I've been trying to glean some insight into the motion of a body that is rotating about an axis through the COM of said body whilst traveling in a orbital-like path about a perpendicular axis outside of the body. To paint the scenario a bit more clearly, I use the example of a single propeller airplane negotiating a turn. Depending on the direction of angular momentum vector of the propeller, the negotiation of that turn will cause the plane to pitch up or down.

Recreating this scene with a gyroscope in hand at arm's length and spinning you're body to recreate the turn, the gyro will pitch up/down until the angular momentum vector of the gyro is parallel/antiparallel to the angular momentum vector of turn.

My question is this: if the magnitude of the angular velocity of both the gyro and of the turn were held constant, would the torque (precession)causing the pitching motion remain constant until the the angular momentum vectors are parallel, or would it increase/decrease until the angular momentum vectors are parallel? What is the proper train of thought one could use to derive the equations of motion of that body?

  • $\begingroup$ Related: Two axes for rotational motion and links therein. $\endgroup$
    – Qmechanic
    Dec 29, 2020 at 9:59
  • $\begingroup$ I’ve seen the Euler rotation theorem and wondered whether or not it would be applicable in my example. I see that, if both rotational axes are perpendicular through the same point within the body, the theorem would be applicable. However, I can’t see if the theorem is (or how the theorem is) applicable when one axis falls within the body whilst the other axis sits outside the body perpendicularly to the spin axis. $\endgroup$ Dec 29, 2020 at 10:12

1 Answer 1


First, the arm's length distance is not material.

There's no need here to go into why that is so. Suffice it to say that for evaluation of gyroscopic effect you can translate all cases to motion where the center of mass remains at the same position. You can set up a gyroscope in a gimbal mounting at center of a lazy susan. Or you can hold your gyroscope in your hand and give it the same freedom to move as if gimbal mounted.

I wil use the aviation terms: Rolling, Pitching, Yawing.

Let the initial orientiation of the spinning be around the rolling axis.
As you have observed, when you introduce yawing motion the gyroscope wheel responds with pitching motion.
As you have observed, the pitching motion brings the gyroscope spin into alignment with the yaw axis.

The response is the most vigorous at the start, when the spin axis is perpendicular to the imposed rotation (here: the yawing).

What you can try is the following:
As initial condition, start with the spin axis already close to alignment with the yaw axis. Then the response will be sluggish.

The thing about acceleration is that the effect of acceleration is cumulative.

In the idealized frictionless case:
The amount of pitching push is the largest at the start of the pitching motion, due to the perpendicular angle.

As the pitching motion progresses: the further the wheel pitches the less oomph the pitching push has, but it will still accumulate.

So in the idealized frictionless case the pitching motion will overshoot.


In the real world there is plenty of friction. I rather expect there is so much friction that the pitching motion will not overshoot at all.

Let's say there is a pitching friction, and that the amount of friction is proportional to the pitching velocity.

Then you expect a gentle pitching motion. As the gyroscope closes in on aligned-with-yaw-axis the pitching velocity will decrease, but it won't quite stop because the slower the pitching velocity, the less friction.

So with friction like that you expect the pitching motion to ease towards alignment with the yaw axis.

Further reading: my 2012 discussion of the mechanics of gyroscopic precession


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.