JEE Main 2020 — Rotational Motion Question with Solution

A uniform rod of length ‘$$l$$’ is pivoted at one of its ends on a vertical shaft of negligible radius. When the shaft rotates at angular speed $$\omega$$ the rod makes an angle $$\theta$$ with it (see figure). To find $$\theta$$ equate the rate of change of angular momentum (direction going into the paper) $$\frac{m{l}^2}{12}{\omega }^2\sin \theta \cos \theta$$ about the centre of mass (CM) to the torque provided by the horizontal and vertical forces F_H and F_V about the CM. The value of $$\theta$$ is then such that :