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When learning the position vector $\:\mathbf r$, I was told it was not a real vector because it is tied to a particular coordinate system. Now I have learned that the derivative of any vector $\:\mathbf C\:$ in a rotating frame and the inertial frame can be related via (equation \ref{9.8} in $^{\prime\prime}$An Introduction to Mechanics$^{\prime\prime}$ by D.Kleppner-R.Kolenkow, CUP 2nd Edition 2014), \begin{equation} \left(\frac{\mathrm d\mathbf C}{\mathrm d\, t}\right)_{in}\!\!\!\!\!=\left(\frac{\mathrm d\mathbf C}{\mathrm d\,t}\right)_{rot}\!\!\!\!\!+\boldsymbol{\Omega\times} \mathbf C \tag{9.8}\label{9.8} \end{equation} The author claims that $\:\mathbf C\:$ can be replaced by $\:\mathbf r$, then the velocity in a rotating frame can be related to the inertial velocity via \begin{equation} \mathbf v_{in}=\mathbf v_{rot}+\boldsymbol{\Omega\times} \mathbf r \end{equation} where $\:\mathbf v_{in}\:$ denotes the velocity in the inertial frame, $\:\mathbf v_{rot}\:$ denotes the velocity in the rotation frame. $\:\boldsymbol \Omega\:$ denotes the angular velocity of the rotating frame.

What confuses me is how we can replace $\:\mathbf C\:$ with $\:\mathbf r\:$ if $\:\mathbf r\:$ is not a real vector. Even if we can, I wonder to which coordinate system this $\:\mathbf r\:$ is tied, and where the origin is.

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Position is a real vector, when phrased properly. If position is defined as the displacement from an "origin" to a point in space, that is a vector. Note that the fact that we had to include the origin aligns with your phrasing "... tied to a particular coordinate system."

The equation for rotation given has a specific coordinate system. $\Omega$ and $r$ are vectors with respect to an "origin" of the rotation (a point along the axis of rotation). It would be possible to come up with an equation for a rotating frame that was not tied to this origin, but the math would be much more complicated and it would not add any insight you could not achieve with a set of coordinate system transformations.

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