Euler's equations

The fundamental equation of motion of a rotating body [see Equation (8.3)],

$$\mbox{\boldmath$\tau$} = \frac{d{\bf L}}{dt},$$ (8.23)

is only valid in an inertial frame. However, we have seen that ${\bf L}$ is most simply expressed in a frame of reference whose axes are aligned along the principal axes of rotation of the body. Such a frame of reference rotates with the body, and is, therefore, non-inertial. Thus, it is helpful to define two Cartesian coordinate systems with the same origins. The first, with coordinates $x$, $y$, $z$, is a fixed inertial frame; let us denote this the fixed frame. The second, with coordinates $x'$, $y'$, $z'$, co-rotates with the body in such a manner that the $x'$-, $y'$-, and $z'$-axes are always pointing along its principal axes of rotation; we shall refer to this as the body frame. Because the body frame co-rotates with the body, its instantaneous angular velocity is the same as that of the body. Hence, it follows from the analysis in Section 6.2 that

$$\frac{d{\bf L}}{dt} = \frac{d{\bf L}}{dt'} + \mbox{\boldmath$\omega$} \times{\bf L}.$$ (8.24)

Here, $d/dt$ is the time derivative in the fixed frame, and $d/dt'$ the time derivative in the body frame. Combining Equations (8.23) and (8.24), we obtain

$$\mbox{\boldmath$\tau$} = \frac{d{\bf L}}{dt'} + \mbox{\boldmath$\omega$} \times{\bf L}.$$ (8.25)

In the body frame, let $\tau$$= (\tau_{x'},\,\tau_{y'},\tau_{z'})$ and $\omega$$= (\omega_{x'},\,\omega_{y'},\,\omega_{z'})$. It follows that

$${\bf L} = ({\cal I}_{x'x'}\,\omega_{x'},\,{\cal I}_{y'y'}\,\omega_{y'},\,{\cal I}_{z'z'}\,\omega_{z'}),$$ (8.26)

where ${\cal I}_{x'x'}$, ${\cal I}_{y'y'}$ and ${\cal I}_{z'z'}$ are the principal moments of inertia. Hence, in the body frame, the components of Equation (8.25) yield

$$\tau_{x'} = {\cal I}_{x'x'}\,\skew{3}\dot{\omega}_{x'} - ({\cal I}_{y'y'}-{\cal I}_{z'z'})\,\omega_{y'}\,\omega_{z'},$$ (8.27)

$$\tau_{y'} = {\cal I}_{y'y'}\,\skew{3}\dot{\omega}_{y'} - ({\cal I}_{z'z'}-{\cal I}_{x'x'})\,\omega_{z'}\,\omega_{x'},$$ (8.28)

$$\tau_{z'} = {\cal I}_{z'z'}\,\skew{3}\dot{\omega}_{z'} - ({\cal I}_{x'x'}-{\cal I}_{y'y'})\,\omega_{x'}\,\omega_{y'},$$ (8.29)

where $\dot{\phantom\omega}=d/dt'$. Here, we have made use of the fact that the moments of inertia of a rigid body are constant in time in the co-rotating body frame. The preceding three equations are known as Euler's equations.

Consider a body that is freely rotating; that is, in the absence of external torques. Furthermore, let the body be rotationally symmetric about the $z'$-axis. It follows that ${\cal I}_{x'x'} = {\cal I}_{y'y'} = {\cal I}_\perp$. Likewise, we can write ${\cal I}_{z'z'} = {\cal I}_\parallel$. In general, however, ${\cal I}_\perp\neq {\cal I}_\parallel$. Thus, Euler's equations yield

$${\cal I}_\perp\,\frac{d\omega_{x'}}{dt'} + ({\cal I}_{\parallel}-{\cal I}_{\perp})\,\omega_{z'}\,\omega_{y'} = 0,$$ (8.30)

$${\cal I}_\perp\,\frac{d\omega_{y'}}{dt'} - ({\cal I}_{\parallel}-{\cal I}_{\perp})\, \omega_{z'}\,\omega_{x'} = 0,$$ (8.31)

$$\frac{d\omega_{z'}}{dt'} = 0.$$ (8.32)

