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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.

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  • $\begingroup$ Assume the radius of circle is $r$ which is constant. Then the tangential component of the acceleration is $a_{t}=r\frac{d\omega}{dt}$ where $\omega=\frac{d\theta}{dt}$ is the angular velocity - and the radial component of the acceleration is $a_{r}=\omega^{2}r$. $a_{r}$ is also known as the centripetal acceleration - it's the fictitious force. You know you're in an non-inertial frame when the origin of your coordinate system is the axis of rotation of the particle you're tracking. A inertial frame is one that moves at constant velocity and doesn't accelerate. $\endgroup$ – Cinaed Simson Jun 17 at 5:12
  • $\begingroup$ And just to be clear, the magnitude of $\hat r=1$, which is constant, hence $\dot r=\ddot r=0$, which yields $F_r=-mr\dot{\theta}^2$ and $F_\theta=mr\ddot{\theta}$ - which are the correct answers. There is no Coriolis force in 2 dimensions - and there's no vector cross product in 2 dimensions either. $F_r=-mr\dot{\theta}^2$ is the centripetal acceleration - a fictitious force. $\endgroup$ – Cinaed Simson Jun 17 at 7:35
  • $\begingroup$ @CinaedSimson This should be an answer, not a comment. $\endgroup$ – J. Murray Jun 17 at 13:45
  • $\begingroup$ @CinaedSimson, what I'm considering is not necessarily circular motion though. The position vector is $\mathbf{r}=r\mathbf{\hat{r}}$. The standard basis vector's magnitude never changes, but $\dot{r}$ may be nonzero because the motion of the particle is arbitrary and so can be radial. $\endgroup$ – Andrew Paul Jun 17 at 15:45
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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.

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  • $\begingroup$ This makes sense, thanks! So then, if we imagine $\dot{\theta}=0$ (in the inertial frame), and let a rotating frame have angular velocity $\omega$, then the shift becomes $0\mapsto -\omega$ and hence the term $2m\dot{r}\dot{\theta}$ is just $2m\boldsymbol{\omega}\times\mathbf{v}$, the Coriolis force observed from the rotating frame, as I now understand it. $\endgroup$ – Andrew Paul Jun 17 at 3:41

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