TITLE: Fastest Rolling Shape?
QUESTION [16 upvotes]: 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!

REPLY [2 votes]: I am going to post a bunch of assumptions which will impose further constraints on the 2nd problem. I think these assumptions maybe natural in view of the thought experiment. 

Without loss of generality, we can assume that the gutter has width 1, so the width of the gutter is parametrized by $[0,1]$. 
The cross-section of the gutter itself is parametrized by some convex function $g: [0,1] \to \mathbb{R}$ with $g(0) = g(1)$. 
The object is not allowed to stick out of the gutter side-ways (else we can just take a heavy rod that rests on the top of the gutter...).
The object is assumed to have uniform density and total mass 1.
The object is obtained by rotation (since it is to rotate) around an axis. 

First, by energy conservation. similar to described in Victor Protsak's answer, to minimize the total time of travel it suffices to minimize
$$ \mathcal{E} = (L + 2 R_0)( 1 + I / R_1^2 ) $$
where $R_0$ is the maximum radius of the rotating body and $R_1$ is radius at which it is rolling, in other words the maximum distance from the axis to a point on the surface of the object that is in contact with the gutter. (Note that it is impossible for a ball of radius 1 to roll inside a gutter of radius 1 without slipping!!! We'll just conveniently forget frictional dissipation for our computations.)
Now, by the definition of the angular momentum and simple re-arrangement argument, it is clear that the cross-section of the object can be obtained as a function over its axis. More precisely, in a cylindrical coordinate system where $z$ is along the axis of rotation, the object 
$$ S = \{ r \leq f(z) \} $$
(If there are any "holes" near the axis, by moving the mass inward closer to the axis it reduces both $I$ and $R$.)
It remains to write $I$ and $R$ as a functional of $f$. Working in the cylindrical coordinate still, the volume of $f$ is
$$ V(f) = \pi \int_0^1 f(z)^2 dz $$
By assumption this is a constant number.
The moment of inertia is 
$$ I(f) = \int_0^1 \int_0^{f(z)} 2\pi r^3 \mu dr dz = \frac{\mu\pi}{2}\int_0^1 f(z)^4 dz $$
where $\mu = 1 / V$ the uniform mass density. 
Lastly, $R_0 = \sup f$ by definition. But $R_1$ is a bit tricky to compute. A bit of thought tells you that $R_1 = \sup_{P} f(z)$, where $P$ is the set of points satisfying $f(z) + g(z) = \sup (f+g)$. 
Observe that $R_1 \leq R_0$. 
To summarize, the task then becomes: given a function $g(z)$ and constants $\alpha,\beta,V$, minimize the functional 
$$\mathcal{E} = (\alpha + \|f\|_\infty)(\beta + \frac{\|f\|^4_4}{R_1(f;g)^2} )$$
under the constraint that $\|f\|_2^2 = V$, where $\|\cdot\|_p$ denotes the Lebesgue $p$-norm over $[0,1]$. 
If we impose the constraint that $R_1 = R_0$, then the problem reduces to the first one which Victor examined, where the solution is a Rolling cylinder. I rather suspect that the cylinder is also the solution in the general case. (I am pretty sure that to minimize $L^4$ while fixing $L^2$ on a compact interval one needs to make the function as flat as possible.)