Question 24 - 2020 (03 Sep Shift 2)

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) $\frac{ml^2}{12}\omega^2\sin\theta\cos\theta$ 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:

(1) $\cos\theta = \frac{2g}{3l\omega^2}$

(2) $\cos\theta = \frac{g}{2l\omega^2}$

(3) $\cos\theta = \frac{g}{l\omega^2}$

(4) $\cos\theta = \frac{3g}{2l\omega^2}$