A solid disk rotates in the horizontal plane at an angular velocity of rad/s with respect to an axis perpendicular to the disk at its center. The moment of inertia of the disk is 0.17 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 0.40 m from the axis. The sand in the ring has a mass of 0.50 kg. After all the sand is in place, what is the angular velocity of the disk

Question is not complete and the complete question is;

A solid disk rotates in the horizontal plane at an angular velocity of 0.067 rad/s with respect to an axis perpendicular to the disk at its center. The moment of inertia of the disk is 0.17kg⋅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 0.40 m from the axis. The sand in the ring has a mass of 0.50 kg. After all the sand is in place, what is the angular velocity of the disk?

0.0456 rad/s

Explanation:

We know that angular momentum is given by;

L = Iω

Where;

L is angular momentum

I is moment of inertia

ω is angular velocity

Initial angular momentum is;

L_d = Iω_d

We have that;

ω_d = 0.067 rad/s

Moment of inertia of disk= 0.17 kg.m²

Thus,

L_d = 0.067 x 0.17 = 0.01139 Kg.m²/s

Now, final angular momentum is given as;

L_f = I_f•ω_f

I_f = I_sand + I_disk

Moment of inertia of sand(I_sand) = mr²

r = 0.4m and m= 0.5kg

So, I_sand = 0.5 x 0.4² = 0.08 kg.m²

Now, I_f = 0.08 + 0.17 = 0.25 kg.m²

Now, from conservation of angular momentum, L_i = L_f

Thus, 0.01139 = 0.18•ω_f

So, ω_f = 0.01139/0.25= 0.0456 rad/s

dang that balls went far