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I am reading the book Classical Mechanics by Taylor, and I have a question about the equations of motion in a non-inertial frame of reference, say $S$.

In particular, if $S$ has an angular velocity $\Omega$ with respect to some inertial frame of reference $S_0$, then the equation of motions in the rotating frame is given by:

$$m \ddot{\textbf{r}} = F + F_\textrm{cor} + F_\textrm{cf} \tag{$\dagger$}$$ where $F$ is the net forced as measured in any reference frame, and $F_{cor}$ and $F_\textrm{cf}$ are Coriolis and centrifugal forces, respectively.

Now, my question is that if I were to "solve" $(\dagger)$ for $\textbf{r}$ in some context, is it correct that I need to expresses the forces in the right hand side of the equation in the frame $S$?

In his discussion of centrifugal force in free fall on earth (pg 345 to 348), Taylor identifies the centrifugal force as $$F_\textrm{cf} = \Omega^2 R_\textrm{earth} \sin(\theta) \hat{\rho}$$, where $\rho$ is the unit radial vector in cylindrical polar coordinates, and since this force, as it is written, is with respect to the frame $S_0$, it needs to be "rewritten" in terms of the frame $S$. Doesn't it?

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Why do you say that the equation for $F_{cf}$ is written w.r.t. $S_0$? In frame $S_0$ there wouldn't be any $F_{cf}$ or $F_{coriolis}$, because it is an inertial frame. So if those forces are non-zero it must be definitely w.r.t. some non-inertial frame $S$.

Response to comment:

Newton's law in its unadorned form is valid in an inertial frame. You obtain equation of motion for non-inertial frame, by transforming the equation in inertial frame, to non-inertial frame. During this transformation, $\Omega$ of non-inertial frame is an essential input (in fact it is one of the things that distinguishes inertial from non-inertial frame). But once the transformation is done, the resulting equation is w.r.t. to the frame into which you have transformed, which is the non-inertial frame. This is also borne out by the fact that transformed equation is not valid anymore in the original inertial frame.

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  • $\begingroup$ I am saying that $F_{cf}$ is written w.r.t. $S_{0}$ because $F_{cf} = m(\Omega \times r) \times \Omega$, where $\Omega$ is the angular velocity vector as seen from the inertial frame $S_0$ $\endgroup$ – JFN Aug 17 '16 at 17:59
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Friend, I am not so experienced as you are. What I am providing as an answer is merely a guesswork.

Visualize the coordinate system in the $S_0$ frame. It is stationary.

Now visualise the same in the $S$ frame rotating w.r.t. $S_0$ frame with $\omega$.

Now visualize, $S$ frame is stationary and $S_0$ frame is rotating w.r.t. $S$ frame with -ve $\omega$.

Say, -ve $\omega = \omega_0$.

So you are using -ve $\omega_0$ in your equation.

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    $\begingroup$ Would you please be so kind as to define your symbols? $\endgroup$ – ZeroTheHero Mar 24 '17 at 4:41

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