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If there where no friction at all, would a spinning wheel held up by one end of the axis spin precess forever without falling down?

Spinning wheel

I just asked another question about the same problem:

Direction of torque precession of a spinning wheel

Since it seems to be a good practice on stackexchange not to ask several questions in one post, I splitted them up into two questions. However if I am wrong, feel free to merge this questions.

Community
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martin
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  • If there were no friction at all and the wheel did fall down, where would the energy in the spinning wheel have gone? – Peter Shor May 05 '12 at 14:05
  • @PeterShor The wheel just continues to spin. – martin May 05 '12 at 14:50
  • So the spinning would speed up to compensate for the loss of gravitational potential energy in the wheel? I suppose that wouldn't violate conservation of energy, and angular momentum isn't locally conserved here anyway, so maybe you do need to use some actual physics to get the right answer. – Peter Shor May 05 '12 at 15:24
  • Actually this is a very good case for having two separate questions. – David Z May 05 '12 at 15:27
  • Related: http://physics.stackexchange.com/q/4844/2451 – Qmechanic May 05 '12 at 18:53
  • I think it must eventually fall down even if there is no friction. If it doesn't, what if I pull down the "free" part of the wheel axis with some extra force? As I am increase the extra force eventually I will apply enough force to push the wheel axis down. – akhmed Jul 19 '17 at 01:39

2 Answers2

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It is spinning forever. As you see, change of angular momentum

$$\frac{\text{d}\vec{L}}{\text{d}t} = \vec{\tau}$$

is always perpendicular to angular momentum itself, which means that angular momentum's direction is changed, while its magnitude is constant. Note the mathematical analogy with velocity and acceleration in case of circular rotation with constant velocity:

$$\frac{\text{d}\vec{v}}{\text{d}t} = \vec{a}_\text{cp}$$

Pygmalion
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  • Hm, this should be also the case for a very very small $\omega$, seems to be very unintuitive for me – martin May 05 '12 at 19:19
  • First step is understanding without thinking about $\omega$. If you want $\omega$ in the picture you should use Euler's equations, because $\vec{L} = I \vec{\omega}$ is valid only for fixed axis rotations (and of course along principal axis of moment of intertia too!). It can become extremely complicated to understand things when Euler's equations are considered. – Pygmalion May 05 '12 at 19:30
  • Not sure I understand the answer. What if I apply more downward force to the "free" part of the wheel axis (say, "increase gravity" and thus wheel weight)? Assume no friction. Would I not be able to push the wheel axis down no matter what extra force I apply? – akhmed Jul 19 '17 at 01:43
  • The question was if wheel keeps spinning, not if axis moves. – Pygmalion Jul 19 '17 at 18:45
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In the steady state, yes it stays forever in the same orbit in terms of the axis of rotation for the gyro.

But when the precession is not constant, things are different, and this is because when you set the gyro up in the first place, it starts at full force and no precession. It doesn't instantly gain a precession $\Omega$ or velocity. That is, it has to accelerate.

But where can it get the energy to do so? Correct me if I'm wrong but the precession acceleration gets it energy from the average height of the gyro. So initially the thing drops until the force is balanced between the double axis rotating gyro, and the gravitational force. Just think of the gyro as having a torque, so that it precesses based off a two-axis rotation.

Torque will accelerate its precession, and once its approaches steady state the gyro will have a torque that is balanced, so that the angular momentum is constant.

joseph h
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Jørgen
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  • Your answer could be improved with additional supporting information. Please [edit] to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers in the help center. – Community Oct 03 '23 at 05:41