Question

A solid disk rotates in the horizontal plane at an angular velocity of 0.0612 rad/s with respect to an axis perpendicular to the disk at its center. The moment of inertia of the disk is 0.134 kg·m2. From above, sand is dropped straight down onto this rotating disk, so that a thin uniform ring of sand is formed at a distance of 0.398 m from the axis. The sand in the ring has a mass of 0.509 kg. After all the sand is in place, what is the angular velocity of the disk?

Answer 1

Initial angular velocity, w1 = 0.0612 m/s
Initial moment of inertia, I1 = 0.134 kg-m^2

Mass of ring m = 0.509 kg
Radius of ring r = 0.398 m
Moment of inertia of ring = mr^2 = 0.509 * 0.398^2 = 0.081 kg-m^2
Total moment of inertia of ring+disk I2 = 0.134 + 0.081 = 0.215 kg-m^2

Let w2 = final angular velocity

By conservation of angular momentum,
I1 * w1 = I2 * w2
0.134 * 0.0612 = 0.215 * w2
w2 = 0.038 rad/s
Angular velocity of the disk, w2 = 0.038 rad/s