From "Theoretical Mechanics of Particles and Continua" by A. Fetter and J. Walecka.

As emphasized in the preceding section, the general expression $(7.11)$ can be applied to the coordinate vector $\mathbf r$ of a moving particle, relating the velocity seen by observers fixed in the inertial frame and in the rotating frame: $$\left(\frac{\text d \mathbf{r}}{\text d t}\right)_{inertial}=\left(\frac{\text d\mathbf{r}}{\text d t}\right)_{body}+\mathbf{\omega}\times\mathbf{r}$$

What does the notation of the subscripts in the above equation mean?

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    $\begingroup$ Can you link to a text or article that uses this notation? $\endgroup$ – rob Feb 27 at 14:15
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    $\begingroup$ The text above the equation explains it. They don't say "body", but you can use process of elimination, right? $\endgroup$ – Aaron Stevens Feb 27 at 14:34

The subscript 'inertial' means 'in an interial frame' (think 'as seen by a non-accelerating observer'). The subscript body means 'in the rotating body's frame'.

For example (using cylindrical basis vectors $\hat{z}, \hat{r},\hat{\varphi}$):

Let's say the earth is rotating counter-clockwise with ${\boldsymbol \omega} = \omega\hat{z}$. And there's a motionless object at ${\boldsymbol r}= r \hat{r}$. The object is motionless so for a inertial observer: $$ \left(\frac{\text d \mathbf{r}}{\text d t}\right)_{\rm inertial}=0 $$

However, an observer on earth should see the object moving clockwise at velocity $\omega r$, and indeed: $$ \left(\frac{\text d \mathbf{r}}{\text d t}\right)_{\rm body} =\left(\frac{\text d \mathbf{r}}{\text d t}\right)_{\rm inertial}-{\boldsymbol \omega}\times{\boldsymbol r}= -\omega r \hat{\varphi} $$


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