Question

The bob of a pendulum at rest is given a sharp hit to impart a horizontal velocity of $\mu =\sqrt{5gl}$, where $l$ is the length of the pendulum . Find the height from the lowermost point at which tension is zero.

• A. never zero for $v=\sqrt{5gl}$
• B. $2l$
• C. $l$
• D. $l/2$

Answer 1

The correct option is B $2l$


Let the tension in the string be zero at point $B$ at height $h$ above $A$.
Applying conservation of energy between $A$ and $B$, taking $A$ as reference
$\Delta U+\Delta KE=0$
$mgh+[\frac{1}{2}m{v}_B^2-\frac{1}{2}m{v}_A^2]=0$
where $h=l+l\cos \theta$

$\therefore$ Velocity at point $B$, $v_B^2=v_A^2-2g(l+l\cos \theta )...\left(i\right)$

When tension is zero, balancing forces along the string,
$\frac{m{v}_B^2}{l}=mg\cos \theta$
$\Rightarrow v_B^2=gl\cos \theta ........\left(ii\right)$

From $\left(i\right)$ and $\left(ii\right)$, we get
$gl\cos \theta =5gl-2gl-2gl\cos \theta$
$\Rightarrow 3gl\cos \theta =3gl$
$\Rightarrow \cos \theta =1$
$\Rightarrow \theta =0^{\circ }$

Therefore, tension is zero at the topmost point.
i.e $h=l+l\cos \theta =2l$