A ball of mass ' \(\mathrm{m}\) ' is attached to the free end of a string of length ' \(l\) '. The ball is moving in horizontal circular path about the vertical axis as shown in the diagram.
The angular velocity ' \(\omega\) ' of the ball will be [ \(\mathrm{T}=\) Tension in the string. \(]\)

The tension in the string can be resolved in two components along the perpendicular axis. The gravitational force is acting downwards and the centrifugal force is acting in \(-\mathrm{x}\) direction
\(
\begin{aligned}
& \mathrm{T} \sin \theta=\mathrm{mr} \omega^2 \\
\therefore \quad & \omega^2=\frac{\mathrm{T} \sin \theta}{\mathrm{mr}}
\end{aligned}
\)
\(\therefore \quad \omega=\sqrt{\frac{\mathrm{T} \sin \theta}{\mathrm{mr}}}\)
From figure, \(\sin \theta=\frac{\mathrm{r}}{l}\)
\(
\begin{aligned}
& \therefore \quad \omega=\sqrt{\frac{\mathrm{Tr}}{\mathrm{mr} l}} \\
& \therefore \quad \omega=\sqrt{\frac{\mathrm{T}}{\mathrm{m} l}} \end{aligned}
\)