The following questions occurred to me.
This is not research mathematics, just idle curiosity.
Apologies if it is inappropriate.
-
Suppose you have a fixed volume V of maleable material,
perhaps clay.
The goal is to form it into a shape S (convex or nonconvex) that would roll down an inclined
plane as fast as possible.
The plane is tilted at θ with respect to the horizontal.
The race track that quantifies "as fast as possible"
is of length L. The shape S must be
entirely behind a starting plane orthogonal to the inclined plane,
and its race is finished when it is entirely ahead of a finishing plane,
L distant, again orthogonal to the inclined plane.
So if S is a disk of radius r, its center of
gravity will have to travel a total distance of L + 2r
to complete the race. S is released at rest, and rolls
under gravity. Assume that the materials of S and of the
inclined plane have a sufficient coefficient of friction μ
such that there is no slippage, just pure rolling.
The stipulation that S start behind and end
ahead of the orthogonal planes suggests that the optimal shape
is an arbitrarily thin (and therefore arbitrarily long) cylinder, to minimize r.
But this does not accord with (my) physical intuition.
What am I missing here? -
A variation on the same problem replaces
the inclined plane with an inclined half-cylinder of radius R, like a rain gutter.
It may be necessary to assume a relation between R and the volume V of material
to make this problem reasonable (so that S fits in the gutter). But certainly this version's
answer could not be an arbitrarily long cylinder!
Best Answer
There is one thing that's way more important than the precise definition of how the race starts and ends: the moment of intertia of $S$, which determines how much energy is wasted on the rotation. Let us assume that $S$ is cylindrically symmetrical with mass $m$, radius $R$, moment of intertia $I$, and let $k=I/mR^2$. Then from the conservation of energy,
$$\frac{I\omega^2}{2}+\frac{mv^2}{2}=mgx\sin\theta,$$
and using that $v=\omega R,$ we get that
$$\frac{dx}{dt}=v=\sqrt{\frac{2g\sin\theta}{k+1}}x^{1/2}.$$
This already makes it clear that in order to minimize the rolling time, $k$ must be as small as possible, but this can also be quantified by solving the separable ODE and finding that
$$T=\sqrt{\frac{2(k+1)d}{g\sin\theta}}\geq \sqrt{\frac{2L}{g\sin\theta}},$$
where $d=L+2R$ is the distance traveled. The conclusion, in the cylindrically symmetrical case of radius $R$, is that $S$ should be as close as possible to the infinitely thin heavy rod with a disk-shaped thin "flap" of radius $R$ attached, so that $k$ is just slightly over $1$. In a practical sense, imagine a wheel pair with very light rims and a very heavy axle. [I like to think of this situation as "anti-flywheel problem": we want to minimize rotation energy accumulated by $S$, so contrary to the usual flywheel, the wheel is light and the axle is heavy.] If your "malleable material" has uniform density and $R$ may vary, one sequence of mathematical solutions approaching the time bound consists of solid cylinders with radii $r_i$ shrinking to 0 and growing widths $\ell_i$, with two $w$-thin disks of radius $R_i$ attached, where
$$\pi(\ell_i r_{i}^2+2w R_i^2)=V, \quad r_i=o(R_i), \quad R_i\to 0.$$
[I've only put in an explicit sequence because of a clarification request in the comments; I actually think that this level of detail hinders understanding]. It's also clear that the cylindrical symmetry restriction is irrelevant.
The second problem is more difficult: I don't have much to say on it other than that, by the same token, a solid ball is better than a disk/cylinder even if it could be fitted in the "gutter".
P.S.: "Physical" intuition may be deceptive: many people assume that a ball will roll according to the same law of motion as a cylinder or an ideal point mass. This actually happened to some of my mathematically oriented classmates in a physics lab assignment, who tried to make up data in order to avoid doing tedious experiments.