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# 12.3: Rotating Reference Frame


Consider a rotating frame of reference which will be designated as the double-primed (rotating) frame to differentiate it from the non-rotating primed (moving) frame, since both of which may be undergoing translational acceleration relative to the inertial fixed unprimed frame as described in Figure $$12.2.1$$.

## Spatial time derivatives in a rotating, non-translating, reference frame

For simplicity assume that $$\mathbf{R}_{fix} = \mathbf{V}_{fix} = 0$$, that is, the primed reference frame is stationary and identical to the fixed stationary unprimed frame. The double-primed (rotating) frame is a non-inertial frame rotating with respect to the origin of the fixed primed frame.

Appendix $$19.4.2C$$ shows that an infinitessimal rotation $$d\theta$$ about an instantaneous axis of rotation leads to an infinitessimal displacement $$d\mathbf{r}^{R}$$ where

$d\mathbf{r}^{R} = d \theta \times \mathbf{r}^{\prime}_{mov} \label{12.7}$

Consider that during a time $$dt$$, the position vector in the fixed primed reference frame moves by an arbitrary infinitessimal distance $$d\mathbf{r}^{\prime}_{mov}$$. As illustrated in Figure $$\PageIndex{1}$$, this infinitessimal distance in the primed non-rotating frame can be split into two parts:

1. $$d\mathbf{r}^{R} = d\theta \times \mathbf{r}^{\prime}_{mov}$$ which is due to rotation of the rotating frame with respect to the translating primed frame.
2. $$(d\mathbf{r}^{\prime\prime}_{rot})$$ which is the motion with respect to the rotating (double-primed) frame.

That is, the motion has been arbitrarily divided into a part that is due to the rotation of the double-primed frame, plus the vector displacement measured in this rotating (double-primed) frame. It is always possible to make such a decomposition of the displacement as long as the vector sum can be written as

$d\mathbf{r}^{\prime}_{mov} = d\mathbf{r}^{\prime\prime}_{rot} + d\theta \times \mathbf{r}^{\prime}_{mov} \label{12.8}$

Since $$d\theta = \omega dt$$ then the time differential of the displacement, Equation \ref{12.8}, can be written as

$\left( \frac{d\mathbf{r}^{\prime}}{dt}\right)_{mov} = \left(\frac{d\mathbf{r}^{\prime\prime}}{dt}\right)_{rot} + \omega \times \mathbf{r}^{\prime}_{mov} \label{12.9}$

The important conclusion is that a velocity measured in a non-rotating reference frame $$\left(\frac{d\mathbf{r}^{\prime}}{dt}\right)_{mov}$$ can be expressed as the sum of the velocity $$\left(\frac{d\mathbf{r}^{\prime\prime}}{dt}\right)_{rot}$$, measured relative to a rotating frame, plus the term $$\omega \times \mathbf{r}^{\prime}_{mov}$$ which accounts for the rotation of the frame. The division of the $$d\mathbf{r}^{\prime}_{rot}$$ vector into two parts, a part due to rotation of the frame plus a part with respect to the rotating frame, is valid for any vector as shown below.

## General vector in a rotating, non-translating, reference frame

Consider an arbitrary vector $$\mathbf{G}$$ which can be expressed in terms of components along the three unit vector basis $$\hat{\mathbf{e}}^{fix}_i$$ in the fixed inertial frame as

$\mathbf{G} = \sum^{3}_{i=1} G^{fix}_i \hat{\mathbf{e}}_i^{fix} \label{12.10}$

Neglecting translational motion, then it can be expressed in terms of the three unit vectors in the non-inertial rotating frame unit vector basis $$\hat{\mathbf{e}}^{rot}_i$$ as

$\mathbf{G} = \sum^{3}_{i=1} (G_i)_{rot} \hat{\mathbf{e}}^{rot}_i \label{12.11}$

Since the unit basis vectors $$\hat{\mathbf{e}}^{rot}_i$$ are constant in the rotating frame, that is,

$\left(\frac{d\hat{\mathbf{e}}^{rot}_i}{dt}\right)_{rot} = 0 \label{12.12}$

then the time derivatives of $$\mathbf{G}$$ in the rotating coordinate system $$\hat{\mathbf{e}}^{rot}_i$$ can be written as

$\left(\frac{d\mathbf{G}}{dt}\right)_{rot} = \sum^3_{i-1} \left(\frac{dG_i}{dt}\right)_{rot} \hat{\mathbf{e}}^{rot}_i \label{12.13}$

The inertial-frame time derivative taken with components along the rotating coordinate basis $$\hat{\mathbf{e}}^{rot}_i$$, Equation \ref{12.11}, is

$\left(\frac{d\mathbf{G}}{dt}\right)_{fix} = \sum^3_{i-1} \left(\frac{dG_i}{dt}\right)_{rot} \hat{\mathbf{e}}^{rot}_i + (G_i)_{rot} \frac{d\hat{\mathbf{e}}^{rot}_i }{dt} \label{12.14}$

Substitute the unit vector $$\hat{\mathbf{e}}^{rot}$$ for $$\mathbf{r}^{\prime}_{mov}$$ in Equation \ref{12.9}, plus using Equation \ref{12.12}, gives that

$\left(\frac{d\hat{\mathbf{e}}^{rot}}{dt}\right)_{fix} = \omega \times \hat{\mathbf{e}}^{rot} \label{12.15}$

Substitute this into the second term of Equation \ref{12.14} gives

$\left(\frac{d\mathbf{G}}{dt}\right)_{fix} = \left(\frac{d\mathbf{G}}{dt}\right)_{rot} + \omega \times \mathbf{G} \label{12.16}$

This important identity relates the time derivatives of any vector expressed in both the inertial frame and the rotating non-inertial frame bases. Note that the $$\omega \times \mathbf{G}$$ term originates from the fact that the unit basis vectors of the rotating reference frame are time dependent with respect to the non-rotating frame basis vectors as given by Equation \ref{12.15}. Equation \ref{12.16} is used extensively for problems involving rotating frames. For example, for the special case where $$\mathbf{G} = \mathbf{r}^{\prime}$$, then Equation \ref{12.16} relates the velocity vectors in the fixed and rotating frames as given in Equation \ref{12.9}.

Another example is the vector $$\mathbf{\dot{\omega}}$$

$\mathbf{\dot{\omega}} = \left(\frac{d\omega}{dt}\right)_{fix} = \left(\frac{d\omega}{dt}\right)_{rot} + \omega \times \omega = \left(\frac{d\omega}{dt}\right)_{rot} = \mathbf{\dot{\omega}} \label{12.17}$

That is, the angular acceleration $$\dot{\omega}$$ has the same value in both the fixed and rotating frames of reference.

This page titled 12.3: Rotating Reference Frame is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Douglas Cline via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.