# 12.3: Rotating Reference Frame

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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:

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