I am very confused about net acceleration and angular acceleration in circular motion [closed]

Question

I have seen in many places that angular acceleration can be zero but net acceleration can't be zero in circular motion. I want to know whose components are tangential and radial acceleration (net or angular acceleration) and what is the significance of angular acceleration then? After studying axial vectors like angular displacement, angular velocity, angular acceleration why shift to net acceleration which is in plane(I think) (Please correct me wherever I am wrong I am new to this concept)

Answer 1

Say you have a movement on a plane, described by the position vector $\mathbf{r} = r \mathbf{\hat{r}}$ (note that this unit vector points from the origin to the particle/object). Using the kinematic derivative rules $\frac{d}{dt}\mathbf{\hat{r}} = \dot{\theta}\mathbf{\hat{\theta}}$ and $\frac{d}{dt}\mathbf{\hat{\theta}} = -\dot{\theta}\mathbf{\hat{r}}$, you get by differentiating position the velocity (chain rule!): $$ \mathbf{v} = \dot{r} \mathbf{\hat{r}} + r\dot{\theta} \mathbf{\hat{\theta}} $$ and differentiating once again, we get acceleration: $$ \mathbf{a} = (\ddot{r} - r\dot{\theta}^2)\mathbf{\hat{r}} + (2\dot{r}\dot{\theta} + r \ddot{\theta})\mathbf{\hat{\theta}} $$

This is the most general case and there's velocity and acceleration in both radial $\mathbf{\hat{r}}$ and tangential $\mathbf{\hat{\theta}}$ directions. When the movement is circular, the radius $r$ has no derivatives of any order (doesn't change), se the above definitions are now: $$ \mathbf{v} = r\dot{\theta} \mathbf{\hat{\theta}} $$ and $$ \mathbf{a} = - r\dot{\theta}^2\mathbf{\hat{r}} + r \ddot{\theta}\mathbf{\hat{\theta}} $$ that is, velocity is always tangent to the (circular) trajectory but acceleration has a radially inward centripetal component! it is also usual to relabel $\dot{\theta} = \omega$ (angular velocity) and $\ddot{\theta} = \dot{\omega} = \alpha$ (angular acceleration). Now you can say: $$ \mathbf{a} = - r\omega^2\mathbf{\hat{r}} + r \alpha\mathbf{\hat{\theta}} $$ That is, acceleration has both radial/centripetal and tangential components, and the tangential component is related to the angular acceleration by $a_{\text{tangent}} = r \alpha$.

Finally, you can let $\alpha = 0$ and there will still be some component to acceleration.