Question

A large horizontal circular platform (M=97.3 kg, r=4.71 m) rotates about a frictionless vertical axle. A student (m=78.94 kg) walks slowly from the rim of the platform toward the center. The angular velocity ω of the system is 2.56 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. answer = 1079.26 kg*m^2

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 1.02 m from the center of the platform.

D. Find the angular speed when the student is 1.02 m from the center of the platform.

Answer 1

here,

mass of student, m = 97.3 kg

radius of platform, r = 4.71 m

Part a:
moment of inertia of platform about z axis
Ip = 0.5*m*r^2
Ip = 0.5 * 97.3 * 4.71^2
Ip = 1079.256 kg.m^2

Part b:
moment of inertia of student about central axis
Ib = m*r^2
Ib = 78.94 * 4.71^2
Ib = 1751.213 kg.m^2

Part c:
moment of inertia of student about central axis
Ib' = m*r^2
Ib' = 78.94 * 1.02^2
Ib' = 82.129 kg.m^2

Part d:
from conservation of angular momentum we have :

(ip+ib) * w1 = (ip+Ib') * w2

final angular speed, w2 = (ip+ib)/(ip+Ib') * w1

final angular speed, w2 = (1079.256+1751.213)/(1079.256+82.129) * 2.56

final angular speed, w2 = 6.24 rad/s