What is the minimum velocity at the lowest po | vedclass

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Question

(C) At the highest point,the forces acting are tension $T$ and weight $mg$,both directed downwards. The centripetal force is provided by these forces:

Vedclass · Accepted Answer

$\sqrt{5gl}$

Vedclass · Answer

(C) At the highest point,the forces acting are tension $T$ and weight $mg$,both directed downwards. The centripetal force is provided by these forces: $T + mg = \frac{mv_1^2}{l}$.For the string to just become slack at the highest point,the tension $T$ must be zero $(T = 0)$.Thus,$mg = \frac{mv_1^2}{l}$,which gives the critical velocity at the top as $v_1 = \sqrt{gl}$.Using the law of conservation of energy between the lowest point (velocity $v_2$) and the highest point (velocity $v_1$):Total energy at bottom = Total energy at top.Taking the lowest point as the reference level for potential energy $(PE = 0)$:$\frac{1}{2}mv_2^2 + 0 = \frac{1}{2}mv_1^2 + mg(2l)$.Substituting $v_1^2 = gl$:$\frac{1}{2}mv_2^2 = \frac{1}{2}m(gl) + 2mgl$.$\frac{1}{2}mv_2^2 = \frac{5}{2}mgl$.$v_2^2 = 5gl$.$v_2 = \sqrt{5gl}$.

What is the minimum velocity at the lowest point of a vertical circle of radius $l$ such that the string just becomes slack at the highest point?

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