A large horizontal circular platform (M=102.5 kg, r=3.71 m) rotates about a fric
ID: 1470186 • Letter: A
Question
A large horizontal circular platform (M=102.5 kg, r=3.71 m) rotates about a frictionless vertical axle. A student (m=60.43 kg) walks slowly from the rim of the platform toward the center. The angular velocity of the system is 3.12 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 1.48 m from the center of the platform.
D.Find the angular speed when the student is 1.48 m from the center of the platform.
Explanation / Answer
A) moment of inertia of platform = 0.5*M*r^2
= 0.5*102.5*3.71^2
= 705.4 kg.m^2
B) moment of inertia of the student = m*r^2
= 60.43*3.71^2
= 831.8 kg.m^2
C) when,the student is at r = 1.48 m
moment of inertia of the student = m*r^2
= 60.43*1.48^2
= 132.4 kg.m^2
D) Apply conservation of angular momentum
I2*w2 = I1*w1
w2 = I1*w1/I2
= (705.4 + 132.4)*3.12/(705.4 + 831.8)
= 1.7 rad/s
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