Question

The bob of the pendulum shown in the figure describes an arc of a circle in a vertical plane. If the tension in the cord is $3$ times of the weight of the bob for the position shown, the velocity of the bob at this position is

• A. $6.47\text{m/s}$
• B. $10\text{m/s}$
• C. $0\text{m/s}$
• D. $15\text{m/s}$

Answer 1

The correct option is A $6.47\text{m/s}$

FBD of the bob:


Let $T$ and $v$ be the tension in the bob and velocity of the bob respectively.
Given, $T=3mg$

From the FBD,
$T=mg\cos \theta +\frac{m{v}^2}{r}$
[where, $r=2\text{m}$] , $[\theta ={30}^o]$
$\therefore 3mg-mg\cos {30}^{\circ }=\frac{m{v}^2}{2}$
$3\times 9.81-9.81\times \frac{\sqrt{3}}{2}=\frac{v^2}{2}$
$v=6.47\text{m/s}$