Will a precessing spinning whell fall down if there is no friction at all?

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?

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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 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 5 '12 at 14:05 @PeterShor The wheel just continues to spin. – martin May 5 '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 5 '12 at 15:24 Actually this is a very good case for having two separate questions. – David Zaslavsky♦ May 5 '12 at 15:27 – Qmechanic♦ May 5 '12 at 18:53

$$\frac{\text{d}\vec{L}}{\text{d}t} = \vec{\tau}$$
$$\frac{\text{d}\vec{v}}{\text{d}t} = \vec{a}_\text{cp}$$
 Hm, this should be also the case for a very very small $\omega$, seems to be very unintuitive for me – martin May 5 '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 5 '12 at 19:30