# The secret behind the spinning, asymmetrically weighted, 2D disk-shaped top?

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

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. –  nijankowski Jun 16 '13 at 13:11
That didn't help –  Mark Jun 17 '13 at 16:51
–  Qmechanic Apr 26 at 19:57

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.

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

I) Here we will give a qualitative rather than quantitative analysis. Consider first the axially symmetric 3D tippe top, which means that there will be two principal axis with maximal moment of inertia. (If we wish, we can for simplicity model the top as a spherically symmetric ball, with a heavy point mass off the geometric center $O$.) Empirically, we observe to a first approximation that:

1. The angular vector $\vec{\omega}$ is mostly vertical and completes a small precession for each revolution of the top. (Let's assume that $\vec{\omega}$ is upwards rather than downwards.)

2. The rotation of the top is around a point $Q$ somewhere between the geometric center $O$ and the center of mass $C$.

3. The point $P$ of contact completes a small circular orbit for each revolution of the top.

4. The top is mostly rolling without sliding. In other words, to a first approximation, it is static friction rather than kinetic friction, and hence the mechanical energy is conserved.

The angular momentum vector $\vec{L}$ will mostly point upwards but tilted a bit to the same (opposite) side as $C$ iff $O$ is above (below) $C$. In other words, the angular momentum vector $\vec{L}$ completes a small precession for each revolution of the top. We will assume that this precession of $\vec{L}$ (and the corresponding net torque needed for this precession) remains very small during the entire inversion process.

Now the torques of gravity and the normal force work in unison to increase the precession of $\vec{L}$. The cancellation of this torque can only come from the friction force acting in the same direction (horizontally speaking) as where the center of mass $C$ is.$^1$

Naturally, due to an imperfect rolling motion, the friction force will then tend to slide the point of contact in the direction of the friction force, thereby raising the center of mass $C$. (Del Campo showed that sliding friction must play an essential role in the inversion process, cf. Ref. 1. Of course sliding friction is also responsible for that the top eventually comes to rest and looses its mechanical energy.)

II) The asymmetric 2D discs$^2$ works similarly, although now the rotation will mainly be around the unstable principal axis with intermediate moment of inertia, leading to the tennis racket/Dzhanibekov effect, as correctly pointed out by ja72 in his answer, see also e.g. this Phys.SE post and links therein. In situations without friction, the tennis racket/Dzhanibekov effect explains the last unresolved part of the solution video.

References:

1. Richard J. Cohen, The tippe top revisited, http://dx.doi.org/10.1119/1.10926 (hat tip: Leonida)

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$^1$ Note that the friction torque argument in this Basic Saucer Physics 101 video seems to lead to the opposite effect where the center of mass $C$ gets lowered (rather than raised).

$^2$ The geometric details of the asymmetric 2D discs is to a large extent irrelevant for the effect.

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Correction to the answer (v1): Del Campo's argument only guarantee that friction is involved, not that it is necessarily sliding friction. –  Qmechanic Apr 28 at 8:21
Also the word is in the last sentence of the answer should be are. –  Qmechanic Apr 30 at 16:19