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 0.825 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 5.76 kg·m2. If by moving the bricks the man decreases the rotational inertia of the system to 2.43 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?

Explanation / Answer

given angular speed of platform, w = 0.825 rev/s = 0.825*2*pi rad/s

initial rotational inertia of the system Ii = 5.76 kg m^2

final rotational inertia of the system, If = 2.43 kg m^2

a. initial angular momentum of the system = Ii*w

final angular momentum of the system = If*wf ( where wf is the final angular speed of the system)

from conservation o f angular momentum

If*wf = Ii*w

wf = Ii*w/If = 1.9555 rev/s

b. new KE of the system = 0.5*If*wf^2 = 183.43251 J

initial KE = 0.5*Ii*w^2 = 77.3855 J

ratio = 2.37037

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