What can go wrong with applying chain rule to angular velocity of circular motion? Lets say I have a circular motion, like this:

I know that:
$$\omega = \frac{\text{d} \phi}{\text{d}t}$$
Mathematically, what I am doing wrong, when I attempt to apply the chain rule in the following way?
$$\omega = \frac{\text{d}\phi}{\text{d}t}=\frac{\text{d}r}{\text{d}t}\frac{\text{d}\phi}{\text{d}r}(=0?)$$
Which result seems wrong because $\frac{\text{d} r}{\text{d} t}$ is zero, since my $r$ coordinate is not changing. So $\omega$ would be $0$ for all circular motion. What am I doing wrong?

This is not a homework problem, but inspired by one.
 A: The meaning of pushing an intermediate variable is to relate the dependencies of change. So, the expression
$$ \frac{d\phi}{dt} = \frac{d \phi}{dr} \frac{dr}{dt}$$
Assumes that the angle $(\phi)$ can be written as some function of the radius from the origin i.e: $\phi(r(t))$ but is that really possible in this case? We can say that there is no 'differentiable' map from $r \to \phi$.
I mean think about it, how would you construct a function which associates the radius , a fixed variable, with the angle which changes with time? It would not even be a function because you would have to span all the angles with a single radius value.
A: The problem is that $r$ is a constant, so its rate of change relative to any other variable must be zero. Thus
$$\frac{\text{d}r}{\text{d}t} = 0$$
and
$$\frac{\text{d}r}{\text{d}\phi} = 0$$
So $\frac{\text{d}\phi}{\text{d}r}$ is undefined. When you try to apply the chain rule in
$$\omega = \frac{\text{d}\phi}{\text{d}t}=\frac{\text{d}r}{\text{d}t}\frac{\text{d}\phi}{\text{d}r} = \frac{\text{d}r}{\text{d}t} / \frac{\text{d}r}{\text{d}\phi}$$
you're attempting to divide zero by zero, which is an indeterminate form.
A: $d\phi/dr$  is undefined. $\phi$ is only defined for 1 value of $r$ and is thus not differentiable. So you end up with $\omega$ being zero x undefined quantity.
A: The reason is that $r(t)$ and $\phi(t)$ are your independent variables, so you are not allowed to use the chain rule for them, because $\phi$ does not depend on $r$. An analogy would be the following:
\begin{equation}
\frac{d}{dx}x=1
\end{equation}
However, if I introduce another independent variable $y$, it would be of course wrong to write
\begin{equation}
\frac{d}{dx}x=\frac{dy}{dx}\frac{d}{dy}x=0
\end{equation}
A: Though $\frac{\mathrm{d}r}{\mathrm{d}t}$ is $0$, $\frac{\mathrm{d}\phi}{\mathrm{d}r}$ also tends to infinity. So this is a zero into infinity form and its limit will result in a finite quantity which will be equal to the angular velocity calculated without chain rule .
A: Since you've got vectors on your diagram, I think it's worth noting that you could write
$$\omega = \frac{\mathrm{d}\phi}{\mathrm{d}t} = \frac{\mathrm{d}\vec{r}}{\mathrm{d}t} \cdot \vec{\nabla}\phi$$
because $\vec{r}$ is a function of $t$.
A: $dr/ dt ≠0$ as direction of $r$ is continuously changing.
$r$ means both direction as well as magnitude, you are only considering the magnitude, which is wrong.
