triton.physics.ndsu nodak.edu A large horizontal circular platform (M-92.5 kg, r

Question

triton.physics.ndsu nodak.edu A large horizontal circular platform (M-92.5 kg, r 4.14 m) rotates about a frictionless vertical axle. A vertical axle. A student (m-72.78 kg) walks siowly from the rim of the platform toward the center. The angular velocity of the system is 3.24 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 2378 12kg"mg Submit Answer Tries 1/5 Previous Tries Find the moment of inertia of the student about the center axis (while standing at the rim) of the platform Submit Answer Tries 0ys Find the moment of inertia of the student about the center axis while the student is standing 1.28 m from the center of the platform. Submit Answer Tries 0ys Find the angular speed when the student is 1.28 m from the center of the platform. Submit Answer Tries o/s Tries o/s Submit Anawer 18 29

Explanation / Answer

Given that

Cicular platform of mass(M) =92.5kg

Radius (r) =4.14m

Mass of the student is (m) =72.78kg

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

=0.5*92.5*(4.14)2+(72.78)(4.14)2

=2040.126kg.m2

Given that

Angular speed (w) =3.24rad/s

Moment inertia when student at a distance d' =1.28m is

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

=0.5*92.5*(4.14)2+(92.5)*(1.28)2=792.706+151.552=944.258kg.m2

Now from the conservation of angular momentum

Iw =I'w'

Therefore the required angualr speed is w' =Iw/I' =(2040.126kg.m2)(3.24)/944.258kg.m2=7.00rad/s

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