I considered a particle in polar coordinates, $(r,\theta)$, with mass $m$. The standard basis vectors in polar coordinates are:
$$\mathbf{\hat{r}}=\cos{\theta}\mathbf{\hat{x}}+\sin{\theta}\mathbf{\hat{y}}$$
And:
$$\boldsymbol{\hat{\theta}}=\frac{\partial\mathbf{\hat{r}}}{\partial\theta}=-\sin{\theta}\mathbf{\hat{x}}+\cos{\theta}\mathbf{\hat{y}}$$
Differentiating the vector $\mathbf{r}$ to the particle twice, we find that:
$$\mathbf{\ddot{r}}=(\ddot{r}-r\dot{\theta}^2)\mathbf{\hat{r}}+(2\dot{r}\dot{\theta}+r\ddot{\theta})\boldsymbol{\hat{\theta}}$$
From which it follows that the radial component of force on this particle is $F_r=m(\ddot{r}-r\dot{\theta}^2)$ and the tangential component is $F_\theta=m(2\dot{r}\dot{\theta}+r\ddot{\theta})$.
I was able to understand three out of four of the terms in this pair of equations by considering the particle undergoing radial and circular motion (in which case $\dot{\theta}=0$ and $\dot{r}=0$, respectively).
Incidentally, however, the $2m\dot{r}\dot{\theta}$ term is the Coriolis force. But isn't this force fictitious and only observable in a non-inertial reference frame? Was I working in a non-inertial reference frame during this derivation? Does what I'm asking even make sense?
I think I primarily need some clarification of how inertial/non-inertial reference frames come into play in this derivation.
Best Answer
It’s a very fair question. The answer is that that’s not the Coriolis force, but it is related.
Suppose you were to now examine the same particle in coordinates rotating about the origin with angular velocity $\omega$, this would perform a shift $\dot\theta\mapsto \dot\theta -\omega$ while leaving $r, \dot r, \ddot\theta$ invariant. As a consequence we would find that $$\mathbf {\ddot r}' = \mathbf{\ddot r} + 2 \omega \left(r \dot\theta ~\hat r-\dot r ~\hat \theta\right) - r \omega^2~\hat r.$$
The first of these terms is the actual Coriolis force $-2m~\vec\omega\times\vec v$. The final term is the similarly fictitious centrifugal force.