13.21: Torque-free rotation of an asymmetric rigid rotor

The Euler equations of motion for the case of torque-free rotation of an asymmetric (triaxial) rigid rotor about the center of mass, with principal moments of inertia \(I_1 \neq I_2 \neq I_3\), lead to more complicated motion than for the symmetric rigid rotor.³ The general features of the motion of the asymmetric rotor can be deduced using the conservation of angular momentum and rotational kinetic energy.

Figure \(\PageIndex{1}\): Rotation of an asymmetric rigid rotor. The dark lines correspond to contours of constant total rotational kinetic energy T, which has an ellipsoidal shape, projected onto the angular momentum L sphere in the body-fixed frame.

Assuming that the external torques are zero then the Euler equations of motion can be written as

\[\begin{align} I_1\dot{\omega}_1 = (I_2 − I_3) \omega_2\omega_3 \label{13.156}\\ I_2\dot{\omega}_2 = (I_3 − I_1) \omega_3\omega_1 \notag \\ I_3\dot{\omega}_3 = (I_1 − I_2) \omega_1\omega_2 \notag \end{align}\]

Since \(L_i = I_i\omega_i\) for \(i = 1, 2, 3\), then Equation \ref{13.156} gives

\[\begin{align}I_2I_3\dot{L}_1 = (I_2 − I_3)L_2L_3 \label{13.157} \\ I_1I_3\dot{L}_2 = (I_3 − I_1)L_3L_1 \notag \\ I_1I_2 \dot{L}_3 = (I_1 − I_2)L_1L_2 \notag \end{align}\]

Multiply the first equation by \(I_1L_1\), the second by \(I_2L_2\) and the third by \(I_3L_3\) and sum, which gives

\[I_1I_2I_3 \left( L_1\dot{L}_1 + L_2\dot{L}_2 + L_3\dot{L}_3 \right) = 0 \label{13.158}\]

The bracket is equivalent to \(\frac{d}{dt} (L^2_1 + L^2_2 + L^2_3)=0\) which implies that the total rotational angular momentum \(L\) is a constant of motion as expected for this torque-free system, even though the individual components \(L_1, L_2, L_3\) may vary. That is

\[L^2_1 + L^2_2 + L^2_3 = L^2 \label{13.159}\]

Note that equation \ref{13.159} is the equation of a sphere of radius \(L\).

Multiply the first equation of \ref{13.157} by \(L_1\), the second by \(L_2\), and the third by \(L_3\), and sum gives

\[I_2I_3L_1\dot{L}_1 + I_1I_3L_2\dot{L}_2 + I_1I_2L_3\dot{L}_3 = 0 \label{13.160}\]

Divide \ref{13.160} by \(I_1I_2I_3\) gives \(\frac{d}{dt}( \frac{L^2_1}{2I_1} + \frac{L^2_2}{ 2I_2} + \frac{L^2_3}{2I_3} )=0\). This implies that the total rotational kinetic energy \(T\), given by

\[\frac{L^2_1}{2I_1} + \frac{L^2_2}{2I_2} + \frac{L^2_3}{2I_3} = T \label{13.161}\]

is a constant of motion as expected when there are no external torques and zero energy dissipation. Note that \ref{13.161} is the equation of an ellipsoid.

Equations \ref{13.159} and \ref{13.161} both must be satisfied by the rotational motion for any value of the total angular momentum \(\mathbf{L}\) and kinetic energy \(T\). Fig \(\PageIndex{1}\) shows a graphical representation of the intersection of the \(L\) sphere and \(T\) ellipsoid as seen in the body-fixed frame. The angular momentum vector \(\mathbf{L}\) must follow the constant-energy contours given by where the \(T\)-ellipsoids intersect the \(L\)-sphere, shown for the case where \(I_3 > I_2 > I_1\). Note that the precession of the angular momentum vector \(\mathbf{L}\) follows a trajectory that has closed paths that circle around the principal axis with the smallest \(I\), that is, \(\hat{\mathbf{e}}_1\), or the principal axis with the maximum \(I\), that is, \(\hat{\mathbf{e}}_3\). However, the angular momentum vector does not have a stable minimum for precession around the intermediate principal moment of inertia axis \(\hat{\mathbf{e}}_2\). In addition to the precession, the angular momentum vector \(\mathbf{L}\) executes nutation, that is a nodding of the angle \(\theta\). For any fixed value of \(L\), the kinetic energy has upper and lower bounds given by

\[\frac{L^2}{2I_3} \leq T \leq \frac{L^2}{2I_1} \label{13.162}\]

Thus, for a given value of \(L\), when \(T = T_{\text{min}} = \frac{L^2}{2I_3}\), the orientation of \(\mathbf{L}\) in the body-fixed frame is either \((0, 0, +L)\) or \((0, 0, −L)\), that is, aligned with the \(\hat{\mathbf{e}}_3\) axis along which the principal moment of inertia is largest. For slightly higher kinetic energy the trajectory of \(L\) follows closed paths precessing around \(\hat{\mathbf{e}}_3\). When the kinetic energy \(T = \frac{L^2_2}{2I_2}\) the angular momentum vector \(L\) follows either of the two thin-line trajectories each of which are a separatrix. These do not have closed orbits around \(\hat{\mathbf{e}}_2\) and they separate the closed solutions around either \(\hat{\mathbf{e}}_3\) or \(\hat{\mathbf{e}}_1\). For higher kinetic energy the precessing angular momentum vector follows closed trajectories around \(\hat{\mathbf{e}}_1\) and becomes fully aligned with \(\hat{\mathbf{e}}_1\) at the upper-bound kinetic energy.

Note that for the special case when \(I_3 > I_2 = I_1\), then the asymmetric rigid rotor equals the symmetric rigid rotor for which the solutions of Euler’s equations were solved exactly in chapter \(13.19\). For the symmetric rigid rotor the \(T\)-ellipsoid becomes a spheroid aligned with the symmetry axis and thus the intersections with the \(L\)-sphere lead to circular paths around the \(\hat{\mathbf{e}}_3\) body-fixed principal axis, while the separatrix circles the equator corresponding to the \(\hat{\mathbf{e}}_3\) axis separating clockwise and anticlockwise precession about \(\mathbf{L}_3\). This discussion shows that energy, plus angular momentum conservation, provide the general features of the solution for the torque-free symmetric top that are in agreement with those derived using Euler’s equations of motion.

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³Similar discussions of the freely-rotating asymmetric top are given by Landau and Lifshitz [La60] and by Gregory [Gr06].