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 rod marks an angle $θ$ with it (see figure). To find $θ$ equate the rate of change of angular momentum (direction going into the paper $\frac{{ml}^{2}}{12} ω^{2} \sin θcos θ$ about the centre of mass (CM) to the torque provided by the horizontal and vertical forces F_v and F_H about the CM. the value of $θ$ is then such that

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$\cos θ = \frac{g}{{lω}^{2}}$

$\cos θ = \frac{2 g}{3 {lω}^{2}}$

$\cos θ = \frac{3 g}{2 {lω}^{2}}$

$\cos θ = \frac{g}{2 {lω}^{2}}$

answer is D.

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Detailed Solution

for equilibrium, net force on the rod should be zero.

$\Rightarrow F_{V} = mg$ ; and, $F_{H} = {mω}^{2} \frac{ℓ}{2} sinθ$

since, about centre of mass,

Torque = rate of change of Angular momentum

$\Rightarrow mg (\frac{ℓ}{2} \sin θ) - {mω}^{2} (\frac{ℓ}{2} \sin θ) (\frac{ℓ}{2} \cos θ) = \frac{{mℓ}^{2}}{12} ω^{2} \sin θcos θ$

$\Rightarrow \cos θ = \frac{3 g}{2 ω^{2} ℓ}$

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