Gyroscopes simply explained with Coriolis pseudotorques

One of the most amusingly counterintuitive encounters with physics is to watch a gyroscope that is supported only at one end as it slowly rotates in a horizontal plane and defies gravity. The configuration is shown in Fig. 1. The gyroscope is modeled as a negligible-mass shaft with a massive disk at one end, and is supported only at the other end. The weight of the disk results in a gravitational torque in the positive x direction. The intuitive expectation is that the unsupported disk will fall. The falling disk, after all, would create angular momentum about the positive x direction, a change in angular momentum required by the gravitational torque.

This analysis is simple and well known, but does not really give an intuitive explanation of the counterintuitive failure of the disk to fall. Such an explanation has, happily, already been given more than half a century ago, by Barker² in the pages of this journal; a much more recent version of this explanation has appeared in which vpython simulations may offer addtional clarity.³ 

Here, I would like to give another explanation, another viewpoint, that is not as direct as those cited above,^2,3 but that may be more appealing than the standard nonintuitive torque/angular momentum approach. It is, in any case, interestingly different.

This viewpoint requires only that we look at the configuration in a frame that is rotating with the precessing gyroscope. In this frame, the configuration differs from that in Fig. 1 only in the absence of precession; the only motion is the rotation of the disk around the shaft. In this reference frame, the gyroscope shaft remains in the z direction. The angular momentum $I ω$ is therefore constant in time and we face a new puzzle: Why doesn't the gravitational torque result in some change?

When acting on a mass point, the $Ω$ dependent terms on the right of Eq. (2) are the accelerations due to “pseudoforces” in the noninertial frame. The first of these is the most familiar, the centrifugal force. This pseudoforce, in the z direction, pushes mass away from the y axis and is opposed by tension in the gyroscope shaft so that one may (correctly) predict that these forces add to zero.¹⁰ We also ignore the last term on the right, since $Ω$ is constant.

This leaves us with only the term $− 2 Ω × v rot$⁠, the Coriolis acceleration, where $v rot$ is the motion due to the $ω$ rotation about the z axis. For mass elements on the disk, the distribution of Coriolis pseudoforces is sketched in Fig. 2. It should now be evident that the Coriolis “pseudotorque” on the disk, in the negative x direction, must balance the gravitational torque in the positive x direction, so that there is no net torque, and the puzzle is solved. But to be sure of this, we should calculate the pseudotorque.