Question

A heavy particle is suspended by a string of length $l$. The particle is given a horizontal velocity $v_0$. The string becomes 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. $v_0=\sqrt{gl[2+\sqrt{3}]}$
• B. $v_0=\sqrt{gl}$
• C. $v_0=\sqrt{gl[5+\sqrt{3}]}$
• D. $v_0=\sqrt{gl\left[\frac{3+\sqrt{3}}{2}\right]}$

Answer 1

The correct option is A $v_0=\sqrt{gl[2+\sqrt{3}]}$

As mentioned in the question that the string becomes slack at some angle which means at some angle tension in the string become zero but still the particle is having some velocity due to which the particle continues to move and then it falls and passes through the point of suspension
The free body diagram of particle at angle $\theta$ from vertical is given as,

Now, taking component of force along the string
$T+mg\cos \theta =\frac{m{v}^2}{l}$
Let, this is the point at which tension in the string become zero then,
$\begin{array}{}{v}^2=gl\cos \theta & \text{(1)}\end{array}$
As from the above free body diagram the height of the particle above center is $l\cos \theta$
Potential energy of particle at an angle $\theta$ from vertical is
$U=mgh=mg[l+l\cos \theta ]$
and kinetic energy, $K.E=\frac{1}{2}m{v}^2$
If we assume lowest point is the reference level, then potential energy is zero at that point.
Kinetic energy at bottom most point is,
$\frac{1}{2}m{v_0}^2$
From conservation of energy,
$\frac{1}{2}m{v_0}^2=mg[l+l\cos \theta ]+\frac{1}{2}m{v}^2$
puting value of $v^2$
$\frac{1}{2}{v_0}^2=gl[1+\cos \theta ]+\frac{1}{2}gl\cos \theta$
$\begin{array}{}\frac{1}{2}{v_0}^2=gl[1+\frac{3}{2}\cos \theta ]& \text{(2)}\end{array}$

After angle $\theta$ the partical will be in projectile motion

From the above diagram,
$\begin{array}{}x=l\sin \theta =v\cos \theta t& \text{(3)}\end{array}$
$\begin{array}{}y=-l\cos \theta =v\sin \theta t-\frac{1}{2}g{t}^2& \text{(4)}\end{array}$
replacing $t$ from eq.(3) in eq.(4)
$\begin{array}{}-l\cos \theta =v\sin \theta \frac{l\sin \theta }{v\cos \theta }-\frac{1}{2}g\frac{l^2{\sin }^2\theta }{v^2{\cos }^2\theta }& \text{(4)}\end{array}$
$-{\cos }^2\theta =si{n}^2\theta -\frac{1}{2}g\frac{l{\sin }^2\theta }{v^2\cos \theta }$
$\frac{1}{2}g\frac{l{\sin }^2\theta }{v^2\cos \theta }={\sin }^2\theta +{\cos }^2\theta$
$1=\frac{1}{2}g\frac{l{\sin }^2\theta }{v^2\cos \theta }$
by using equation 1
$1=\frac{1}{2}\frac{{\sin }^2\theta }{{\cos }^2\theta }$
$\tan \theta =\sqrt{2}$

$\cos \theta =\frac{1}{\sqrt{3}}$
replace this in eq.(2) we get
$\frac{1}{2}{v_0}^2=gl[1+\frac{3}{2}\frac{1}{\sqrt{3}}]$
$v_0=\sqrt{2gl[1+\frac{\sqrt{3}}{2}]}$
$v_0=\sqrt{gl[2+\sqrt{3}]}$