A solid disk rotates in the horizontal plane at an angular velocity of 0.0609 ra

Question

A solid disk rotates in the horizontal plane at an angular velocity of 0.0609 rad/s with respect to an axis perpendicular to the disk at its center. The moment of inertia of the disk is 0.120 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 0.386 m from the axis. The sand in the ring has a mass of 0.514 kg. After all the sand is in place, what is the angular velocity of the disk? (Can you please draw a picture)

Explanation / Answer

Initial angular velocity w1 =0.0609 m/s
Initial moment of inertia I1 = 0.120 kg-m^2

Mass of ring m = 0.514 kg
Radius of ring r = 0.386 m
Moment of inertia of ring = mr^2 =0.514 *0.386 ^2 = 0.0765 kg-m^2
Total moment of inertia of ring+disk I2 = I1 + 0.0765 kg-m^2
Or I2 = 0.120 + 0.0765 = 0.1965
Let w2 = final angular velocity

By conservation of angular momentum,
I1 * w1 = I2 * w2
Or 0.120 * 0.0609 =0.1965 * w2
Or w2 = 0.120 * 0.0609/0.1965
= 0.03719 rad/s

Ans: 0.03719 rad/s

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