Question

A large horizontal circular platform (M=91 kg, r=3.99 m) rotates about a frictionless vertical axle. A student (m=77.93 kg) walks slowly from the rim of the platform toward the center. The angular velocity ω of the system is 3.8 rad/s when the student is at the rim.

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

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

Find the moment of inertia of the student about the center axis while the student is standing 1.09 m from the center of the platform

Find the angular speed when the student is 1.09 m from the center of the platform.

Answer 1

This is a conservation of angular momentum problem.
Angular momentum = I * ω

For the platform, I = ½*Mr² = ½ * 91 * 3.99^2 = 724.36 kg.m²

For the student, I = mr² = 77.93 * 3.99^2 = 1240.65 kg.m²
Total I = 1965.01 kg.m²

Initial angular momentum = 1965.01*3.8 = 7467.05 kg.m²

When the student is 1.09 m from the center, I = 77.93 * 1.09² = 92.58 kg.m²
Total momentum for platform and student is I = 724.36 + 92.58 = 816.94 kg.m²

Final angular momentum = 816.94 * ωf
816.94 * ωf = 7467.05
ωf = 9.14 rad/s