A block follows the path as shown in the figu | vedclass

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(D) To complete a full vertical circle,the block must maintain contact with the track at the highest point of the loop.Let $v_{top}$ be the velocity o

Vedclass · Accepted Answer

$h \ge 5r/2$

Vedclass · Answer

(D) To complete a full vertical circle,the block must maintain contact with the track at the highest point of the loop.Let $v_{top}$ be the velocity of the block at the highest point of the circular path of radius $r$.At the highest point,the forces acting on the block are gravity $(mg)$ acting downwards and the normal force $(N)$ from the track acting downwards.The equation of motion is: $mg + N = \frac{mv_{top}^2}{r}$.For the block to complete the circle,the normal force must be at least zero $(N \ge 0)$.Setting $N = 0$,we get the minimum velocity at the top: $mg = \frac{mv_{top}^2}{r} \Rightarrow v_{top}^2 = gr$.Using the law of conservation of mechanical energy between the starting point (at height $h$) and the highest point of the loop (at height $2r$):Initial Energy $(E_i)$ = Final Energy $(E_f)$$mgh = mg(2r) + \frac{1}{2}mv_{top}^2$Substituting $v_{top}^2 = gr$:$mgh = 2mgr + \frac{1}{2}m(gr)$$mgh = 2mgr + 0.5mgr = 2.5mgr$$h = 2.5r = 5r/2$.Since the block must have at least this much energy to complete the circle,the condition is $h \ge 5r/2$.

$A$ block follows the path as shown in the figure from height $h$. If the radius of the circular path is $r$,then the relation that holds good to complete a full circle is

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