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A uniform rod length $2 l$ rocks to and from on the top of a rough semicircular fixed cylinder of radius a. The period of small oscillation, of rod is

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$2π \sqrt{\frac{l}{3g}}$

$2π \sqrt{\frac{a}{3g}}$

$\frac{π l}{\sqrt{3ga}}$

$\frac{2π l}{\sqrt{3ga}}$

answer is A.

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Detailed Solution

$I α = τ = m g a \sin θ$

$\frac{m {(2 l)}^{2}}{12} α = m g a θ α = (\frac{3 g a}{l^{2}}) θ = ω^{2} θ T = \frac{2 π}{ω} = \frac{2 π l}{\sqrt{3 g a}}$

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