One end of a string of length R is tied to a stone of mass m and the other end to a small pivot on a frictionless vertical board. The stone is whirled in a vertical circle with the pivot as the centre. The minimum speed the stone must have, when it is at the topmost point on the circle, so that the string does not slack is given by

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$\sqrt{g R}$

$\sqrt{2 m g R}$

$\sqrt{m g R}$

$\sqrt{2 g R}$

answer is A.

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Detailed Solution

Referring to Fig, when the stone is at the topmost point A, the net force towards the centre is
$T_{A} + m g = \frac{m v_{A}^{2}}{R}$
When the stone is at the lowermost point $B$, the net force towards the centre is
$T_{B} - m g = \frac{m v_{B}^{2}}{R}$
The relation between $v_{A}$ and $v_{B}$ can be obtained from the principle of conservation of energy. Let the gravitational potential energy be zero at the lowermost point B. Then
$KE$ at $B + PE$ at $B = KE$ at $A + PE$ at A
or $\frac{1}{2} m v_{B}^{2} + 0 = \frac{1}{2} m v_{A}^{2} + m g \times A B$
$= \frac{1}{2} m v_{A}^{2} + m g \times 2 R$
or $v_{B}^{2} = v_{A}^{2} + 4 g R$
Now, when the stone is at A, the string will not slack if the whole of centripetal force is provided by the weight mg, i.e. $T_{A} = 0$ . Putting $T_{A} = 0$ , we have
$m g = \frac{m v_{A}^{2}}{R} or v_{A} = \sqrt{g R}$