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Let us consider two reference frames: $S$ and $S'$. $S$ is an inertial frame and $S'$ is a non-inertial frame as it is rotating wrt $S$ with an angular velocity $\omega$ about a fixed axis. The origins of $S$ and $S'$: $O$ and $O'$, respectively, coincide.

An observer in $S'$ writes the Newton's second law in $S'$:

$$m\textbf{a}' = \textbf{F} + \textbf{F}_c + \textbf{F}_{cen} + \textbf{F}_{Euler}$$

where, $\textbf{F}$ is the applied force, $\textbf{F}_c$ is the coriolis force, $\textbf{F}_{cen}$ is the centrifugal force and $\textbf{F}_{Euler}$ force. I am excluding their mathematical expressions for brevity.

Question

If the observer wants to write an expression for $\textbf{F}$ for solving a physics problem, then should he write $\textbf{F}$ as a function of the coordinates of $S'$; i.e., $\textbf{F} = \textbf{F}(\textbf{r}')$ ?

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  • $\begingroup$ Using the coordinates of a frame is the definition of a frame. What is the alternative in your question? $\endgroup$
    – safesphere
    Nov 6 '17 at 0:46
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If the observer can only measure the position $\mathbf{r}'$, velocity $\dot{\mathbf{r}}'$ and acceleration $\ddot{\mathbf{r}}'$ then the equations of motion are derived from the kinematics

$$ \mathbf{r} = \mathbf{r}_{S'} + \mathrm{R}_{S'} \mathbf{r}' $$

$\mathbf{r}$ coordinates of particle in inertial frame, $\mathbf{r}_{S'}=0$ location of S' relative to inertial frame, $\mathrm{R}_{S'}$ $3×3$ rotation matrix of S'.

Take the total derivative of the above, and note that $\dot{\mathrm{R}}_{S'} = {\boldsymbol \omega} \times \mathrm{R}_{S'}$.

$$ \mathbf{v} = \mathbf{v}_{S'} + {\boldsymbol \omega} \times \mathrm{R}_{S'} \mathbf{r}' + \mathrm{R}_{S'} \dot{\mathbf{r}}' $$

again differentiate and collect similar terms

$$ \mathbf{a} = \underbrace{ \mathbf{a}_{S'}}_{\text{Frame accel.}} + \underbrace{ {\boldsymbol \alpha} \times \mathrm{R}_{S'} \mathbf{r}'}_{\text{Euler accel.}} + \underbrace{ \mathrm{R}_{S'}\ddot{\mathbf{r}}'}_{\text{Observ. accel.}} + \underbrace{ {\boldsymbol \omega} \times ({\boldsymbol \omega} \times \mathrm{R}_{S'} \mathbf{r}')}_{\text{Centripetal accl.}}+\underbrace{2 {\boldsymbol \omega} \times \mathrm{R}_{S'}}_{\text{Coriolis accel.}} \dot{\mathbf{r}}' $$

Considering that $\mathbf{F} = m \,\mathbf{a}$ you can convert the above into "fictitious forces" and solve for $\mathbf{F}' = m \ddot{\mathbf{r}}'$. It is easy to consider the instant where the axes align $\mathrm{R}_{S'}=[1]$

$$ \mathbf{F}' = m \ddot{\mathbf{r}}' = \mathbf{F} - m \mathbf{a}_{S'} - {\boldsymbol \alpha} \times m \mathbf{r}' - {\boldsymbol \omega}\times({\boldsymbol \omega} \times m \mathbf{r}') -2 {\boldsymbol \omega} \times m \dot{\mathbf{r}}'$$

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  • $\begingroup$ Edit: Fixed the notation where a dot represents the derivatives of location $\mathbf{r}'$ as measured in the moving frame. $\endgroup$ Nov 6 '17 at 14:16

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