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A particle originally at rest at the highest point of a smooth vertical circle is slightly displaced. It will leave the circle at a vertical distance h below the highest point, such that h is equal to

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$\frac{R}{4}$

$\frac{R}{2}$

$\frac{R}{3}$

answer is D.

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Detailed Solution

$h = R - R \cos θ; v = \sqrt{2 gh} = \sqrt{2 gR (1 - \cos θ)}$

$mg \cos θ - N = \frac{{mv}^{2}}{R}$

when it leaves circle : N = 0

$mg \cos θ = \frac{{mv}^{2}}{R} \Rightarrow \cos θ = \frac{2}{3}$

$h = R - R \cos θ = \frac{R}{3}$

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