A particle starting from rest moves atong the circumference of a circle of radius ' \(r\) ' with angular acceleration ' \(\alpha\) '. The magnitude of the average velocity in time it completes the small angular displacement ' \(\theta\) ' is

Using kinematic equation,
\(\begin{aligned}
\theta & =\omega_0 \mathrm{t}+\frac{1}{2} \alpha \mathrm{t}^2 \\
\theta & =\frac{1}{2} \alpha \mathrm{t}^2 \\
\therefore \quad \mathrm{t} & =\left(\frac{2 \theta}{\alpha}\right)^{1 / 2}...(i)
\end{aligned}\)
\(\ldots\left(\because\right.\) particle was at rest, \(\left.\omega_0=0\right)\)
Angular displacement of the particle \(=\mathrm{r} \theta\)...(ii)
\(\begin{aligned}
\therefore \quad & \text { Average velocity }=\frac{\text { Angular displacement }}{\text { time }} \\
& \mathrm{V}_{\text {average }}=\frac{\mathrm{r} \theta}{\mathrm{t}}=\frac{\mathrm{r} \theta}{\left(\frac{2 \theta}{\alpha}\right)^{1 / 2}}=\frac{\mathrm{r}}{\sqrt{2}} \sqrt{\alpha \theta}
\end{aligned}\)