Question

A particle is moving along a vertical circle of radius r. the velocity of particle at p will be

Answer 1

Let the speed at the point P be $u$

For the critical point C we can write, $\frac{m{v}^2}{R}=mg$

Thus, ${v}^2 = Rg$

Applying conservation of energy,

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

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

$2gR = {u}^2$

$u = \sqrt{2gR}$

Answer 2

Answer:

u= root over 2gr is the correct answer for this question