Mathematical Tools for Physics

7—Operators and Matrices 170

Add and subtract the same thing on the right side of the equation (add zero) to get

~r×F~=~r×

d~p

dt

+

d~r

dt

×~p−

d~r

dt

×~p

=

d

dt

(

~r×~p

)

−

d~r

dt

×~p

Now recall that~pism~v, and~v=d~r/dt, so the last term in the preceding equation is zero because you’re taking
the cross product of a vector with itself. This means that when I added and subtracted the term from the right
side above I was really adding and subtracting zero.
~r×F~ is the torque applied to the point massmand~r×~pis the mass’s angular momentum about the
origin. Now if I have many masses and many forces I simply put an index on this torque equation and add the
resulting equations over all the masses in the rigid body. The sums on the left and the right provide the definitions
of torque and of angular momentum.

~τtotal=

∑

k

~rk×F~k=

d

dt

∑

k

(

~rk×~pk

)

=

dL~

dt

~v 2 (out)

~r 2 ×m 2 ~v 2

m 2

~r 2

~ω

~r 1

~r 1 ×m 1 ~v 1

~v 1 (in)

m 1

For a specific example, attach two masses to the ends of a light rod and attach that rod to a second,
vertical one as sketched — at an angle. Now spin the vertical rod and figure out what the angular velocity
and angular momentum vectors are. Since I said that I’m spinning it along the vertical rod, that defines the
direction of the angular velocity vector~ωto be upwards in the picture. (Viewed from above everything is rotating
counter-clockwise.) The angular momentum of one point mass is~r×~p=~r×m~v. The mass on the right has
a velocity pointingintothe page and the mass on the left has it pointingout. Take the origin to be where the