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) $\left(\dfrac{\sin\theta}{2 + 3\sin\theta}\right)^{1/2}$
Concept. The string goes slack when tension is zero, so gravity's radial component alone gives the centripetal force; energy is conserved from the bottom.
Step 1. At P. $mg\sin\theta = \dfrac{mv^2}{l} \Rightarrow v^2 = gl\sin\theta$.
Step 2. Energy. $v_0^2 = v^2 + 2gl(1 + \sin\theta) = gl(2 + 3\sin\theta)$.
Step 3. Ratio. $\dfrac{v}{v_0} = \left(\dfrac{\sin\theta}{2 + 3\sin\theta}\right)^{1/2}$.
Answer. $\left(\dfrac{\sin\theta}{2 + 3\sin\theta}\right)^{1/2}$ (Option D).
Solution by Sreeraj P, M.Sc Physics