Question

A heavy particles is suspended by a string of length l from a fixed point O. The particle is given a horizontal velocity $v_0$ the string slack at some angle and the particle proceeds on a parabola. Find the value of $v_0$ if the particle passes through the point of suspension.

• A. $v={\lfloor gl(2+\sqrt{3})\rfloor }^{1/2}$
• B. $v={\lfloor 2gl(2+\sqrt{3})\rfloor }^{1/2}$
• C. $v={\lfloor gl(1+\sqrt{3})\rfloor }^{1/2}$
• D. $v={\lfloor gl(2+\sqrt{3})\rfloor }^{1/3}$

Answer 1

The correct option is A $v={\lfloor gl(2+\sqrt{3})\rfloor }^{1/2}$

Let at P, the string slacks when it makes an angle $\theta$ with the vertical, hence at the point P the centripetal force is only due to the component of the gravitational force.
$mgcos\theta =\frac{m{v}^2}{l}$ where v = velocity of the particle at P.
$v^2=gl.cos\theta$ (i)
Conserving energy at initial point and at P we get
$\frac{1}{2}m{v}_0^2=\frac{1}{2}m{v}^2+mg(1+cos\theta )$ (ii)
From (i) and (ii) we get $\frac{1}{2}m{v}_0^2=\frac{1}{2}mglcos\theta +mgl(1+cos\theta )$
$\Rightarrow v_0^2=2gl+\frac{3g}{cos\theta }\Rightarrow v_0^2=gl[2+3cos\theta ]$ (iii)
Now particle will pass point of suspension if $lsin\theta =\left(vcos\theta \right)t$ (iv)
And $lcos\theta =(vsin\theta -\frac{1}{2}g{t}^2)$ (v) Eliminating t from (iv) and (v) we get.
$-lcos\theta =\left(vsin\theta \right)\left[\frac{lsin\theta }{vcos\theta }\right]-\frac{1}{2}g{\left[\frac{lsin\theta }{vcos\theta }\right]}^2$ Substituting $v^2=glcos\theta$, and simplifying $tan\theta =\sqrt{2}$
Therefore $cos\theta =\frac{1}{\sqrt{3}}$ (vi)
Substituting the value of $cos\theta$ in equation (iii), we get $v={\lfloor gl(2+\sqrt{3})\rfloor }^{1/2}$