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 $m$ and 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 $$T=\sqrt{\frac{2(k+1)d}{g\sin\theta}},$$$T=\sqrt{\frac{2(k+1)d}{g\sin\theta}}\geq \sqrt{\frac{2L}{g\sin\theta}},$$where d=L+2R is the distance traveled. The conclusion, in the cylindrically symmetrical case , of radius R, is that S should be as close as possible to a the infinitely thin heavy rod with a disk-shaped thin "flap" of radius R attached, so that k is just slightly over 1. In a practical sense, you can think of imagine a wheel pair with very light rims and a very heavy axle. [I like to think of this situation as "anti-flywheel problem": we want to minimize rotation energy accumulated by S, so contrary to the usual flywheel, the wheel is light and the axle is heavy.] If your "malleable material" has uniform density and R may vary, a one sequence of mathematical solutions approaching the lower time bound will consist consists of solid cylinders of with radii r_i shrinking to 0 and increasing growing widths (to preserve \ell_i, with two w-thin disks of radius R_i attached, where$$\pi(\ell_i r_{i}^2+2w R_i^2)=V, \quad r_i=o(R_i), \quad R_i\to 0.$$[I've only put in an explicit sequence because of a clarification request in the volume constraint)comments; I actually think that this level of detail hinders understanding]. It's also clear that the cylindrical symmetry restriction is irrelevant. The second problem is more difficult: I don't have much to say on it other than that, by the same token, a solid ball is better than a disk/cylinder even if it could be fitted in the "gutter". P.S.: "Physical" intuition may be deceptive: many people assume that a ball will roll according to the same law of motion as a cylinder or an ideal point mass. This actually happened to some of my mathematically oriented classmates in a physics lab assignment, who tried to make up data in order to avoid doing tedious experiments. 1 There is one thing that's way more important than the precise definition of how the race starts and ends: the moment of intertia of S, which determines how much energy is wasted on the rotation. Let us assume that S is cylindrically symmetrical with mass m and radius R, and let k=I/mR^2. Then from the conservation of energy,$$\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}},
where $d=L+2R$ is the distance traveled. The conclusion, in the cylindrically symmetrical case, is that $S$ should be as close as possible to a heavy rod, so that $k$ is just slightly over $1$. In practical sense, you can think of a wheel pair with very light rims and a very heavy axle. If your "malleable material" has uniform density, a sequence of mathematical solutions approaching the lower bound will consist of solid cylinders of radii shrinking to 0 and increasing widths (to preserve the volume constraint). It's also clear that the cylindrical symmetry restriction is irrelevant.