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:
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:
- [JEE MAIN 2020]
- A
$\cos \theta=\frac{g}{2 \ell \omega^{2}}$
- B
$\cos \theta=\frac{3 g}{2 \ell \omega^{2}}$
- C
$\cos \theta=\frac{2 g}{3 \ell \omega^{2}}$
- D
$\cos \theta=\frac{g}{\ell \omega^{2}}$
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- [AIIMS 2019]
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