Question

A bob of a simple pendulum of mass $m$ is projected horizontally with some speed from its lowest point such that the string will slack at a position where it makes an angle ${60}^{\circ }$ with the upward vertical direction. Find the tension in the string at the lowest position.

• A. $\frac{9}{2}mg$
• B. $\frac{5}{2}mg$
• C. $0$
• D. $\frac{3}{2}mg$

Answer 1

The correct option is A $\frac{9}{2}mg$

At angle $\theta ={60}^{\circ }$, $T=0$ in the string at point $B$


Applying mechanical energy conservation from $A$ to $B$, taking datum for $PE$ at $A$
$K{E}_A+P{E}_A=K{E}_B+P{E}_B$
$\frac{1}{2}m{u}^2+0=\frac{1}{2}m{v}^2+mg(l+l\cos {60}^{\circ }$
$\Rightarrow v=\sqrt{u^2-2g(l+l\cos {60}^{\circ })}$
$\Rightarrow v=\sqrt{u^2-3gl}...\left(i\right)$

From FBD of the pendulum bob at $B$ when $T^{\prime }=0$, the equation of dynamics towards centre is:
$mg\cos {60}^{\circ }+T^{\prime }=\frac{m{v}^2}{l}$
$\Rightarrow \frac{mg}{2}=\frac{m(u^2-3gl)}{l}$
$\Rightarrow \frac{m{u}^2}{l}=\frac{7}{2}mg....\left(ii\right)$

Now applying the equation of dynamics towards centre at point $A$:
$T-mg=\frac{m{u}^2}{l}.....\left(iii\right)$
From Eq. $\left(ii\right)$ and $\left(iii\right)$
$T=\frac{7}{2}mg+mg$
$\therefore T=\frac{9}{2}mg$ at point $A$.