Question

The bob of a pendulum at rest is given a sharp hit to impart a horizontal velocity $u$. If $l$ is the length of the pendulum, find the tension in the string when $\theta ={90}^{\circ }$ from the vertical. Given: $u=\sqrt{3gl}$

• A. $\sqrt{2}mg$
• B. $3mg$
• C. $2mg$
• D. $mg$

Answer 1

The correct option is D $mg$


Using energy conservation between points $A$ and $B$.
$K{E}_A+U_A=K{E}_B+U_B$
$\frac{1}{2}m{u}^2+0=\frac{1}{2}m{v}^2+mgl$
$\Rightarrow \frac{1}{2}m\left(3gl\right)=\frac{1}{2}m{v}^2+mgl$
$\Rightarrow \frac{1}{2}mgl=\frac{1}{2}m{v}^2$
$\Rightarrow m{v}^2=mgl$

Balancing forces in the horizontal direction at $B$,
$T=\frac{m{v}^2}{l}=\frac{mgl}{l}=mg$
Option D is correct.