Newtonian motion of a particle confined to a smooth surface Recently, I've been considering a model wherein a lone particle with constant mass m is confined to a surface $F$: $\mathbb{R}^m \to \mathbb{R}^n$ , where $m < n$. I declare this surface to be strictly stationary, even resisting rigid motions. We then declare the particle's initial position to be $w(t_0) = F(u(t_0))$ with $w$ in $\mathbb{R}^n$ and $u$ in $\mathbb{R}^m$, and we declare its initial velocity to be $v(t_0) = w'(t_0) = F'(u(t_0))*u'(t_0)$, where $F'$ is the Jacobian of $w$ with respect to $u$. So, $w(t_0+dt) = F(u(t_0) + u'(t_0)dt)$, to ensure our particle is still confined to $F$ at $t_0+dt$.
However, the velocity vector cannot be unchanged from $t_0$ to $t_0+dt$, since the span of $w'(t_0)$ does not generally coincide with that of $F'(u(t_0) + u'(t_0)dt)$, i.e., the "old" velocity is not tangent to $F$ at the "new" position. To meet this requirement of tangency, I have proposed (axiomatically) that the true velocity of the particle at $t_0+dt$ is the projection of the "old" velocity $v(t_0)$ onto the span of the "new" Jacobian matrix $F'(u(t_0) + u'(t_0)dt)$, as to minimize any instantaneous acceleration. In other words, $$w'(t_0 + dt) := A(A^T A)^{-1}A^Tw'(t_0)$$ where $A = F'(u(t_0) + u'(t_0)dt)$; the $^T$ and $^{-1}$ postfixes denote matrix transposition and inversion, respectively.
Given this axiom, it is clear that $w'(t_0) \neq w'(t_0+dt)$, implying that some instantaneous acceleration $$a = \frac{w'(t_0+dt) - w'(t_0)}{dt}$$ is imposed on the particle.
These are my primary questions:


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*Where does this acceleration come from? Following an energy-conserving (Newtonian) model, some force should be present to impart any supposed acceleration to our particle. This would imply that something in the system loses energy. Would we achieve conservation of energy by simply subtracting this energy from the momentum of the particle in question?

*Is my proposed axiom one which is physically (in a Newtonian sense) valid? Am I going down the wrong rabbit hole when I assume that minimization of instantaneous acceleration is a correct way to describe the motion of a particle confined to a surface?
I would greatly appreciate any sort of insight into how I could continue with and/or improve upon this model. If this model is totally invalid, or if there is a better way to approach this problem, any point in a better direction would be equally appreciated.
P.S. This is my first ever post to any StackExchange site, so I apologize for any formatting/communication errors. I will be more than happy to clarify any uncertainties regarding my model as needed. Thanks!
P.P.S. (edit) My intentions are that the particle not be acted upon by any external force (not even friction) other than that of constrain. Additionally, I intend for the particle to be bounded to the surface strictly, ie it is located on the surface at any given time.
As a final thought, why doesn't a force orthogonal to the particle's current velocity affect its kinetic energy? I suspect it has something to do with centripetal acceleration. I'd prefer to use linear functions to describe the particle's motion, but if trig functions are the only practical option then so be it.
 A: If $m$ < $n$, then your $F$ is a function from $m$-dimensional flat space to $n$ dimensional flat space, which maps $\mathbb{R}^m$ onto a $m$ dimensional surface in $\mathbb{R}^n$.
If the motion of a particle in $\mathbb{R}^n$ is restricted to this surface, then obviously a force is required to keep it on that surface. Without such a force, it follows a straight path in $\mathbb{R}^n$, thus potentially moving off the surface defined by $F$.
As an example consider $m=1$ and $n=2$ with $F$ describing the path of a road and the particle being a car. Then in order to stay on the road, transverse forces (exerted via static friction between road and tyres) act when you're turning a corner (and your stomach knows about them too!).
From this example, you see that the force required is orthogonal to the (tangent to the) path (i.e. the velocity vector in $\mathbb{R}^n$). This should be true in general. Such an orthogonal force does not change the kinetic energy
(this follows trivially from the definition $E_{\text{kin}}=\tfrac12m\mathbf{v}^2$, when $\dot{E}=m\mathbf{v}\cdot\dot{\mathbf{v}}$, which vanishes if $\dot{\mathbf{v}}$ is orthogonal to $\mathbf{v}$).
In Lagrangian mechanics, it is straightforward to constrain the motion to a sub-space (like your surface) and the force(s) required to enforce this are known as force of constrain (look for constraint forces in the above link). They must be provided by the mechanics of the system (possibly with the aid of gravity).
A simple example is a ball rolling down a hill (with potentially a curved shape not simply an inclined plane). Then the ball cannot simply drop vertically as a free particle in 3D would do. Instead the hill (pushing from below) and gravity (pushing downward) constrain the ball to stay on the surface of the hill. However, depending on the shape of the hill, the force of constraint (required to keep the ball on the surface) may exceed that provided by gravity: in this case the ball will leave the surface and follow the trajectory of a gravitating particle moving through 3D space (a parabola) until it hits ground again...
