Question

A large horizontal circular platform (M=105.1 kg, r=3.54 m) rotates about a frictionless vertical axle. A student (m=54.08 kg) walks slowly from the rim of the platform toward the center. The angular velocity ω of the system is 2.24 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 1.7 m from the center of the platform.

Find the angular speed when the student is 1.7 m from the center of the platform.

Answer 1

a)
L = 1/2 mr^2
= 0.5*105.1*3.54^2
= 658.5 kgm^2

b)
the initial moment of inertia when the student is at the rim is
Ii = 1/2 mr^2 for the platform + mr^2 for the student.
I = 0.5*105.1*3.54^2 + 54.08*3.54^2
= 1336.25 kgm^2

c)
the final moment of inertia with the student at radius 1.7 m is
If = 1/2 * 105.1*3.54^2 + 54.08*1.7^2
= 815 kgm^2

d)
The angular moment is conserved. It is
L = I*ω
The initial angular moment is
Li = 1336.25 kgm^2*2.24 rad/s
= 2993.2 kgm^2s^-1
and so
2993.2 kgm^2s^-1 = If*ωf
2993.2 = 815 *ωf
ωf =3.7 rad/s