With the apparatus shown above, you can demonstrate the relationships between the speed and direction of the precession of a rotating object and the speed and direction of its rotation, show different patterns of nutation, and show the effect of displacement of the spin axis from that of the protruding rod by loading it with masses set along a different axis.

The apparatus shown above is an Ealing model 13-2209 large demonstration gyroscope. It is an air gyroscope whose rotor is a solid steel ball, nickel plated for protection. The ball rests in a cup, at the center of which is a small hole through which air is introduced by means of the pump shown at left rear. The resulting thin layer of air between the ball and the cup provides a nearly frictionless bearing on which the ball can rotate. This makes it possible to demonstrate some phenomena that would be difficult or impossible to show with other arrangements. The ball has been drilled out at one point to accept a hub, which holds an aluminum shaft. This shaft provides a means by which to spin the rotor with your fingers, and for attaching masses and other accessories. The ball is four inches in diameter (~100 mm; radius ≈ 50 mm) and has a mass of 4 kg. An o-ring placed around the hub acts as a cushion in case the ball rotates in such a way that the hub makes contact with the base. Please note: The ball is not case hardened, and it can be damaged if it is dropped. Also, the aluminum shaft is not strong enough to be used as a handle. In lifting the ball from the cradle to the base and back, use both hands on the ball itself.

Shown at right below the base of the apparatus, from left to right, are the following:

• A 10-g mass. This mass has on each face a yellow and a red stripe to make the rotation more obvious. It slips onto the aluminum shaft, and is held in place by o-rings within its bore.

• A 130-g mass. This mass has on each face a pair of yellow swirl stripes to make the rotation more obvious. It slips onto the aluminum shaft, and is held in place by o-rings within its bore. Around the outside it has an o-ring, which acts as a cushion in case it falls and makes contact with the base.

• A 5/64″ Allen wrench, for fixing the attachment shown to its right, to the aluminum shaft.

• A sleeve, which slips over the aluminum shaft and sets the height of the attachment shown to its right.

• An attachment for displacing the principal axis about which the ball spins, by means of masses set on its shaft. The 5/64″ Allen wrench placed in a set screw in the middle of the attachment shows the approximate location of the new principal axis.

• Two steel 10-g masses and one aluminum 5-g mass. These may go on the attachment or on the main shaft, as desired. Below the masses are o-rings, either for keeping them from sliding off the attachment, or for setting them at a particular height.

• Above the attachment and masses is a plastic ruler, in case you wish to measure the position of one of the masses.

• The wooden block with the hole in the center is a cradle for the ball while it is not in use.

• Below the apparatus at left are three discs with various patterns on them, to help show certain aspects of the motion of the gyroscope

With this apparatus you can demonstrate precession under constant torque, torque-free precession, nutation, and an interesting axis switching phenomenon that occurs when you add off-axis masses to shift the main symmetry axis. These are described below.

1) Precession under constant torque

Because the ball has been drilled out to accept the hub for the aluminum shaft, even with the shaft installed it has a light spot where the hub is. As a result, its center of mass lies slightly below the geometric center of the ball, and one of the ball’s three principal axes – the one that runs along the central axis of the rod, is not equal to the two principal axes that run perpendicular to it and to each other. (All three axes meet at the center of mass of the ball-rod system.) Since the system pivots about the center of the ball, when you set the ball in the cup and start the pump, if the shaft is not vertical, gravity exerts a torque about the pivot at the center of mass, and the shaft rises to the vertical position. When you displace the shaft downward, the ball oscillates, with the shaft swinging back and forth about the vertical position.

The torque exerted by gravity on a spinning top causes it to precess. (See demonstrations 28.54 – Bicycle wheel as a top, 28.57 – Bicycle wheel precession, and 28.60 – Maxwell’s top.) This torque is τ = r × mg, where m is the mass of the top and g is the acceleration of gravity. It points outward from the pivot point on which the top sits, and rotates about it as the top precesses. It also equals dL/dt, where L is the angular momentum of the top. We will call the angle that L makes with the vertical θ and the angle between L and L + ΔL, in the circle traced by L as it precesses, φ. ω_p = Δφ/Δt, and analysis shows that ω_p = τ/(L sin θ) and τ = rmg sin θ, which gives ω_p = rmg/L, or

ω_p = rmg/Iω.

