A thin uniform rod $(X)$ of mass $M$ and length $L$ is pivoted at a height $\left(\frac{L}{3}\right)

Question

A thin uniform rod $(X)$ of mass $M$ and length $L$ is pivoted at a height $\left(\frac{L}{3}\right)$ as shown in the figure. The rod is allowed to fall from a vertical position and lie horizontally on the table. The angular velocity of this rod when it hits the table top, is $\_\_\_\_$ .
(g = gravitational acceleration)

Professor PC Thomas Classes · Answer

The center of mass (CM) of the rod is at a distance of $L/2$ from either end. The distance between the pivot and the CM is $d = |L/2 - L/3| = L/6$. The moment of inertia of the rod about the pivot is found using the parallel axis theorem: $I = I_{CM} + Md^2 = \frac{ML^2}{12} + M(\frac{L}{6})^2 = \frac{ML^2}{12} + \frac{ML^2}{36} = \frac{ML^2}{9}$. As the rod falls from a vertical to a horizontal position, the CM descends by a vertical distance $h = d = L/6$. By conservation of mechanical energy, the loss in potential energy equals the gain in rotational kinetic energy: $Mgh = \frac{1}{2}I\omega^2 \implies Mg(\frac{L}{6}) = \frac{1}{2}(\frac{ML^2}{9})\omega^2 \implies \frac{gL}{6} = \frac{L^2}{18}\omega^2 \implies \omega^2 = \frac{3g}{L} \implies \omega = \sqrt{\frac{3g}{L}}$.