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In the The Feynman Lectures on Physics, Volume I, Chapter 20, Section 3,The Gyroscope found here, Feynman wrote

We note that when the wheel is precessing, the particles that are going around the wheel are not really moving in a plane because the wheel is precessing (see Fig. 20–4). As we explained previously (Fig. 19–4), the particles which are crossing through the precession axis are moving in curved paths, and this requires application of a lateral force. This is supplied by our pushing on the axle, which then communicates the force to the rim through the spokes.

Can someone please explain how the forces $F$ applied in Fig. 20-2 are lateral? I would have thought that the lateral force needed to be applied along the y-axis instead of the z-axis in order to supply the force needed to make the particle travel in the curve shown in Fig. 20-4 similar to Fig. 19-4. I can't figure out what this has to do with the Coriolis-effect which Fig. 19.4 is demonstrating.

-Thanks in advance!

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  • $\begingroup$ Figure 19.4 looks down on a turntable. So the rotational axis (positive y axis) points up, out of the page. lateral forces are applied in the x-z plane. Right? $\endgroup$ – docscience Feb 28 '20 at 18:33
  • $\begingroup$ Yep, I agree with that. Do you know how the other axes relate? $\endgroup$ – eball Feb 29 '20 at 21:21
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I can’t see the numbered figures, so let me walk you through a slightly different physical example: a helicopter rotor.

Imagine the rotor is turning counterclockwise views from above. A rotor blade goes from “ahead” to “left” to “behind” to “right”.

Now imagine you provide an impulse down on a blade tip at the front, up on a blade at the back. Normally, you’d expect the rotor to move down at the front, up at the back: tilt forward.

But note what’s really happening: the impulse is a force for a time, which is a velocity change. Only after a time will it become a position change.

So when the rotor is turning, the blade is moving right to left at the front. The impulse adds a small downward component to the large right-left speed: the blade tip starts moving on the diagonal. It doesnt move down right at the front. It starts moving down, but only gets down by the time the rotor blade has moved to the left: the turning rotor tilts to the left, not forward.

If you want a rotating rotor, gyroscope or other object to rotate in a particular way, you have to exert the torque 90 degrees before in the rotation.

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  • $\begingroup$ Thanks for the explanation. Can you explain what you mean when you say: "the impulse is a force for a time, which is a velocity change. Only after a time will it become a position change." -Thanks! $\endgroup$ – eball Feb 28 '20 at 18:04
  • $\begingroup$ That’s basic mechanics: F=ma means that F delta t = m delta v $\endgroup$ – Bob Jacobsen Feb 28 '20 at 18:06
  • $\begingroup$ Are you suggesting that for every rotation rate, no matter how small or how large, the timing always comes out in such a way that the effects of the exerted force occur 90 degrees later? For instance, let the helicopter rotor rotate at an anguler velocity of one revolution per minute. With such a low angular velocity, will the response still be 90 degrees delayed? So: we have that the explanation you offered fails very visibly at low rotation rates. Unfortunately, this explanation fails for every rotation rate, it's just that at high rotation rate the failure is harder to spot. $\endgroup$ – Cleonis Feb 28 '20 at 18:53
  • $\begingroup$ @cleonis Timing is not relevant: the motion is constrained to be circular rotation, so it’s 90 degrees of shift, not some number of seconds. Clearly the behavior at low rate is different from high rate; the answer mentions that. The exact answer is $L+\tau \delta t$, but the Question seems to need a conceptual answer, not a mathematical one. $\endgroup$ – Bob Jacobsen Feb 28 '20 at 20:13
  • $\begingroup$ @BobJacobsen The point is, this is so elementary that failure in one case implies complete failure. The 90-degrees-delayed explanation is like proposing a Universal law of gravity while acknowledging that it fails for the planet Mercury. But if it fails visibly for Mercury it is in fact failing for all planets, the failure just happens to be below detection level. The fact that the 90 degrees delayed idea fails for slow rotation rates carries the implication that it fails for all rotation rates. No wiggle room: any proposed mechanism is either correct for all cases, or totally wrong. $\endgroup$ – Cleonis Feb 28 '20 at 21:12
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image credit: Feynman Lectures on Physics book I chapter 20

Image source: Feynman Lectures on Physics book I, chapter 20

The reasoning that Feynman presents is very indirect.

First: the discussion that Feynman offers here is for the case of a fast spinning wheel. When a wheel is spinning fast the gyroscopic effects are significantly larger than the inertia of the non-spinning wheel. For the purpose of simplicity Feynman treats the inertia of the non-spinning wheel as negligable compared to the gyroscopic effects of the fast spinning motion.

So: if the wheel would not be spinning you would need a bit of torque around the x-axis to get the reorientation depicted in fig. 20-2, but as I said, Feynman treats that one as negligable.


Feynman points out the following: the spinning wheel is subjected to rotation around the x-axis. The result of that is a change of the direction of the angular momentum of the spinning wheel. In the picture the length of the vector for the angular momentum is not to scale. Think of the angular momentum vector as a very very long vector. The vector $\Delta L$ represent the change from previous angular momentum to new angular momentum. Think of that vector $\Delta L$ as a very very long vector. (This is why Feynman only considers the corresponding torque, everything else by comparison negligable.)

Feynman then reasons as follows: the torque that corresponds to the vector $\Delta L$ is a torque around the z-axis. So: as you were reorienting the wheel from initial angular momentum to new angular momentum you must have been exerting the forces $F$ and $-F$ which both lie in the xy-plane.

Notice that here Feynman doesn't attempt to explain. He is laying out: if you apply the formulas as presented earlier in the chapter this is the result you get.

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  • $\begingroup$ Thank you for the explanation. If both $L$ and $\Delta L$ are very very long vectors, how do they compare to one another? I would have thought $L$ >> $\Delta L$. Also, can you explain how Fig. 20-4 compares to Fig. 19.-4? I think I'm missing Feynman's point -Thanks! $\endgroup$ – eball Feb 29 '20 at 21:31
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    $\begingroup$ @eball Here is my recommendation: buy a quality gyroscope and start feeling what it does. Compare different rotation rates, from slow to as fast you are able to spin it up. There is no substitute for feeling with your own hands what it does. $\endgroup$ – Cleonis Feb 29 '20 at 22:31
  • $\begingroup$ @eball I exaggerated. Indeed, since this example is for $\Delta \Theta$ a small value you get that $\Delta L$ << $L$ I should have concentrated on emphasizing that with a very large rotation rate the torque associated with gyroscopic effect is so much larger than the torque associated with the plain, non-gyroscopic inertia of the gyro wheel, that the plain inertia is by comparison negligable. $\endgroup$ – Cleonis Feb 29 '20 at 22:38
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    $\begingroup$ @eball I find the narratives to both fig 19-4 and fig 20-4 quite convoluted; unnecessarily complicated. I don't think they're helpful. $\endgroup$ – Cleonis Feb 29 '20 at 22:43
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    $\begingroup$ @eball The stackexchange comment sections are by design not intended for conversation, and I endorse that design. For questions that require follow-up a threaded forum is a more suitable format. My preferred explanation of gyroscopic precession is in an answer I wrote in 2012, to a question titled, What determines the direction of precession of a gyroscope. (I link only now because I want to avoid giving the impression of vanity linking.) (And yeah, that looong series of comments to Bob Jacobson is very much against my standard policy.) $\endgroup$ – Cleonis Mar 5 '20 at 20:14

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