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Consider a ring rolling without slipping along a horizontal surface. Regardless of the speed of the ring, it is continuously in contact with the surface.
Let's deform the ring slightly so that it becomes an ellipse with small eccentricity of $e\rightarrow 0$.

Now, consider the deformed ring rolling without slipping along the surface. At low speed, the ring still keeps contact with the surface continuously. However, at some critical speed the ring will jump.

I am interested in how to find this speed?

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How about (1) assume continuous contact and rolling without slipping; (2) figure the vertical acceleration of the CoM as a function of velocity; and (3) the minimum of the above falls below $-g$ you've started skipping? You'll need the arc length for an ellipse which I believe means getting into special functions in a hurry. –  dmckee Feb 5 '13 at 17:26
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Is the ring deformation due to a force? Is this force rotating also or fixed w.r.t. the plane. If you are interested in the kinematics of a rolling ellipse, then it's a great question, but needs a little refinement. –  ja72 Feb 5 '13 at 23:10
    
One of the problem with solving this, is that if the ellipse is purely rolling, to find the horizontal transverse distance, the perimeter of the ellipse needs to be calculated, which involves an elliptical integral. –  ja72 Feb 6 '13 at 14:30
    
@ja72 But if the eccentricity goes to zero, then perhaps things go easier? –  Martin Gales Feb 6 '13 at 15:01
    
@MartinGales, yes. The $e\ll1$ condition could be used to produce an approximation. –  ja72 Feb 6 '13 at 15:10
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1 Answer

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The polar shape of the ellipse is $r(\varphi)$, with $\varphi=0$ at major radius $r_{\mbox{major}}=a$ , and $\varphi=\frac{\pi}{2}$ at the minor radius $r_{\mbox{minor}}=a\left(1-e\right)$ , where $e$ is the eccentricity.

$$r(\varphi) = \frac{a\,(1-e)}{\sqrt{e(e-2)\cos^2\varphi+1}} \approx a\,\left(1-e\,\sin^{2}\varphi\,\right)$$

The angle between the contact normal and the polar location of the contact point $$\tan\alpha = \mbox{-}\frac{\frac{{\rm d}}{{\rm d}\varphi}r(\varphi)}{r(\varphi)} \alpha = 2e\,\sin\varphi\cos\varphi$$

The angular position of the ellipse as a function of the contact point location angle $\varphi$ $$ \theta = \varphi+\alpha(\varphi) = \varphi+2e\,\sin\varphi\cos\varphi$$

The angular velocity of the ellipse $$\omega = \dot{\varphi}+\left(\frac{{\rm d}}{{\rm d}\varphi}2e\,\sin\varphi\cos\varphi\right)\dot{\varphi} = \dot{\varphi}+\left(4e\,\cos^{2}\varphi-2e\right)\dot{\varphi} = \left(4e\,\cos^{2}\varphi-2e+1\right)\dot{\varphi}$$

The angular acceleration of the ellipse $$\dot{\omega} = \frac{{\rm d}\omega}{{\rm d}t}=\frac{\partial\omega}{\partial\varphi}\dot{\varphi}+\frac{\partial\omega}{\partial\dot{\varphi}}\ddot{\varphi}$$ $$ = \left(\mbox{-}8e\,\sin\varphi\cos\varphi\right)\dot{\varphi}^{2}+\left(4e\,\cos^{2}\varphi-2e+1\right)\ddot{\varphi}$$

The above equations are used to solve for $\dot{\varphi}(\omega)=$ and $\ddot{\varphi}(\dot{\omega})=$

The vertical position of the ellipse as a function of the contact point location angle $\varphi$ $$ y = r\,\cos\alpha = a\,\left(1-e\,\sin^{2}\varphi\,\right)$$ since $\cos\alpha\sim1$ .

The vertical speed of the ellipse is

$$\dot{y} = \left(\frac{\partial}{\partial\varphi}a\,\left(1-e\,\sin^{2}\varphi\,\right)\right)\dot{\varphi} = \left(\mbox{-}2a\, e\,\sin\varphi\cos\varphi\right)\dot{\varphi} $$

$$\dot{y} = \frac{\left(\mbox{-}2a\, e\,\sin\varphi\cos\varphi\right)}{\left(4e\,\cos^{2}\varphi-2e+1\right)}\omega$$ And the vertical acceleration

$$\ddot{y} = \left(\frac{\partial}{\partial\varphi}\dot{y}\right)\dot{\varphi}+\left(\frac{\partial}{\partial\dot{\varphi}}\dot{y}\right)\ddot{\varphi} = \left(2a\, e\,\left(1-2\cos^{2}\varphi\right)\right)\dot{\varphi}^{2}+\left(\mbox{-}2a\, e\,\sin\varphi\cos\varphi\right)\ddot{\varphi}$$

$$ = \mbox{-}2a\, e\,\frac{\left(2\cos^{2}\varphi+2e-1\right)\dot{\varphi}^{2}+\sin\varphi\cos\varphi\,\dot{\omega}}{4e\,\cos^{2}\varphi-2e+1}$$

Before I expand even more, I look at the peak acceleration is at $\varphi=0$ or where $\dot{\varphi}=\frac{1}{2e+1}\omega$ and $\ddot{\varphi}=\frac{1}{2e+1}\dot{\omega}$

$$ \ddot{y} = \mbox{-}2a\, e\,\frac{\left(2+2e-1\right)\dot{\varphi}^{2}+0}{2e+1} = \mbox{-}2a\, e\,\dot{\varphi}^{2} = \mbox{-}2a\, e\,\left(\frac{1}{2e+1}\omega\right)^{2}$$

$$ \boxed{\ddot{y}=\mbox{-}\frac{2a\, e\,\omega^{2}}{\left(2e+1\right)^{2}}}$$

When this acceleration equals gravity with $\ddot{y}=\mbox{-}\, g$ then the ellipse jumps. This occurs at the critical angular velocity of

$$\omega_{C}=\left(2e+1\right)\sqrt{\frac{g}{2a\, e}}$$

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Many thanks for this work! You probably meant $\ddot{y}=\mbox{-}\, g$. The result does not need to depend on $m$? –  Martin Gales Feb 7 '13 at 15:21
    
Oops! .... good catch. –  ja72 Feb 7 '13 at 15:29
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Really, great work! –  Martin Gales Feb 7 '13 at 15:35
    
The polar equation for the ellipse is not exact, but approximate for small $e$. The exact is $r(\varphi)=\frac{a b}{\sqrt{(b^2-a^2)\cos^2\varphi+a^2}}$ where $a$,$b$ are the major and minor radiuses. In this case, $b=a (1-e)$ is used. –  ja72 Feb 7 '13 at 17:32
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