Question:
A driver having a moment of inertia of $6.0 \mathrm{~kg}_{-} \mathrm{m}^{2}$ about an axis through its centre of mass rotates at an angular speed of $2 \mathrm{rad} / \mathrm{s}$ about the axis. If he folds his hands and feet to decrease the moment of inertia to $5.0 \mathrm{~kg}_{-} \mathrm{m}^{2}$, what will be the new angular speed?
Solution:
Since, $\tau_{\text {ext }}=0$
$\therefore L_{i}=L_{f}$
$6(2)=5^{\omega_{f}}$
$\omega_{f}=2.4 \mathrm{rad} / \mathrm{sec}$