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A sleeve $A$ can slide freely along a smooth rod bent in the shape of a half circle of radius $R$ . The system is set in rotation with a constant angular velocity $ω$ about a vertical axis $O O'$ . Find the angle $θ$ corresponding to the steady position of the sleeve.

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$\cos^{- 1} \frac{3 g}{2 ω^{2} R}$

$\cos^{- 1} \frac{g}{ω^{2} R}$

$\cos^{- 1} \frac{2 g}{3 ω^{2} R}$

$\cos^{- 1} \frac{g}{4 ω^{2} R}$

answer is A.

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Detailed Solution

If $θ$ is the angle corresponding to the steady position, then

$N \sin θ = m ω^{2} r = m ω^{2} (R \sin θ) \Rightarrow N = m ω^{2} R - - - - - (i) a n d N \cos θ = m g - - - - - (i i)$

Question Image — FBD of sleeve in rotating frame

Solving equations (i) and (ii), we get

$\cos θ = \frac{g}{ω^{2} R}$

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