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Q.
A particle is moving along a vertical circle of radius $R$. At $P$, what will be the velocity of particle (assume critical condition at $C)$ ?
Work, Energy and Power
Solution:
At critical condition at $C$
$v=\sqrt{g R}$
Using mechanical energy conservation between points $P$ and $C$ (taking P.E. $=0$ at $P$ )
$U_{i}+k_{i}=U_{f}+k_{f}$
$0+\frac{1}{2} m v^{2}=m g R\left(1-\cos 60^{\circ}\right)+\frac{1}{2} m(\sqrt{g R})^{2}$
$\frac{1}{2} m v^{2}=\frac{m g R}{2}+\frac{m g R}{2}$
$v=\sqrt{2 g R}$
