Question

Starting from rest, a particle rotates in a circle of radius $R=\sqrt{2}m$ with an angular acceleration $\alpha =\frac{\pi }{4}rad/s^2$. The magnitude of a average velocity of the particle over the time it rotates quarter circle is:

• A. $1.5m/s$
• B. $2m/s$
• C. $1m/s$
• D. $1.25m/s$

Answer 1

Answer: (C)

$1m/s$

Given, $Radius=\sqrt{2}$ Angular Acceleration $\alpha =\frac{\pi }{4}$

Let a particle moves from P to Q during t time of period in a circular Path of radius R .where R = √2m

Angular acceleration,$\alpha =\frac{\pi }{4}rad/s^2$

Initial angular velocity, ${\omega }_0=0$

So, use formula, $\theta =\frac{1}{2}\alpha .t^2$

Since, particle rotates a quarter of circle " so, $\theta =\frac{\pi }{2}$

so, $\frac{\pi }{2}=\frac{1}{2}\frac{\pi }{4}{t}^2\Rightarrow t^2=4\Rightarrow t=2sec$

so, $v_{av}=\frac{\text{total displacement}}{\text{total time taken}}$

$v_{av}=\frac{\sqrt{2}R}{2}$

$v_{av}=\frac{\sqrt{2}.\sqrt{2}m}{2}$

$v_{av}=1m/s$