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A rough L – shaped rod is located in a horizontal plane and a sleeve of mass m in inserted in the rod. The rod is rotated with a constant angular velocity $ω$ in the horizontal plane. The lengths $l_{1} and l_{2}$ are shown in figure. The normal reaction and frictional force acting on the sleeve when its starts slipping are $(μ : coefficient of friction between rod and sleeve)$

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$N = m ω^{2} l_{1}$

$ω^{4} = \frac{μ^{2} g^{2}}{l_{2}^{2} - μ^{2} l_{1}^{2}}$

$N = m \sqrt{g^{2} + ω^{2} l_{1}^{2}}$

$f = μ N$

answer is B, C, D.

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Detailed Solution

$N_{1}^{2} + N_{2}^{2} = N^{2} N_{2} = m g m ω^{2} \times \sin ϕ = f f = m ω^{2} l_{2} N_{1} = m ω^{2} l_{1}$

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