A frictionless track ABCDE ends in a circular | vedclass

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(D) To successfully complete a vertical circular loop of radius $R$,the minimum velocity $v_{top}$ required at the highest point of the loop is $\sqrt

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$2$

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(D) To successfully complete a vertical circular loop of radius $R$,the minimum velocity $v_{top}$ required at the highest point of the loop is $\sqrt{gR}$.Using the principle of conservation of mechanical energy between point $A$ (height $h$) and the top of the loop (height $2R$):$mgh = mg(2R) + \frac{1}{2}mv_{top}^2$Substituting $v_{top}^2 = gR$:$mgh = 2mgR + \frac{1}{2}mgR$$mgh = \frac{5}{2}mgR$$h = \frac{5}{2}R$Given $h = 5 \, cm$:$5 = \frac{5}{2}R$$R = 2 \, cm$.Thus,the maximum value of $R$ 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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