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 ω the…

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