Question

A particle is suspended from a fixed point by a string of length $5\text{m}$. It is projected from equilibrium position with such a velocity that the string slackens after the particle has reached a height $8\text{m}$ above the lowest point. Choose the correct option(s)
[Take $g=10{\text{m/s}}^2$]

• A. Velocity of particle just before the string slackens is $\sqrt{30}\text{m/s}$
• B. Velocity of particle just before the string slackens is $\sqrt{20}\text{m/s}$
• C. Particle can rise further to a vertical height of $1.96\text{m}$
• D. Particle can rise further to a vertical height of $0.96\text{m}$

Answer 1

Answer: (A), (D)

Particle can rise further to a vertical height of $0.96\text{m}$

Let the speed of the particle just before the string slackens be $v$.


From the figure, angle made by the vertical at the point where the string slackens
$\theta ={\cos }^{-1}\left(\frac{3}{5}\right)$
On applying equation of dynamics towards the centre of vertical circular path, at the point where the string just slackens:
$T+mg\cos \theta =\frac{m{v}^2}{r}$
String will slacken, means tension in the string become zero. Substituting $T=0$ in the above equation
$mg\cos \theta =\frac{m{v}^2}{r}$
$\Rightarrow v=\sqrt{rg\cos \theta }$
$\Rightarrow v=\sqrt{5\times 10\times \frac{3}{5}}$
$\because \cos \theta =\frac{3}{5}$
$\Rightarrow v=\sqrt{30}\text{m/s}$

Just after the string slackens, particle starts projectile motion with velocity of projection equal to velocity of the particle at that instant.
Maximum height of projectile,
$H=\frac{v^2{\sin }^2\theta }{2g}$
$\Rightarrow H=\frac{30\times {\left(\frac{4}{5}\right)}^2}{2\times 10}$
$\Rightarrow H=0.96\text{m}$
Hence, particle can rise further to a vertcial height of $0.96\text{m}$.
$\Rightarrow \left(a\right)\&\left(d\right)$ options are correct.