A body attached to a string of length l describes a vertical circle such that it is just able to cross the highest point.Find the minimum velocity at the bottom of the circle.

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Solution

The tendency of the string to become slack is maximum when the particle is at the topmost point of the circle.
At the top, the tension of the string is $\frac{m v_{T}^{2}}{R} - m g$
where $v_{T}$ is the speed of the body at the top.
For $v_{T}$ to be minimum $T \approx 0$ $V_{T} = \sqrt{g} R$
If $v_{B}$ be the critical velocity of the particle at the bottom, then
by conservation of energy:
$M g (2 R) + \frac{1}{2} m v_{T}^{2} = 0 + \frac{1}{2} m v_{B}^{2}$
Since $v_{T} = \sqrt{g} R$ ,
$2 m g R + \frac{1}{2} m g R = \frac{1}{2} m v_{B}^{2}$
$v_{B} = \sqrt{5 g R}$

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