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Q. In a simple pendulum, the breaking strength of the string is double the weight of the bob. The bob is released from rest when the string is horizontal. The string breaks when it makes an angle $\theta$ with the vertical :

Work, Energy and Power

Solution:

Let speed of bob at $B$ is $v$
Tension at $\theta$ angle is
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$T =m g \cos \theta+\frac{m v^{2}}{l} $
$[$ here $v=\sqrt{2 g l \cos \theta}]$
$2 m g =m g \cos \theta+\frac{m v^{2}}{l} $
$m v^{2} =2 m g l-m g l \cos \theta $
$3 \cos \theta =2$
$\theta =\cos ^{-1}\left(\frac{2}{3}\right)$