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

Last updated

May 11, 2024
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- 28.1: Introduction to Euler’s Equations
- 28.3: Using Energy and Angular Momentum Conservation

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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

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

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

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