momentum the sum of these two parallel vectors, and it isnotin the direction of the angular
velocity.
Now make this quantitative and apply it to a general rigid body. There are two basic pieces to the problem:
the angular momentum of a point mass and the velocity of a point mass in terms of its angular velocity. The
position of one point mass is described by its displacement vector from the origin,~r. Its angular momentum is
then~r×~p, where~p=m~v. If the rigid body has an angular velocity vector~ω, the linear velocity of a mass at
coordinate~ris~ω×~r.

~ω

θ ~r

~v

rsinθ

~r

~ω

dm

The total angular momentum of a rotating set of massesmkat respective coordinates~rkis the sum of all
the individual pieces of angular momentum

L~=

∑

k

~rk×mk~vk, and since ~vk=~ω×~rk,

L~=

∑

k

~rk×mk

(

~ω×~rk

) (2)

If you have a continuous distribution of mass then using an integral makes more sense. For a given distribution of
mass, this integral (or sum) depends on the vector~ω. It defines a function having a vector as input and a vector
~Las output. Denote the function byI, so~L=I(~ω).

~L=

∫

dm~r×

(

~ω×~r

)

=I(~ω) (3)