A man stands on a platform that is rotating (without friction) with an angular s

Question

A man stands on a platform that is rotating (without friction) with an angular speed of 1.4 rev/s; his arms are outreached and he holds a weight in each hand. The rotational inertia of the system of man, weights, and platform about the central axis is 11.00 kg m2. If by moving the weights the man decreases the rotational inertia of the system to 4.84 kg m2, the resulting angular speed of the platform is 2.00×10^1 rad/s. What is the ratio of the new kinetic energy of the system to the original kinetic energy?

Explanation / Answer

kinetic energy = 1/2I2

1.4rev/s = 1.4*2rad/s=8.79rad/s

initial kinetic energy = 1/2I2 = 425.58

final knetic energy = 1/2I2 = 968

ratio = 968/425.58 = 2.275

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