Euler's Equations

The fundamental equation of motion of a rotating body [see Eq. (528)],

$${\bf T} = \frac{d{\bf L}}{dt},$$ (573)

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. Since 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 Sect. 8.2 that

$$\frac{d{\bf L}}{dt} = \frac{d{\bf L}}{dt'} + \mbox{\boldmath$\omega$}\times{\bf L}.$$ (574)

Here, $d/dt$ is the time derivative in the fixed frame, and $d/dt'$ the time derivative in the body frame. Combining Eqs. (573) and (574), we obtain

$${\bf T} = \frac{d{\bf L}}{dt'} + \mbox{\boldmath$\omega$}\times{\bf L}.$$ (575)

Now, in the body frame let ${\bf T}= (T_{x'},\,T_{y'},T_{z'})$ and $\mbox{\boldmath$\omega$}= (\omega_{x'},\,\omega_{y'},\,\omega_{z'})$. It follows that ${\bf L} = (I_{x'x'}\,\omega_{x'},\,I_{y'y'}\,\omega_{y'},\,I_{z'z'}\,\omega_{z'})$, where $I_{x'x'}$, $I_{y'y'}$ and $I_{z'z'}$ are the principal moments of inertia. Hence, in the body frame, the components of Eq. (575) yield

$$T_{x'} = I_{x'x'}\,\dot{\omega}_{x'} - (I_{y'y'}-I_{z'z'})\,\omega_{y'}\,\omega_{z'},$$ (576)

$$T_{y'} = I_{y'y'}\,\dot{\omega}_{y'} - (I_{z'z'}-I_{x'x'})\,\omega_{z'}\,\omega_{x'},$$ (577)

$$T_{z'} = I_{z'z'}\,\dot{\omega}_{z'} - (I_{x'x'}-I_{y'y'})\,\omega_{x'}\,\omega_{y'},$$ (578)

where $\dot{~}=d/dt$. 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 above equations are known as Euler's equations.

Consider a rigid body which is constrained to rotate about a fixed axis with constant angular velocity. It follows that $\dot{\omega}_{x'}=\dot{\omega}_{y'} = \dot{\omega}_{z'}=0$. Hence, Euler's equations, (576)-(578), reduce to

$$T_{x'} = - (I_{y'y'}-I_{z'z'})\,\omega_{y'}\,\omega_{z'},$$ (579)

$$T_{y'} = - (I_{z'z'}-I_{x'x'})\,\omega_{z'}\,\omega_{x'},$$ (580)

$$T_{z'} = - (I_{x'x'}-I_{y'y'})\,\omega_{x'}\,\omega_{y'}.$$ (581)

These equations specify the components of the steady (in the body frame) torque exerted on the body by the constraining supports. The steady (in the body frame) angular momentum is written

$${\bf L} = (I_{x'x'}\,\omega_{x'},\,I_{y'y'}\,\omega_{y'},\, I_{z'z'}\,\omega_{z'}).$$ (582)

It is easily demonstrated that ${\bf T}= \mbox{\boldmath$\omega$}\times{\bf L}$. Hence, the torque is perpendicular to both the angular velocity and the angular momentum vectors. Note that if the axis of rotation is a principal axis then two of the three components of $\omega$ are zero (in the body frame). It follows from Eqs. (579)-(581) that all three components of the torque are zero. In other words, zero external torque is required to make the body rotate steadily about a principal axis.

Suppose that the body is freely rotating: i.e., there are no external torques. Furthermore, let the body be rotationally symmetric about the $z'$-axis. It follows that $I_{x'x'} = I_{y'y'} = I_\perp$. Likewise, we can write $I_{z'z'} = I_\parallel$. In general, however, $I_\perp\neq I_\parallel$. Thus, Euler's equations yield

$$I_\perp\,\frac{d\omega_{x'}}{dt} + (I_{\parallel}-I_{\perp})\,\omega_{z'}\,\omega_{y'} = 0,$$ (583)

$$I_\perp\,\frac{d\omega_{y'}}{dt} - (I_{\parallel}-I_{\perp})\, \omega_{z'}\,\omega_{x'} = 0,$$ (584)

$$\frac{d\omega_{z'}}{dt} = 0.$$ (585)

Clearly, $\omega_{z'}$ is a constant of the motion. Equation (583) and (584) can be written

$$\frac{d\omega_{x'}}{dt} + {\mit\Omega}\,\omega_{y'} = 0,$$ (586)

$$\frac{d\omega_{y'}}{dt} - {\mit\Omega}\,\omega_{x'} = 0,$$ (587)

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

$$\omega_{x'} = \omega_\perp\,\cos({\mit\Omega}\,t),$$ (588)

$$\omega_{y'} = \omega_\perp\,\sin({\mit\Omega}\,t),$$ (589)

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 makes 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

$$\omega_{x'} = \omega\,\sin\alpha\,\cos({\mit\Omega}\,t),$$ (590)

$$\omega_{y'} = \omega\,\sin\alpha\,\sin({\mit\Omega}\,t),$$ (591)

$$\omega_{z'} = \omega\,\cos\alpha,$$ (592)

where

$${\mit\Omega} =\omega\,\cos\alpha \left(\frac{I_\parallel}{I_\perp}-1\right).$$ (593)

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

$$L_{x'} = I_\perp\,\omega\,\sin\alpha\,\cos({\mit\Omega}\,t),$$ (594)

$$L_{y'} = I_\perp\,\omega\,\sin\alpha\,\sin({\mit\Omega}\,t),$$ (595)

$$L_{z'} = I_\parallel\,\omega\,\cos\alpha.$$ (596)

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 makes a constant angle $\theta$ with the symmetry axis, where

$$\tan\theta = \frac{I_\perp}{I_\parallel}\,\tan\alpha.$$ (597)

Note that the angular momentum vector, the angular velocity vector, and the symmetry axis all lie in the same plane: i.e., ${\bf e}_{z'}\cdot{\bf L}\times \mbox{\boldmath$\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., $I_\perp< I_\parallel$), 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., $I_\perp>I_\parallel$).