Question

A particle suspended by a light inextensible thread of length $l$ is projected horizontally from its lowest position with a velocity $\sqrt{7gl/2}$. The string will slack after swinging through an angle $\theta$ with the horizontal. Then, $\theta$ is

• A. ${30}^{\circ }$
  
• B. ${90}^{\circ }$
  
• C. ${45}^{\circ }$
  
• D. ${37}^{\circ }$
  

Answer 1

The correct option is A ${30}^{\circ }$


Let the speed of the particle at the point where the string slacks be $v$

Applying conservaiton of mechanical energy

$\frac{1}{2}m{v}^2+mg\left(h\right)=\frac{1}{2}m{u}^2$

$h=l+lsin\theta =l(1+sin\theta )$


$v^2=u^2-2gh$

$v^2=u^2-2gl(1+sin\theta ).......\left(1\right)$

Writing Newton's second law at the point where string slacks,

$T+mgsin\theta =\frac{m{v}^2}{l}$

Condition for the string to slack is $T=0$

$mgsin\theta =\frac{m{v}^2}{l}$

$mgsin\theta =\frac{m}{l}[u^2-2gl(1+sin\theta \left)\right]$

Given, $u^2=\frac{7gl}{2}$

we get sin $\theta =\frac{1}{2}$

or, $\theta ={30}^{\circ }$

Why this Question? Note: Whenever a velocity $\sqrt{2gl}<u<\sqrt{5gl}$ is imparted to a particle at the lowest point of the verticle circular path, the particle takes a parabolic trajectory and a point is always obained where the tension in the string becomes zero, or the string slacks.