A bucket tied at the end of a 1.6, m long str | vedclass

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(A) For the water not to spill at the highest point of a vertical circle,the gravitational force must provide the necessary centripetal force.At the h

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$4$

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(A) For the water not to spill at the highest point of a vertical circle,the gravitational force must provide the necessary centripetal force.At the highest point,the condition for the minimum speed $V$ is given by:$\frac{mV^2}{r} = mg$Where $m$ is the mass of the water,$V$ is the velocity,$r$ is the radius of the circle $(1.6\, m)$,and $g$ is the acceleration due to gravity $(10\, m/sec^2)$.Canceling $m$ from both sides,we get:$V^2 = rg$$V = \sqrt{rg}$Substituting the given values:$V = \sqrt{1.6 	imes 10} = \sqrt{16} = 4\, m/sec$.Thus,the minimum speed required is $4\, m/sec$.

$A$ bucket tied at the end of a $1.6\, m$ long string is whirled in a vertical circle with constant speed. What should be the minimum speed so that the water from the bucket does not spill,when the bucket is at the highest position? (Take $g = 10\, m/sec^2$)

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