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(A) The particle starts from rest,so initial angular velocity $\omega_0 = 0$.Using the equation of rotational motion: $	heta = \omega_0 t + \frac{1}{

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The particle starts from rest,so initial angular velocity $\omega_0 = 0$.Using the equation of rotational motion: $	heta = \omega_0 t + \frac{1}{2} \alpha t^2$.Since $\omega_0 = 0$,we have $	heta = \frac{1}{2} \alpha t^2$,which gives $t = \sqrt{\frac{2 	heta}{\alpha}}$.The displacement of the particle for an angular displacement $	heta$ is the chord length $d = 2r \sin(	heta/2)$. For small $	heta$,$\sin(	heta/2) \approx 	heta/2$,so $d \approx r 	heta$.The average velocity is defined as $v_{avg} = \frac{	ext{displacement}}{	ext{time}} = \frac{r 	heta}{t}$.Substituting $t$: $v_{avg} = \frac{r 	heta}{\sqrt{2 	heta / \alpha}} = r 	heta \sqrt{\frac{\alpha}{2 	heta}} = r \sqrt{\frac{\alpha 	heta^2}{2 	heta}} = r \sqrt{\frac{\alpha 	heta}{2}}$.Thus,the correct option is $r \sqrt{\frac{\alpha 	heta}{2}}$,which matches option $A$ if written as $r \left(\frac{\alpha 	heta}{2}ight)^{1/2}$.

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

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