The speed of the precession is inversely proportional to the rotational speed of the top, and the top precesses in the direction of the torque exerted by gravity. (ΔL points in the direction of τ.)

You can demonstrate this type of precession by setting the black, striped 10-g mass on the shaft. When it is centered anywhere below 7.3 cm above the top of the hub, the center of mass of the system is below the pivot point, and the torque from gravity acts to bring the shaft vertical. With the ball sitting as shown, this torque points toward the front of the table. If you spin the shaft clockwise as viewed from above, L points downward along the line of the shaft. ΔL points toward the front of the table, so the tip of the shaft moves toward the rear of the table, and the ball precesses counterclockwise as viewed from the top. If you spin the shaft counterclockwise, L points upward along the shaft. ΔL points toward the front of the table, so the tip of the shaft moves toward the front of the table, and the ball precesses clockwise. The ball precesses in the opposite direction to the direction in which it is spinning.

If you move the 10-g mass to a point farther than 7.3 cm from the top of the hub, this raises the center of mass of the system above the pivot point, and gravity pulls the shaft downward. With the ball sitting as shown, the torque now points toward the rear of the table. This reverses the direction of ΔL. Now if you spin the shaft clockwise, the tip moves toward the front of the table, and the ball precesses clockwise, and if you spin the shaft counterclockwise, the tip moves toward the rear of the table, and the ball precesses counterclockwise. The ball now precesses in the same direction to that in which it is spinning, as does a conventional top.

If you set the 10-g mass so that it is centered at the point 7.3 cm from the top of the hub, this puts the center of mass at the pivot, and gravity cannot exert a torque. If you now spin the shaft, the ball does not precess. (You may have to make small adjustments to the position of the mass to find the exact spot where there is no precession.)

As the equation above would indicate, the faster you spin the shaft, the more slowly the ball precesses. One can watch the precession of a spinning bicycle wheel or top increase in speed as friction slows its spinning. In this demonstration, the friction is low enough that it would take a long time for this to happen. You can illustrate the inverse relationship between the rate of spin and the speed of the precession either by starting the ball spinning at different speeds, or by starting it spinning at high speed and then occasionally slowing it a bit by lightly grabbing the shaft and then releasing it.

Because the Earth is somewhat oblate and its rotation axis is tilted with respect to its orbit about the Sun, the Sun exerts a torque about the Earth in the direction that would bring its rotation axis perpendicular to its orbit. The Moon’s orbit lies at angle of 5° to that of the Earth around the Sun, so it, too, exerts a torque in the same direction. (The bulge is more strongly attracted to the Sun or the Moon on the side that faces it than on the opposite side, hence the torque. Because the Moon is so much closer than the Sun is to the Earth, the torque it exerts on the Earth is a little over twice as great as that due to the Sun.) This causes the Earth’s axis to precess at a rate of 50.3 arc seconds per year (~0.14 arc second per day). The period of this precession is about 25,800 years. (See demonstrations 92.06 -- Celestial globe, and 92.30 -- Cutaway telescope.)

2) Torque-free precession

If an object is set in motion so that it is spinning about its main axis of symmetry, but is also rotating in some other way so that its total angular momentum L points in some other direction, this creates a situation in which L is not parallel to ω. Absent an external torque, conservation of angular momentum requires that the direction of L remain fixed. As a result, the spin axis precesses about L. This results in the characteristic wobbling motion of a spinning coin, for example, as it falls after having been released with a flip. A complete analysis is somewhat complicated, but we can simplify it by considering the motion of a symmetrical object whose precessional motion is not great, and for which rhe angular displacements about the axes perpendicular to the main symmetry axis are small.

Our spinning object has spin angular momentum L_s = I_sω_s, along its main symmetry axis. Let us say that L_s is close to the z axis, sitting at an angle θ_x ≪ 1 from the x axis, and at an angle θ_y ≪ 1 from the y axis. For small angular displacements, to first order the moments of inertia about the principal axes, I_xx, I_yy and I_zz, are constant, and the products of inertia, I_xy, I_xz, I_yx, I_yz, I_zx and I_zy, are zero. Also, for small angles we can treat the rotations separately and add them. Since rotation about either the x or y axis gives L_s a component along the other axis, the angular momenta about the x and y axes are

L_x = I_xx(dθ_x/dt) + L_s sin θ_y and

L_y = I_yy(dθ_y/dt) + L_s sin θ_x.

