Physics - Rotation Question with Solution | TestHub

By the principle of conservation of mechanical energy, the loss in potential energy of the rod is equal to the gain in its rotational kinetic energy.

The initial height of the center of mass of the rod is $h_i=\frac{l}{2}$ and the final height is $h_f=0$.

Loss in potential energy $=mg\left(\frac{l}{2}\right)$

Gain in rotational kinetic energy $=\frac{1}{2}I{\omega }^2$

The moment of inertia of the rod about the hinged end A is $I=\frac{m{l}^2}{3}$.

Equating the two energies:

$mg\left(\frac{l}{2}\right)=\frac{1}{2}\left(\frac{m{l}^2}{3}\right){\omega }^2$

$mgl=\frac{m{l}^2}{3}{\omega }^2$

${\omega }^2=\frac{3g}{l}$

$\omega =\sqrt{\frac{3g}{l}}$

Answer: $\sqrt{\frac{3g}{l}}$