A particle starting from rest moves atong the circumference of a circle of radius ' $r$ ' with angular…

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
  1. $\frac{r^2}{2 \alpha \theta}$
  2. $\frac{\mathrm{r}}{2 \alpha \theta}$
  3. $\frac{\mathrm{r} \alpha \theta}{2}$
  4. $\frac{\mathrm{r}}{\sqrt{2}} \sqrt{\alpha \theta}$

Solution

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) $\begin{aligned} \therefore \quad & \text { Average velocity }=\frac{\text { Angular displacement }}{\text { time }} \\ & \mathrm{V}_{\text {average }}=\frac{\mathrm{r} \theta}{\mathrm{t}}=\frac{\mathrm{r} \theta}{\left(\frac{2 \theta}{\alpha}\right)^{1 / 2}}=\frac{\mathrm{r}}{\sqrt{2}} \sqrt{\alpha \theta} \end{aligned}$ *

Asked in: MHT CET 2024 (16 May Shift 1)

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