Question 222: Work, Energy and Power & Circular Motion

Question

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 time it completes the small angular displacement ' $	heta$ ' is

Accepted Answer

Using kinematic equation,
$$\begin{aligned}
& 	heta=\omega_0 t+\frac{1}{2} \alpha t^2 \
& 	heta=\frac{1}{2} \alpha t^2 \quad \ldots\left(\because 	ext { particle was at rest, } \omega_0=0ight)\
	herefore \quad & t=\left(\frac{2 	heta}{\alpha}ight)^{1 / 2}\quad	ext{.... (i)}
\end{aligned}$$
Angular displacement of the particle $=r 	heta\quad	ext{.... (iii)}$
$$\begin{aligned}
& 	herefore \quad 	ext { Average velocity }=\frac{	ext { Angular displacement }}{	ext { time }} \
& \mathrm{v}_{	ext {avenge }}=\frac{\mathrm{r} 	heta}{\mathrm{t}}=\frac{\mathrm{r} 	heta}{\left(\frac{2 	heta}{\alpha}ight)^{1 / 2}}=\frac{\mathrm{r}}{\sqrt{2}} \sqrt{\alpha 	heta}
\end{aligned}$$

Answer

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

Answer

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

Answer

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

Answer

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

Work, Energy and Power & Circular Motion

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 time it completes the small angular displacement ' \theta ' is

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