27.3: Free Motion of a Symmetrical Top

As a warm up in using Euler’s angles, we’ll redo the free symmetric top covered in the last lecture. With no external torques acting the top will have constant angular momentum $\vec{L}$.

Figure $\PageIndex{1}$:

We’ll take $\vec{L}$ in the fixed Z direction. The axis of the top is along $x_{3}$.

Taking the $x_{1}$ axis along the line of nodes ON (Figure $\PageIndex{1}$) at the instant considered, the constant angular

$$\begin{align*} \vec{L} &=\left(I_{1} \Omega_{1}, I_{1} \Omega_{2}, I_{3} \Omega_{3}\right) \\[4pt] &=\left(I_{1} \dot{\theta}, \quad I_{1} \dot{\phi} \sin \theta, \quad I_{3}(\dot{\phi} \cos \theta+\dot{\psi})\right) \end{align*}$$

Figure $\PageIndex{1}$: Free motion of symmetric top: Constant $vec{L}$ along fixed Z

Remember, this new $x_{1}$ axis (Figure $\PageIndex{1}$) is perpendicular to the Z axis we’ve taken $\vec{L} \text { along, so } L_{1}=I_{1} \dot{\theta}=0$, and $\theta$ is constant, meaning that the principal axis $x_{3}$ describes a cone around the constant angular momentum vector $\vec{L}$. The rate of precession follows from the constancy of $L_{2}=I_{1} \dot{\phi} \sin \theta$. Writing the absolute magnitude of the angular momentum as L, $L_{2}=L \sin \theta$ (remember L is in the Z direction, and $x_{1}$ is momentarily along ON ) so the rate of precession $\dot{\phi}=L / I_{1}$. Finally, the component of $\vec{L}$ along the $x_{3}$ axis of symmetry of the top is $L \cos \theta=I_{3} \Omega_{3}$, so the top’s spin along its own axis is $\Omega_{3}=\left(L / I_{3}\right) \cos \theta$.