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(D) Let the initial speed be $u = \sqrt{4 g \ell}$. The string becomes slack when the tension $T = 0$. Let this occur at an angle $	heta$ with the ve

Vedclass · Accepted Answer

$\sqrt{\frac{8}{27} g \ell}$

Vedclass · Answer

(D) Let the initial speed be $u = \sqrt{4 g \ell}$. The string becomes slack when the tension $T = 0$. Let this occur at an angle $	heta$ with the vertical.At this point,the radial component of gravity provides the centripetal force: $mg \cos 	heta = \frac{mv^2}{\ell} \Rightarrow v^2 = g \ell \cos 	heta$.Using the conservation of energy between the lowest point and the point where the string becomes slack:$\frac{1}{2}mu^2 = \frac{1}{2}mv^2 + mg\ell(1 + \cos 	heta)$Substituting $u^2 = 4g\ell$ and $v^2 = g\ell \cos 	heta$:$\frac{1}{2}m(4g\ell) = \frac{1}{2}m(g\ell \cos 	heta) + mg\ell(1 + \cos 	heta)$$2g\ell = \frac{1}{2}g\ell \cos 	heta + g\ell + g\ell \cos 	heta$$1 = \frac{3}{2} \cos 	heta \Rightarrow \cos 	heta = \frac{2}{3}$.Thus,$v^2 = g\ell(\frac{2}{3}) = \frac{2}{3}g\ell$.At the highest point of the trajectory (where the vertical velocity component is zero),the velocity of the bob is purely horizontal and equal to $v_x = v \cos 	heta$.$v_{top} = v \cos 	heta = \sqrt{\frac{2}{3}g\ell} 	imes \frac{2}{3} = \sqrt{\frac{2}{3} 	imes \frac{4}{9} g\ell} = \sqrt{\frac{8}{27} g\ell}$.

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