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(D) For a particle moving in a vertical circle,the critical condition at the highest point $C$ is that the velocity $v_C = \sqrt{g R}$.We use the law

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$\sqrt{2 g R}$

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(D) For a particle moving in a vertical circle,the critical condition at the highest point $C$ is that the velocity $v_C = \sqrt{g R}$.We use the law of conservation of mechanical energy between points $P$ and $C$.Let the potential energy at point $P$ be $U_P = 0$.The height of point $C$ relative to point $P$ is $h = R - R \cos 60^{\circ} = R(1 - 0.5) = 0.5 R$.Applying conservation of energy: $U_P + K_P = U_C + K_C$$0 + \frac{1}{2} m v_P^2 = m g (0.5 R) + \frac{1}{2} m v_C^2$Substituting $v_C = \sqrt{g R}$:$\frac{1}{2} m v_P^2 = 0.5 m g R + \frac{1}{2} m (g R)$$\frac{1}{2} m v_P^2 = 0.5 m g R + 0.5 m g R = m g R$$v_P^2 = 2 g R$$v_P = \sqrt{2 g R}$

$A$ particle is moving along a vertical circle of radius $R$. At point $P$,what will be the velocity of the particle? (Assume the critical condition at the highest point $C$).

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