A rod of length 50 cm is pivoted at one end. It is raised such that it makes an angle of 30$$^{\circ}$$ from the horizontal as shown and released from rest. Its angular speed when it passes through the horizontal (in rad s$$^{-1}$$) will be: ($$g = 10$$ ms$$^{-2}$$)

$$E_i = E_f \implies U_i + K_i = U_f + K_f$$

$$Mg\left(\frac{L}{2}\sin\theta\right) + 0 = 0 + \frac{1}{2}I\omega^2$$

$$Mg\frac{L}{2}\sin30^\circ = \frac{1}{2}\left(\frac{1}{3}ML^2\right)\omega^2 \implies g\sin30^\circ = \frac{1}{3}L\omega^2$$

$$10 \times \frac{1}{2} = \frac{1}{3} \times 0.5 \times \omega^2 \implies 5 = \frac{1}{6}\omega^2 \implies \omega^2 = 30 \implies \omega = \sqrt{30}\text{ rad/s}$$