Question

A particle rotates along a circle of radius under root 2 with angular acceleration pi/4 starting from rest calculate the magnitude of average velocity of the particle over the time it rotate a quarter circle

Answer 1

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,$\alpha=\frac{\pi}{4}rad/s^2$
as you can see , initial angular velocity ,$\omega_0=0$

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

question said " particle rotates a quarter of circle " so, $\theta=\frac{\pi}{4}$

so, $\frac{\pi}{2}=\frac{1}{2}\frac{\pi}{4}t^2$
t² = 4 => t = 2 sec

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

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

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

$v_{av}=1m/s$

Answer 2

A particle rotates along a circle of radius under root 2 with angular acceleration pi/4 starting from rest calculate the magnitude of average velocity of the particle over the time it rotate a quarter circle

Report by R

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

as you can see , initial angular velocity ,\omega_0=0

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

question said " particle rotates a quarter of circle " so, \theta=\frac{\pi}{4}

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

t² = 4 => t = 2 sec

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