Question

A particle is initially at rest, moves along in a circle of radius $R=2\text{m}$ with an angular acceleration $\alpha =\frac{\pi }{8}{\text{rad/sec}}^2$. The magnitude of average velocity of the particle over the time it moves by half of the circle is

• A. $4\text{m/s}$
• B. $3\text{m/s}$
• C. $2\text{m/s}$
• D. $1\text{m/s}$

Answer 1

The correct option is D $1\text{m/s}$

Given, radius of circle $R=2\text{m}$, angular acceleration $\alpha =\frac{\pi }{8}{\text{rad/s}}^2$
Since, particle was initially at rest, ${\omega }_0=0$
Using the formula,
$\theta ={\omega }_{0}t+\frac{1}{2}\alpha t^2$ where, $\theta$ is the angular displacement
$\Rightarrow \theta =\frac{1}{2}\alpha t^2=\frac{\pi t^2}{16}...\left(i\right)$
Also, we have for half circle, $\theta =\pi ...\left(ii\right)$
Using $\left(i\right)\text{\&}\left(ii\right)$ we have
$\pi =\frac{\pi t^2}{16}\Rightarrow t^2=16\Rightarrow t=4\text{sec}$
Now, average velocity is given by
$v_{avg}=\frac{\text{total displacement}}{\text{total time taken}}$
$\Rightarrow v_{avg}=\frac{2R}{4}=\frac{4}{4}=1\text{m/s}$