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(B) Let $T_A$ and $T_B$ be the tensions at the highest point $A$ and the lowest point $B$,respectively.At the highest point $A$,the forces acting on t

Vedclass · Accepted Answer

$1: 2$

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(B) Let $T_A$ and $T_B$ be the tensions at the highest point $A$ and the lowest point $B$,respectively.At the highest point $A$,the forces acting on the particle are tension $T_A$ (downward) and weight $mg$ (downward). The net centripetal force is:$T_A + mg = \frac{mv_A^2}{r}$Given $v_A = \sqrt{7gr}$,we have:$T_A + mg = \frac{m(7gr)}{r} = 7mg$$T_A = 6mg$By the law of conservation of energy between points $A$ and $B$:$\frac{1}{2}mv_A^2 + mg(2r) = \frac{1}{2}mv_B^2$$\frac{1}{2}m(7gr) + 2mgr = \frac{1}{2}mv_B^2$$3.5mgr + 2mgr = 0.5mv_B^2$$5.5mgr = 0.5mv_B^2 \Rightarrow v_B^2 = 11gr$At the lowest point $B$,the forces are tension $T_B$ (upward) and weight $mg$ (downward). The net centripetal force is:$T_B - mg = \frac{mv_B^2}{r}$$T_B = mg + \frac{m(11gr)}{r} = 12mg$The ratio of tensions is:$\frac{T_A}{T_B} = \frac{6mg}{12mg} = \frac{1}{2}$Thus,the ratio is $1: 2$.

$A$ particle attached to a string of length $r$ is moving in a vertical circular path. If the speed of the particle at the highest point is $\sqrt{7gr}$,then the ratio of the tension in the string at the highest point to the tension at the lowest point is ($g=$ acceleration due to gravity).

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