A particle of mass m is released from a heigh | vedclass

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Question

$N=\frac{m v&rsquo;^{2}}{r}-m g$

$\frac{m v&rsquo;^{2}}{r}=m g$

$v^{\prime}=\sqrt{g r}$

$0+0=-m g(h-2 R)+\frac{1}{2} m g r$

$=-m g h+2 m g

Vedclass · Accepted Answer

$2.5\, R$

Vedclass · Answer

$N=\frac{m v&rsquo;^{2}}{r}-m g$

$\frac{m v&rsquo;^{2}}{r}=m g$

$v^{\prime}=\sqrt{g r}$

$0+0=-m g(h-2 R)+\frac{1}{2} m g r$

$=-m g h+2 m g R+\frac{1}{2} m g R$

$h-2 R+\frac{R}{2}=\frac{5 R}{2}$

$=2.5 R$

$A$ 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

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