A large horizontal circular platform (M=98.4 kg, r=3.29 m) rotates about a frict

Question

A large horizontal circular platform (M=98.4 kg, r=3.29 m) rotates about a frictionless vertical axle. A student (m=79.21 kg) walks slowly from the rim of the platform toward the center. The angular velocity of the system is 3.06 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 1.53 m from the center of the platform. Find the angular speed when the student is 1.53 m from the center of the platform.

Explanation / Answer

Here ,

For the moment of inertia of platform , I = 0.5 * M * r^2

I = 0.5 * 98.4 * 3.29^2

I = 532.5 Kg.m^2

the moment of inertia of platform is 532.5 Kg.m^2

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Now , for the student at the rim

I1 = m * r^2

I1 = 79.21 * 3.29^2

I1 = 827.4 Kg.m^2

the moment of inetia of student is 827.4 Kg.m^2

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when the student is 1.53 m away from the centre of platform

I2 = m * r^2

I2 = 79.21 * 1.53^2

I2 = 185.4 Kg.m^2

the moment of inertia of student when it is 1.53 m is 185.4 Kg.m^2

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Using conservation of angular momentum

(I + I1) * wi = (I + I2) * wf

( 532.5 + 827.4) * 3.06 = ( 532.5 + 185.4 ) * wf

wf = 5.796 rad/s

the angular speed of the platform when student is 1.53 m from the centre is 5.796 rad/s

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