How fast do you have to spin an egg to have it standing?

Question

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 doesn't work, assume its center of mass is shifted a bit, if that doesn't 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?

Answer 1

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 $F(M_{ring}, r_a/r_b; \omega_s) = 0$?

Answer 2

In my opinion 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 geometric 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 it's still way over my head). As you already pointed out, this behaviour is not possible on a frictionless table. For some readings see:

• http://en.wikipedia.org/wiki/Tippe_top

• https://arxiv.org/abs/chao-dyn/9501008

• https://www.jstor.org/stable/2102374

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

https://www.wobsta.de/uni/tippetop/literature.shtml.html

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

Answer 3

I think that if you could spin the egg perfectly on its 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 its sharp edge. Also, this can be done with a football, so the matter of "one side of the egg is bigger" isn't relevant.