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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 ) $\frac{ m \ell^{2}}{12} \omega^{2} \sin \theta \cos \theta$ 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:
$\cos \theta=\frac{g}{2 \ell \omega^{2}}$
$\cos \theta=\frac{3 g}{2 \ell \omega^{2}}$
$\cos \theta=\frac{2 g}{3 \ell \omega^{2}}$
$\cos \theta=\frac{g}{\ell \omega^{2}}$
Solution
$F_{V}=m g$
$F_{H}=m \omega^{2} \frac{\ell}{2} \sin \theta$
$mg \frac{\ell}{2} \sin \theta- m \omega^{2} \frac{\ell}{2} \sin \theta \frac{\ell}{2} \cos \theta=\frac{ m \ell^{2}}{12} \omega^{2} \sin \theta \cos \theta$
$\cos \theta=\frac{3}{2} \frac{g}{\omega^{2} \ell}$ $….(ii)$
