27.3: Free Motion of a Symmetrical Top

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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 \(\vec{L}\).

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

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

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 \(\Omega_{3}=\left(L / I_{3}\right) \cos \theta\).