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.

Answer 1

I w

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