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It seems that the equation for transforming between two different expressions of the relative acceleration of a vector involves absolute angular velocities, which I find surprising.

Consider two non-inertial coordinate systems $\alpha$ and $\beta$, whose absolute angular velocity vectors are $\boldsymbol{\Omega}^\alpha$ and $\boldsymbol{\Omega}^\beta$, respectively.

The absolute velocity and acceleration of $\boldsymbol{x}$ (i.e. the motion as seen by an observer attached to the inertial system), is $\dot{\boldsymbol{x}}$ and $\ddot{\boldsymbol{x}}$, respectively.

We denote the velocity and acceleration relative to $\alpha$, i.e. that seen by an observer attached to the system $\alpha$, as $\dot{\boldsymbol{x}}|_\alpha$ and $\ddot{\boldsymbol{x}}|_\alpha$, respectively, and similarly for $\beta$. Note that $\dot{\boldsymbol{x}}|_\alpha$ is the time derivative of $\boldsymbol{x}$, holding the basis vectors of $\alpha$ fixed.

$\dot{\boldsymbol{x}}$ can be expressed in terms of $\dot{\boldsymbol{x}}|_\alpha$ and $\dot{\boldsymbol{x}}|_\beta$ as follows:

\begin{equation} \begin{split} \dot{\boldsymbol{x}} &= \dot{\boldsymbol{x}}|_\alpha + \boldsymbol{\Omega}^\alpha\times\boldsymbol{x}\\ \dot{\boldsymbol{x}} &= \dot{\boldsymbol{x}}|_\beta + \boldsymbol{\Omega}^\beta\times\boldsymbol{x}\\ \end{split} \end{equation}

and $\ddot{\boldsymbol{x}}$ can be expressed as follows: \begin{equation} \begin{split} \ddot{\boldsymbol{x}} &= \ddot{\boldsymbol{x}}|_\alpha + 2\boldsymbol{\Omega}^\alpha\times\dot{\boldsymbol{x}}|_\alpha + \dot{\boldsymbol{\Omega}}^\alpha\times\boldsymbol{x} + \boldsymbol{\Omega}^\alpha\times(\boldsymbol{\Omega}^\alpha\times\boldsymbol{x})\\ \ddot{\boldsymbol{x}} &= \ddot{\boldsymbol{x}}|_\beta + 2\boldsymbol{\Omega}^\beta\times\dot{\boldsymbol{x}}|_\beta + \dot{\boldsymbol{\Omega}}^\beta\times\boldsymbol{x} + \boldsymbol{\Omega}^\beta\times(\boldsymbol{\Omega}^\beta\times\boldsymbol{x}) \end{split} \end{equation}

We can use the above equations to express $\dot{\boldsymbol{x}}|_\alpha$ in terms of $\dot{\boldsymbol{x}}|_\beta$: \begin{equation} \dot{\boldsymbol{x}}|_\alpha = \dot{\boldsymbol{x}}|_\beta + \boldsymbol{\Omega}^{\beta/\alpha}\times\boldsymbol{x} \end{equation}

and $\ddot{\boldsymbol{x}}|_\alpha$ in terms of $\ddot{\boldsymbol{x}}|_\beta$: \begin{equation} \ddot{\boldsymbol{x}}|_\alpha = \ddot{\boldsymbol{x}}|_\beta + 2\boldsymbol{\Omega}^{\beta/\alpha}\times\dot{\boldsymbol{x}}|_{\beta} +\boldsymbol{\Omega}^{\beta/\alpha}\times\boldsymbol{x} +\boldsymbol{\Omega}^{\beta/\alpha}\times(\boldsymbol{\Omega}^{\beta/\alpha}\times\boldsymbol{x}) + (\boldsymbol{\Omega}^\beta\times\boldsymbol{\Omega}^\alpha)\times\boldsymbol{x} \end{equation}

where $\boldsymbol{\Omega}^{\beta/\alpha} = \boldsymbol{\Omega}^\beta - \boldsymbol{\Omega}^\alpha$.

The last equation (whose fairly lengthy derivation I've skipped) is the one I'm referring to in the title. Say that we know the motion relative to $\beta$, and the relative angular acceleration between $\alpha$ and $\beta$ ($\boldsymbol{\Omega}^{\beta/\alpha})$. Then this equation directly gives the relative acceleration in terms of $\alpha$ (without the need to transform from $\beta$ to an inertial frame, and from there to $\alpha$).

Although as it turns out, the equation also requires knowledge of the absolute angular velocities of the coordinate systems $\alpha$ and $\beta$ ($\boldsymbol{\Omega}^\alpha$ and $\boldsymbol{\Omega}^\beta$), since these appear as irreducible terms in this equation. Note that this is not the case for the equation relating the relative velocities.

Am I missing something here, and does this in fact make physical sense? How can the term $(\boldsymbol{\Omega}^\beta\times\boldsymbol{\Omega}^\alpha)\times\boldsymbol{x}$ be understood from a physical point of view?

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