Question

A solid disk rotates in the horizontal plane at an angular velocity of 0.0647 rad/s with respect to an axis perpendicular to the disk at its center. The moment of inertia of the disk is 0.199 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 0.420 m from the axis. The sand in the ring has a mass of 0.499 kg. After all the sand is in place, what is the angular velocity of the disk?

Answer 1

$initial\: \: angular \: \: velocity\: \: (\omega ) = 0.0647rad/s.$

Conservatiof of angular momentum,

$I_{1}\omega _{1}=(I_{1 }+I_{2})\omega 2$

$\omega 2 =\frac{I_{1}\omega _{1}}{I_{2}+I_{1}} = 0.0449rad/s$