4.2: Angular Velocity and Eulerian Angles

Let $Oxyz$ be a set of space-fixed axis, and let $Ox_{0}y_{0}z_{0}$ be the body-fixed principal axes of a rigid body.

The orientation of the body-fixed principal axes $Ox_{0}y_{0}z_{0}$ with respect to the space-fixed axes $Oxyz$ can be described by the three Euler angles: $\theta$, $\phi$, and $\psi$. These are illustrated in Figure IV.1a. Those who are not familiar with Euler angles or who would like a reminder can refer to their detailed description in Chapter 3 of my notes on Celestial Mechanics.

We are going to examine the motion of a body that is rotating about a non-principal axis. If the body is freely rotating in space with no external torques acting upon it, its angular momentum $\bf{L}$ will be constant in magnitude and direction. The angular velocity vector $\bf{\omega}$, however, will not be constant, but will wander with respect to both the space-fixed and body-fixed axes, and we shall be examining this motion. I am going to call the instantaneous components of $\omega$ relative to the body-fixed axes $\omega_{1} , \omega_{2}, \omega_{3}$, and its magnitude $\omega$. As the body tumbles over and over, its Euler angles will be changing continuously. We are going to establish a geometrical relation between the instantaneous rates of change of the Euler angles and the instantaneous components of $\omega$. That is, we are going to find how $\omega_{1} , \omega_{2}$ and $\omega_{3}$ are related to $\dot{\theta} , \dot{\phi}$ and $\dot{\psi}$.

I have indicated, in Figure VI.2a, the angular velocities $\boldsymbol{\dot{\theta}} , \boldsymbol{\dot{\phi}}$ and $\boldsymbol{\dot{\psi}}$ as vectors in what I hope will be agreed are the appropriate directions.

It should be clear that $\omega_1$ is equal to the $x_{0}$-component of $\boldsymbol{\dot{\phi}}$ plus the $x_{0}$-component of $\boldsymbol{\dot{\theta}}$ and that $\omega_2$ is equal to the $y_{0}$-component of $\boldsymbol{\dot{\phi}}$ plus the $y_{0}$-component of $\boldsymbol{\dot{\theta}}$ and that $\omega_3$ is equal to the $z_{0}$-component of $\boldsymbol{\dot{\phi}}$ plus $\boldsymbol{\dot{\psi}}$

Let us look at Figure IV.2b

We see that the $x_{0}$ and $y_{0}$ components of $\boldsymbol{\dot{\theta}}$ are $\dot{\theta} \cos \psi$ and $- \dot{\theta} \sin \psi$ respectively. The $x_{0}$, $y_{0}$ and $z_{0}$ components of $\boldsymbol{\dot{\phi}}$ are, respectively:

• $\dot{\phi} \sin \theta \sin \psi$,
• $\dot{\phi} \sin \theta \sin \psi$, and
• $\dot{\phi} \cos \theta$.

Hence we arrive at

$$\omega_{1} = \dot{\phi} \sin \theta \sin \psi + \dot{\theta} \cos \psi . \tag{4.2.1}\label{eq:4.2.1}$$

$$\omega_{2} = \dot{\phi} \sin \theta \cos \psi - \dot{\theta} \sin \psi . \tag{4.2.2}\label{eq:4.2.2}$$

$$\omega_{3} = \dot{\phi} \cos \theta + \dot{ \psi} \tag{4.2.3}\label{eq:4.2.3}$$