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 $\mathrm{\omega }$ the rod marks an angle $\mathrm{\theta }$ with it (see figure). To find $\mathrm{\theta }$ equate the rate of change of angular momentum (direction going into the paper $\frac{{\mathrm{ml}}^2}{12}{\mathrm{\omega }}^2\sin \mathrm{\theta cos}\mathrm{\theta }$ about the centre of mass (CM) to the torque provided by the horizontal and vertical forces F_v and F_H about the CM. the value of $\mathrm{\theta }$ is then such that