Question

The bob of a stationary pendulum is given a sharp hit to impart it a horizontal speed of \[\sqrt{3 gl}\] . 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,

\[\frac{1}{2} \text{ m} \nu^2 - \frac{1}{2}\text{m}u^2 = - \text{mgh}\]

\[ \nu^2 = \text{u}^2 - 2\text{g} \left( \text{l + l} \cos \theta \right) . . . (\text{i})\]

\[\text{Again}, \frac{\text{m} \nu^2}{l} = \text{mg} \cos \theta\]

\[ \nu^2 = \text{lg} \cos \theta . . . . . . . . . . . . . . . . . . . (\text{ii})\]

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

\[gl \cos \theta = 3gl - 2gl - 2gl \cos \theta\]
\[3 \cos \theta = 1\]
\[\theta = \cos^{- 1} \left( \frac{1}{3} \right)\]
\[ = \cos^{- 1} \left( - \frac{1}{3} \right)\]