Question

The bob of a simple pendulum at rest position is given a velocity V in horizontal direction so that bob describes verticle circle of radius equal to length of pendulum L . If the tension in string is 4 times weight of of bob when the string is horizontal , the velocity of bob :

Answer 1

The bob has a velocity ${}^{\prime }{v}^{\prime }$ as in lowest point and it describe a vertical circle of radius $=L.$

Tension in the string when it makes angle $\theta$ with vertical is given by :

$T=\frac{m{v}^2}{L}-mg(2-3\cos \theta )$

Given that , at horizontal $(\theta ={90}^o),T=4mg$, so

Gmg $=\frac{m{V}^2}{L}-mg(2-3\cos {90}^o)$

$=\frac{m{V}^2}{L}-2mg$

$\Rightarrow 6mg=\frac{m{V}^2}{L}\Rightarrow V=\sqrt{6gL}$