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 3.69 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 7.7 kg*m^2. When the man pulls the weights inward toward his body, the moment of inertia decreases to 2.4 kg*m^2. What is the resulting angular speed of the platform? 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

given

w1 = 3.69 rev/s = 3.69*2*pi rad/s = 23.18 rad/s

I1 = 7.7 kg.m^2

I2 = 2.4 kg.m^2

a) Let w2 is the final angular velcoity

Apply conservation of angular momentum

I2*w2 = I1*w1

w2 = I1*w1/I2

= 7.7*3.69/2.4

= 11.84 rev/s or 74.35 rad/s

b) KEi = 0.5*I1*w1^2

= 0.5*7.7*23.18^2

= 2068.7 J

c)
KEf = 0.5*I2*w2^2

= 0.5*2.4*74.35^2

= 6633.5 J

d) change in kinetic energy, delta_KE = KEf - KEi

= 6633.5 - 2068.7

= 4564.8 J

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