A bob of heavy mass $m$ is suspended by a light string of length $l$. The bob is given a horizontal velocity $v_0$ as shown in figure. If the string gets slack at some point $P$ making an angle $\theta$ from the horizontal, the ratio of the speed $v$ of the bob at point $P$ to its initial speed $v_0$ is:
Answer: (D) $\left(\dfrac{\sin\theta}{2 + 3\sin\theta}\right)^{1/2}$
At P the bob is at height $l\sin\theta$ above O, i.e. $l(1 + \sin\theta)$ above the lowest point.
The string becomes slack when the tension becomes zero. Then the radial component of gravity alone provides the centripetal force:
$$mg\sin\theta = \frac{mv^2}{l} \;\Rightarrow\; v^2 = gl\sin\theta$$
Energy conservation from the lowest point to P:
$$v_0^2 = v^2 + 2gl(1 + \sin\theta) = gl\sin\theta + 2gl + 2gl\sin\theta = gl(2 + 3\sin\theta)$$
$$\frac{v}{v_0} = \left(\frac{\sin\theta}{2 + 3\sin\theta}\right)^{1/2}$$
Solution by Sreeraj P, M.Sc Physics