Question

A small bob hangs at one end of a massless string of length $l$ and the other end of the string is fixed. The bob is given a sharp hit to impart it a horizontal speed of $\sqrt{3gl}$. Find the angle $\left(\theta \right)$ made by the string with the vertical when the string becomes slack.

• A. $\theta ={\cos }^{-1}\left(\frac{2}{3}\right)$
• B. $\theta ={\cos }^{-1}\left(\frac{3}{4}\right)$
• C. $\theta ={\cos }^{-1}\left(\frac{1}{3}\right)$
• D. $\theta ={\cos }^{-1}\left(\frac{5}{3}\right)$

Answer 1

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

Horizontal speed of the bob at point $A$ is,
$u=\sqrt{3gl}$
Let $m$ be the mass of the bob.


At point $B$, string becomes slack i.e $(T=0)$ in the string $\&$ angle made by the string with vertical is $\theta$. Let speed of the bob be $v$

Applying Newton's $2^{nd}$ law towards the centre of the vertical circle at point $B$:
$mg\cos \theta +T=\frac{m{v}^2}{l}...\left(i\right)$
(putting $T=0$ in Eq. $i)$
$\Rightarrow mg\cos \theta =\frac{m{v}^2}{l}$
$\therefore v^2=lg\cos \theta ....\left(ii\right)$

Height to which bob rises above point $A$
$h=l\cos \theta +l=l(1+\cos \theta )...\left(iii\right)$
Applying mechanical energy conservation from $A$ to $B$:
$\Rightarrow \text{Loss in KE}=\text{Gain in PE}$
$\frac{1}{2}m{u}^2-\frac{1}{2}m{v}^2=mgh$
$\Rightarrow v^2=u^2-2gh...\left(iv\right)$

From Eq. $\left(ii\right),\left(iii\right),\&\left(iv\right)$,
$lg\cos \theta =3gl-2gl-2gl\cos \theta$
$\Rightarrow 3gl\cos \theta =gl$
$\Rightarrow \cos \theta =\frac{1}{3}$
$\therefore \theta ={\cos }^{-1}\left(\frac{1}{3}\right)$