If you take a hardboiled egg and put it on a table and start to spin it, if you spin it fast enough it will start to spin in an upright position. What is the angular velocity needed for this transition to occur?

Is energy conserved in the transition? Must there be some imperfection in the egg for the transition to occur?

Assume a prolate spheroid egg of constant density. If that doesnt work, assume its center of mass is shifted abit, if that dont work, assume an egg.

Edit: One is allowed to use simpler shapes of the same symmetry type, to avoid messy integrals.

And most importantly, can you explain without equations, why does the egg prefer to spin in an upright position ?

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    $\begingroup$ How do you assume an egg? ;-) $\endgroup$ – Sklivvz Dec 23 '10 at 16:06
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    $\begingroup$ sounds like the tippy-top problem, which can be solved as a function of surface friction, so I assume that is needed here too. But the problem is just too complicated for me to work on... $\endgroup$ – Jeremy Dec 23 '10 at 19:29
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    $\begingroup$ I've just tried and failed to observe this phenomenon with a raw egg. I would hard boil it, but unfortunately it's the expensive kind of egg and it belongs to my roommate, who's out of town until after Christmas. I like the question - hopefully someone can answer it in an accessible way. $\endgroup$ – Mark Eichenlaub Dec 24 '10 at 4:15
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    $\begingroup$ here's a decent video: youtube.com/watch?v=Q_H2kWGanDA $\endgroup$ – Mark Eichenlaub Dec 24 '10 at 5:12
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    $\begingroup$ @Sklivvz --assume a spherical egg (a variant on the spherical cow joke) $\endgroup$ – Gordon Feb 11 '11 at 6:37

In my oppinion this problem is equivalent to the Tippe_top, with the difference of the egg not turning around 180 degrees but 90 degrees. The underlying physics should be the same, since the center of mass and the geomtric center are not being the same. The egg should be a little bit easier to calculate, since its geometric shape is not that horrible. (However its still way over my head). As you already pointed out, this behaviour is not possible on a frictionless table. For some readings see:




The last two links are quite an extensive discussion about the involved physics. A more basic introduction to this topic is


Unfortunatelly, this website is only available in German. However you might take the above links as as starting point for furhter readings.


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  • $\begingroup$ Hi @ftiaronsem, the last link is dead. Can you fix it? $\endgroup$ – Urb Nov 24 at 22:45

I think one suited approach to your problem is to look at the stability of the movement with respect to the energy.

Energy stability analysis

The rotation energy is given by $$E_{rot} = \frac{1}{2}\Theta{(\mathbf{\omega},\mathbf{\omega})}$$

where $\Theta$ is the (tensorial) moment of inertia with respect to the (vectorial) rotation $\mathbf{\omega}$.

The mass of a prolate, homogeneous ellipsoid is given by $M = \frac{4}{3}\rho r_a^2r_b$ where $\rho$ is the mass density and $r_b$ is assumed to be the axis longer (or, say different) than $r_b$.

Then, we can write down the principal moments of inertia for this ellipsoid as $$\Theta_b = \frac{M}{5}\cdot 2 r_a^2$$ and $$\Theta_a = \frac{M}{5}\cdot (r_a^2 + r_b^2)$$

If we now chose $\theta$ to be the angle between $\mathbf{\omega}$ and the $z$-axis, we find the rotation energy as $$E_{rot} = \frac{M}{10}\left( 2r_a^2 \cos^2\theta \omega^2 + (r_a^2 + r_b^2) \sin^2\theta \omega^2 \right)$$

with $\omega$ now being the scalar value of rotation (I could not make it bold).

Now, we have to turn to the potential energy in the gravitational field. I suppose, this should take the form $$E_{pot} = Mg\left( r_b \cos\theta + r_a \sin\theta \right)$$

If you now analyze the energy $$E = E_{rot}+ E_{pot}$$

with respect to $\theta$ and the parameters $r_a$ and $r_b$, you will find out that $\partial_\theta E$ will vanish at $\theta = 0$ but it will be a maximum (remember, $\theta\geq 0$) assuming $r_b > r_a$.

That means, that the movement will be unstable assuming a rotating prolate body. Or, in other words, you cannot find some $\omega_s$ to have the egg at resting rotation in this model.

Other approaches

One further ansatz to calculate a stable rotating egg, one could assume that the rotating egg is not homogeneous. A simple model could be to assume that a small ring of mass is attached to the "smaller belt" leading to $$\Theta_b = \frac{M}{5}\cdot 2 r_a^2 + \frac{M_{ring}}{2}r_a^2\quad, \qquad \Theta_a = \frac{M}{5}\cdot (r_a^2 + r_b^2) + \frac{M_{ring}}{2\cdot 2\pi r_a}r_a^2$$

Now, can you show, when $E$ has a minimum in $\theta = 0$ leading to a stability condition of the form $F(M_{ring}, r_a/r_b; \omega_s) = 0$?



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  • $\begingroup$ @Robert Experimentally, it turns out that the egg is stable when horizontal for low rotation rates, and only becomes unstable and stands up when the rotation rate is above some cutoff value. $\endgroup$ – Mark Eichenlaub Dec 26 '10 at 8:27
  • $\begingroup$ @Mark: Thank you for the information. If I interpret it correctly, this underlines the assumption that the egg cannot be seen as homogeneous. I think my added-ring-of-mass model might be simple but should explain the situation qualitatively. Greets $\endgroup$ – Robert Filter Dec 26 '10 at 15:57
  • $\begingroup$ @Robert Filter: I think the issue here is that the prolate ellipsoid doesn't have a "vertical" asymmetry. I think this is what makes the egg stand up (when the egg is horizontal, the center of mass is not in the middle of the egg. When we spin it the different lengths of the sides will make the egg oscillate unevenly. $\endgroup$ – Sklivvz Dec 26 '10 at 23:29
  • $\begingroup$ @Sklivvz: Thank you for your thoughts. I think even in this case, the moments of inertia along the axis of symmetry will be smaller making the rotation still unstable for the case of a homogeneous egg (Remember, $\Theta = \int r^2 dm$). I think I pretty much answered the question under the assumptions given by kalle43 :) $\endgroup$ – Robert Filter Dec 27 '10 at 8:33
  • $\begingroup$ I am not satisfied by this explanation. Why would the energy be a function of theta? What if were rotating on a frictionless table, then it cant loose energy. $\endgroup$ – TROLLHUNTER Dec 28 '10 at 14:13

I think that if you could spin the egg perfectly on it's edge, then it wouldn't stand up. It stands because of a wobble as it spins. This wobble changes the stability of the egg's rotation, causing it to stabilize itself. If you do this enough times, you'll notice that the egg will sometimes stand on it's sharp edge. Also, this can be done with a football, so the matter of "one side of the egg is bigger" isn't relevant.

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