A frictionless track ABCDE ends in a circular | vedclass

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Question

(D) For a body to successfully complete a vertical circular loop of radius $R$ starting from rest at a height $h$,the minimum height required is given

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(D) For a body to successfully complete a vertical circular loop of radius $R$ starting from rest at a height $h$,the minimum height required is given by the conservation of energy and the condition for the normal force to be non-negative at the top of the loop.At the top of the loop,the minimum velocity $v$ required is $\sqrt{gR}$.Using the conservation of energy between point $A$ and the top of the loop:$mgh = mg(2R) + \frac{1}{2}mv^2$$mgh = 2mgR + \frac{1}{2}m(gR)$$h = 2R + 0.5R = 2.5R = \frac{5}{2}R$Given $h = 5\, cm$,we have:$5 = \frac{5}{2}R$$R = 2\, cm$.Thus,the maximum radius $R$ for which the body can complete the loop is $2\, cm$.

$A$ frictionless track $ABCDE$ ends in a circular loop of radius $R$. $A$ body slides down the track from point $A$ which is at a height $h = 5\, cm$. The maximum value of $R$ for the body to successfully complete the loop is ........... $cm$.

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