If the tension in the string is

T

, resolving horizontally and vertically and using F=ma (where a is the acceleration towards the centre):

T \cos θ = mg

T \sin θ = ml \sin θ ω^{2}

Eliminating

T

l ω^{2} \cos θ = g

and hence the angle

θ

\cos^{- 1} \frac{g}{l ω^{2}}

If we increase the angular velocity then $\cos θ$ decreases and angle $θ$ increases and hence the mass rises up and the radius of the circle it moves on also increases.

As $\cos θ \leq 1$ we have

$\frac{g}{l ω^{2}} \leq 1$

so

$ω \geq \sqrt{\frac{g}{l}}$

and hence the smallest angular speed with which the ball can whirl in a circle on this string is

$\sqrt{\frac{g}{l}} .$