An acrobatic physics professor stands at the center of turntable, holding his ar

Question

An acrobatic physics professor stands at the center of turntable, holding his arms extended horizontally, with a 5.0 kg dumbbell in each hand. He is set rotating about a vertical axis, making one revolution in 2.0 seconds. His moment of inertia, without dumbbells, is 3.0 kg * m^2 when his arms are outstretched and drops to 2.2 kg * m^2 when his arms are pulled in close to his chest. The dumbbells are 1.0 m from thesis initially and 0.2 m from it at the end. Find the professor's new angular velocity if he pulls the dumbbells close to his chest, and compare the final kinetic energy with the initial. please show work

Explanation / Answer

apply the law of conservation of angular momentum:

I2w2 = I1w1

Hence, the new angular velocity is,

w2 = I1w1/I2

= [I1cm+2mr12]w1/ [I2cm+2mr22]

= [3+2(5)(1)2][1 rev/2 s] /  [2.2+2(5)(0.2)2]

= 2.5 rev/s

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the initial kinetic energy is,

Ki = 0.5I1w12 = 0.5*[3+2(5)(1)2][{(1 rev/2 s)(2*pi / rev)}2] = 64 J

the final kinetic energy is,

Kf = 0.5I2w22 = 0.5*[2.2+2(5)(0.2)2][{(2.5)(2*pi / rev)}2] = 320 J

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