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) $${{m{l^2}} \over {12}}{\omega ^2}\sin \theta \cos \theta$$ about the centre of mass (CM) to the torque provided by the horizontal and vertical forces FH and FV about the CM. The value of $\theta$ is then such that :