pervect said:
In a more realistic case, w2 and w3 might have small values, and the motion will be more complicated, but it will depend on the exact inital conditions, there is nothing in the physics I can see to give it a preferred direction of precession.
I agree, it is not clear from Euler's equations that there should be a preferred precession direction. In the two books I've looked at (Landau's and Goldstein's), they draw the picture so that the total angular momentum makes an acute angle with the spin angular velocity. But they do not say that it must be this way, so I suspect that is just a drawing convention. It makes it easier to visualize the various projections. But this acute angle is in fact what I observe every time I toss the spinning plate. I don't know how the initial conditions could be made different so that it would precess the other way. I toss it slightly different each time, but it does the same thing.
This is the traditional analysis, but there's an easier (IMO) way to cary out the problem. This is to simply note that L (the total angular momentum) must remain constant.
Yes. That is what I was using when I added up the torques of all the mass elements and said that the total must be zero. Actually, I did the calculation for a thin ring. That is sufficient since the disk is made up of concentric rings. This calculation gives the correct precession rate, but apparently the wrong direction. Here is how I am going about it:
The total angular velocity is \vec{\Omega}_{tot} = \vec{\Omega}_{pr} + \vec{\Omega}_{sp}. The velocity of a particular mass element is
\dot{\vec{r}} = \vec{\Omega}_{tot} \times \vec{r}.
The acceleration is (omitting gravity)
\ddot{\vec{r}} = \dot{\vec{\Omega}}_{tot} \times \vec{r} + \vec{\Omega}_{tot} \times \dot{\vec{r}}.
Using \dot{\vec{\Omega}}_{pr} = 0 and \dot{\vec{\Omega}}_{sp} = \vec{\Omega}_{pr} \times \vec{\Omega}_{sp} and substituting for \dot{\vec{r}},
\ddot{\vec{r}} = (\vec{\Omega}_{pr} \times \vec{\Omega}_{sp}) \times \vec{r} + \vec{\Omega}_{tot} \times (\vec{\Omega}_{tot} \times \vec{r}).
Apply BAC-CAB a couple of times and note that \vec{\Omega}_{sp} \cdot \vec{r} = 0 to get
\ddot{\vec{r}} \ = \ (\vec{\Omega}_{pr} \cdot \vec{r})(\vec{\Omega}_{pr} + 2\vec{\Omega}_{sp}) - \Omega_{tot}^2 \vec{r}.
then form the cross product of this with \vec{r} dm to get the torque on the mass element
d\vec{N} = (\vec{\Omega}_{pr} \cdot \vec{r})\vec{r} \times (\vec{\Omega}_{pr} + 2\vec{\Omega}_{sp}) dm, and integrate around the ring. To do the integral, I took the ring to lie in the xy-plane (principal axes) and precession angular velocity to make an angle theta with the z-body axis, so that it has a component \Omega_{pr} \sin \theta along the positive x-axis. In other words, I chose the x- and y-axes this way, since you are free to do that for a symmetrical top. It turned out that in order for the integral to vanish, \vec{\Omega}_{pr} + 2\vec{\Omega}_{sp} must be proportional to \hat{e}_x, which means that the z-component of the precession is {\Omega}_{pr} \cos \theta = -2\Omega_{sp}. So it looks like the precession should be in the direction opposite what I observe.
edit: actually, I'm pretty sure my choice of axes to perform the integral has nothing to do with principal axes. They were just chosen to make the integration convenient, and happen to coincide with the principal axes. And by the way, my calculations of the velocity and acceleration are in the non-rotating "lab" frame. So the components of the angular velocity differ from those used in Euler's equations.
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