First slide

Vertical circular motion

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

Moderate

A

$\sqrt{2 gl}$
B

${\sqrt{3 gl}}^{}$
C

$\sqrt{\frac{gl}{2}}$
D

$\sqrt{\frac{gl}{3}}$

Solution

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

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App