By symmetry, L_xx = L_yy, and since both are perpendicular to L_z, we will designate them L_⟂. Since for small angles, sin θ ≈ θ and cos θ ≈ 1, these become

L_x = L_⟂(dθ_x/dt) + L_sθ_y and

L_y = L_⟂(dθ_y/dt) + L_sθ_x.

For small θ_x and θ_y, L_s is close to the z axis, and

L_z = I_sω_s.

Since there is no torque, dL/dt = 0, L_s and ω_s are constant, and differentiating the equations above gives

I_⟂(d²θ_x/dt²) + L_s(dθ_y/dt) = 0 and

I_⟂(d²θ_y/dt² - L_s(dθ_x/dt) = 0.

Substituting ω_x for dθ_x/dt and ω_y for dθ_y/dt gives

I_⟂(dω_x/dt) + L_sω_y = 0 and

I_⟂(dω_y/dt) + L_sω_x = 0.

Rearranging the second of these gives (dω_y/dt) = (L_s/I_⟂)ω_x, so differentiating the first one and then substituting this into the result gives

(I_⟂²/L_s)(d²ω_x/dt²) + L_sω_x = 0,

which gives

(d²ω_x/dt²) + γ²ω_x = 0,

where γ = L_s/I_⟂ (= (I_s/I_⟂)ω_s). This is identical to the equations on the pages for demonstrations that show simple harmonic motion, and a solution is

ω_x = A sin (γt + φ),

where A and φ are constants that depend on initial conditions. From the equation above for rotation about the x axis, ω_y = -(I_⟂/L_s)(dω_x/dt). Differentiating the solution above, then, gives

ω_y = (I_⟂/I_sω_s)Aγ cos (γt + φ), or

ω_y = A cos (γt + φ).

To find the position as a function of time, we can integrate these to put them in terms of θ_x and θ_y. This gives

θ_x = (A/γ) cos (γt + φ) + θ_x0 and

θ_y = -(A/γ) sin (γt + φ) + θ_y0.

θ_x0 and θ_y0 are constants of integration. To satisfy our small-angle approximation, A/γ ≪ 1. These equations describe the motion of a body whose spin axis rotates about some fixed line. If we take this line as the z axis, then both θ_x0 and θ_y0 equal zero. If at t = 0, θ_x = θ₀ and θ_y = 0, and A/γ = θ₀ and φ = 0, these equations become

θ_x = θ₀ cos γt and

θ_y = θ₀ sin γt.

These equations describe torque-free precession, with the spin axis precessing about the z axis at an angle of θ₀, with a frequency γ = ω_sI_s/I_⟂. For an oblate spheroid, I_s > I_⟂, and γ > ω_s. For a thin coin, the example mentioned above, I_s = 2I_⟂ and γ = 2ω_s. As it falls, a spinning coin wobbles twice as fast as it spins.

In all of the above, we have considered the rotating object as viewed from an external (laboratory) reference frame. In some cases it can be advantageous to view the motion of an object in the frame of reference of the object itself. The Earth, which is an oblate spheroid, exhibits torque-free precession. Since an observer on Earth is spinning with the earth, the rate of precession he would observe is γ′ = γ - ω_s, or

γ′ = ω_s[(I_s - I_⟂)/I_⟂].

For the Earth, the ratio in brackets is approximately 1/306, which gives a precession period of about 306 days. Because the Earth is elastic, however, and its motion is affected by deformation of the mantle and by currents in the liquid core, oceans and atmosphere, the observed period is about 430 days, and the motion is somewhat irregular. It is known as the Chandler Wobble, after S. C. Chandler, who discovered it in 1891. (It had been predicted earlier by Leonhard Euler in 1765.) The rotation axis wanders about the pole in a circle having a radius of about 10 m.

You can demonstrate torque-free precession with the striped 10-g mass set on the shaft. (This helps show the rate of spin.) Set the ball spinning, and then give the shaft a kick with your hand. From whichever direction you strike the shaft, it precesses in the same direction as the one in which it is spinning, and at a similar rate to that of the spin. The shaft sweeps out a cone centered about the original rotation axis. Some people refer to this cone as a wobble cone.

