The angular velocity of a stone whirled in a horizontal circle of radius $r$ at angle $\theta$ is:

1. From the two force components, $v^2=rg\tan\theta$. 2. For circular motion $v=\omega r$, so $v^2=\omega^2r^2$. 3. Equating: $\omega^2r^2=rg\tan\theta$, giving $\omega^2=\dfrac{g\tan\theta}{r}$. 4. Taking the square root gives $\omega=\sqrt{\dfrac{g\tan\theta}{r}}$. 5. The revolutions per second follow as $\dfrac{\omega}{2\pi}$. Omitting the square root, a recorded error, leaves the wrong units entirely. _Source: NECTA ACSEE 2023 Physics 131/1, Question 3: Mechanics (Uniform Circular Motion)_