Q 11-06-241JEE MainJEE Main 2017 (2 Apr)Easy
A slender uniform rod of mass $M$ and length $l$ is pivoted at one end so that it can rotate in a vertical plane (see figure). There is negligible friction at the pivot. The free end is held vertically above the pivot and then released. The angular acceleration of the rod when it makes an angle $\theta$ with the vertical is:
Answer: (B) $\dfrac{3g}{2l}\sin\theta$
Gravity acts at the centre, a distance $l/2$ from the pivot. When the rod makes angle $\theta$ with the vertical, the lever arm of $Mg$ is $\frac{l}{2}\sin\theta$:
$$\tau = Mg\,\frac{l}{2}\sin\theta$$
About the end, $I = \dfrac{Ml^2}{3}$, so
$$\alpha = \frac{\tau}{I} = \frac{Mg\frac{l}{2}\sin\theta}{\frac{Ml^2}{3}} = \frac{3g}{2l}\sin\theta$$
Solution by Sreeraj P, M.Sc Physics