26.2: Precession of a Symmetrical Top

A more interesting case is the free rotation (zero external torque) of a symmetrical top, meaning $I_{1}=I_{2} \neq I_{3}$.

Figure $\PageIndex{1}$

We can take any pair of orthogonal axes, perpendicular to the body’s symmetry axis, as the $x_{1}, x_{2}$ axes. We’ll choose $x_{2}$ following Landau, as perpendicular to the plane containing $\vec{L}$ and the momentary position of the $x_{3}$ axis, so in the diagram here $x_{2}$ is perpendicularly out from the paper/screen, towards the viewer.

This means the angular momentum component $L_{2}=0$ and therefore $\Omega_{2}=0 . \text { Hence } \vec{\Omega}$ is in the same plane as $\vec{L}, x_{3}, \text { and so the velocity } \vec{v}=\vec{\Omega} \times \vec{r}$ of every point on the axis of the top is perpendicular to this plane (into the paper/screen). The axis of the top $O x_{3}$ must be rotating uniformly about the direction of $\vec{L}$.

The spin rate of the top around its own axis is

$$\begin{equation} \Omega_{3}=L_{3} / I_{3}=\left(L / I_{3}\right) \cos \theta \end{equation}$$

The angular velocity vector $\vec{\Omega}$ can be written as a sum of two components, one along the body’s axis $O x_{3}$ and one parallel to the angular momentum $\vec{L}$ (these components are shown dashed in the figure)

$$\begin{equation} \vec{\Omega}=\vec{\Omega}_{\text {precession }}+\vec{\Omega}_{3} \end{equation}$$

The component along the body’s axis $O x_{3}$ does not contribute to the precession, which all comes from the component along the (fixed in space) angular momentum vector.

The speed of precession follows from

$$\begin{equation} \Omega_{\text {precession }} \sin \theta=\Omega_{1} \end{equation}$$

and

$$\begin{equation} \Omega_{1}=L_{1} / I_{1}=\left(L / I_{1}\right) \sin \theta \end{equation}$$

so

$$\begin{equation} \Omega_{\text {precession }}=L / I_{1} \end{equation}$$

Note also the ratio of precession rate to spin around axis is

$$\begin{equation} \Omega_{\text {precession }} / \Omega_{3}=\left(I_{3} / I_{1}\right) \sec \theta \end{equation}$$

This means the precession rate and the spin are very comparable, except when $\theta$ is near $\pi / 2$, when the precession becomes much faster. Remember this is the body’s precession with no external torque, and is clearly completely different—much faster precession—than the familiar case of a fast spinning top under gravity.