A bucket filled with water is tied to a rope | vedclass

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(C) For the water not to spill out of the bucket at the highest point,the velocity at the highest point must be at least $\sqrt{gR}$.Using the princip

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$5 ~m/s$

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(C) For the water not to spill out of the bucket at the highest point,the velocity at the highest point must be at least $\sqrt{gR}$.Using the principle of conservation of energy between the lowest point $(v_L)$ and the highest point $(v_H)$:$\frac{1}{2}mv_L^2 = \frac{1}{2}mv_H^2 + mg(2R)$$v_L^2 = v_H^2 + 4gR$Since $v_H = \sqrt{gR}$,we have $v_L^2 = gR + 4gR = 5gR$.$v_L = \sqrt{5gR}$Substituting the given values $g = 10 ~m/s^2$ and $R = 0.5 ~m$:$v_L = \sqrt{5 	imes 10 	imes 0.5} = \sqrt{25} = 5 ~m/s$.

$A$ bucket filled with water is tied to a rope of length $0.5 ~m$ and is rotated in a circular path in a vertical plane. The minimum velocity it should have at the lowest point of the circle so that the water does not spill is,$(g = 10 ~m/s^2)$.

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