A thin uniform rod \(A B\) of mass ' \(\mathrm{m}\) ' and length ' \(l\) ' is hinged at one end A to the ground level. Initially the rod stands vertically and is allowed to fall freely to the ground in the vertical plane. The angular velocity of the rod when its end B strikes the ground is ( \(g\) = acceleration due to gravity)

We know,
Loss in P.E = Gain in rotational K.E.
When the centre of mass of the rod falls through a distance \(\frac{\mathrm{L}}{2}\),
Loss in P.E \(=m g \frac{L}{2}\) ...(i)
and,
Gain in Rotational K.E \(=\frac{1}{2} \mathrm{I} \omega^2=\frac{1}{2}\left[\frac{\mathrm{ML}^2}{3}\right] \omega^2\) ...(ii)
Equating (i) and (ii)
\(
\begin{aligned}
& \frac{\mathrm{MgL}}{2}=\frac{1}{2} \frac{\mathrm{ML}^2}{3} \omega^2 \\
& \omega^2=\frac{3 \mathrm{~g}}{\mathrm{~L}} \\
& \omega=\sqrt{\frac{3 \mathrm{~g}}{\mathrm{~L}}}=\sqrt{\frac{3 \mathrm{~g}}{l}} \quad \ldots(\because \text { here } \mathrm{L}=l)
\end{aligned}
\)
\(\ldots(\because\) here \(\mathrm{L}=l)\)