Question

The simple pendulum bob is given a horizontal velocity $\sqrt{gl}$ at the bottom. The angle with the vertical through which the ball swings before its velocity becomes zero is

• A. ${120}^{\circ }$
• B. ${60}^{\circ }$
• C. ${450}^{\circ }$
• D. ${30}^{\circ }$

Answer 1

The correct option is B ${60}^{\circ }$

When pendulum bob reaches point $B$ making an angle $\theta$ with vertical, such that its velocity becomes zero,
Gain in height from bottom$h=l-l\cos \theta =l(1-\cos \theta )$

Applying energy conservation from point $A$ to point $B$ as shown in figure:


$\text{Loss in KE of bob}=\text{Gain in PE of bob}$
$\frac{1}{2}m{u}^2-0=mgl(1-\cos \theta )$
$\Rightarrow 1-\cos \theta =\frac{u^2}{2gl}$
$\Rightarrow \cos \theta =1-\frac{u^2}{2gl}=1-\frac{gl}{2gl}$
(given $u=\sqrt{gl}$)
$\Rightarrow \cos \theta =1-\frac{1}{2}=\frac{1}{2}$
$\therefore \theta ={60}^{\circ }$