Question

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. Then

• A. $\theta ={\cos }^{-1}\left(\frac{1}{3}\right)$
• B. $\theta ={60}^{\circ }$
• C. $\theta ={\cos }^{-1}\left(\frac{2}{3}\right)$
• D. $\theta =0^{\circ }$

Answer 1

The correct option is C $\theta ={\cos }^{-1}\left(\frac{2}{3}\right)$


Let the speed of bob at point B be $v$

From the FBD:
$T-mg\cos \theta =\frac{m{v}^2}{l}\dots \left(i\right)$ (where $l$ is the length of the string)

From conservation of energy,
$mgl=mgl(1-\cos \theta )+\frac{1}{2}m{v}^2$
{Bob is released from horizontal }
$\Rightarrow v=\sqrt{2gl\cos \theta }\dots \left(ii\right)$

$T=2mg$ (Given)
$\Rightarrow 2mg=mg\cos \theta +\frac{m{v}^2}{l}$
(from (i))
$\Rightarrow m{v}^2=2mgl-mgl\cos \theta$
$\Rightarrow m\left(2gl\cos \theta \right)=2mgl-mgl\cos \theta$
(substituting from (ii))
$\Rightarrow 3mgl\cos \theta =2mgl$
$\Rightarrow \cos \theta =\frac{2}{3}$
$\therefore \theta ={\cos }^{-1}\left(\frac{2}{3}\right)$