Dtyjlui

A perfectly rough plane is inclined at an angle $alpha$ to the horizontal. What is the least eccentricity of an elliptic cylinder that resists rolling down the plane?

At equilibrium, the weight $W$, the normal reaction $R$ and the friction $F$ should be concurrent (at the point of contact). The inclination $alpha$ is the angle between the radius vector and the corresponding normal.

Unit outward normal vector:

$$mathbf{n}=
frac{(bcos t,asin t)}{sqrt{a^2sin^2 t+b^2cos^2 t}}$$

Unit radius vector:

$$hat{mathbf{r}}=
frac{(acos t,bsin t)}{sqrt{a^2cos^2 t+b^2sin^2 t}}$$

Now

begin{align}
cos alpha &= mathbf{n} cdot hat{mathbf{r}} \
&= frac{ab}{sqrt{(a^2 sin^2 t+b^2 cos^2 t)(a^2cos^2 t+b^2sin^2 t)}} \
&= absqrt{frac{8}{a^4+6a^2b^2+b^4-(a^2-b^2)^2 cos 4t}}
end{align}

The upper and lower bounds are given by $t=npi/2$ and $t=(2n+1)pi/4$ respectively.

That is,

$$frac{2ab}{a^2+b^2} le cos alpha le 1$$

$$cos^{-1} frac{2ab}{a^2+b^2} ge alpha ge 0$$

Equivalently

$$frac{2sqrt{1-e^2}}{2-e^2} le cos alpha$$

On solving,

$$e^2 ge frac{2sin alpha}{1+sin alpha}$$