A student sits on a freely rotating stool holding two weights, each of mass 2.90

Question

A student sits on a freely rotating stool holding two weights, each of mass 2.90 kg. When his arms are extended horizontally, the weights are 1.09 m from the axis of rotation and he rotates with an angular speed of 0.756 rad/s. The moment of inertia of the student plus stool is 2.90 kg.m2 and is assumed to be constant. The student pulls the weights inward horizontally to a position 0.299 m from the rotation axis. (a) (a) Find the new angular speed of the student. (b) Find the kinetic energy of the rotating system before and after he pulls the weights inward.

Explanation / Answer

Here ,

Ii (inital interia) = 2.90 + 2*2.9*1.09^2

Ii = 9.791 Kg.m^2

If (final interia) = 2.90 + 2*2.9*0.299^2

If = 3.42 Kg.m^2

Now , using conservation of anglular momentum

If*wf = Ii*Wi

3.42 * wf = 9.791 * 0.756

wf = 2.17 rad/s

the new angular speed is 2.17 rad/s

b)

as KE = 0.5*I*w^2

KE(before) = 0.5 *9.791 * 0.756^2

KE(Before ) = 2.8 J

Kinectic energy before is 2.8 J

Now ,

KE(after ) = 0.5*3.42 *2.17^2

KE(after) = 8.05 J

Kinetic energy after is 8.05 J

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