9.3: Differential Rotations

Now consider small rotations from one frame to another; using the small angle assumption to ignore higher-order terms gives

\begin{align} R \, &\simeq \, \begin{bmatrix} 1 & \delta \phi & -\delta \theta \\[4pt] -\delta \phi & 1 & \delta \psi \\[4pt] \delta \theta & -\delta \psi & 1 \end{bmatrix} \\[4pt] &= \, \begin{bmatrix} 0 & \delta \phi & -\delta \theta \\[4pt] -\delta \phi & 0 & \delta \psi \\[4pt] \delta \theta & -\delta \psi & 0 \end{bmatrix} + I_{3 \times 3}, \end{align}where \(I_{3 \times 3}\) donotes the identity matrix. \(R\) comprises the identity plus a part equal to the (negative) cross-product operator \( (-\delta \vec{E} \times) \), where \(\delta \vec{E} = [\delta \psi, \, \delta \theta, \, \delta \phi]\), the vector of differential Euler angles, ordered with the axes \([x, \, y, \, z].\) Small rotations are completely decoupled; their order does not matter. Since \(R^{-1} = R^T\), we also have \(R^{-1} = I_{3 \times 3} + \delta \vec{E} \times\);

\begin{align} \vec{x}_b \, &= \, \vec{x} - \delta \vec{E} \times \vec{x} \\[4pt] \vec{x} \, &= \, \vec{x}_b + \delta \vec{E} \times \vec{x}_b. \end{align}

We now fix the point of interest on the body, instead of in inertial space, calling its location in the body frame \(\vec{r}\) (radius). The differential rotations occur over a time step \(\delta t\), so that we can write the location of the point before and after the rotation, with respect to the first frame as follows:

\begin{align} \vec{x} (t) \, &= \, \vec{r} \\[4pt] \vec{x} (t + \delta t) \, &= \, R^T \vec{r} \, = \, \vec{r} + \delta \vec{E} \times \vec{r} \end{align}

Dividing by the differential time step gives

\begin{align} \dfrac{\delta \vec{x}}{\delta t} \, &= \, \dfrac{\delta \vec{E}}{\delta t} \times \vec{r} \\[4pt] &= \, \vec{\omega} \times \vec{r}, \end{align}

where the rotation rate vector \(\vec{\omega} \simeq d \vec{E} / dt\) because the Euler angles for this infinitesimal rotation are small and decoupled. This same cross-product relationship can be derived in the second frame as well:

\begin{align} \vec{x}_b (t) \, &= \, R \vec{r} \, = \, \vec{r} - \delta \vec{E} \times \vec{r} \\[4pt] \vec{x}_b (t + \delta t) \, &= \, \vec{r}, \end{align}

such that

\begin{align} \dfrac{\delta \vec{x}_b}{\delta t} \, &= \, \dfrac{\delta \vec{E}}{\delta t} \times \vec{r} \\[4pt] &= \, \vec{\omega} \times \vec{r}. \end{align}

On a rotating body whose origin point is fixed, the time rate of change of a constant radius vector is the cross-product of the rotation rate vector \(\vec{\omega}\) and the radius vector itself. The resultant derivative is in the moving body frame. In the case that the radius vector changes with respect to the body frame, we need an additional term:

\[ \dfrac{d \vec{x}_b}{dt} \, = \, \vec{\omega} \times \vec{r} + \dfrac{\partial \vec{r}}{\partial t}. \]

Finally, allowing the origin to move as well gives

\[ \dfrac{d \vec{x}_b}{dt} \, = \, \vec{\omega} \times \vec{r} + \dfrac{\partial \vec{r}}{\partial t} + \dfrac{d \vec{x}_o}{dt}. \]

This result is often written in terms of body-referenced velocity \(\vec{v}\):

\[ \vec{v} \, = \, \vec{\omega} \times \vec{r} + \dfrac{\partial \vec{r}}{\partial t} + \vec{v}_o, \]

where \(\vec{v}_o\) is the body-referenced velocity of the origin. The total velocity of the particle is equal to the velocity of the reference frame origin, plus a component due to rotation of this frame. The velocity equation can be generalized to any body-referenced vector \(\vec{f}\):

\[ \dfrac{d \vec{f}}{dt} \, = \, \dfrac{\partial \vec{f}}{\partial t} + \vec{\omega} \times \vec{f}. \]