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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.

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Solution

## 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, $\frac{1}{2}m{\nu }^{2}-\frac{1}{2}m{u}^{2}=-mgh\phantom{\rule{0ex}{0ex}}{\nu }^{2}={u}^{2}-2g\left(l+l\mathrm{cos}\theta \right)...\left(i\right)\phantom{\rule{0ex}{0ex}}\mathrm{Again},\frac{m{\nu }^{2}}{l}=mg\mathrm{cos}\mathrm{\theta }\phantom{\rule{0ex}{0ex}}{\nu }^{2}=lg\mathrm{cos}\mathrm{\theta }...................\left(\mathrm{ii}\right)$ Using equation (i) and (ii) and the value of u, we get, $gl\mathrm{cos}\mathrm{\theta }=3gl-2gl-2gl\mathrm{cos}\mathrm{\theta }\phantom{\rule{0ex}{0ex}}3\mathrm{cos}\mathrm{\theta }=1\phantom{\rule{0ex}{0ex}}\mathrm{\theta }={\mathrm{cos}}^{-1}\left(\frac{1}{3}\right)\phantom{\rule{0ex}{0ex}}={\mathrm{cos}}^{-1}\left(-\frac{1}{3}\right)$

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