Gyroscopic Precession

Let us now study the motion of a rotationally symmetric rigid top which is free to turn about a fixed point (without friction), but which is subject to a gravitational torque (see Fig. 44). Let the $z'$-axis coincide with the symmetry axis. Let the principal moment of inertia about the symmetry axis be $I_\parallel$, and let the other principal moments both take the value $I_\perp$. Let the $z$-axis run vertically upward, and let the common origin, $O$, of the fixed and body frames coincide with the fixed point about which the top turns. Suppose that the center of mass of the top lies a distance $l$ along its axis for $O$, and that the mass of the top is $m$. Let the symmetry axis of the top subtend an angle $\theta$ (which is an Eulerian angle) with the upward vertical.

Figure 44: A symmetric top.

Consider an instant in time at which the Eulerian angle $\psi$ is zero. This implies that the $x'$-axis is horizontal, as shown in the diagram. The gravitational force, which acts at the center of mass, thus exerts a torque $m\,g\,l\, \sin\theta$ in the $x'$-direction. Hence, the components of the torque in the body frame are

$$T_{x'} = m\,g\,l\,\sin\theta,$$ (626)

$$T_{y'} = 0,$$ (627)

$$T_{z'} = 0.$$ (628)

The components of the angular velocity vector in the body frame are given by Eqs. (607)-(609). Thus, Euler's equations (576)-(578) take the form:

$$m\,g\,l\,\sin\theta = I_\perp\,(\ddot{\theta}- \cos\theta\,\sin\theta\,\dot{\phi}^{\,2}) + L_\psi\,\sin\theta\,\dot{\phi},$$ (629)

$$0 = I_\perp\,(2\,\cos\theta\,\dot{\theta}\,\dot{\phi} +\sin\theta\,\ddot{\phi}) - L_\psi\,\dot{\theta},$$ (630)

$$0 = \dot{L}_\psi,$$ (631)

where

$$L_\psi = I_\parallel\,(\cos\theta\,\dot{\phi}+\dot{\psi})= I_\parallel\,{\mit\Omega},$$ (632)

where ${\mit\Omega} = \omega_{z'}$ is the angular velocity of the top. Multiplying Eq. (630) by $\sin\theta$, we obtain

$$\dot{L}_\phi = 0,$$ (633)

where

$$L_\phi = I_\perp\,\sin^2\theta\,\dot{\phi} + L_\psi\,\cos\theta.$$ (634)

According to Eqs. (631) and (633), the two quantities $L_\psi$ and $L_\phi$ are constants of the motion. These two quantities are the angular momenta of the system about the $z'$- and $z$-axis, respectively. They are conserved because the gravitational torque has no component along either the $z'$- or the $z$-axis.

If there are no frictional forces acting on the top then the total energy, $E=K+U$, is also a constant of the motion. Now,

$$E = \frac{1}{2}\,\left(I_\perp\,\omega_{x'}^{\,2} + I_\perp\... ...+ I_\parallel\,\omega_{z'}^{\,2}\right) + m\,g\,l\,\cos\theta.$$ (635)

When written in terms of the Eulerian angles (with $\psi=0$), this becomes

$$E=\frac{1}{2}\left(I_\perp\,\dot{\theta}^{\,2} + I_\perp\,\s... ...{\,2} + L_\psi^{\,2}/I_\parallel\right) + m\,g\,l\,\cos\theta.$$ (636)

Eliminating $\dot{\phi}$ between Eqs. (634) and (636), we obtain the following differential equation for $\theta$:

$$E = \frac{1}{2}\,I_\perp\,\dot{\theta}^{\,2} + \frac{(L_\phi... ...c{1}{2}\frac{L_\psi^{\,2}}{I_\parallel} + m\,g\,l\,\cos\theta.$$ (637)

Let

$$E' = E - \frac{1}{2}\frac{L_\psi^{\,2}}{I_\parallel},$$ (638)

and $u=\cos\theta$. It follows that

$$\dot{u}^{\,2} = 2\,(E' -m\,g\,l\,u)\,(1-u^2)\,I_\perp^{-1} - (L_\phi-L_\psi\,u)^2\,I_\perp^{-2},$$ (639)

or

$$\dot{u}^{\,2} = f(u),$$ (640)

where $f(u)$ is a cubic polynomial. In principal, the above equation can be integrated to give $u$ (and, hence, $\theta$) as a function of $t$:

$$t = \int\frac{du}{\sqrt{f(u)}}.$$ (641)

Fortunately, we do not have to perform the above integration (which is very ugly) in order to discuss the general properties of the solution to Eq. (640). It is clear, from Eq. (641), that $f(u)$ needs to be positive in order to obtain a physical solution. Hence, the limits of the motion in $\theta$ are determined by the three roots of the equation $f(u)=0$. Since $\theta$ must lie between $0$ and $\pi/2$, it follows that $u$ must lie between 0 and 1. It can easily be demonstrated that $f\rightarrow\pm\infty$ as $u\rightarrow\pm\infty$. It can also be shown that the largest root $u_3$ lies in the region $u_3>1$, and the two smaller roots $u_1$ and $u_2$ (if they exist) lie in the region $-1\leq u\leq +1$. It follows that, in the region $-1\leq u\leq 1$, $f(u)$ is only positive between $u_1$ and $u_2$. Figure 45 shows a case where $u_1$ and $u_2$ lie in the range 0 to 1. The corresponding values of $\theta$--$\theta_1$ and $\theta_2$, say--are then the limits of the vertical motion. The axis of the top oscillates backward and forward between these two values of $\theta$ as the top precesses about the vertical axis. This oscillation is called nutation. Incidentally, if $u_1$ becomes negative then the nutation will cause the top to strike the ground (assuming that it is spinning on a level surface).

Figure 45: The function $f(u)$.

If there is a double root of $f(u)=0$ (i.e., if $u_1 = u_2$) then there is no nutation, and the top precesses steadily. However, the criterion for steady precession is most easily obtained directly from Eq. (629). In the absence of nutation, $\dot{\theta}=\ddot{\theta}=0$. Hence, we obtain

$$m\,g\,l = -I_\perp\,\cos\theta\,\dot{\phi}^{\,2}+ L_\psi\,\dot{\phi},$$ (642)

or

$${\mit\Omega}= \frac{m\,g\,l}{I_\parallel\,\dot{\phi}} + \frac{I_\perp}{I_\parallel}\,\cos\theta\,\dot{\phi}.$$ (643)

The above equation is the criterion for steady precession. Since the right-hand side of Eq. (643) possesses the minimum value $2\,\sqrt{m\,g\,l\,I_\perp\,\cos\theta}/I_\parallel$, it follows that

$${\mit\Omega} > {\mit\Omega}_{\rm crit} = \frac{2\,\sqrt{m\,g\,l\,I_\perp\,\cos\theta}}{I_\parallel}$$ (644)

is a necessary condition for obtaining steady precession at the inclination angle $\theta$. For ${\mit\Omega}>{\mit\Omega}_{\rm crit}$, there are two roots to Eq. (643), corresponding to a slow and a fast steady precession rate for a given inclination angle $\theta$. If ${\mit\Omega}\gg {\mit\Omega}_{\rm crit}$ then these two roots are approximately given by

$$(\dot{\phi})_{\rm slow} \simeq \frac{m\,g\,l}{I_\parallel\,{\mit\Omega}},$$ (645)

$$(\dot{\phi})_{\rm fast} \simeq \frac{I_\parallel\,{\mit\Omega}}{I_\perp\,\cos\theta}.$$ (646)

The slower of these two precession rates is the one which is generally observed.