You can show the relationship between the speed of precession and the rate of spin as described above for precession under torque.

3) Nutation

As noted above, and shown in the demonstrations linked above, when we release the free end of the axis of a spinning gyroscope or spinning wheel (while the other end sits on a pivot), gravity exerts a torque about the pivot that should cause the gyroscope or wheel to fall, but instead it precesses about the pivot. In order for the gyroscope or wheel to precess, it must have some component of angular momentum about the (vertical) pivot axis, so when we release it, the shaft falls to a level little bit lower than where it started. At first, however, it drops to a level lower than where it should sit during steady precession, rises again almost to its original level, and goes through several similar cycles until when the motion damps out, the gyroscope or wheel precesses at a level slightly below where it started. This oscillatory motion is called nutation. For many systems, the nutation is fast enough, and it damps out quickly enough, that it is difficult to observe. The low friction afforded by the bearing in the air gyroscope, and the mass distribution of the rotor with the 130-g mass attached, however, make this motion obvious and persistent.

For the following analysis, we consider a wheel (not yet spinning) with its axle horizontal and one end resting on a pivot that sits at the origin. The z axis is vertical, and the axle lies along or very close to the y axis. The analysis will be similar to that above for section 2), torque-free precession. Since the center of mass of the rotor in this case is displaced from the origin by a distance l, however, we must use the parallel axis theorem to find the angular momentum about the x and z axes. If the moment of inertia of the wheel through a vertical axis is I_zz, then the moment of inertia about the z axis is I_zz + Ml². If we call this sum I_p, then L_z = I_pω_z By symmetry, we obtain a similar expression for the moment of inertia about the x axis, and L_x = I_pω_x. These are exact when the wheel sits on the y axis, and hold to first order for small angles with respect to the y axis.

Now we set the wheel spinning at a speed ω_s. If its moment of inertia about the spin axis is I_s, then L_s = I_sω_s.

For small displacements from the y axis, the angular momentum has one term for motion about the z axis, of I_pω_z, and a component from the change in direction of L_s. For small angular displacements about the z and x axes, θ_z ≪ 1 and θ_x ≪ 1, and as noted above, we can treat the rotations separately and add them. For rotation about the x axis we obtain

L_x = I_pω_x, L_y = L_s cos θ_x ≈ L_s and L_z = L_s sin θ_x ≈ L_sθ_x.

Similarly, for rotation about the z axis we obtain

L_x = -L_s sin θ_z ≈ -L_xθ_z, L_y = L_s cos θ_z ≈ L_s and L_z = I_pω_z.

These equations indicate that to first order, small rotations about the x and z axes leave L_y unchanged, but contribute to both L_x and L_z, giving:

L_x = I_pω_x - L_sθ_z,

L_y = L_s

L_z = I_pω_z + L_sθ_x.

The instantaneous torque about the origin is τ_x = -lW, where l is as above (the length of the axle from the wheel to the origin) and W is the weight of the wheel. If L_s is constant, and if dθ_z/dt = ω_z and dθ_x/dt = ω_x, since τ = dL/dt, we have:

I_p(dω_x/dt) - L_sω_z = -lW

dL_s/dt = 0, and

I_p(dω_z/dt) + L_sω_x = 0.

The last of these gives (dω_z/dt) = -L_xω_x/I_p. Differentiating the first and substituting this into the result gives:

(d²ω_x/dt) + (L_s²/I_p²)ω_x = 0.

Setting γ = L_s/I_p = ω_sI_s/I_p gives (d²ω_x/dt²) + γ²ω_x = 0, for which a solution is

ω_x = A cos(γt + φ),

where A and φ are constants as described above. In similar fashion to the analysis for part 2), further substitutions and then integration to obtain angular position as a function of time gives:

θ_x = B sin(γt + φ) + C and

θ_z = (lW/L_s)t + B cos(γt + φ) + D,

where B, C and D are constants of integration.

Say that we release the wheel axle from rest along the y axis at t = 0 (with the wheel spinning). At t = 0, (θ_x)₀ = 0 and (dθ_x/dt)₀ = 0. From the above equations we have

B sin φ + C = 0 and Bγ cos φ = 0.

