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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 ?

EricB
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    You should say something about your understanding of the problem, including where you're stuck. This helps answerers tailor responses to your skill set, without wasting time telling you things you already know. – Blue Apr 17 '18 at 04:15
  • Is the glass $2$-dimensional or $3$-dimensional? – Blue Apr 17 '18 at 04:19
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    I feel a 2 dimensional glass or 3D will come to rest at the same angle. I feel a vertical line at the resting position will pass thru the center of gravity and will intersect the parabola at 2 places. I feel calculus must be applied to find where the length of the curve to one side of the vertical line at rest will be equal to the length of the 2 curves on the other side of the line – EricB Apr 17 '18 at 04:23
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    Keeping the parabola of height x inches vertical at zero degrees, I feel it would be first easier to find the center of gravity by finding the horizontal line at height y inches that slices the glass- such that length of the 2 upper edges would equal the length of the bottom portion of the parabola. Then when the glass is at the resting angle- the vertical line where it touches must pass through that same point – EricB Apr 17 '18 at 04:44
  • You should edit your question to add in your thoughts about the question. This site is not for posting bare questions. – user21820 Apr 21 '18 at 06:04

1 Answers1

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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 $h$.

amd
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  • thank you. Do you think my premises was correct about the length of the curves on both sides of the vertical line at rest? It may be a different and perhaps more cumbersome approach, but was just wondering about my assumption? – EricB Apr 17 '18 at 14:30
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    @EricB Yes, splitting the total mass in two is a valid approach, but the vertical mass distributions of a parabolic wire and that of a paraboloid of revolution are quite different. You’d have to find a plane that bisects the surface or use a variable density for the wire, one that depends on the line that you’re trying to find. – amd Apr 17 '18 at 16:07