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# 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.

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