A particle originally at rest at highest point of a smooth vertical circle is slightly displaced. It

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5.Work, Energy, Power and Collision

hard

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

A

$h = R$

B

$h = \frac{R}{3}$

C

$h = \frac{R}{2}$

D

$h = \frac{{2R}}{3}$

Solution

$h=R-R \cos \theta, v=\sqrt{2 g h}=\sqrt{2 g R(1-\cos \theta)}$

$m g \cos \theta-N=\frac{m v^{2}}{R}$

When it leaves circle: $N=0$ $m g \cos \theta=\frac{m v^{2}}{R} \Rightarrow \cos \theta=\frac{2}{3}$

$h=R-R \cos \theta=\frac{R}{3}$

Standard 11

Physics

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