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