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) $(m l^{2} / 12) ω^{2} \sin θ \cos θ$ 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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$\cos θ = 3 g / 2 l ω^{2}$

$\cos θ = g / 2 l ω^{2}$

$\cos θ = 2 g / 3 l ω^{2}$

$\cos θ = g / l ω^{2}$

answer is A.

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Detailed Solution

$F_{v} = m g F_{H} = m ω^{2} \frac{l}{2} \sin θ ∴ T_{n e t} a b o u t C O M = F_{v} . \frac{l}{2} \sin θ - F_{H} \frac{l}{2} \cos θ = \frac{m l^{2}}{12} ω^{2} \sin θ \cos θ m g (\frac{l}{2} \sin θ) - m ω^{2} (\frac{l}{2} \sin θ) (\frac{l}{2} \cos θ) = m ω^{2} (\frac{l^{2}}{12}) (\sin θ) (\cos θ) \Rightarrow \frac{g l}{2} - \frac{ω^{2} l^{2}}{4} \cos θ = \frac{l^{2}}{12} ω^{2} \cos θ \frac{g l}{2} = ω^{2} l^{2} \cos θ (\frac{1}{12} + \frac{1}{4}) \frac{g l}{2} = \frac{ω^{2} l^{2} \cos θ}{3} \cos θ = \frac{3 g}{2 ω^{2} l} \int (\frac{M}{L} d x) (x \sin θ) ω^{2} \cos θ = \frac{M g L}{2} \sin θ \cos θ = \frac{3 g}{2 l ω^{2}}$