A student sits on a freely rotating stool holding two weights, each of mass 3.00

Question

A student sits on a freely rotating stool holding two weights, each of mass 3.00 kg. When his arms are extended horizontally, the weights are 1.03 m from the axis of rotation and he rotates with an angular speed of 0.747 rad/s. The moment of inertia of the student plus stool is 3.00 kg · m2 and is assumed to be constant. The student pulls the weights inward horizontally to a position 0.305 m from the rotation axis.

(a) Find the new angular speed of the student. rad/s

(b) Find the kinetic energy of the rotating system before and after he pulls the weights inward. J (before) and J (after)

Explanation / Answer

This is nothing but a simple application of Angular Moment Conservation for the system

Where

I1= (2*3*1.032)+3

I2= (2*3*.3052)+3

I1w1=I2w2

(a) .747((2*3*1.032)+3) = .wnew((2*3*.3052)+3)

wnew = 1.96 rad/s

(b)

Initial Kinetic Energy = .5* I1*w12 = .5*.7472 * ((2*3*1.032)+3) = 2.612 Joule

Final Kinetic Energy = .5* I2*w22= .5* 1.962 * ((2*3*.3052)+3) = 6.83 Joule

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