Question

A man stands on a platform that is rotating (without friction) with an angular speed of 1.29 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 4.66 kg.m^2. If by moving the bricks the man decreases the rotational inertia of the system to 1.03 kg-m^2, (a) what is the resulting angular speed of the platform and (b) what ?s the ratio of the new kinetic energy of the system to the original kinetic energy?

Answer 1

Apply conservation of angular momentum initial and final positions of the person

L_i = L_f

I_i w_i = I_f w _f)

4.66 kg m^2 ( 1.29 rev/s) = 1.03 kg m^2 ( W_f)

w_f = 5.836 rev/s ( 2 pi rad/ rev ) = 36.65 rad/s