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The derivation of the precession rate of a wheel hanging from a rope, as shown below usually involves taking the total torque acting on the system and equating it to the change in angular momentum. Gyroscopic precession

This gives $$\omega_{precession}= \frac{\tau}{L_{spin}}$$

as shown in this answer.

But this doesn't give an intuitive idea why the wheel rotates and why it rotates in a specific direction. After thinking about this, I got an intuitive explanation for why the wheel rotates and have answered it here

Is there a way to derive the precession rate from this argument? If yes, How?

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2 Answers 2

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If $\tau$ is high then the precession rate is high because the analogous pair of forces acting on the rim are high.

If $L_{spin}$ is low, then the speed of the rim particles going around in the circle are low. So for a given $\tau$, the velocities of the particles will be more amenable to change, thereby increasing the precession rate.

If you want to draw the full analogy then you need to understand $\omega_{precession}$ as a sort of "rate of steering" of the velocity vectors of the rim particles.

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About deriving precession rate from the model presented in the answer you wrote

I recommend starting with dividing the wheel in four quadrants. What happens in the top half is symmetric to what happens in the bottom half, and what happens in the leading half is symmetric to what happens in the trailing half. Concentrating on a quadrant reduces complexity, and once you have worked out one quadrant you have in effect worked out all four quadrants.

Let's say we do the quadrant that is part of the top half and part of the leading half.

As you moving along the length of that quadrant, starting at the top:
We are looking at the velocity component in the direction parallel to the horizontal. As you move along the rim from top point to the leading point: the velocity component in the horizontal direction becomes ever slower. The magnitude of that velocity component (as a function of position along the rim) can be expressed mathematically.

The magnitude of the torque as you move along the rim (from top point to leading point) is proportional to the height above the height of the hub. So you can express the magnitude of the torque as a function of position along the rim.

But it's not clear how to integrate the effects along the rim of the wheel to a resultant effect. Neither the magnitude nor the direction of the effect is uniform along the rim.


There's an additional problem. When a wheel is in precessing motion there are transient internal stresses the entire time.

Here is an argument to make the existence of those internal stresses plausible.
I assume you have seen the skill of swinging around a piece of pizza dough, and throwing into the air. You never see any precessing motion. The reason for that is: dough is flexible, so any internal bending forces immediately bend the dough, and the energy is dissipated.

In order to display sustained precessing motion a body must be quite rigid.


The only way to derive the rate of precession, it seems to me, is to find a way such that it isn't necessary to express the force at every point along the rim.

A way must be found such that the forces of internal stresses drop out of the calculation. Or better still: to do what existing derivations do: set up such that the forces of internal stresses do not enter the calculation in the first place.

Here I need to discuss the following: if it is assumed the wheel is rigid isn't it the case that the internal stresses drop out of the calculation automatically?

I think they don't drop out automatically. It may be that an internal stress shifts a force to another part of the wheel where it has either less or more leverage.


If you try the strategy I outlined above (writing a mathematical expressing for the force at every point along the rim) and it doesn't work out, then I think that's due to not taking internal stresses into account.

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  • $\begingroup$ "Here is an argument to make the existence of those internal stresses plausible. I assume you have seen the skill of swinging around a piece of pizza dough, and throwing into the air. You never see any precessing motion. The reason for that is: dough is flexible, so any internal bending forces immediately bend the dough, and the energy is dissipated." Cool example! Very illustrative, too. $\endgroup$
    – Evan
    Aug 11, 2021 at 19:02
  • $\begingroup$ Yeah Ok . Thanks for the detailed answer. $\endgroup$
    – Sophile
    Aug 12, 2021 at 9:26
  • $\begingroup$ @AbhinavPB Several of the ideas I pointed out in this answer are ideas that have used before. I presented a quadrant strategy in an answer I wrote in 2012 about gyroscopic precession. Rather than trying to attribute force the strategy uses just motion. The wheel is rigid, so I know the motion everywhere. That way the internal stresses do not enter the calculation. A calculation based on that quadrant strategy does not use the vector form of the angular momentum, nor the vector form of the torque. I ended up with the standard relation. $\endgroup$
    – Cleonis
    Aug 12, 2021 at 14:39

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