Question

The bob of the pendulum shown in the figure describes an arc of circle in a vertical plane. If the tension in the chord is $2.5$ times the weight of the bob for the position shown. Find the velocity of the bob in that position. (Take $g=10{\text{m/s}}^2$)

• A. $\sqrt{42}\text{m/s}$
• B. $33\text{m/s}$
• C. $\sqrt{33}\text{m/s}$
• D. $6\text{m/s}$

Answer 1

The correct option is C $\sqrt{33}\text{m/s}$

It is given that, $T=2.5mg$


By writing the force equation of the bob at that position we get,
$T-mg\cos \theta =\frac{m{v}^2}{r}$, where $mg\cos \theta$ is the component of weight along the line of tension and $v$ is the velocity of the bob at that point.

For the angle
$\theta ={30}^{\circ }$, we get
$2.5mg-mgcos{30}^{\circ }=\frac{m{v}^2}{r}$
$\Rightarrow \frac{5}{2}mg-\frac{\sqrt{3}}{2}mg=\frac{m{v}^2}{2}$

$\Rightarrow v^2=g(5-\sqrt{3})$, $\sqrt{3}\approx 1.7$

$v=\sqrt{3.3\times g}=\sqrt{33}$
Hence option C is the correct answer