A thin uniform rod X of mass M and length L is pivoted at a height $$(\frac{L}{3})$$ 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 is the acceleration due to gravity)

Moment of Inertia about the Pivot ($$I_P$$):

The rod is pivoted at a distance of $$L/3$$ from the bottom. The center of mass (CM) of the rod is at its midpoint, $$L/2$$ from the bottom.

The distance ($$d$$) between the pivot and the center of mass is:

$$d = \frac{L}{2} - \frac{L}{3} = \frac{L}{6}$$

Using the Parallel Axis Theorem:

$$I_P = I_{cm} + Md^2$$

$$I_P = \frac{1}{12}ML^2 + M\left(\frac{L}{6}\right)^2$$

$$I_P = \frac{1}{12}ML^2 + \frac{1}{36}ML^2 = \frac{1}{9}ML^2$$

Change in Potential Energy ($$\Delta PE$$):

When the rod falls from the vertical position to the horizontal position, the center of mass descends by a vertical height equal to its initial distance from the pivot.

$$\Delta h = d = \frac{L}{6}$$

$$\Delta PE = Mg\Delta h = Mg\left(\frac{L}{6}\right)$$

Conservation of Energy:

$$\Delta PE = \Delta KE_{rot}$$

$$Mg\left(\frac{L}{6}\right) = \frac{1}{2} I_P \omega^2$$

$$Mg\frac{L}{6} = \frac{1}{2} \left(\frac{1}{9}ML^2\right) \omega^2$$

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