Question:
A boy is seated in a revolving chair revolving at an angular speed of 120 revolutions per minute. Two heavy balls form part of the revolving system and the boy can pull the balls closer to himself or may push them apart. If by pulling the balls closer, the boy decreases the moment of inertia of the system by $6 \mathrm{~kg}_{-} m^{2}$ to $2 \mathrm{~kg}^{-} \mathrm{m}^{2}$, what will be the new angular speed?
Solution:
Since, ${ }^{\tau_{\text {ext }}}=0$
$\therefore L_{i}=L_{f}$
$6(120)=(2)\left(\omega_{f}\right)$
$\omega_{f}=360$
$\mathrm{rev} / \mathrm{sec}$
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