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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.

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  • $\begingroup$ If there were no friction at all and the wheel did fall down, where would the energy in the spinning wheel have gone? $\endgroup$ – Peter Shor May 5 '12 at 14:05
  • $\begingroup$ @PeterShor The wheel just continues to spin. $\endgroup$ – martin May 5 '12 at 14:50
  • $\begingroup$ 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. $\endgroup$ – Peter Shor May 5 '12 at 15:24
  • $\begingroup$ Actually this is a very good case for having two separate questions. $\endgroup$ – David Z May 5 '12 at 15:27
  • $\begingroup$ Related: physics.stackexchange.com/q/4844/2451 $\endgroup$ – Qmechanic May 5 '12 at 18:53
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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:

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

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  • $\begingroup$ Hm, this should be also the case for a very very small $\omega$, seems to be very unintuitive for me $\endgroup$ – martin May 5 '12 at 19:19
  • $\begingroup$ 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. $\endgroup$ – Pygmalion May 5 '12 at 19:30
  • $\begingroup$ 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? $\endgroup$ – akhmed Jul 19 '17 at 1:43
  • $\begingroup$ The question was if wheel keeps spinning, not if axis moves. $\endgroup$ – Pygmalion Jul 19 '17 at 18:45

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