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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 $h \ge 5r/2$.

List-$I$ (Speed at lowest point)	List-$II$ (Possible situation)
$(P) v_0 = \sqrt{5 g \ell}$	$(1)$ Tension at lowest point $= 6 mg$
$(Q) v_0 = \sqrt{g \ell}$	$(2)$ String will slack for a finite time
$(R) v_0 = 2 \sqrt{g \ell}$	$(3)$ Bob will oscillate
$(S) v_0 = 3 \sqrt{g \ell}$	$(4)$ Tension at highest point $= 4 mg$

$A$ 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

Similar Questions

$A$ stone tied to a string is rotated in a vertical circle. The minimum speed with which the string has to be rotated at the lowest point is:

$A$ mass $m$ is attached to a thin wire and whirled in a vertical circle. The wire is most likely to break when

$A$ stone of mass $m$ is moving in a vertical circle of radius $20 \ cm$. What is the difference in kinetic energy between the lowest and highest points?

$A$ body of mass $m$ attached at the end of a string is just completing the loop in a vertical circle. The apparent weight of the body at the lowest point in its path is ($g =$ gravitational acceleration).

Vedclass Test Series

Exam Paper Generator

Online Exam Module