A particle of mass m is released from a height H on a smooth curved surface which ends into a vertic

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5.Work, Energy, Power and Collision

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$A$ particle of mass $m$ is released from $a$ height $H$ on $a$ smooth curved surface which ends into a vertical loop of radius $R$, as shown The minimum value of $H$ required so that the particle makes a complete vertical circle is given by

A

$5 \,R$

B

$4\, R$

C

$2.5\, R$

D

$2\, R$

Solution

$N=\frac{m v’^{2}}{r}-m g$

$\frac{m v’^{2}}{r}=m g$

$v^{\prime}=\sqrt{g r}$

$0+0=-m g(h-2 R)+\frac{1}{2} m g r$

$=-m g h+2 m g R+\frac{1}{2} m g R$

$h-2 R+\frac{R}{2}=\frac{5 R}{2}$

$=2.5 R$

Standard 11

Physics

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