# Rotational motion

PROBLEM 2:

A playground whirl consists of a uniform disk of mass 10M and radius R. It is spinning at a rate of 2.0 rmp. A child mass M, originally at rest, steps on to the whirl. She does not slide back off.

What is the final angular velocity of the whirl and the child (assuming the axis is frictionless)?

What fraction of the initial kinetic energy is lost? Where did it go?

PROBLEM: 3

The whirl is problem 2 is now connected to a motor that keeps it rotating at a constant 20.0 rpm even with the child aboard. She is initially at the center. After awhile she begins walking from the center. Draw a free body diagram of the child when she is a distance r from the center. If the coefficient of friction is 0.25, how far can she walk from the center before sliding off. How fast is she going at the maximum distance?

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#### Solution Preview

Please refer to the attachment.

Angular velocity = ω = 2Πν = 2Π(2/60) = 4Π/60 = Π/15 rad/sec

Moment of inertia of the whirl = I = ½ (10M)R2 = 5MR2

Initial angular momentum of the whirl = Iω = (5MR2)(Π/15) = (MR2)(Π/3)

As soon as the child steps on to the whirl, the angular velocity of the whirl changes so as to conserve net angular momentum of the whirl and the girl.

Angular momentum of the child before it steps on to the whirl = 0

Moment of inertia of the child after it steps on the whirl = MR2

Angular momentum of the child after it steps on to the whirl = MR2ω' where ω' is the angular velocity of the ...

#### Solution Summary

The expert examines the rotational motion and kinetic energy lost. A step by step solution provided.