A thin uniform rod $(X)$ of mass $M$ and length $L$ is pivoted at a height $\left(\frac L3\right)$ 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 ______ .
( $\text{g}$ = gravitational acceleration)
Answer: (B) $\sqrt{\dfrac{3g}{L}}$
The pivot is at the level of the table top, $L/3$ from the lower end. The centre of mass is $\dfrac L2 - \dfrac L3 = \dfrac L6$ above the pivot. When the rod lies on the table it is horizontal at the pivot's level, so the centre of mass falls by $\dfrac L6$.
Moment of inertia about the pivot:
$$I = \frac{ML^2}{12} + M\left(\frac L6\right)^2 = \frac{ML^2}{12} + \frac{ML^2}{36} = \frac{ML^2}{9}$$
Energy conservation:
$$Mg\frac L6 = \frac12\cdot\frac{ML^2}{9}\omega^2\Rightarrow\omega^2 = \frac{3g}{L}$$
$$\omega = \sqrt{\frac{3g}{L}}$$
Solution by Sreeraj P, M.Sc Physics