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(B) To complete a vertical loop,the body must maintain a minimum velocity at the highest point $C$ such that the tension $T_C$ in the string (or norma

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$\sqrt{5gR}$

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(B) To complete a vertical loop,the body must maintain a minimum velocity at the highest point $C$ such that the tension $T_C$ in the string (or normal force) is at least zero.At the highest point $C$,the forces acting on the body are gravity $(mg)$ acting downwards and tension $(T_C)$ acting downwards. The centripetal force is provided by the sum of these forces:$T_C + mg = \frac{mv_C^2}{R}$For the minimum velocity,we set $T_C = 0$,which gives:$mg = \frac{mv_C^2}{R} \Rightarrow v_C = \sqrt{gR}$Now,we apply the law of conservation of mechanical energy between the lowest point $A$ and the highest point $C$. Let $v_0$ be the velocity at point $A$:Total Energy at $A$ = Total Energy at $C$$\frac{1}{2}mv_0^2 = \frac{1}{2}mv_C^2 + mg(2R)$Substituting $v_C^2 = gR$:$\frac{1}{2}mv_0^2 = \frac{1}{2}m(gR) + 2mgR$$\frac{1}{2}mv_0^2 = \frac{5}{2}mgR$$v_0^2 = 5gR$$v_0 = \sqrt{5gR}$

What is the minimum velocity with which a body of mass $m$ must enter a vertical loop of radius $R$ so that it can complete the loop $?$

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