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(D) Using the rotational kinematic equation for angular displacement:$	heta = \omega_0 t + \frac{1}{2} \alpha t^2$Since the particle starts from rest

Vedclass · Accepted Answer

$\frac{r}{\sqrt{2}} \sqrt{\alpha 	heta}$

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(D) Using the rotational kinematic equation for angular displacement:$	heta = \omega_0 t + \frac{1}{2} \alpha t^2$Since the particle starts from rest,$\omega_0 = 0$,so:$	heta = \frac{1}{2} \alpha t^2$Solving for time $t$:$t = \sqrt{\frac{2 	heta}{\alpha}} \quad ...(i)$The linear distance (arc length) covered by the particle for an angular displacement $	heta$ is:$s = r 	heta \quad ...(ii)$The average velocity is defined as the total displacement divided by the total time. For a small displacement $	heta$,the magnitude of displacement is approximately the arc length $s$:$V_{	ext{average}} = \frac{s}{t} = \frac{r 	heta}{\sqrt{\frac{2 	heta}{\alpha}}}$$V_{	ext{average}} = r 	heta \cdot \sqrt{\frac{\alpha}{2 	heta}} = r \sqrt{\frac{\alpha 	heta^2}{2 	heta}} = \frac{r}{\sqrt{2}} \sqrt{\alpha 	heta}$

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

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