Find the least eccentricity of an ellipse which can rest on the plane.

Question

A perfectly rough plane is inclined at an angle $\alpha$ to the horizontal.Find the least eccentricity of an ellipse which can rest on the plane.

Any ideas on how to solve this problem?

Answer 1

This problem has nothing to do with friction, it is a problem of kinematics alone. My result differs from the formula obtained in the quoted source, sailing under the section heading "friction".

Let the ellipse be parametrized by $$E:\quad t\mapsto\bigl(x(t),y(t)\bigr):=(a\sin t,-b\cos t)\ ,$$ and let $\ell$ be the supporting line at $(x,y):=\bigl(x(t),y(t)\bigr)$ for some $t\in\bigl[0,{\pi\over2}\bigr]$, see the figure. Then the upward normal direction is given by $\tilde n=(-\dot y, \dot x)=(-b\sin t, a\cos t)$, and the unit normal by $n={\tilde n\over |\tilde n|}$, where $$|\tilde n|=\dot s=\sqrt{\dot x^2+\dot y^2}=\sqrt{a^2\cos^2 t+b^2\sin^2 t}=:\sqrt{g(t)}\ .$$

It follows that the height $p$ of the center of $E$ with respect to $\ell$ computes to $$p(t)=- n\cdot(x,y)=-{\tilde n\cdot(x,y)\over|\tilde n|}={ab\over\sqrt{a^2\cos^2 t+b^2\sin^2 t}}= ab\>g^{-1/2}\ .\tag{1}$$ We now consider $\ell$ as fixed and let $E$ roll on $\ell$ instead. While "time" $t$ is running the height $p$ undergoes a "sinusoidal" movement of period $\pi$ described by $(1)$.

In reality $\ell$ is not horizontal, but slanted with an angle $\alpha$ with respect to the gravitational horizontal (drawn in red in the figure). Therefore the gravitational height $h$ of the center of $E$ satisfies the differential equation $$\dot h=\dot p\>\cos\alpha-\dot s\>\sin\alpha\ .$$ Here the first term is the projection of $\dot p$ to the gravitational vertical, and the second term comes from $E$ rolling without friction along $\ell$ downwards: The point of contact moves along $\ell$ with speed $\dot s$, and this produces a downward drift $\dot s\>\sin\alpha$ for the quantity $h$.

When $0<\alpha\ll 1$ the function $t\mapsto h(t)$ is approximatively $=p(t)$. It has a local minimum shortly after $t=0$ and a local maximum shortly before $t={\pi\over2}$. As $\alpha$ increases these local extrema move to each other and at a certain critical value $\alpha_*$ coalesce to an inflection point with horizontal tangent. For even larger $\alpha$ the condition $$\dot h(t)=0\tag{2}$$ has no real solutions anymore. We now proceed to determine the critical value $\alpha_*$.

Using $(1)$ we obtain $$\dot h=-{1\over2}ab\> g^{-3/2}\dot g\>\cos\alpha-g^{1/2}\>\sin\alpha = g^{1/2}\cos\alpha\left(-{ab\>\dot g\over 2g^2}-\tan\alpha\right)\ .$$ In view of $(2)$ this means that we have to analyse under which conditions the equation $$-{ab\>\dot g(t)\over 2 g^2(t)}=\tan\alpha\tag{3}$$ has real solutions $t$. To this end write $$g(t)={a^2+b^2\over 2}\bigl(1+\rho \cos(2t)\bigr)$$ with ${\displaystyle\rho:={a^2-b^2\over a^2+b^2}\geq0}$. It follows that $$\dot g(t)=-(a^2+b^2)\rho\>\sin(2t)\ ,$$ so that $(3)$ amounts to $${2ab \rho\over a^2+b^2}\ {\sin u\over (1+\rho \cos u)^2}=\tan\alpha\ ,\tag{4}$$ where we have put $2t=:u$. Here the left hand side has a certain maximal value $\mu>0$ in the interval $0\leq u\leq\pi$. If $0\leq \tan\alpha<\mu$ the equation $(4)$ has two real solutions in this interval, and if $\tan\alpha>\mu$ the equations $(4)$, resp. $(2)$, have no real solutions. It follows that $\alpha_*=\arctan \mu$.

It is possible to compute $\mu$ explicitly, but the result is not inspiring: Let $f(u):=\sin u(1+\rho\cos u)^{-2}$. Then $$f'(u)={\cos u(1+\rho\cos u)-2\rho\sin^2 u\over(1+\rho\cos u)^3}\ ,$$ so that we have to solve the equation $\rho\cos^2 u-\cos u-2\rho=0$ for $\cos u$. This gives $$\cos u={1-\sqrt{1+8\rho^2}\over 2\rho}=:c\ .$$ In this way we finally obtain $$\mu={2ab \rho\over a^2+b^2}\>{\sqrt{1-c^2}\over(1+\rho c)^2}\ .$$