Physics - Rotation Question with Solution | TestHub

The moment of inertia of the rod about the pivot at its end is $I=\frac{1}{3}M{L}^2$. When the rod is released from a horizontal position, its center of mass falls by a vertical distance of $h=\frac{L}{2}$.

By conservation of mechanical energy, the loss in potential energy equals the gain in rotational kinetic energy: $$Mgh=\frac{1}{2}I{\omega }^2$$. Substituting the values, $$Mg\frac{L}{2}=\frac{1}{2}\left(\frac{1}{3}M{L}^2\right){\omega }^2$$. Simplifying, $$\frac{MgL}{2}=\frac{1}{6}M{L}^2{\omega }^2$$. This gives $${\omega }^2=\frac{3g}{L}$$. Therefore, the angular speed is $$\omega =\sqrt{\frac{3g}{L}}$$.