A man stands at the center of a platform that rotates without friction with an a

Question

A man stands at the center of a platform that rotates without friction with an angular speed of 1.21 rev/s. His arms are outstretched, and he holds a heavy weight in each hand. The moment of inertia of the man, the extended weights, and the platform is 11.6 kg*m^2. When the man pulls the weights inward toward his body, the moment of inertia decreases to 4.4 kg*m^2. What is the kinetic energy of the system before the man pulls the weights to his body?

What is the kinetic energy of the system after the man pulls the weights to his body??

What is the change in kinetic energy of the system?

Explanation / Answer

Rotational kinetic energy is given by:

KE = 0.5*I*w^2

Initial KE will be

KEi = 0.5*I1*w1^2

w1 = 1.21 rev/sec = 7.6 rad/sec

Using given values:

KEi = 0.5*11.6*7.6^2 = 335.01 J

Final KE will be

using momentum conservation

L1 = L2

I1*w1 = I2*w2

w2 = I1*w1/I2 = 11.6*7.6/4.4 = 20.04 rad/sec

KEf = 0.5*I2*w1^2

Using given values:

KEf = 0.5*4.4*20.04^2 = 883.52 J

Change in KE = 335.01 - 883.52 = -548.51 J

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