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

Physics Asked on April 12, 2021

Lets say I have a circular motion, like this:

enter image description here

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.

7 Answers

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}

Correct answer by Ruben Campos Delgado on April 12, 2021

$dphi/dr$ 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.

Answered by user256872 on April 12, 2021

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 .

Answered by Naman Parikh on April 12, 2021

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

Answered by Nandani Kumari on April 12, 2021

The meaning of pushing an intermediate variable is to relate the dependencies of change. So, the expression

$$ frac{dphi}{dt} = frac{d phi}{dr} frac{dr}{dt}$$

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.

Answered by Buraian on April 12, 2021

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.

Answered by PM 2Ring on April 12, 2021

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 $t$.

Answered by jezza on April 12, 2021

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