A diver having a moment of inertia of $6.0 \mathrm{~kg}-\mathrm{m}^2$ about an axis through its cent

Question

A diver 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 this 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?

Professor PC Thomas Classes · Answer

Since there is no external torque acting on the diver, the angular momentum remains conserved.
$
\begin{gathered}
I_1 \omega_1=I_2 \omega_2 \
6.0 	imes 2=5.0 	imes \omega_2 \
\omega_2=\frac{12.0}{5.0}=2.4 \mathrm{rad} / \mathrm{s}
\end{gathered}
$