Question

A pendulum bob of mass m is suspended at rest. A constant horizontal force $F=\frac{mg}{2}$ starts acting on it. The maximum angular deflection of the string is

• A. ${90}^{\circ }$
• B. ${53}^{\circ }$
• C. ${37}^{\circ }$
• D. ${60}^{\circ }$

Answer 1

The correct option is B ${53}^{\circ }$


Let angular deflection be $\theta$ when velocity is $v$,
then by work energy theorem $W=\Delta KE$
$\frac{1}{2}m{v}^2=-mgh+Flsin\theta$

$\frac{1}{2}m{v}^2=-mg(l-lcos\theta )+Flsin\theta$ $(\because h=l-lcos\theta )$

At maximum angular deflection, $v=0$
$0=-mgl(1-cos\theta )+\frac{mgl}{2}sin\theta \Rightarrow 2-2cos\theta =sin\theta \Rightarrow 4+4co{s}^2\theta -8cos\theta =si{n}^2\theta =1-co{s}^2\theta$
$\Rightarrow 5co{s}^2\theta -8cos\theta +3=0\Rightarrow 5co{s}^2\theta -5cos\theta -3cos\theta +3=0$
$\Rightarrow 5cos\theta (cos\theta -1)-3(cos\theta -1)=0\Rightarrow (5cos\theta -3)(cos\theta -1)=0$
$\Rightarrow cos\theta =\frac{3}{5}orcos\theta =1\Rightarrow \theta ={53}^{\circ }or\theta =0$