Question

The bob of a pendulum initially at rest is given a sharp hit to impart a horizontal velocity of $\sqrt{3.5gl}\text{m/s}$ where $l$ is the length of the pendulum in metres. Find the height (in $\text{metres}$) from the bottom, at which tension in the string becomes zero. Take $g=10{\text{m/s}}^2$.

• A. $l$
• B. $\frac{3l}{2}$
• C. $2l$
• D. Tension in the string will never be zero.

Answer 1

The correct option is B $\frac{3l}{2}$

Let $\theta$ be the angle with the vertical as shown in figure, at which tension in the string is zero.
FBD at angle $\theta$ with the vertical:


Applying equation of dynamics towards centre of circle:
$T+mg\cos \theta =\frac{m{v}^2}{l}$
Putting $T=0$,
$gl\cos \theta =v^2...\left(i\right)$
or $v^2=gl\cos \theta$

At bottommost position, velocity of the particle was $u=\sqrt{3.5gl}$
Since work done by tension $W_T=0$, applying mechanical energy conservation from bottommost position to given position :
$\Delta KE+\Delta PE=0$
$(\frac{1}{2}m{v}^2-\frac{1}{2}m{u}^2)+\left(mgl\right[1-\cos \theta ]-0)=0$
Taking datum for $PE=0$ at bottom
$u^2-v^2=2gl[1+\cos \theta ]$
$3.5gl-gl\cos \theta =2gl+2gl\cos \theta$
$1.5gl=3gl\cos \theta$
$\Rightarrow \cos \theta =\frac{1}{2}$

Hence, height from the bottom at which tension in the string becomes zero is given by:
$h=l+l\cos \theta =l[1+cos\theta ]$
$\therefore h=l(1+\frac{1}{2})=\frac{3l}{2}\text{m}$