Physics - Rotation Question with Solution | TestHub

When the rod is released from a horizontal position, its center of mass is at a height of $L/2$ above its lowest point (when it's vertical). As it swings to the vertical position, its potential energy is converted into rotational kinetic energy. Initial potential energy $P{E}_i=mg(L/2)$. Initial kinetic energy $K{E}_i=0$.

Final potential energy $P{E}_f=0$ (taking the vertical position as reference). Final kinetic energy $K{E}_f=\frac{1}{2}I{\omega }^2$.

The moment of inertia of a uniform rod of mass $m$ and length $L$ about an axis through one end is $I=\frac{1}{3}m{L}^2$. By conservation of mechanical energy: $P{E}_i+K{E}_i=P{E}_f+K{E}_f$.$$mg\frac{L}{2}+0=0+\frac{1}{2}(\frac{1}{3}m{L}^2){\omega }^2$$ $$mg\frac{L}{2}=\frac{1}{6}m{L}^2{\omega }^2$$ $$3g=L{\omega }^2⟹\omega =\sqrt{\frac{3g}{L}}$$