A large horizontal circular platform (M=95.3 kg, r=3.74 m) rotates about a frict

Question

A large horizontal circular platform (M=95.3 kg, r=3.74 m) rotates about a frictionless vertical axle. A student (m=61.31 kg) walks slowly from the rim of the platform toward the center. The angular velocity of the system is 3.93 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 platformFind the moment of inertia of the student about the center axis while the student is standing 1.52 m from the center of the platform.Find the angular speed when the student is 1.52 m from the center of the platform.

Explanation / Answer

moment of inertia of platform = M r^2 /2 = 95.3 x 3.74^2 /2 = 666.51 kg m^2

when student was at rim,

moment of inertia of student = m r^2 = 61.31 x 3.74^2 = 857.58 kg m^2

when student is at r = 1.52 m,

moment of inertia of student = 61.31 x 1.52^2 = 141.65 kg m^2

using angular momentum conservation,

initial angular moentum = final angular momentum

Iiwi = Ifwf

( 666.51 + 857.58) x 3.93 = ( 666.51 + 141.65) wf

wf = 7.41 rad/s

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