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 :
Option: 1 $\cos \theta =\frac{2g}{3l\omega ^{2}}$
Option: 2 $\cos \theta =\frac{g}{2l\omega ^{2}}$
Option: 3 $\cos \theta =\frac{g}{l\omega ^{2}}$
Option: 4 $\cos \theta =\frac{3g}{2l\omega ^{2}}$

Answers (1)

$\begin{array}{l} \mathrm{F}_{\mathrm{V}}=\mathrm{mg} \\ \mathrm{F}_{\mathrm{H}}=\mathrm{m} \omega^{2} \frac{\ell}{2} \sin \theta \end{array}$

$\begin{array}{l} \therefore \tau_{\mathrm{net}} \text { about } \mathrm{COM}=\mathrm{F}_{\mathrm{v}} \cdot \frac{\ell}{2} \sin \theta-\mathrm{F}_{\mathrm{H}}( \frac{\ell}{2} \cos \theta) =\frac{\mathrm{m} \ell^{2}}{12} \omega^{2} \sin \theta \cos \theta \end{array}$

$\begin{array}{l} \Rightarrow \operatorname{mg}\left(\frac{\ell}{2} \sin \theta\right)-\operatorname{m\omega }^{2}\left(\frac{\ell}{2} \sin \theta\right)\left(\frac{\ell}{2} \cos \theta\right)=\operatorname{m\omega }^{2}\left(\frac{\ell^{2}}{12}\right)(\sin \theta)(\cos \theta) \\ \\ \Rightarrow \frac{g \ell}{2}-\frac{\omega^{2} \ell^{2}}{4} \cos \theta=\frac{\ell^{2}}{12} \omega^{2} \cos \theta \\ \\ \Rightarrow \frac{g \ell}{2}=\omega^{2} \ell^{2} \cos \theta\left(\frac{1}{12}+\frac{1}{4}\right) \\ \\ \Rightarrow \frac{g \ell}{2}=\frac{\omega^{2} \ell^{2} \cos \theta}{3} \\ \\ \Rightarrow \cos \theta=\frac{3 g}{2 \omega^{2} \ell} \end{array}$

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