I'm a psychologist and I'm interested in perception of gyroscopic forces. For that reason, I'm having to think about many of the classic issues relating to gyroscopes in reverse. That is, instead of predicting what imposed forces will do to a gyroscope assembly, I must think in terms what forces will be imposed by the assembly on a perceiver, in order to make predictions about how it will be perceived.


Imagine a gyroscope mounted on a rod, such that the rod could be wielded by someone. Held perfectly level, there would be no precession forces because the person's hand is applying a torque that counteracts gravity. Thus, the direction of angular momentum is not changed.

Wielded, rod-mounted gyroscope.

Now imagine the wielder rotates his wrist 90 degrees in a forehand direction. This applies a torque about the axis of the wrist, sweeping the rod away from our view as shown in the figure. This reorients the gyroscope by 90 degrees as well, and the angular momentum causes the end of the rod to "lift" somewhat during this rotation, because it is in the direction of precession. That is, less torque is required by the wielder to counteract gravity during this rotation when the gyroscope is spinning, than if it were not spinning.


How do I characterize this apparent "lifting" force mathematically? (Is "force" even the right way to describe it?) That is, if I have a flywheel of a known mass and radius, spinning at a known rate, how do I calculate the force that it will apply to the palm of the person wielding it on a rod of known length, through a 90 degree rotation?

  • $\begingroup$ Have a look at this video and the following one. m.youtube.com/watch?v=GeyDf4ooPdo $\endgroup$
    – Farcher
    Commented Feb 2, 2018 at 14:52
  • $\begingroup$ Thanks for posting these -- they're a great illustration of the issue. They don't contain the information I'm after (which is how I should treat the angular momentum vector change, in terms of its imposed forces), but hopefully other readers or commenters find it helpful. $\endgroup$
    – PhiloT
    Commented Feb 2, 2018 at 15:04
  • $\begingroup$ The only force that you need to apply is equal to the weight of the gyroscope as is explained in the videos. $\endgroup$
    – Farcher
    Commented Feb 2, 2018 at 15:15
  • $\begingroup$ That is my understanding too, however in applying the force, the resulting torque will be redirected by the angular momentum vector. I'm essentially wondering how I can calculate what the force of the angular momentum vector will be at the hand during the application of force. Can I just treat the angular momentum (pseudo)vector like the same as a torque vector? And if so, since the magnitude doesn't change, in which units do I measure this reorientation? $\endgroup$
    – PhiloT
    Commented Feb 2, 2018 at 15:34
  • $\begingroup$ The following discussion (here on stackexchange) of gyroscopic precesssion was contributed by me. (It's an abbreviated version of the discussion on my own website.) When you change the orientation of a (fast) spinning wheel, the wheel responds to that motion in a particular way. The discussion I'm linking to explains why it responds in the way it does. That will allow you to work out the answer to the question you raised. No need for vector calculus, in this case. $\endgroup$
    – Cleonis
    Commented Feb 16, 2018 at 23:58

2 Answers 2


No. Torque is the time rate of change of angular momentum $$\vec{\tau} = \frac{\operatorname{d}\vec{L}}{\operatorname{d}t},$$ just like force is the time rate of change of linear momentum $$\vec{F} = \frac{\operatorname{d}\vec{p}}{\operatorname{d}t}.$$ Thus, when $\vec{L}$ is large it takes a large torque to make a significant difference in it.

Note that the angular momentum does not in any way redirect the force. $\vec{F}=m\vec{a}$ still applies. What's happening is there's gravity pulling down, and the hand pushing up. Therefore there is no net force. Because they act at different locations, though, there is a net torque that acts perpendicular to the angular momentum. Just like when $\vec{F}$ is always perpendicular to $\vec{p}$ you get circular motion, this produces circular precession.

  • $\begingroup$ I think my point of difficulty is in what it means to change the angular momentum. Specifically, the difference between changing the direction versus the magnitude. My understanding is that it takes the same amount of torque to reorient the axis of a spinning gyroscope regardless of its angular momentum, but that the direction in which the torque must be applied changes. So, if it takes 5 nm to rotate a the axis of a gyroscope while it's off, it takes 5 nm while it's on too, but in a different direction. Is that correct? If so, how do I use L to calculate the difference in direction? $\endgroup$
    – PhiloT
    Commented Feb 2, 2018 at 16:24
  • $\begingroup$ "My understanding is that it takes the same amount of torque to reorient the axis of a spinning gyroscope regardless of its angular momentum" This is not true. It takes the same amount of "angular impulse", for lack of a better term for time integrated torque. Any torque kept perpendicular to $\vec{L}$ can cause the axis of rotation to rotate, it's just that the magnitude of the torque controls the rate of rotation (given constant, perpendicular, $\tau$ and $L$ then $\Omega_p = \tau/L$). $\endgroup$ Commented Feb 2, 2018 at 17:09
  • $\begingroup$ This is very helpful. Thank you. What is omega in that last equation? Is that the "time integrated torque" you mentioned? $\endgroup$
    – PhiloT
    Commented Feb 2, 2018 at 17:23
  • $\begingroup$ "same amount of "angular impulse"" addendum: same amount of angular impulse to reorient a given angular momentum. More angular momentum requires a greater angular impulse to reorient. $\endgroup$ Commented Feb 2, 2018 at 17:24

How do I characterize this apparent "lifting" force mathematically? (Is "force" even the right way to describe it?)

Let me address this specific subquestion.
This is about the distinction we need to make between 'force' and 'inertia'.

First a very simple case, to establish the basics.
You have some wooden workpiece on a bench, and you want to hammer a wooden peg into a hole, the fit is tight.

If you just tap the peg the wood-wood friction is strong enough to bring your mallet to a dead stop. The tap does exert a force on the peg, but a tap is not enough.

As we know, we say that the mallet exerts a force when it strikes the peg because the mallet has inertia; a force is required to bring the moving mallet to a stop. The peg exerts that force upon the mallet. The faster you swing the mallet, the stronger the force that is required to bring it to a dead stop. At some point the wood-wood friction between the peg and the wall of the hole is not enough to stop the mallet completely, and you get to hammer the peg in.

We call it a 'force' when there is force pair. When object A exerts a force upon object B then object B exerts an equal and opposite force upon A.

Inertia itself cannot be categorized as a force because there is no 'equal and opposite pair'. The definition of inertia is that a force is required to change the velocity of an object.

A complete picture features both inertia and a force pair. Of course, in everyday language we just say that the mallet strikes are forcing the peg into the hole

Spinning gyroscope wheel
Ultimately, the way a spinning gyroscope wheel responds to motion that is imposed upon it arises from inertia. (As to how that happens: I added a link to a discussion of gyroscopic precession (here on physics.stackexchange) in a comment to your question.)
With you physically holding the axle of the gyroscope wheel the wheel is indeed exerting a force upon you everytime you are in the process of changing the orientation of the spinning wheel.

Of course, there is that surprising thing that the direction of the spinning wheel response is not opposite to the direction of the force you imposed upon it. So yeah, you question: 'is "force" even the right way to describe it?' is well founded.

The difference in direction goes into internal stresses of the gyroscope wheel. A non-rigid wheel that can flex from side to side will not have the same behavior as a rigid gyroscope wheel; try to reorient it and it will deform violently.


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.