Question

A particle is suspended from a fixed point by a string of length $5\text{m}$. It is projected from the equilibrium position with such a velocity that the string becomes slack after the particle has reached a height of $8\text{m}$ above the lowest point. Find the velocity of the particle, just before the string becomes slack.(take $g=10{\text{m/s}}^2$)

• A. $\sqrt{30}\text{m/s}$
• B. $\sqrt{40}\text{m/s}$
• C. $\sqrt{50}\text{m/s}$
• D. $\sqrt{60}\text{m/s}$

Answer 1

The correct option is A $\sqrt{30}\text{m/s}$

For finding the velocity of the particle just before the slackness we have the condition of tension at point $B$, $T_B=0$

Here the angle $\theta ={37}^{\text{o}}$, as $\sin \theta =\frac{3}{5}$
Thus, by the FBD of the bob at the point $B$ we get

$\frac{m{v}^2}{l}=mg\cos {53}^o$ $(T=0)$
$v=\sqrt{gl\cos {53}^{\text{o}}}=\sqrt{30}\text{m/s}$

Hence option A is the correct answer