posted by Nik on .
A thin rod of length L and mass M rotates about a vertical axis through its center with angular velocity w. The rod makes an angle B with the rotation axis. Determine the angular momentum of the rod.
I w = 2 (M/L) integral from 0 to L/2 of (x sin B)^2 dx * w
= 2 w (M sin^2 B/L) integral 0 to L/2 of x^2 dx
= 2 w (M sin^2 B / L) x^3/3 at x = L/2 - at x = 0
= (2/3) w (M sin^2 B / L) L^3/8
= (1/12) w M L^2 sin^2 B