Apparently one of the reasons why all manhole covers are shaped like discs is because for any other shape, the manhole cover would fall through its own hole. As stated this is not necessarily a mathematics question, since there are many physical restrictions and requirements for the design of a manhole cover which are not easy to state in mathematical terms.
Here I attempt to formulate a mathematical question which models the above. Let $D \subset\mathbb{R}^2$ be a compact region whose interior is path-connected and whose boundary consists of finitely many smooth arcs. This is our manhole. By a manhole cover we mean a region in $\mathbb{R}^3$ defined by $D$ and two additional parameters $r, \delta > 0$. First we dilate $D$ by a factor of $r > 1$: that is, define $D(r)$ by $(x,y) \in D(r)$ if and only if $(x/r, y/r) \in D$. The manhole cover $C(r, \delta)$ is then obtained from $D(r)$ by having a uniform thickness of $\delta > 0$. It is understood that $\delta$ is small compared to the diameter of $D(r)$.
A manhole cover $C(r, \delta)$ is said to fall through its own hole if, up to a translation and rotation, $C(r, \delta)$ can pass through $D$ without the boundaries intersecting.
Here whether or not $C(r, \delta)$ falls through its own hole depends not only on $D$ but also the choice of $r$ and $\delta$.
My question is: what choices of $r, \delta$ (may depend on the diameter of $D$) ensure that the circle is the only shape that falls through its own hole?
