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Q.

A particle originally at rest at the highest point of a smooth vertical circle is slightly displaced. At what height below the highest point will the particle leave the sphere?

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a

R

b

R/4

c

R/2

d

R/3

answer is D.

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Detailed Solution

From geometry,

$\begin{array}{l} h = R - Rcos θ; v = \sqrt{2 gh} \\ = \sqrt{2 gR (1 - \cos θ)} \end{array}$

Force along the radius

$mgcos θ - N = \frac{{mv}^{2}}{R}$

When it leaves circle: N = 0

$\begin{array}{l} \Rightarrow mgcos θ = \frac{{mv}^{2}}{R} = \frac{m \times 2 gR (1 - \cos θ)}{R} \\ \Rightarrow \cos θ = \frac{2}{3} \end{array}$

Thus we get, $h = R - Rcos θ = R / 3$

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A particle originally at rest at the highest point of a smooth vertical circle is slightly displaced. At what height below the highest point will the particle leave the sphere?