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When you spin an asymmetrically weighted, 2D disk-shaped top, the heavy part actually rises to the top. Why is this?

  1. http://www.youtube.com/watch?v=h0SZZTBQmEs

  2. http://www.youtube.com/watch?v=tDr26U49_VA

In the second video it shows that this is because of friction but trying this on a frictionless surface won't do much: The disk will go up and down and still the heavy part will be on the top.

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Here is an article on the tippe top physics.umd.edu/lecdem/services/refs_scanned_WIP/… The same arguments will hold for the disk too. –  Leonida Jun 16 '13 at 13:11
That didn't help –  neomhmd Jun 17 '13 at 16:51
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up vote 4 down vote accepted

I think the solution has more to do with the tennis racket effect (see: http://physics.stackexchange.com/a/17507/392).

Let me clarify the disk with hole in it has two stable axes of rotation and one unstable one. The unstable one is through the hole and the stable one is across (below in green) and normal to the disk.

Axes Def

I have confirmed that without friction (and from the videos in the link above) when the disk is spun on the unstable axis, it will perdiodically flip. This is what caused it to flip when the disk was falling without friction.

The additional nuiance here is the once it is on the "upsidedown" oriention and friction is present then the unstable axis becomes stable.

If the hole spans from the center to the edge of the disk, then the center of gravity is at $$ \vec{c} = (0,-\frac{R}{6},0)$$ where $R$ is the outside radius of the disk. The principal moments of inertia about the center of mass are $$ \begin{aligned} I_{XX} & = m \left( \frac{\ell^2}{12} + \frac{29 R^2}{144} \right) \approx 0.2 m R^2 \\ I_{YY} & = m \left( \frac{\ell^2}{12} + \frac{5 R^2}{16} \right) \approx 0.31 m R^2 \\ I_{ZZ} & = m \left( \frac{37 R^2}{72} \right) \approx 0.51 m R^2 \end{aligned} $$

where $\ell$ is the thickness of the disk. Since $I_{YY}-I_{XX} = \frac{8}{37} I_{ZZ}$ this means that the y direction is the medium inertia value, x the minimum and z the maximum. Hence the instability about the y axis according to the Tennis Racket Effect.

I am working to qualify the above statement and I am going to update this post with my findings.

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may you please explain why the axis through the hole is unstable, I would say it should be the one on the x-axis because it is not symmetrical. –  Mhmd Dec 31 '13 at 12:43
@user689 Any rigid body has three principal moments of inertia of different values. Rotation about the smallest and largest direction is stable, whilst rotation about the medium value is unstable. This is the tennis racket effect. –  ja72 Dec 31 '13 at 16:17
I have added a little more about the mass moments of inertia on the post. –  ja72 Dec 31 '13 at 16:58
The tennis racket effect/theorem can be accurately understood by the Poinsot-Binet Energy Ellipsoid. More reference is in Goldstein –  Torsten Hĕrculĕ Cärlemän Dec 31 '13 at 17:24
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