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.

• I think this is a good question, but I'm not entirely sure what you're asking. Is this about an object falling on to a surface (e.g. the ground) and stopping immediately, or is it about an object sliding on a surface described by $y=x^2$? In the latter case, is the object constrained to stay on the surface, or can it "float" up off the surface? And is that supposed to be a 2D surface in 3D space, i.e. a paraboloid, or is it a 1D "surface" in 2D space, which is literally what you show in the image? Nov 5 '13 at 6:57
• I tried to clarify the example of the falling ball. My question is about $y=x^2$. I just used the falling ball example because it seemed like a more intuitive example of how velocity can change the strength of the normal force. Eventually I want to extend this to a 2D surface in 3D space, but for now, I would like at least to understand it for a 1D curve in a 2D space. I don't think the conservation of energy solution would work as well in 3D, which is really why I am not using it.
– 0xFE
Nov 5 '13 at 21:34
• Can you clarify what you are looking for? The force from the surface to the ball? Is that what? Please ask your question, you told a story, it was really annoying to have to guess your question. Nov 6 '13 at 15:22
• My question is at the end in bold. Yeah, I'm looking for the mathematically how the force from the surface to the ball changes when the velocity of the ball changes. Sorry about the long story. I tried to keep it short, but I probably could have made it shorter.
– 0xFE
Nov 6 '13 at 21:47

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 $$\boxed{R = \vec{n} \cdot \vec{F} = \frac{m v^2}{\rho} = \frac{m v^2 f''}{\left( 1 + f'^2 \right)^\frac{3}{2}}}$$

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

• I'm going to try it to make sure I understand it and that it works properly, but this looks like exactly what I was looking for.
– 0xFE
Nov 6 '13 at 4:00
• -1. My answer is much better and simpler. Nov 6 '13 at 12:21
• @user1708 I understand you being proud of your answer, but it is bad manners in Phys.SE to go around and put down all other answers because you do not like them. It is up to the OP to decide which is the best suited answer. Nov 6 '13 at 14:25
• @user1708 in my defense, I am presenting a complete answer with all the bits and pieces so as to answer all the questions the OP might have. Oh, and it is correct too. Nov 6 '13 at 14:27
• Nov 6 '13 at 16:16

All you need is Newtons law and the fact that the force from the surface is the normal force since you have no friction. Then draw a diagram and see that $N_x/N_y=df/dx=v_y/v_x$. Then solve $F=d\vec{p}/dt=m\vec{g}+\vec{N}$. If you need $\vec{v}$ aswell use energy conservation and initial condition.

Post solution. ;)

• This solution is incomplete because you haven't specified how the velocity related components enter into $F$. On the bottom of the curve $N$ is more than the static case (which you show above) because it has to force the particle along a curved path of some osculating radius. The centripetal force is added to gravity to yield $N$. Nov 6 '13 at 14:53
• a) F=mv is not mentioned in the post, and b) F=mv is not correct, F=ma is correct. Nov 6 '13 at 16:18
• N here isnt for the static case, N is the normal force, meaning the total force from the surface to the mass. F=mv, ja72's law? LOL. F=mdv/dt Nov 6 '13 at 16:25
• And that leads to the other problem in this post, the fact that the terms used are not defined or explained. Can you explain how the normal force is $F=m v$? (PS. Try to use math formatting in the comments also) Nov 6 '13 at 16:27
• F=ma is the net force on the object. That is standard newtons law. Nov 6 '13 at 16:29