28.2: Free Rotation of a Symmetric Top Using Euler’s Equations

This is a problem we’ve already solved, using Lagrangian methods and Euler angles, but it’s worth seeing just how easy it is using Euler’s equations.

For \(I_{1}=I_{2}\), the third equation gives immediately \(\Omega_{3}=\text { constant. }\).

Then, writing for convenience

\begin{equation}
\Omega_{3}\left(I_{3}-I_{1}\right) / I_{1}=\omega
\end{equation}

the first two equations are

\begin{equation}
\dot{\Omega}_{1}=-\omega \Omega_{2}, \quad \dot{\Omega}_{2}=\omega \Omega_{1}
\end{equation}

These equations can be combined to give

\begin{equation}
d\left(\Omega_{1}+i \Omega_{2}\right) / d t=i \omega\left(\Omega_{1}+i \Omega_{2}\right), \text { sо }\left(\Omega_{1}+i \Omega_{2}\right)=A e^{i \omega t}
\end{equation}

That is, \(\left(\Omega_{1}, \Omega_{2}\right)\) moves around a circle centered at the origin with constant angular velocity. So \(\begin{equation}\Omega_{1}^{2}+\Omega_{2}^{2}=|A|^{2}\end{equation}\) stays constant, and \(\Omega_{3}\) is constant, the angular velocity vector has constant length and rotates steadily about the axis \(x_{3}\).

From

\begin{equation}
L_{1}=I_{1} \Omega_{1}, L_{2}=I_{2} \Omega_{2}, L_{3}=I_{3} \Omega_{3}
\end{equation}

it follows that the angular momentum vector also precesses at a steady rate about \(x_{3}\). This is, remember, in the body frame—we know that in the fixed space frame, the angular momentum vector is constant! It follows that, as viewed from the outside, the \(x_{3}\) axis precesses around the fixed angular momentum vector at a steady rate.

Of course, the rate is the same as that found using Euler’s angles, recall from the previous lecture that

\begin{equation}
\vec{L}=\left(I_{1} \Omega_{1}, I_{1} \Omega_{2}, I_{3} \Omega_{3}\right)=\left(I_{1} \dot{\theta}, \quad I_{1} \dot{\phi} \sin \theta, \quad I_{3}(\dot{\phi} \cos \theta+\dot{\psi})\right)
\end{equation}

so in precession

\begin{equation}
L_{3}=L \cos \theta=I_{3}(\dot{\phi} \cos \theta+\dot{\psi}) \text { and } \dot{\phi}=L / I_{1}
\end{equation}

so

\begin{equation}
\dot{\psi}=L \cos \theta\left(\dfrac{1}{I_{3}}-\dfrac{1}{I_{1}}\right)=-\Omega_{3}\left(I_{3}-I_{1}\right) / I_{1}
\end{equation}