What is the angular acceleration of a particle moving with constant angular velocity?

Hint: In this question, we have to remember that the angular acceleration is the ratio of change in angular velocity and time taken by the particle from moving one point to another. We can solve this question by putting $\omega = $constant

Complete step by step solution:
We know that the angular velocity of a particle moving on a circular path of radius \[r\], then the angular velocity $\omega $ is given as-
$\omega = \dfrac{{d\theta }}{{dt}}$ (i)
Where $d\theta $ is the angular displacement of the particle after time ${dt}$.
Now, let the angular acceleration is $\alpha $ of the particle. So, angular acceleration
$\alpha = \dfrac{{d\omega }}{{dt}}$ (ii)
Where $d\omega $ is the change in the velocity and ${dt}$ is the time taken.
Now, according to the question, the angular velocity is constant. So,
$\omega = $constant
Then $d\omega = 0$
Now, from equation (i)
$\alpha = 0$

Hence, the angular acceleration is zero when the angular velocity is constant for a particle.

Additional Information:
We know that in circular motion, a particle is moving in a circular motion, while in linear motion, a particle is moving in a linear path. In circular motion, the particle will have an angular displacement while in linear motion, the particle will have linear displacement. In circular motion, the particle has angular velocity and angular acceleration while in linear motion, the particle has linear velocity and linear acceleration.

Note:
In this question, we have to remember that angular velocity and angular acceleration are related with circular motion. We have to remember that the angular acceleration is the ratio between change in angular velocity and the time. We have given in the question that the angular velocity is constant. So, change in angular velocity will be zero.