A man stands in the center of a platform that is rotating (without friction) wit

Question

A man stands in the center of a platform that is rotating (without friction) with an angular speed of 1.2 revolutions per second, his arms are outstretched and he holds a brick in each hand. The rotational inertia of the system consists of the mans body, the bricks and the platform about a central vertical axis and has a total value of 6 kilograms - meter^2. If by moving the bricks in towards his side the man decreases the rotational inertia of the system to only 2 kilograms - meter^2 what will be the resulting angular speed of the spinning man now?

Explanation / Answer

from the conservation of angular momentum,

(angular momentum, L=I)

I11=I22

therefore,

6*1.2=2*final

hence, final = 3.6 revolution/s

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