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(D) To complete a full vertical circle,the tension at the highest point must be at least zero $(T \ge 0)$. At the highest point,the forces acting are

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$v = \sqrt{5gl}$

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(D) To complete a full vertical circle,the tension at the highest point must be at least zero $(T \ge 0)$. At the highest point,the forces acting are gravity $(mg)$ and tension $(T)$,providing the centripetal force: $T + mg = \frac{mv_h^2}{l}$. Setting $T = 0$ for the minimum velocity at the top,we get $mg = \frac{mv_h^2}{l}$,which implies $v_h = \sqrt{gl}$. Using the law of conservation of energy between the lowest point (velocity $v$) and the highest point (velocity $v_h$): $\frac{1}{2}mv^2 = \frac{1}{2}mv_h^2 + mg(2l)$. Substituting $v_h^2 = gl$: $\frac{1}{2}mv^2 = \frac{1}{2}m(gl) + 2mgl = \frac{5}{2}mgl$. Solving for $v$,we get $v = \sqrt{5gl}$.

$A$ body of mass $m$ is suspended from a string of length $l$. What is the minimum horizontal velocity that should be given to the body in its lowest position so that it may complete one full revolution in a vertical plane with the point of suspension as the center of the circle?

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