Which shape rolls the farthest? Let $K$ be a $2$-dimensional convex body of homogenous material and diameter $\le 1$.
Releasing it from an unstable position, it rolls on a horizontal line $L$.
Assume there is sufficient friction to preclude slippage.

Q1. Which shape rolls the farthest, where the rolling distance is determined by the
farthest point of contact on $L$?

For example, it seems that a thin ellipse of diameter $1$ would roll nearly
its full length (but not flip over and roll farther):


Q2. Same question but permitting inhomogenous material, i.e., the object can
be "loaded" to be heavier in some internal locations.

Can a rolling distance of $1$ be exceeded?
 A: You can solve this using simply energy considerations. Due to the non slip condition, your system is conservative and has only one degree of freedom. You can parametric this DOF by the angle $\theta$ of the orientation and setting its origin as the initial orientation. You'll have only two sorts of energies: kinetic $K\geq 0$ and potential $U=Mgh(\theta)$ with $M$ the mass of the object, $h$ the height of the center of mass of the object with respect to the ground depending on its orientation, which depends only on the geometry of the object (for example for a circle, $h=R$ its radius).
Technically, if you rigorously release it from the unstable position, it will just stay put. I'll therefore assume that it is slightly tilted. Energetically, it's assuming that you have slightly less energy than the energy of the reference unstable position.
All the evolution of the object will be determined by the graph $h$. Since you start by an unstable equilibrium, this means you start right next to a local maximum $h(0)$ and you go downhill. You will continue going until you reach the first point of equal initial height at orientation $\theta_f$. If at $\theta_f$ $h$ is also stationary, you'll asymptotically reach this new equilibrium point (heteroclitic orbit), if not, you'll backtrack and come back to your initial position.
For the case of the ellipse, $h$ has two equally spaced points where its maximum is obtained (and two minima), so this confirms intuition that it only rolls over. For low energy, it will rock back and forth between the two maxima and in the limit when you reach the maximum potential energy from below, it links the two unstable postions asymptotically.
With this approach, there is no real difference between Q1 and Q2. The furthest you can roll is given by the perimeter of the object. Your object can do a full turn iff $h$ has only one global maximum, ie the starting position. This is possible for example for an egg-like object (note that it can arbitrarily close to a circle) when it's homogenous or even a circular object with an internal off-center heavy component.
If you are fixing the perimeter, it's done. Fixing the diameter is mathematically more challenging. Based on experimenting with polygons, I'd say the convex object with the greatest perimeter with fixed diameter would be the circle, but I'm not sure, and certainly didn't prove it. In this case, using the two previous examples, this would solve your problem.
Even the answer is incomplete, I hope it helps.
