Eulerian Angles

The fixed frame and the body frame share the
same origin. Hence, we can transform from one to the other by
means of an appropriate *rotation* of our coordinate axes.
In general, if we restrict ourselves to rotations about one of the Cartesian
axes, *three* successive rotations are required to
transform the fixed frame into the body frame. There are, in fact, many different
ways to combined three successive rotations in order to achieve this. In the following, we shall describe the
most widely used method, which is due to Euler.

We start in the fixed frame, which has coordinates , , , and
unit vectors , , . Our first rotation
is counterclockwise (looking down the axis) through an angle about the -axis. The new frame has coordinates , , , and
unit vectors , , . According
to Equations (A.1277)-(A.1279), the transformation of coordinates can be represented as
follows:

Clearly, is the precession rate about the axis, as seen in the fixed frame.

The second rotation is counterclockwise (looking down the axis) through
an angle about the -axis. The new frame has coordinates
, , , and unit vectors
,
,
. By analogy with Equation (528), the transformation
of coordinates can be represented as follows:

(530) |

The third rotation is counterclockwise (looking down the axis) through
an angle about the -axis. The new frame is the body frame, which has coordinates
, , , and unit vectors , , . The transformation of coordinates can be represented as
follows:

(532) |

Clearly, is

The full transformation between the fixed frame and the body frame
is rather complicated. However, the following results can easily be
verified:

It follows from Equation (534) that . In other words, is the angle of inclination between the - and -axes. Finally, since the total angular velocity can be written

Equations (529), (531), and (533)-(535) yield

The angles , , and are termed *Eulerian angles*. Each has a clear physical interpretation: is the angle of precession
about the axis in the fixed frame, is minus the angle of precession about the
axis in the body frame, and is the angle of inclination
between the and axes. Moreover, we can
express the components of the angular velocity vector in the body frame entirely in terms of the Eulerian angles, and their time derivatives [see Equations (537)-(539)].

Consider a rigid body which is constrained to rotate about a fixed axis with
the constant angular velocity . Let the fixed angular velocity
vector point along the -axis. In the previous section, we saw that
the angular momentum and the torque were both steady in the body frame.
Since there is no precession of quantities in the body frame, it follows that
the Eulerian angle is constant. Furthermore, since the angular velocity vector is
fixed in the body frame, as well as the fixed frame [as can be seen by applying Equation (502)
to instead of ], it must subtend a constant angle with the
axis. Hence, the Eulerian angle is also constant.
It follows from Equations (537)-(539) that

(540) | |||

(541) | |||

(542) |

which implies that . In other words, the precession rate, , in the fixed frame is equal to . Hence, in the fixed frame, the constant torque and angular momentum vectors found in the body frame

Consider a rotating device such as a flywheel or a propeller. If the device
is *statically balanced* then its center of mass lies on the axis
of rotation. This is desirable since, otherwise, gravity, which effectively
acts at the center of mass, exerts a varying torque about the
axis of rotation as the device rotates, giving rise to unsteady rotation. If the device is
*dynamically balanced* then the axis of rotation is also
a principal axis, so that, as the device rotates its angular momentum vector,
,
remains parallel to the axis of rotation. This is desirable since, otherwise,
the angular momentum vector is not parallel to the axis of rotation, and, therefore, precesses around it. Since is equal to the
torque, a precessing torque must also be applied to the device (at right-angles
to both the axis and ). The result is a reaction
on the bearings which can give rise to violent vibration and
wobbling, even when the device is statically balanced.

Consider a freely rotating body which is rotationally symmetric about one axis (the -axis). In the absence of an external torque, the
angular momentum vector is a constant of the motion [see Equation (456)]. Let point along the -axis. In the
previous section, we saw that the angular momentum vector subtends a
constant angle with the axis of symmetry: *i.e.*, with the -axis. Hence, the time derivative
of the Eulerian angle is zero. We also saw that the angular momentum
vector, the axis of symmetry, and the angular velocity vector are coplanar.
Consider an instant in time at which all of these vectors lie in the -
plane. This implies that . According to the
previous section, the angular velocity vector subtends a constant
angle with the symmetry axis. It follows that
and
. Equation (537) yields
. Hence, Equation (538) yields

Finally, Equations (539), together with (525) and (543), yields

(545) |

(546) |

As an example, consider the free rotation of a thin disk. It is easily
demonstrated (from the perpendicular axis theorem)
that

(547) |

(548) |

(549) |

(550) | |||

(551) |

Thus, the symmetry axis precesses in the fixed frame at approximately twice the angular speed of rotation. This precession is manifest as a wobbling motion.

It is known that the axis of rotation of the Earth is very slightly inclined to
its symmetry axis (which passes through the two geographic poles). The angle is approximately seconds of an arc
(which corresponds to a distance of about on the Earth's surface).
It is also known that the ratio of the terrestrial moments of inertia
is about
, as determined from the
Earth's oblateness--see Section 12.7. Hence, from (521), the precession rate of the
angular velocity vector about the symmetry axis, as viewed in a geostationary reference frame, is

(552) |

(553) |

(554) |

(555) |