Question

A particle rotates along a circle of radius $R=\sqrt{2}$ m with an angular acceleration $\alpha =\frac{\pi }{4}rad/s^2$ starting from rest. Calculate the magnitude of average velocity of the particle over the time it rotates a quarter circle.

Answer 1

$\text{Step 1: Time required to cover quarter circle (t)}$ $\text{[Refer Fig.]}$

Suppose the particle goes from A to B as shown in the figure.

Angular displacement, $\theta =\frac{\pi }{2}$

Initial angular speed, ${\omega }_o=0$

Since Angular acceleration $\left(\alpha \right)$ is constant

$\therefore$ Applying equation of motion in rotation

$\theta ={\omega }_{o}t+\frac{1}{2}\propto t^2$

$\Rightarrow \frac{\pi }{2}=0+\frac{1}{2}\frac{\pi }{4}{t}^2$

$\Rightarrow t=2s$

$\text{Step 2: Average velocity calculation}$

$|\overrightarrow{V_{avg}}|=\frac{\text{Displacement}}{\text{time}}$

From figure, Displacement(shortest distance between A and B) $=\sqrt{R^2+R^2}$ $=\sqrt{2}R$

$\Rightarrow |\overrightarrow{V_{avg}}|=\frac{\sqrt{2}R}{2s}=\frac{\sqrt{2}\sqrt{2}}{2}m/s$$=1m/s$

Hence the magnitude of average velocity of the particle over the time it rotates a quarter circle is $1m/s$.