Q 11-06-202JEE MainJEE Main 2019 (9 Jan, Shift 2)Medium
A rod of length $50\ \text{cm}$ 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 $\text{rad s}^{-1}$) will be $(g = 10\ \text{m s}^{-2})$
Answer: (B) $\sqrt{30}$
The centre of mass is at $L/2$ from the pivot, so it falls through $\dfrac{L}{2}\sin30^\circ = \dfrac{L}{4}$.
The moment of inertia about the end is $\dfrac{mL^2}{3}$. Energy conservation:
$$mg\frac{L}{4} = \frac12\cdot\frac{mL^2}{3}\,\omega^2 \;\Rightarrow\; \omega^2 = \frac{3g}{2L} = \frac{3(10)}{2(0.5)} = 30$$
$$\omega = \sqrt{30}\ \text{rad s}^{-1}$$
Solution by Sreeraj P, M.Sc Physics