Gyroscopic Precession

Consider a gyroscope that consists of a symmetric top of mass $M$ that rotates about its symmetry axis at the constant angular velocity $\omega$. (The rotation axis is obviously a principal axis of rotation.) The top is free to rotate (without friction) about a fixed pivot point $P$. Suppose that the axis of the top subtends a constant angle $\theta$ with the vertical. Let us set up a Cartesian coordinate system such that the $z$-axis is vertical, and the $x$-$y$ plane is horizontal. Suppose that the projection of the axis of the top on to the $x$-$y$ plane subtends an (instantaneous) angle $\phi$ with the $x$-axis. Let $C$ be the center of mass of the top (which lies on the symmetry axis), and let the distance $PC$ be $l$. Finally, let $I$ be the (principal) moment of inertia of the top about its symmetry axis. See Figure 1.10.

Figure 1.10: A gyroscope.

It follows, from Figure 1.10, that the angular velocity of the top is

$$\mbox{\boldmath$\omega$} = \omega\,\sin\theta\,\cos\phi\,{\bf e}_x+ \omega\,\sin\theta\,\sin\phi\,{\bf e}_y+ \omega\,\cos\theta\,{\bf e}_z.$$ (1.230)

Hence, the angular momentum of the top is

$${\bf L} = I\,\omega\,\sin\theta\,\cos\phi\,{\bf e}_x+ I\,\omega\,\sin\theta\,\sin\phi\,{\bf e}_y+ I\,\omega\,\cos\theta\,{\bf e}_z,$$ (1.231)

where use has been made of Equation (1.191). Now, the weight of the top exerts a torque $\tau$ that is of magnitude $M\,g\,l\,\sin\theta$ and whose direction is specified in the figure. Here, $g$ is the acceleration due to gravity. Hence,

$$\mbox{\boldmath$\tau$} = -M\,g\,l\,\sin\theta\,\sin\phi\,{\bf e}_x+M\,g\,l\,\sin\theta\,\cos\phi\,{\bf e}_y.$$ (1.232)

The angular equation of motion of the top, (1.180), is

$$\frac{d {\bf L}}{dt} = \mbox{\boldmath$\tau$} .$$ (1.233)

Assuming that $\theta$ and $\omega$ are constants, whereas $\phi$ is time-varying, the $x$-, $y$-, and $z$-components of the previous equation are

$$-I\,\omega\,\sin\theta\,\sin\phi\,\frac{d\phi}{dt} = - M\,g\,l\,\sin\theta\,\sin\phi,$$ (1.234)

$$I\,\omega\,\sin\theta\,\cos\phi\,\frac{d\phi}{dt} = M\,g\,l\,\sin\theta\,\cos\phi,$$ (1.235)

$$=0,$$ 0 (1.236)

respectively. Hence, we deduce that the gravitational torque acting on the top causes its axis of rotation to precess about the vertical at the rate

$$\frac{d\phi}{dt} = \frac{M\,g\,l}{I\,\omega},$$ (1.237)

while maintaining a constant inclination to the vertical, and a constant spin rate. Note that the precession is in the same direction as the vertical component of the top's angular velocity. Interestingly enough, the precession rate is independent of the angle of inclination of the rotation axis to the vertical.