Q 11-06-172JEE MainJEE Main 2020 (9 Jan, Shift 1)Medium
One end of a straight uniform $1$ m long bar is pivoted on a horizontal table. It is released from rest when it makes an angle $30^\circ$ from the horizontal. Its angular speed when it hits the table is given as $\sqrt{n}$ rad s$^{-1}$, where $n$ is an integer. The value of $n$ is ______. (Take $g = 10$ m s$^{-2}$)
Numerical value type. Enter your answer.
Answer: 15
The centre of mass falls through $\dfrac{L}{2}\sin30^\circ = 0.25$ m. The bar rotates about the pivot with $I = \dfrac{mL^2}{3}$:
$$mg(0.25) = \frac{1}{2}\cdot\frac{m(1)^2}{3}\,\omega^2$$
$$\omega^2 = 6\times0.25\times10 = 15 \Rightarrow \omega = \sqrt{15}\ \text{rad s}^{-1}$$
Solution by Sreeraj P, M.Sc Physics