A boy stands over the center of a horizontal platform which is rotating freely with a speed of 2 revolutions per second about a vertical axis through the center of the platform and straight up through the boy. He holds 2 kg masses in each of his hands close to his body. The combined moment of inertia of the system is 1 kg x metre². The boy now stretches his arms so as to hold the masses far from his body. In this situation, the moment of inertia of the system increases to 2 kg x metre². The kinetic energy of the system in the latter case as compared with that in the previous case will

Easy

Solution

Since, external torque on system is zero, angular momentum is conserved.
$L = I ω$
So, if I is doubled. $ω$ is halved
Now, Rotational K.E. $= \frac{L^{2}}{2 I}$
Clearly, rotational KE decreases (becomes half).

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