The bob of a stationary pendulum receives a sharp hit, imparting to...

Let the angle rotated by the string from the lowest point be $\theta and the velocity of the bob at this position be$v$.<br><br>Using the principle of conservation of mechanical energy between the lowest point and the position at angle$\theta$:<br><br>$\dfrac{1}{2}mv_0^2 = \dfrac{1}{2}mv^2 + mgl(1 - \cos\theta)$<br><br>Given$v_0 = \sqrt{3gl}$, substituting this into the equation:<br><br>$\dfrac{1}{2}m(3gl) = \dfrac{1}{2}mv^2 + mgl(1 - \cos\theta)$<br><br>$v^2 = 3gl - 2gl(1 - \cos\theta) = gl(1 + 2\cos\theta)$<br><br>The equation of motion for the bob along the radial direction is:<br><br>$T - mg\cos\theta = \dfrac{mv^2}{l}$<br><br>Substituting the expression for$v^2$:<br><br>$T = mg\cos\theta + \dfrac{m}{l}[gl(1 + 2\cos\theta)]$<br><br>$T = mg\cos\theta + mg(1 + 2\cos\theta) = mg(1 + 3\cos\theta)$<br><br>The string becomes slack when the tension$T$becomes zero:<br><br>$mg(1 + 3\cos\theta) = 0$<br><br>$1 + 3\cos\theta = 0$<br><br>$\cos\theta = -\dfrac{1}{3}$<br><br>$\theta = \cos^{-1} $\left(-\dfrac{1}{3}\right)$