What can go wrong with applying chain rule to angular velocity of circular motion?

Question

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?

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This is not a homework problem, but inspired by one.

Answer 1

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}

Answer 2

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 .

Answer 3

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.

Answer 4

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.

Answer 5

$d\phi/dr$ 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.

Answer 6

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 $t$.

Answer 7

$dr/ dt ≠0$ as direction of $r$ is continuously changing. $r$ means both direction as well as magnitude, you are only considering the magnitude, which is wrong.