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.

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

Answer 1

The correct option is B ${\cos }^{-1}\left(\frac{-1}{3}\right)$

Let the string make an angle $\theta$ with the vertical finally. FBD at this position is given below.


$u=\sqrt{3gl}$ (given)
Applying energy conservation at $A$ and $B$, taking $A$ as a reference
$\frac{1}{2}m{v}^2+mgh=\frac{1}{2}m{u}^2$
where $h$ is the vertical height of point $B$.
$h=l+l\cos \theta$

$\Rightarrow v^2=u^2-2g(l+l\cos \theta )$
$\Rightarrow v^2=3gl-2gl(1+\cos \theta )\dots \dots \left(1\right)$

From the FBD at point $B$:
$T+mg\cos \theta =\frac{m{v}^2}{R}$
Again, as tension is zero (for slack string)
$\frac{m{v}^2}{l}=mg\cos \theta$
$v^2=lg\cos \theta \dots \dots \left(2\right)$

From (1) and (2) equation,
$3gl-2gl(1+\cos \theta )=gl\cos \theta$
$3gl\cos \theta =gl\Rightarrow \cos \theta =\frac{1}{3}$
$\Rightarrow \theta ={\cos }^{-1}\left(\frac{1}{3}\right)$

So, the angle rotated before the string becomes slack
$={180}^{\circ }-{\cos }^{-1}\left(\frac{1}{3}\right)={\cos }^{-1}\left(\frac{-1}{3}\right)$