A solid disk rotates in the horizontal plane at an angular velocity of 0.0653 rad/s with respect to an axis perpendicular to the disk at its center. The moment of inertia of the disk is 0.156 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 0.391 m from the axis. The sand in the ring has a mass of 0.511 kg. After all the sand is in place, what is the angular velocity of the disk?

Answer 1

I think they're trying to get to a conservation of angular momentum question, i.e.
Initial momentum = I omega = .156*.0653
Sand added is zero as it has no angular component

Final is
(.156 omegaf) + (1/2 mr^2 omegaf) (we are assuming the sand is a thin walled cylinder, m and r given)
Set initial equal to final and solve for omegaf