From the following equation from hyperphysics, as rotational speed $\omega$ decreases, precession speed $\omega_p$ increases. However, from what we observe physically, as rotational speed of the top decreases, not only does precession speed increase, we also see that $\phi$ increases. I believe this discrepancy is due to the equation assuming zero friction slowing down the spin.

How would one modify / derive the equation to take into account friction slowing the spin leading to an increase in $\phi$?

Precessing Top


1 Answer 1


As you point out:
In the case of a spinning top that is in the process of decaying three things are happening hand in hand.

  • The spinning rate decays
  • The center of mass of the spinning top loses height
  • The precessing speed increases.

As you suspect:
the equations presented on the Hyperphysics website are approximations, they are not exhaustive. I'll get back to that further on in this answer.

A larger precessing speed has a corresponding larger rotational kinetic energy. The height loss of the center of mass of the spinning top is the source of that energy. That is: gravitational potential energy gets converted to kinetic energy.

This conversion process is discussed in the Feynman Lectures, Book I, chapter 20, Rotation in space. In section 20-3 Feynman discusses the onset of the precessing motion, right after the gyroscope has been released. The onset of precessing motion is summerized as: "So you see it has to go down a little, in order to go around."

(Feynman's discussion uses angular momentum considerations. The angular velocity of the precessing motion has a corresponding angular momentum vector. If the precessing speed changes then the corresponding change of the angular momentum must be acounted for.)

This discussion by Feynman has been experimentally verified with a tabletop experiment. This was done by Svilen Kostov and Daniel Hammer. The article which they describe their corroboration of Feynman's discussion is named after that key assertion It has to go down a little, in order to go around.

The actual mechanism of the conversion of pitching motion to precessing motion is discussed by me in an 2012 answer here on physics.stackexchange, to a question titled "What determines the direction of precession of a gyroscope?"

To find the rate at which the spinning top's center of mass loses height:
It should be possible to use either energy considerations or angular momentum considerations. I expect the energy approach to be the shortest.

You would need to find an expression for the rotational kinetic energy of the precessing motion. To reach the required precessing speed the center of mass needs to drop a corresponding height.

An exhaustive derivation of gyroscopic precession uses Euler angles throughout.

It may well be that nobody has ever worked out that problem. In engineering gyroscopes are always equipped with some driving mechanism (usually an electric motor) to sustain a constant spin rate.

  • $\begingroup$ Phenomenal! Thx for the huge amount of detail and reference! $\endgroup$
    – Rufus
    Commented Jun 10, 2019 at 4:59
  • $\begingroup$ @Rufus About providing those references: to my knowledge they are unique. To my knowledge all other discussions of gyroscopic precesssion give only the approximation, but presenting that as exhaustive discussion. So there is mismatch between what you observe and what the discussion says. The question "How does gyroscopic precession work? is a regularly recurring question here on physics.stackexchange, presumably due to the mismatch I mentioned. Out of curiosity: If I wouldn't have provided any of the references, would you have been skeptical? $\endgroup$
    – Cleonis
    Commented Jun 10, 2019 at 5:31
  • $\begingroup$ I was looking for the equation that allows me to find the rate at which it "dipped down". Your line "The height loss of the center of mass of the spinning top is the source of that energy." was something that I hadn't thought of. The link "It has to go down a little in order to go around" further gave me the equation I needed and also helped solidify the concept. $\endgroup$
    – Rufus
    Commented Jun 10, 2019 at 6:01

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