When a top is spun, it will precess in some direction, either clockwise or counterclockwise. It's possible to find out which way using $\boldsymbol{\tau} = d\mathbf{L}/dt$ and $\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F}$, where $\mathbf{F}$ is the force of gravity.

However, I was never totally satisfied with this because I couldn't "see" exactly why the conclusion was true. Intuitively, I don't get, in my gut, why a downward force $\mathbf{F}$ can push the rotation axis to the left or right.

In principle, one can explain this by just applying non-rotational mechanics to different pieces of the gyroscope. I think this would really help me visualize what's going on. Is there such an explanation?

Note: I already know about the math of rotational mechanics, so please don't rewrite it. I am not interested in any answer that contains a $\boldsymbol{\tau}$, $\mathbf{L}$, $\boldsymbol{\omega}$, or $\times$ anywhere in it.

  • 1
    $\begingroup$ In the book "Feynman's Tips on Physics" by Feynman, Gottlied & Leighton there is a transcription of a lecture of Feynman's where he explains the action of precession without mathematics. The entire book - a great read. $\endgroup$
    – docscience
    Aug 25, 2015 at 10:47
  • $\begingroup$ Conservation of angular momentum, a very important physics concept, provides a great insight in the this behavior. Unfortunately, you don't want that. $\endgroup$
    – Bill N
    Aug 25, 2015 at 14:31
  • $\begingroup$ There is an answer from 2012 (submitted by me) that addresses this question. The 2012 question: What determines the direction of precession of a gyroscope? This 2012 discussion does not involve the concept of angular momentum. The dynamics of gyroscopic precession is made transparent by capitalizing on symmetry. This discussion also allows the reader to understand why at high rates of spin the expression with the vector cross product is a good approximation, whereas at slow rate of spin the vector cross product expression is a failure. $\endgroup$
    – Cleonis
    Jun 12, 2021 at 13:32

1 Answer 1


The reason why a gyroscope does behave in this strange way is that if you try to rotate it's axis in some direction, the "endpoints" of this axis have to be pushed perpendicular to what our first intuition would say.

In order to verify why the axis starts rotating in this strange way, let's make some simplifications: the gyroscope consists of two identical rotating particles and the particles rotate much fasten than we rotate the axis.

To visualize how the particles move, I've made the following drawing: the circle depicts a sphere on which the particles revolve, the center of the particles is always the center of the sphere and the axis of rotation precesses counterclockwise. The rotation axis is initially in up-down direction. Blue and red correspond to different particles, continuous and non-continuous lines correspond to the particle being on our side or the back side of the sphere. Blue starts at B and red at I.

On the figure, you can notice that because of the curvatures of the trajectories, the particles on our side must be pushed up while those of the back side must be pushed down (when the axis hasn't yet rotated much). This force has to be compensated - in order to keep the axis rotating counterclockwise, the axis has to be pushed away from the downside and pulled from the upside.

Movement of the particles

Going back to the gyroscope, if the axis is tilted in the beginning, gravitation "tries" to make it fall down, but instead this axis will rotate in the direction perpendicular to it. Given that information, you can probably figure out yourself in which direction the gyroscope will preceed. I think it should do this in the same direction as it rotates, if I visualized this process correctly with a pencil.

  • $\begingroup$ Aha! This is the answer I've been looking for for years. The last step took me a while since I needed to add in support forces (if there's nothing but those two masses then they'll just fall down, not precess) but then the right direction popped right out! $\endgroup$
    – knzhou
    Aug 26, 2015 at 0:38
  • $\begingroup$ (If anybody in the future is reading this answer, I think the easiest way to make the last jump is to imagine the two masses constrained to be on a ring, with the ring itself constrained to be on a second ring (like a gimbal), and that second ring attached to the gyroscope's support.) $\endgroup$
    – knzhou
    Aug 26, 2015 at 0:43

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