homework and exercises - elliptical ring rolling on a horizontal surface

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?

Answer

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

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