Parabola wine glass tips over - how does it come to rest?

Question

A wine glass with no base nor stem is the shape of a simple parabola y=x^2. The glass is perfectly balanced vertically upright (at 0 degrees). It gets tipped over sideways and wobbles a moment and comes to rest. At what angle is the wine glass resting? Is this angle a function of the glass height or will the angle be the same regardless of how tall the glass is ?

Answer 1

The glass will be stable when there’s no net moment. This will occur when the center of mass is directly above the point of contact with the ground. Equivalently, the points of (meta)stability on the curve are those at which the inward normal points toward the center of mass. By symmetry, we can restrict our attention to the $x$-$z$ plane, in which the cross-section of the glass is the parabola $z=x^2$. For a fixed height $h$, the center of mass will be at a fixed point $(0,\bar z)$. A normal to the parabola at $x$ is $(2x,-1)$, which is parallel to $(x,x^2)-(0,\bar z)$ when $$2x(x^2-\bar z) + x = 0.$$ We always have $x=0$ as a solution, which corresponds to the initial metastable balance. In addition, when $\bar z\gt\frac12$ we have two other real solutions, $x=\pm\sqrt{2\bar z-1}$, from which we can compute the tilt angle $\arctan\left(2\sqrt{2\bar z-1}\right)$. These two additional solutions correspond to a ring on the surface along which the glass will happily roll without tilting its axis relative to the ground plane.

To compute the $z$-coordinate of the center of mass, we need to be careful to account for the entire paraboloid of revolution. Assuming a uniform density $\rho$ and taking advantage of the rotational symmetry we have, using a standard formula, $$\bar z = {\int z\,2\pi\rho x\,ds \over \int 2\pi\rho x\,ds} = {\int_0^{\sqrt h}x^3\sqrt{1+4x^2}\,dx \over \int_0^{\sqrt h}x\sqrt{1+4x^2}\,dx}.$$ According to Mathematica, for $h\gt0$ this evaluates to $$\bar z = {(6 h-1) (4 h+1)^{3/2}+1 \over 10 \left((4 h+1)^{3/2}-1\right)},$$ which is very close to a linear function of $h$.