Question

The bob of a stationary pendulum is given a sharp hit to impart it a horizontal speed of $\sqrt{3gl}$. Find the angle rotated by the string before it becomes slack.

Answer 1

Suppose the string becomes slack at point P.
Let the bob rise to a height h.
h = l + l cos θ




From the work-energy theorem,
$$\begin{gathered} \frac{1}{2}m{\nu }^2-\frac{1}{2}m{u}^2=-mgh \\ {\nu }^2=u^2-2g\left(l+l\cos \theta \right)...\left(i\right) \\ \mathrm{Again},\frac{m{\nu }^2}{l}=mg\cos \mathrm{\theta } \\ {\nu }^2=lg\cos \mathrm{\theta }...................\left(\mathrm{ii}\right) \end{gathered}$$

Using equation (i) and (ii) and the value of u, we get,

$$\begin{gathered} gl\cos \mathrm{\theta }=3gl-2gl-2gl\cos \mathrm{\theta } \\ 3\cos \mathrm{\theta }=1 \\ \mathrm{\theta }={\cos }^{-1}\left(\frac{1}{3}\right) \\ ={\cos }^{-1}\left(-\frac{1}{3}\right) \end{gathered}$$