A solid disk rotates in the horizontal plane at an angular velocity of 0.078 rad

Question

A solid disk rotates in the horizontal plane at an angular velocity of 0.078 rad/s with respect to an axis perpendicular to the disk at its center. The moment of inertia of the disk is 0.074 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 0.55 m from the axis. The sand in the ring has a mass of 0.59 kg. After all the sand is in place, what is the angular velocity of the disk?

A playground carousel is free to rotate about its center on frictionless bearings, and air resistance is negligible. The carousel itself (without riders) has a moment of inertia of 136 kg·m2. When one person is standing at a distance of 1.79 m from the center, the carousel has an angular velocity of 0.601 rad/s. However, as this person moves inward to a point located 0.714 m from the center, the angular velocity increases to 0.798 rad/s. What is the person's mass?

Explanation / Answer

1)

here

Mass of ring m = 0.47 kg
Radius of ring r = 0.46 m

Moment of inertia of ring = m * r^2 = 0.59 * 0.55^2 = 0.17847 kgm^2

Total moment of inertia

I2 = I1 + 0.17847

I2 = 0.074 + 0.17874 = 0.25274 kg m^2

then conservation of angular momentum,

I1 * w1 = I2 * w2

0.074 * 0.078 = 0.25274 * w2

w2 = 0.022837 rad/s

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