Question

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

Answer 1

Answer:

The average velocity of a particle will be 1 m/s.

Explanation:

Given that,

Radius $r=\srrt{2}\ m$

Angular acceleration $\alpha=\dfrac{\pi}{4}$

According to figure,

The angle rotate A to B

$\theta=\dfrac{1}{2}\alpha t^{2}$

$\dfrac{\pi}{2}=\dfrac{1}{2}\times\dfrac{\pi}{4}\times t^{2}$

$t^{2}=4$

$t=2\s$

The average velocity is the ratio of the total distance and total time.

The total distance from A to B is

$D=\sqrt{(\sqrt{2})^2+(\sqrt{2})^2}$

$D= 2\ m$

The average velocity will be

$v_{avg}=\dfrac{D}{T}$

$v_{avg}=\dfrac{2}{2}$

$v_{avg}=1\ m/s$

Hence, The average velocity of a particle will be 1 m/s.