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 makes an angle $θ$ with it (see figure). To find $θ$ equate the rate of change of angular momentum (direction going into the paper) $\frac{m l^{2}}{12} ω^{2} \sin θ \cos θ$ 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 $θ$ is then such that :

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$\cos θ = \frac{2 g}{3 l ω^{2}}$

$\cos θ = \frac{g}{l ω^{2}}$

$\cos θ = \frac{g}{2 l ω^{2}}$

$\cos θ = \frac{3 g}{2 l ω^{2}}$

answer is A.

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Detailed Solution

$\begin{array}{l} F r o m a r o t a t i n g f r a m e r o d w i l l a p p e a r i n e q u i l i b r i u m . N e t t o r q u e a b o u t s u s p e n s i o n p o i n t m u s t b e z e r o . \\ τ_{C . F} = τ_{mg} \\ N o w, t o r q u e d u e t o g r a v i t y = τ_{mg} = mg \frac{ℓ}{2} \sin θ \\ t o r q u e d u e t o c e n t r i f u g a l f o r c e = τ_{C . F} = \int F_{C F} (x \sin θ) = \int dm (xcos θ) ω^{2} xsin θ = \int \frac{m}{ℓ} d x x^{2} ω^{2} \cos θ \sin θ = \frac{m ℓ^{2}}{3} ω^{2} \cos θ \sin θ \\ ∴ mg \frac{ℓ}{2} \sin θ = \frac{m ℓ^{2} ω^{2} \cos θ \sin θ}{3} \\ \Rightarrow \cos θ = \frac{3 g}{2 ℓ ω^{2}} \end{array}$