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My textbook gives the following derivation of the angular momentum of gyroscopic precession:

The magnitude of the torque is $$\tau = r M g\sin\theta.$$ Thus $$dL = r M g\sin\theta\, dt.$$ The angle the top precesses through in time $dt$ is $$d\phi=\frac{dL}{L\sin\theta} =\frac{rMg\sin\theta}{L\sin\theta}dt=\frac{rMg}{L}dt.$$ The precession angular velocity is $\omega_P = \frac{d\phi}{dt}$ and from this equation we see that $$\omega_P = \frac{rMg}{L}= \frac{rMg}{I \omega}.$$

(source)

The derivation on Wikipedia is similar.

I am confused by the use of differentials here. I can see that in the first two lines, the authors have used the fact that $\tau = \frac{dL}{dt}$. But why, in the third line, is $d\phi = \frac{dL}{L\sin\theta}$?

There are many questions on this site seeking a conceptual understanding of precession. I think I understand the concept (how the angular momentum vector changes under the influence of gravity), but I am confused by this derivation.

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2 Answers 2

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The length of an arc on a circumference $\delta \ell$ is related to the circle's radius $r$ and angle subtended $\delta\theta$ by: $$ \delta \theta = \frac{\delta\ell}{r} \qquad \Rightarrow \qquad \mathrm{d}\theta = \frac{\mathrm{d}\ell}{r},$$ where on the right I took the limit of the $\delta$ quantities becoming infinitesimally small.

The circumference here is the precession orbit.
In this case, the small angle subtended is $\mathrm{d}\theta$ (usually called $\mathrm{d}\Omega$), the arc length is the small increase in angular momentum causes by the torque $\mathrm{d} L$ $(= \Gamma \mathrm{d}t)$, and the radius of the orbit is the (projected) angular momentum associated with the gyroscope spinning $L\sin\theta = I\omega\sin\theta$.

Visual aid:

enter image description here

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  • $\begingroup$ "the radius of the orbit is the (projected) angular momentum associated with the gyroscope spinning" What does this mean? Can you show, in equations, why $$\frac{dl}{r} = \frac{dL}{L\sin\theta}$$? I understand the projection and the geometry, but not the relationship between change in momentum and $l$ used here. $\endgroup$
    – Max
    May 24, 2020 at 5:14
  • $\begingroup$ I added a picture to make it clearer. $\endgroup$
    – SuperCiocia
    May 24, 2020 at 5:24
  • $\begingroup$ For one thing, you haven't distinguished between $\phi$ (the angle of precessional rotation) and $\theta$ (the angle between the center of mass's position vector and the axis of rotation), although this may be my fault because I made a similar typo in an earlier version of my question. Also, your drawing shows that $d\phi$ is proportional to $dL / L\sin\theta$, but I looking for a chain of equations establishing that $$\frac{dl}{r} = \frac{dL}{L\sin\phi}$$. Also, you have written $dL = \Gamma dt$ but I believe $\Gamma$ should be $\tau$ unless this is an alternative symbol for torque. $\endgroup$
    – Max
    May 24, 2020 at 5:36
  • $\begingroup$ $\Gamma$ should be $\tau$, and the $\theta$ confusion stems for your typo yes. The equations are the ones I write at the beginning of my answer. The arc length is exactly equal to radius times angle subtended. In L space, the arc length of the precession orbit is dL, the angle subtended d$\phi$, and the radius $L\sin\theta$. $\endgroup$
    – SuperCiocia
    May 24, 2020 at 6:01
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I recommend that you first do a derivation for a more symmetrical case.

In the source you linked to the derivation takes on a more general case right away. Instead you can first consider the case where the spin axis of the gyro wheel is at right angles to the vertical.

Gyroscope with spin axis at right angles to vertical

Image source: wikimedia

Once you have a derivation for that most symmetrical case you can proceed to generalize.


Incidentally, note that the description of gyroscopic precession in the source that you linked to is (like most sources) not complete.

"The top precesses around a vertical axis, since the torque is always horizontal and perpendicular to L. If the top is not spinning, it acquires angular momentum in the direction of the torque, and it rotates around a horizontal axis, falling over just as we would expect."

What is not addressed is what happens at intermediate rates of spinning. When the gyroscope is spinning fast enough then what we see with our eyes is that the gyroscope goes in precessing motion instead of falling over; when the gyroscope is spinning too slow it will simply fall over.

The description seems to suggest a hard either-or outcome. Either the gyroscope will start precessing, or it wiil fall over. But we know that in the real world it is a continuum; in the range from non-spinning to spinning fast there is nowhere a hard cut-off.

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  • $\begingroup$ In the symmetrical case, $\theta = 90^{\circ}$ and the expression in question simplifies to $d\phi = \frac{dL}{L}$. I can see, from @SuperCiocia's answer, why this is true conceptually, but I would like to see a chain of equalities showing that $$d\phi = \frac{ds}{r} = \,\frac{~\cdots~}{\cdots} \,= \frac{dL}{L\sin\theta}$$. I don't require *absolute* mathematical rigor (dropping a second-order term or whatever is fine), but to me the picture relating $dL$ and $L$ unconvincing. (1/2) $\endgroup$
    – Max
    May 24, 2020 at 12:11
  • $\begingroup$ For one thing, how do we know that $d\vec L$ is orthogonal to $\vec L$? $\vec L = \vec r \times m \vec v_t$, so it looks like we're assuming that the torque due to gravity alters only the orientation of $\vec r$ and $\vec v_t$, and that it changes both orientations at the same rate (maintaining their orthogonality) without changing their magnitudes. But clearly, the length of $\vec r$ does change when the gyroscope falls over. So is there a way to state quantitatively our assumption that the gyroscope won't fall over, yielding the chain of equalities I've described above? (2/2) $\endgroup$
    – Max
    May 24, 2020 at 12:11
  • $\begingroup$ @Max I have to say: I recommend against trying to learn from that openstax source. I suspect the derivation is fudged. Other than that I noticed two unrelated errors. How bicycles stay upright has been the subject of quite a bit of study. The general findings: it's more due to properly designed steering geometry; gyroscopic effect plays a minor role, if any. The description of how nutation comes about is nonsensical. $\endgroup$
    – Cleonis
    May 24, 2020 at 17:14
  • $\begingroup$ @Max I really do not understand your problem. The equations you show in the first comment in this thread do not pertain to a "conceptual" and "qualitative" model. It is exact. It is the mathematical definition of a (2D) angle. With $r$ and $ds$ as the radius and the arc-length, you are in physical space (like, real life position). But you can define an angular-momentum space where the same equation holds, but now the radius and arc lengths have different physical meanings, namely the angular momentum causing that specific orbital motion, and the differential increase in angular momentum. $\endgroup$
    – SuperCiocia
    May 27, 2020 at 22:55
  • $\begingroup$ The picture shows that $dL$ is proportional to $ds$, i.e. $$dL = \alpha ds$$ and that $L\sin\theta$ is proportional to $r$, i.e. $$L\sin\theta = \beta r$$. I would like to see a mathematical expression showing that both proportions are the same, i.e. that $\alpha = \beta$, so that they cancel out when taking $ds / r$. The constants $\alpha$ and $\beta$ would be the transformations defining your "angular-momentum space." The answers so far have simply assumed that $\alpha = \beta$ in this transformation. $\endgroup$
    – Max
    May 27, 2020 at 23:56

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