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Next: Euler angles Up: Rigid body rotation Previous: Principal axes of rotation


Euler's equations

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

$\displaystyle \mbox{\boldmath$\tau$}$$\displaystyle = \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

$\displaystyle \frac{d{\bf L}}{dt} = \frac{d{\bf L}}{dt'} +$   $\displaystyle \mbox{\boldmath$\omega$}$$\displaystyle \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

$\displaystyle \mbox{\boldmath$\tau$}$$\displaystyle = \frac{d{\bf L}}{dt'} +$   $\displaystyle \mbox{\boldmath$\omega$}$$\displaystyle \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

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

    $\displaystyle \tau_{x'}$ $\displaystyle = {\cal I}_{x'x'}\,\skew{3}\dot{\omega}_{x'} - ({\cal I}_{y'y'}-{\cal I}_{z'z'})\,\omega_{y'}\,\omega_{z'},$ (8.27)
    $\displaystyle \tau_{y'}$ $\displaystyle = {\cal I}_{y'y'}\,\skew{3}\dot{\omega}_{y'} - ({\cal I}_{z'z'}-{\cal I}_{x'x'})\,\omega_{z'}\,\omega_{x'},$ (8.28)
and   $\displaystyle \tau_{z'}$ $\displaystyle = {\cal I}_{z'z'}\,\skew{3}\dot{\omega}_{z'} - ({\cal I}_{x'x'}-{\cal I}_{y'y'})\,\omega_{x'}\,\omega_{y'},$     (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

    $\displaystyle {\cal I}_\perp\,\frac{d\omega_{x'}}{dt'} + ({\cal I}_{\parallel}-{\cal I}_{\perp})\,\omega_{z'}\,\omega_{y'}$ $\displaystyle = 0,$ (8.30)
    $\displaystyle {\cal I}_\perp\,\frac{d\omega_{y'}}{dt'} - ({\cal I}_{\parallel}-{\cal I}_{\perp})\, \omega_{z'}\,\omega_{x'}$ $\displaystyle = 0,$ (8.31)
and   $\displaystyle \frac{d\omega_{z'}}{dt'}$ $\displaystyle = 0.$     (8.32)

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

    $\displaystyle \frac{d\omega_{x'}}{dt'} + {\mit\Omega}\,\omega_{y'}$ $\displaystyle = 0,$ (8.33)
and   $\displaystyle \frac{d\omega_{y'}}{dt'} - {\mit\Omega}\,\omega_{x'}$ $\displaystyle = 0,$     (8.34)

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

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

    $\displaystyle \omega_{x'}$ $\displaystyle = \omega\,\sin\alpha\,\cos({\mit\Omega}\,t'),$ (8.37)
    $\displaystyle \omega_{y'}$ $\displaystyle = \omega\,\sin\alpha\,\sin({\mit\Omega}\,t'),$ (8.38)
and   $\displaystyle \omega_{z'}$ $\displaystyle = \omega\,\cos\alpha,$     (8.39)

where

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

    $\displaystyle L_{x'}$ $\displaystyle = {\cal I}_\perp\,\omega\,\sin\alpha\,\cos({\mit\Omega}\,t'),$ (8.41)
    $\displaystyle L_{y'}$ $\displaystyle = {\cal I}_\perp\,\omega\,\sin\alpha\,\sin({\mit\Omega}\,t'),$ (8.42)
and   $\displaystyle L_{z'}$ $\displaystyle = {\cal I}_\parallel\,\omega\,\cos\alpha.$     (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

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


next up previous
Next: Euler angles Up: Rigid body rotation Previous: Principal axes of rotation
Richard Fitzpatrick 2016-03-31