A large horizontal circular platform (M=95.8 kg, r=4.19 m) rotates about a frict

Question

A large horizontal circular platform (M=95.8 kg, r=4.19 m) rotates about a frictionless vertical axle. A student (m=51.21 kg) walks slowly from the rim of the platform toward the center. The angular velocity of the system is 3.9 rad/s when the student is at the rim.

A. Find the moment of inertia of platform through the center with respect to the z-axis.

B. Find the moment of inertia of the student about the center axis (while standing at the rim) of the platform.

C. Find the moment of inertia of the student about the center axis while the student is standing 1.59 m from the center of the platform.

D. Find the angular speed when the student is 1.59 m from the center of the platform.

Explanation / Answer

Angular momentum = Moment of Inertia * angular velocity
Momentum is conserved, so the product of Moment of Inertia * angular velocity is constant.

Initial Moment of Inertia * Initial angular velocity = Final Moment of Inertia * Final angular velocity

As the student walks toward the center, the moment of inertia decreases, so the angular velocity must increase.

Initial Moment of Inertia of platform = ½ * mass * radius^2 = ½ * 95.8 * 4.19^2
This does not change!

Initial Moment of Inertia of student = mass * radius^2 = 51.21 * 4.19^2
Total initial Moment of Inertia = ½ * 95.8 * 4.19^2 + 51.21 * 4.19^2



Final Moment of Inertia of student = 51.21 * 1.59^2
Total final Moment of Inertia = ½ * 95.8 * 4.19^2 + 83.3 * 1.59^2


Initial Moment of Inertia * Initial angular velocity = Final Moment of Inertia * Final angular velocity

(½ * 95.8 * 4.19^2 +  51.21 * 4.19^2) * 2.70 = (½ *  95.8 * 4.19^2  + 83.3 * 1.59^2) * f

Solve for f = rad/s

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