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.