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

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

To determine the angular momentum of the rod, we can use the formula:

Angular momentum (L) = moment of inertia (I) * angular velocity (ω).

In this case, the rod is rotating about a vertical axis through its center. Let's assume that the axis of rotation coincides with the origin (0,0). Also, let's consider the rod to have a uniform mass distribution.

The moment of inertia (I) of a thin rod rotating about an axis perpendicular to its length and passing through its center is given by the formula:
I = (1/12) * m * L^2,

where m is the mass of the rod and L is its length.

Given that the mass of the rod is M and its length is L, we can substitute these values into the moment of inertia formula:
I = (1/12) * M * L^2.

The angular velocity of the rod is given as ω.

Now we can calculate the angular momentum by multiplying the moment of inertia with the angular velocity:
L = I * ω,
= ((1/12) * M * L^2) * ω.

Therefore, the angular momentum of the rod is ((1/12) * M * L^2) * ω.