Question

A man stands on a platform that is rotating (without friction) with an angular speed of 1.61 rev/s; his arms are outstretched and he holds a brick in each hand. The rotational inertia of the system consisting of the man, bricks, and platform about the central axis is 7.06 kg·m². If by moving the bricks the man decreases the rotational inertia of the system to 1.96 kg·m², (a) what is the resulting angular speed of the platform and (b) what is the ratio of the new kinetic energy of the system to the original kinetic energy?

Answer 1

a) From the conservation of momentum:

I₁w₁ = I₂w₂

7.06 x 1.61 = 1.96 x w₂

w₂ = 5.8 rev/s

b)

Initial kinetic energy = (1/2)I₁w₁² = (1/2) x 7.06 x (1.61)²

Final kinetic energy = (1/2)I₂w₂² = (1/2) x 1.96 x 5.8²

Hence ratio = (1/2) x 1.96 x 5.8² / ((1/2) x 7.06 x (1.61)² ) = 3.6