27.5: Steady Precession

Under what conditions will a top, spinning under gravity, precess at a steady rate? The constancy of $L_{3}$, $L_{Z}$ mean that $\Omega_{3}=\dot{\phi} \cos \theta+\dot{\psi}$, and $\Omega_{\mathrm{pr}}=\dot{\phi}$ are constants.

The $\theta$ Lagrange equation is

$$I_{1}^{\prime} \ddot{\theta}=I_{1}^{\prime} \dot{\phi}^{2} \sin \theta \cos \theta-I_{3}(\dot{\phi} \cos \theta+\dot{\psi}) \dot{\phi} \sin \theta+M g \ell \sin \theta$$

For constant $\theta$, $\ddot{\theta}=0$, so, with $\Omega_{3}=\dot{\phi} \cos \theta+\dot{\psi}$, and $\Omega_{\mathrm{pr}}=\dot{\phi}$.

$$I_{1}^{\prime} \Omega_{p r}^{2} \cos \theta-I_{3} \Omega_{3} \Omega_{p r}+M g \ell=0 \label{eq3}$$

Since Equation $\ref{eq3}$ is a quadratic equation for the precession rate, there are two solutions in general: on staring at a precessing top, this is a bit surprising! We know that for the top, when it’s precessing nicely, the spin rate $\Omega_{3}$ far exceeds the precession rate $\Omega_{p r}$. Assuming $I_{1}^{\prime}, I_{3}$ to be of similar size, this means the first term in the quadratic is much smaller than the second. If we just drop the first term, we get the precession rate

$$\Omega_{\text {precess }(\mathrm{slow})}=\dfrac{M g \ell}{I_{3} \Omega_{3}}, \quad\left(\Omega_{3} \gg \Omega_{\text {precess }}\right)$$

Note that this is independent of angle—the torque varies as $\sin \theta$, but so does the horizontal component of the angular momentum, which is what’s changing.

This is the familiar solution for a child’s fast-spinning top precessing slowly. But this is a quadratic equation, there’s another possibility: in this large $\Omega_{3}$ limit, this other possibility is that $\Omega_{p r} \text { is itself of order } \Omega_{3}$, so now in the equation the last term, the gravitational one, is negligible, and

$$\Omega_{\text {precess }(\text { fast })} \cong I_{3} \Omega_{3} / I_{1}^{\prime} \cos \theta$$

This is just the nutation of a free top! In fact, of course, both of these are approximate solutions, only exact in the limit of infinite spin (where one goes to zero, the other to infinity), and a more precise treatment will give corrections to each arising from the other. Landau indicates the leading order gravitational correction to the free body nutation mode.