A single wire ACB passes through a smooth rin | vedclass

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(A) Let $T$ be the tension in the wire. Since the ring is smooth,the tension is the same throughout the wire.Resolving the forces acting on the ring $

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$\sqrt{rg}$

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(A) Let $T$ be the tension in the wire. Since the ring is smooth,the tension is the same throughout the wire.Resolving the forces acting on the ring $C$ along the vertical ($y$-axis) and horizontal ($x$-axis) directions:Along the vertical direction,the sum of the vertical components of tension balances the weight of the ring:$T \cos 30^{\circ} + T \cos 60^{\circ} = mg$$T \left( \frac{\sqrt{3}}{2} + \frac{1}{2} \right) = mg$$T \left( \frac{\sqrt{3} + 1}{2} \right) = mg$ $...(i)$Along the horizontal direction,the sum of the horizontal components of tension provides the necessary centripetal force:$T \sin 30^{\circ} + T \sin 60^{\circ} = \frac{mv^2}{r}$$T \left( \frac{1}{2} + \frac{\sqrt{3}}{2} \right) = \frac{mv^2}{r}$$T \left( \frac{\sqrt{3} + 1}{2} \right) = \frac{mv^2}{r}$ $...(ii)$Dividing equation $(ii)$ by equation $(i)$:$\frac{T \left( \frac{\sqrt{3} + 1}{2} \right)}{T \left( \frac{\sqrt{3} + 1}{2} \right)} = \frac{mv^2/r}{mg}$$1 = \frac{v^2}{rg}$$v^2 = rg$$v = \sqrt{rg}$

$A$ single wire $ACB$ passes through a smooth ring at $C$ which revolves at a constant speed in a horizontal circle of radius $r$ as shown in the figure. The speed of revolution is

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