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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{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}}

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\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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