A large horizontal circular platform (M=98.2 kg, r=4.81 m) rotates about a frict

Question

A large horizontal circular platform (M=98.2 kg, r=4.81 m) rotates about a frictionless vertical axle. A student (m=58.09 kg) walks slowly from the rim of the platform toward the center. The angular velocity ? of the system is 3.18 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 1.61 m from the center of the platform.

Find the angular speed when the student is 1.61 m from the center of the platform.

Explanation / Answer

Given that

Cicular platform of mass(M) =98.2kg

Radius (r) =4.81m

Mass of the student is (m) =58.09kg

The moment of inertia when a student at rim is I =(1/2)Mr2+mr2

=0.5*98.2*(4.81)2+(58.09)(4.81)2

=2480kg.m2

Given that

Angular speed (w) =3.18rad/s

Moment inertia when student at a distance d' =1.61m is

I' =(1/2)Mr2+Md'2

=0.5*98.2*(4.81)2+(98.2)*(1.28)2=1135.982+160.890=1296.872kg.m2

Now from the conservation of angular momentum

Iw =I'w'

Therefore the required angualr speed is w' =Iw/I' =(2480kg.m2)(3.18)/1296.872kg.m2=6.08rad/s

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