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A ring of mass $M$ and radius $R$ is hanged from a point on the rim. The ring displaced slightly along its plane, and free to oscillate. The length of equivalent simple pendulum $l$ is

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$l = 2 R$

$l = \sqrt{2} R$

none of these

$l = R$

answer is B.

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Detailed Solution

The restoring torque (for small displacement),

$τ_{r e s t} = - m g (R θ) ∴ α = \frac{τ_{r e s t}}{I} = \frac{m g R}{2 m R^{2}} (- θ) = \frac{g}{2 R} ∴ T = 2 π \sqrt{\frac{2 R}{g}}$

The length of equivalent pendulum

$l = 2 R$

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