Question

A particle is moving along a vertical circle of radius $R$. The velocity of particle at $P$ will be : (assume critical condition at $C$)

• A. $\sqrt{gR}$
• B. $\sqrt{2gR}$
• C. $\sqrt{3gR}$
• D. $\sqrt{\frac{3}{2}gR}$

Answer 1

The correct option is B $\sqrt{2gR}$

Let,

$R=$ radius of circle

$m=$ mass of particle

$v=$ critical speed of particle

$\theta ={60}^0$

Critical condition at $C$,

$mg=\frac{m{v}^2}{R}$

$v^2=gR$

$v=\sqrt{gR}$

Lets assume the speed of particle at point $P$ is $u$

Applying conservation of energy,

$2mgR+\frac{1}{2}m{v}^2=\frac{1}{2}m{u}^2+mgR(1+cos\theta )$

$4gR+gR=u^2+3gR$

$u^2=2gR$

$u=\sqrt{2gR}$

The correct option is B.