Question

A heavy particle hanging from a string of length $l$ is projected horizontally with speed $\sqrt{gl}$ The speed of the particle at the point where the tension in the string equals weight of the particle is

• A. $\sqrt{2gl}$
• B. $\sqrt{3 g l}$
• C. $\sqrt{gl/2}$
• D. $\sqrt{g l / 3}$

Answer 1

Answer: (D)

$T-m g \cos \theta=\frac{m v^{2}}{R}$
Given $T=m g$

$m g-m g \cos \theta=\frac{m v^{2}}{R}$
$g(1-\cos \theta)=\frac{v^{2}}{R}\,\,\,(i)$
$C.O.M.E$ at $A$ and $B ; \Delta K+\Delta U=0$
$\left(\frac{1}{2} m v^{2}-\frac{1}{2} m u^{2}\right)+m g(R-R \cos \theta)=0$
$\Rightarrow v^{2}-u^{2}=-2 g R(1-\cos \theta)$
$\Rightarrow v^{2}-(\sqrt{g l})^{2}= -2 v^{2}$
$3 v^{2}=g l$
$\Rightarrow v=\sqrt{\frac{g l}{3}}$