If B does not equal zero, then φ = π/2 and C = -B. The equation above for θ_z then becomes

θ_z = (lW/L_s)t - B sin γt + D.

At t = 0, (θ_z)₀ = 0 and (dθ_z/dt)₀ = 0, which gives (dθ_z)₀ = (lW/Ls) - Bγ = 0, or B = (lW)/(L_s). Inserting this into the equations above for θ_x and θ_z gives

θ_x = (lW/γL_s)(cos γt - 1) and

θ_z = (lW/γL_s)(γt - sin γt).

According to these equations, as the wheel precesses, the tip of its axle follows the path of a cycloid. As noted above, in most systems, the motion is damped so that each oscillation is smaller than the previous one, until the tip of the spin axis precesses at a level above the lowest points of its travel, but somewhat below where it started.

With the 130-gram mass set on the shaft about 2/3 of the way out from the hub, the path of the tip resembles a cycloid. If you set the mass closer to the hub, the tip makes a loop near the top of its travel, and if you set the mass closer to the outer end of the shaft, the cusps of the cycloid broaden.

4) The effect of shifting the principal axis

A technician at the Ealing Corporation dropped a gyroscope rotor, and when he tried it in a base, he noticed that the shaft, which had bent, exhibited interesting behavior. This behavior was attributed to the displacement of the principal axis and change in symmetry that had occurred when the shaft bent. To provide a means to demonstrate this, Ealing developed an accessory that allows you to shift the principal axis away from the line of the shaft. The photograph below shows this accessory installed for use.

Also shown is the Allen wrench, sitting in the head of a set screw embedded in the base of the holder. The sleeve between the hub and the base of the holder sets the height of the accessory so that the lines of the Allen wrench and the rod holding the masses go through the center of the ball.

Set the sleeve and accessory on the shaft as shown, and use the Allen wrench to tighten the set screw to fix it to the shaft. Set one of the steel 10-g masses and the aluminum 5-g mass on the shaft, and slip an o-ring down the shaft to hold them in place, as shown. Adding these masses shifts the principal axis of the rotor to the line along which the Allen wrench lies when it is inserted in the accessory.

Without the Allen wrench attached, holding the main shaft vertical, set the rotor spinning and then release it. The rotor spins about the main shaft, with the rod with the masses going in a circle around it. After a few rotations, the shaft begins to wobble at an angle that puts the rod with the weights roughly verticle. After a few more rotations, the shaft flips back to vertical. As the rotor spins, these two axes continue periodically to exchange places. It appears that each axis in turn wobbles (goes around in a circle in the same direction as the spin), and then “dewobbles” (spins in place). Throughout this motion, the new principal axis moves in a circle about the vertical line through the center of the rotor. You can make this more obvious by now inserting the Allen wrench as shown and setting the rotor spinning, again with the shaft vertical.

With the Allen wrench in place, it may help to add the other steel 10-g mass, though the rotor should behave well either way.

To show that the principal axis is now at the location described above, you can move the Allen wrench so that it is vertical and set the rotor spinning about it. Now when you release the Allen wrench, it will precess in a circle, and the two other axes will revolve about it as it does so. The periodic wobbling and “dewobbling” will not occur.

1) Kleppner, Daniel and Kolenkow, Robert. An Introduction to Mechanics, Second Edition (Cambridge, UK: Cambridge University Press, 2014), Chapters 7 and 8.
2) Goldstein, Herbert; Poole, Charles P. and Safko, John L. Classical Mechanics, Third Edition (San Francisco: Addison-Wesley, 2002), Chapter 5; torque-free precession of Earth, pp. 207-208; precession of Earth under torque, pp. 223-228.
3) Feynman, Richard P.; Leighton, Robert B. and Sands, Matthew. The Feynman Lectures on Physics, Volume 1 (Menlo Park, California: Addison-Wesley Publishing Company, 1963), pp. 20-5 to 20-7.
4) Physics LibreTexts -- 13.20: Torque-free rotation of an inertially-symmetric rigid rotor
5) Encyclopedia Britannica (britannica.com). Seth Carlo Chandler
6) The Ealing Corporation. 13-2209/13-2258 Large and Student Demonstration Gyroscopes Instruction Manual (Cambridge, Massachusetts: The Ealing Corporation, 1972).