Why does Norton's thought experiment require a specific dome shape? Norton's dome famously shows that Newtonian mechanics (as a mathematical model) is non-deterministic. To do this, a very specific potential is chosen. But why does this not work with any dome shape, e.g. $U(x) = -x^2$?
One could argue as follows: For any dome-shaped potential, we can impart the exact amount of kinetic energy to the particle such that it will come to rest at the peak of the dome. Now, due to T-symmetry, the time-inverted path must also be allowed within Newtonian mechanics. That is, the particle is initially at rest, and then starts moving after some amount of time. Why is this not equivalent to Norton's dome?
 A: This is addressed (briefly) in the Wikipedia article you linked, as well as in Norton's online exposition of the problem.
For other dome shapes, if we give a particle just enough energy to reach the top, it takes an infinite amount of time for the particle to ascend the dome. The particle never actually gets to the top and comes to rest. If we time-reverse this motion to consider a particle moving down the dome, we get a particle that has been moving for an infinite amount of time near the peak instead of a particle that started at rest on top of the dome.
A: Norton's dome has been debated extensively, e.g. in this Phys.SE post, so we will not repeat this here. Let us just mention that Norton chose his explicit counterexample very specifically to make it simple and to let the apparent paradox stand out, but it is easy to generalize. E.g. if we stick with a power law, then we can choose
$$ z~=~-C s^p, \qquad C~>~0, \qquad 1~<~p~<~2, \tag{1}$$
where $z\equiv-h$ is the vertical height and $s\equiv r$ is the arc length.
Then Norton's initial value problem (IVP) is
$$\begin{align} \ddot{s}(t)~=~&Ks(t)^{p-1}, \cr
s(t\!=\!0)~=~&0, \cr
 \dot{s}(t\!=\!0)~=~&0, \cr 
t~\geq~&0,\cr
K~\equiv~&pgC. 
\end{align}\tag{2}$$
and (besides the trivial solution $s\equiv 0$), the 2nd branch solution becomes of the form
$$ s~\propto~t^{\frac{2}{2-p}} \tag{3}$$
Norton choose $p=\frac{3}{2}$. OP's suggestion $p=2$ does not work.
