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