newtonian mechanics - Normal force of ball sliding on concave surface

Imagine a ball is sliding along a surface shaped like $y=x^2$. Like ,

but please ignore the fact that the center of the ball is on the surface instead of the edge.

When the ball is stationary, I can calculate the normal force fairly easily. $F_\text{net}$ in the direction normal to the surface is 0. The normal force is normal to the surface, so it is the force that is exactly strong enough to cancel the normal component of the force of gravity.

The normal force comes out to $<\frac{-a}{(1 + a^2)}, \frac{1}{(1 + a^2)}>$, where $a=f_{\text{surface}}'(x)$ (or in this specific case: $a=2x$)

I initially expected this to be true even after the ball begins sliding down the surface, but I found that it was not. Intuitively, I understand that a velocity with a non-zero component in the normal direction affects the strength of the normal force. This seems like the same mechanism by which the normal force is able to stop a falling object as it hits the ground, for example.

In the falling ball example, the relationship between velocity and normal force seems a bit sticky mathematically. Assuming neither the ground nor the object compress, the normal force would have to be infinite initially, right? It would only be infinite for an infinitesimal amount of time (so the integral would still be finite). However, in that case, the $v(t)$ curve is discontinuous, while it seems like $v(t)$ would be continuous for a ball sliding on $y=x^2$, so hopefully the relationship is less obnoxious in the continuous case.

Is there a mathematical formulation of this relation between normal force and velocity? Can I add a velocity parameter to my equation for $F_\text{net}(x)$ (making it $F_\text{net}(x, v)$) to take this relationship into account?

Note: I also see a conservation of energy solution that sidesteps the normal force issue altogether, but I'd like to understand the forces aspect.

Answer

When a particle follows a path then at any instant there is tangential vector $\vec{e}$ and a normal vector $\vec{n}$ allowing for the velocity to be defined as $$\vec{v} = v \vec{e}$$ and the acceleration $$ \vec{a} = \dot{v} \vec{e} + \frac{v^2}{\rho} \vec{n} $$ where $\rho$ is the radius of curvature of the path.

For a path defined by a function $f(x)$ with derivatives $f'(x)$ and $f''(x)$ the radius of curvature is

$$ \rho(x) = \frac{\left(1+{f'(x)}^2\right)^\frac{3}{2}}{f''(x)} $$

and the direction vectors are

$$ \begin{aligned} \vec{e}(x) &= \begin{pmatrix} \frac{1}{\sqrt{1+{f'(x)}^2}} \\ \frac{f'(x)}{\sqrt{1+{f'(x)}^2}} \end{pmatrix} & \vec{n}(x) &= \begin{pmatrix} - \frac{f'(x)}{\sqrt{1+{f'(x)}^2}} \\ \frac{1}{\sqrt{1+{f'(x)}^2}} \end{pmatrix} \end{aligned} $$

The last part for you, is the fact that gravity only affects the tangential acceleration $\dot{v} = \vec{g} \cdot \vec{e}$ (with $\cdot$ the vector dot product) $$ \dot{v} = -\frac{g f'(x)}{\sqrt{1+{f'(x)}^2}} $$

with total acceleration $$ \vec{a} = \dot{v} \vec{e} + \frac{v^2}{\rho} \vec{n} \\\vec{a} = \begin{pmatrix} - \frac{g f'}{1+f'^2} - \frac{v^2 f' f''}{\left(1+f'^2\right)^2} \\ - \frac{g f'^2}{1+f'^2} + \frac{v^2 f''}{\left(1+f'^2\right)^2} \end{pmatrix} $$

The total force applied is $\vec{F} = m \vec{a}$ and the reaction force is the component normal to motion or $$\boxed{R = \vec{n} \cdot \vec{F} = \frac{m v^2}{\rho} = \frac{m v^2 f''}{\left( 1 + f'^2 \right)^\frac{3}{2}}}$$

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note: this area of study is called differential geometry (Frenet formulas). Look it up.