Question

A particle is suspended from a fixed point by a string of length $3\text{m}$. It is projected from equilibrium position with such a velocity that the string slackens after the particle has reached a height of $5\text{m}$ above the lowest point. Find the velocity of the particle, just before the string slackens. Take $g=9.81{\text{m/s}}^2$.

• A. $10\text{m/s}$
• B. $4.42\text{m/s}$
• C. $0\text{m/s}$
• D. $2\text{m/s}$

Answer 1

The correct option is B $4.42\text{m/s}$


At point B, Tension $T=0$
FBD of particle at point B:


From the frame of reference of particle, there will be a centrifugal force $\frac{m{v}^2}{l}$
$\therefore T+mg\cos \theta =\frac{m{v}^2}{l}$
$\Rightarrow 0+mg\cos \theta =\frac{m{v}^2}{l}$
$\Rightarrow v=\sqrt{gl\cos \theta }$
From the first figure, $\cos \theta =\frac{2}{3}$
$\Rightarrow v=\sqrt{9.81\times 3\times \frac{2}{3}}$
$\Rightarrow v=4.42\text{m/s}$