# The acceleration of circulation motion

We know that in circular motion the position vector is $r\hat{r}$. Then the velocity is the time derivative of it. So it gives $$dv/dt = r d\hat{r}/dt + \frac{dr}{dt} .\hat{r}.$$ now I can't understand why $dr/dt$ isn't zero. The $r$ or radii isn't changing throughout the circle. So should the velocity be only $rd\hat{r}/dt$?

Saying that the position is $r\hat r$ does not only describe circular motion. This is just a general position vector.
Therefore, your time derivative is just a general velocity vector. The first term represents velocity tangent to the position vector, the second parallel to it. Therefore, if you are undergoing circular motion, the second term is $0$.
Based on your comments, I think I see your issue. You are thinking "I see $r\hat r$ is used in the case of circular motion, so it only describes circular motion". The correct way to see it is "$r\hat r$ is always true, so I can apply it to circular motion."
You are indeed right to say that the term $\frac{dr}{dt}$ is zero. Indeed the velocity vector is given just by $r\frac{d\hat{r}}{dt}$.
You can see why this is true too if you consider the dot product of the velocity vector and the radius vector. Consider the displacement or position vector to be $r\hat{r}=r(\cos(\omega t)i+sin(\omega t)j$.
Now we know the derivative is $r\frac{d\hat{r}}{dt}$, which is equal to the derivative of the displacement vector with $r$ as constant, you get: $$r\frac{d\hat{r}}{dt}=r(-\sin(\omega t)i+cos(\omega t)j$$ Now if you take the dot product of the two vectors you should get zero as is true in circular motion, that is the velocity vector is perpendicular to the radius vector. It is indeed true as the dot product yields : $$(-\sin(\omega t)cos(\omega t) + \sin(\omega t)cos(\omega t))=0$$ As was to be shown.