Clearly, $\omega_{z'}$ is a constant of the motion. Equation (8.30) and (8.31) can be written

$$\frac{d\omega_{x'}}{dt'} + {\mit\Omega}\,\omega_{y'} = 0,$$ (8.33)

$$\frac{d\omega_{y'}}{dt'} - {\mit\Omega}\,\omega_{x'} = 0,$$ (8.34)

where ${\mit\Omega}= ({\cal I}_{\parallel}/{\cal I}_\perp-1)\,\omega_{z'}$. As is easily demonstrated, the solution to these equations is

$$\omega_{x'} = \omega_\perp\,\cos({\mit\Omega}\,t'),$$ (8.35)

$$\omega_{y'} =\omega_\perp\,\sin({\mit\Omega}\,t'),$$ (8.36)

where $\omega_\perp$ is a constant. Thus, the projection of the angular velocity vector onto the $x'$-$y'$ plane has the fixed length $\omega_\perp$, and rotates steadily about the $z'$-axis with angular velocity ${\mit \Omega }$. It follows that the length of the angular velocity vector, $\omega=(\omega_{x'}^2+\omega_{y'}^2+\omega_{z'}^2)^{1/2}$, is a constant of the motion. Clearly, the angular velocity vector subtends some constant angle, $\alpha$, with the $z'$-axis, which implies that $\omega_{z'} = \omega\,\cos\alpha$ and $\omega_\perp = \omega\,\sin\alpha$. Hence, the components of the angular velocity vector are

$$\omega_{x'} = \omega\,\sin\alpha\,\cos({\mit\Omega}\,t'),$$ (8.37)

$$\omega_{y'} = \omega\,\sin\alpha\,\sin({\mit\Omega}\,t'),$$ (8.38)

$$\omega_{z'} = \omega\,\cos\alpha,$$ (8.39)

where

$${\mit\Omega} =\omega\,\cos\alpha \left(\frac{{\cal I}_\parallel}{{\cal I}_\perp}-1\right).$$ (8.40)

We conclude that, in the body frame, the angular velocity vector precesses about the symmetry axis (i.e., the $z'$-axis) with the angular frequency ${\mit \Omega }$. Now, the components of the angular momentum vector are

$$L_{x'} = {\cal I}_\perp\,\omega\,\sin\alpha\,\cos({\mit\Omega}\,t'),$$ (8.41)

$$L_{y'} = {\cal I}_\perp\,\omega\,\sin\alpha\,\sin({\mit\Omega}\,t'),$$ (8.42)

$$L_{z'} = {\cal I}_\parallel\,\omega\,\cos\alpha.$$ (8.43)

Thus, in the body frame, the angular momentum vector is also of constant length, and precesses about the symmetry axis with the angular frequency ${\mit \Omega }$. Furthermore, the angular momentum vector subtends a constant angle $\theta$ with the symmetry axis, where

$$\tan\theta = \frac{{\cal I}_\perp}{{\cal I}_\parallel}\,\tan\alpha.$$ (8.44)

The angular momentum vector, the angular velocity vector, and the symmetry axis all lie in the same plane; that is, ${\bf e}_{z'}\cdot{\bf L}\times$   $\omega$$=0$, as can easily be verified. Moreover, the angular momentum vector lies between the angular velocity vector and the symmetry axis (i.e., $\theta<\alpha$) for a flattened (or oblate) body (i.e., ${\cal I}_\perp< {\cal I}_\parallel$), whereas the angular velocity vector lies between the angular momentum vector and the symmetry axis (i.e., $\theta>\alpha$) for an elongated (or prolate) body (i.e., ${\cal I}_\perp>{\cal I}_\parallel$). See Figure 8.2.