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

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

Answer 1

Answer: (D)

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