In addition to Lubos Motl's correct answer, I would like to make a few comments related to Norton's dome:
First a brief derivation of Norton's equation of motion (7). I prefer to call the (non-negative) arc length $r$ for $s$, and the vertical height $-h$ for $z$. Like Lubos Motl, I will introduce a proportionality factor $K$ for dimensional reasons, so that the equation for Norton's dome reads
$$ z~=~-\frac{2K}{3g}s^{3/2}. \tag{1}$$
Here the constant $(g/K)^2$ has dimension of length. Equation (1) is only supposed to be valid for sufficiently small (but finite) arc lengths $s\geq 0$. Since there is no friction, we have mechanical energy conservation$^1$
$$ 0~=~\frac{E}{m}~=~\frac{\dot{s}^2}{2}+gz.\tag{2}$$
In the first equality of (2), we used the initial conditions
$$ s(t\!=\!0)~=~0, \qquad \dot{s}(t\!=\!0)~=~0.\tag{3}$$
We assume that $t\mapsto s(t)$ is twice differentiable wrt. time $t\geq 0$. (In detail, at the initial time $t=0$ we assume that the function is one-sided twice differentiable from right.) Differentiation of eq. (2) wrt. time $t$ leads to
$$ \dot{s}\ddot{s}~\stackrel{(2)}{=}~-g\dot{z}.\tag{4}$$
Division on both sides of eq. (4) with $\dot{s}$ yields$^2$
$$ \ddot{s}~\stackrel{(4)}{=}~-g\frac{\dot{z}}{\dot{s}}~=~-g\frac{dz}{ds}~\stackrel{(1)}{=}~K\sqrt{s}~.\tag{5}$$
Equation (5) is the sought-for equation of motion. Alternatively, combining eqs. (1) and (2) yield the following first-order ODE
$$\dot{s} ~\stackrel{(1)+(2)}{=}~\sqrt{\frac{4K}{3}} s^{\frac{3}{4}}. \tag{6}$$
Norton's initial value problem (IVP) is
$$
\begin{align} \ddot{s}(t)~=~&K\sqrt{s(t)}, \cr
s(t\!=\!0)~=~&0, \cr
\dot{s}(t\!=\!0)~=~&0, \cr
t~\geq~&0.
\end{align}\tag{7}
$$
The IVP (7) has two solution branches$^3$
$$s(t) ~=~\frac{K^2}{144}t^4\qquad\text{and}\qquad s(t) ~=~0~, \tag{8} $$
as can be easily checked. The failure to have local uniqueness of the ODE (7), which leads to indeterminism of the classical system, can from a mathematical perspective be traced to that the square root $\sqrt{s}$ in eq. (7) fails to be Lipschitz continuous at $s=0$.
Alternatively, from mechanical energy conservation (6), one can consider the IVP
$$\begin{align} \dot{s}(t) ~=~&\sqrt{\frac{4K}{3}} s(t)^{\frac{3}{4}}, \cr s(t\!=\!0)~=~&0,\cr t~\geq~&0.\end{align}\tag{9}$$
Not surprisingly, the IVP (9) has the same two solution branches (8), and thus also demonstrates failure to have local uniqueness.
$^1$ I imagine that the point particle is sliding with no friction. (The rolling ball in Norton's figure is slightly misleading and presumably only for illustrative purposes.). A more complete derivation would check that the point particle doesn't loose contact with the doom. If one would like to avoid such an analysis, one may for simplicity assume that the dome is a two-sided constraint.
$^2$ Division with $\dot{s}$ is only valid if $\dot{s}\neq 0$. Now recall that the mechanical energy $E=0$ is zero. If $\dot{s}=0$ then $z=0$ and hence $s=0$ must be zero, cf. eqs. (1) and (2). Hence the division-by-zero issue is limited to the tip of the dome. Ultimately, it turns out that the $\dot{s}=0$ branch does not lead to new solutions not already included in eq. (8), nor alters Norton's IVP (7).
$^3$ For each solution $s$, which is defined for non-negative times $t\geq 0$, let us for convenience extend in a trivially fashion $s(t<0):=0$ for negative times $t<0$. Then if we time-translated a solution $t\mapsto s(t)$ into the future, we get another solution $t\mapsto s(t-T)$ for some moduli parameter $T\geq 0$. Therefore strictly speaking, the first branch in eq. (8) generates a 1-parameter solution with a moduli parameter $T\geq 0$. So in fact, the IVP (7) has infinitely many solutions! Note that the second trivial solution branch (8) can be viewed as the $T\to \infty$ moduli limit of the first solution branch (8).
