Question

A solid disk rotates in the horizontal plane at an angular velocity of .067rad/s with respect to an axis perpendicular to the disk at its center. The moment of inertia of the disk is .10 kg*m2. From above, sand is dropped straight down onto this rotating disk, so that a think uniform ring has a mass of .5 kg. After all the sand is in place, what is the angular velocity of the disk?

Answer #1

Given is:-

Initial angular velocity w_{1} = 0.067 m/s

Initial moment of inertia I_{1} = 0.10
kg-m^{2}

Mass of ring m = 0.50 kg

Radius of ring r = 0.40 m

Now,

Moment of Inertia of the sand ring is

or

Total moment of inertia of the ring and the disk is

Let w_{2} be the final angular veloctiy

Thus

By conservation of angular momentum

by putting all the values in above quation, we get

or

thus the final angular velocity of system will be 0.0372 rad/s

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