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.

Fig. 1.

A gyroscope with a shaft supported only at the end opposite the massive disk. Gravitational torque in the positive x direction drives a precession of the shaft with an angular velocity Ω in the positive y direction.

Fig. 1.

A gyroscope with a shaft supported only at the end opposite the massive disk. Gravitational torque in the positive x direction drives a precession of the shaft with an angular velocity Ω in the positive y direction.

Close modal

But the disk does not fall. Rather, the gyroscope, shaft, and disk1 rotate (“precess”) with angular velocity Ω. The standard explanation is given in terms of the gyroscope angular momentum J=Iω, where I is the gyroscope moment of inertia (“rotational inertia”) about its shaft, and ω is the angular velocity of the disk about the shaft. The gravitational torque drives a rate of change of angular momentum that results not in the disk falling, but in J changing direction

τgrav=dJ/dt=Ω×J=ΩωIx̂.
(1)

Here and below, boldface symbols indicate vectors, nonboldface equivalents are their magnitudes, and symbols like x̂ denote unit vectors in coordinate directions.

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 Iω is therefore constant in time and we face a new puzzle: Why doesn't the gravitational torque result in some change?

The answer starts with the expression for the acceleration arot of a mass point in a frame rotating at angular velocity Ω, our new reference frame

arot=ainΩ×(Ω×R)2Ω×vrotdΩdt×R.
(2)

Here, R is the vector from the coordinate origin to a mass point, ain is the acceleration of that mass point in the inertial frame, and vrot is the velocity of the mass point in the rotating frame. This equation has appeared many times in this journal4–7 and is given in many mechanics texts.8,9

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 2Ω×vrot, the Coriolis acceleration, where vrot 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.

Fig. 2.

The Coriolis forces on the disk in the rotating frame. The figure shows the sign difference in the Coriolis force on the upper and lower half of the disk.

Fig. 2.

The Coriolis forces on the disk in the rotating frame. The figure shows the sign difference in the Coriolis force on the upper and lower half of the disk.

Close modal

To facilitate the calculation, we introduce cylindrical coordinates r,θ,z illustrated in Fig. 3, to locate each mass element dm. In these coordinates, the Coriolis acceleration in Eq. (2) gives us the pseudoforce

dF=2Ω×vrotdm=2Ωŷ×(ωrθ̂)dm=2Ωωrsinθẑdm.
(3)
Fig. 3.

The coordinates for calculating the pseudotorque.

Fig. 3.

The coordinates for calculating the pseudotorque.

Close modal

To calculate the torque about the coordinate origin, for comparison with Eq. (1), we denote the distance from the coordinate origin to the center of the disk by Rc, so that the vector R from the origin to a mass point is

R=Rcẑ+r(x̂cosθ+ŷsinθ).
(4)

We now integrate R×dF to get the total pseudotorque

τpseudo=r(x̂cosθ+ŷsinθ)×dF=2Ωωr2(x̂sin2θŷsinθcosθ)dm.
(5)

The term involving Rc has disappeared due to the cross product. The integration over θ makes the ŷ term vanish and gives a factor of 1/2 for the x̂ term. (The average of sin2θ over all theta is 1/2.) The result is then

τpseudo=Ωωx̂r2dm=ΩωIx̂.
(6)

This pseudotorque in the rotating frame does, indeed, perfectly balance the gravitational torque in Eq. (1).

1.

For the gyroscope to rotate this way, it must initially be put into motion carefully, otherwise in addition to the horizontal precession there will be vertical nutational oscillations.

2.
Ernest F.
Barker
, “
Elementary analysis of the gyroscope
,”
Am. J. Phys.
28
(
10
),
808
810
(
1960
).
3.
Harvey
Kaplan
and
Andrew
Hirsch
, “
Gyroscopic motion: Show me the forces!
,”
Phys. Teach.
52
(
1
),
30
33
(
2014
).
4.
Jacob
Neuberger
, “
Coriolis force revisited
,”
Am. J. Phys.
49
(
8
),
782
784
(
1981
), Eq. (1).
5.
Harold A.
Daw
, “
Coriolis lecture demonstration
,”
Am. J. Phys.
55
(
6
),
1010
1014
(
1987
), Eq. (2).
6.
Guy
Vandegrift
, “
On the derivation of Coriolis and other noninertial accelerations
,”
Am. J. Phys.
63
(
7
),
663
(
1995
), Eq. (7).
7.
L. Filipe
Costa
and
José
Natário
, “
The Coriolis field
,”
Am. J. Phys.
84
(
5
),
388
395
(
2016
), Eq. (10).
8.
L. D.
Landau
and
E. M.
LIfschitz
,
Mechanics
, 3rd ed. (
Elsevier Butterworth-Beinemann
,
Burlington, MA
,
1976
), Vol.
1
, Eq. (39.7) in p.
128
.
9.
Matthew J.
Benacquista
and
Joseph D.
Romano
,
Classical Mechanics
(
Springer
,
Cham
,
2018
), Eq. (1.73) in p.
22
.
10.

This statement is a bit of an oversimplification. The centrifugal pseudoforceforce acts on the mass distributed in the disk. The cancelation of the stress in the shaft and the centrifugal force on the disk therefore involve stresses in the disk. Note that the symmetry of this pseudoforce about the xz plane means that there can be no contribution to the torque that “holds up” the disk.