Intrinsic relations between two solutions

Question

Recently I've encountered a problem stated as follows:
A smooth bowl has the shape of a paraboloid. The equation of the cross-section with the $(x, z)$-plane is $x^2= 2R_Az$, where $R_A$ is the curvature radius at point $A$. One releases at a point with height $z=h$ a point mass with an initial velocity equal to $0$. What is in this case the magnitude of the force exerted by the particle on the bowl when it passes point $A$? Neglect air friction.

I've come up with two solutions to this problem, one needs some naive knowledge of curvatures and one doesn't:

Solution #1: By Law of conservation of energy, we have $$mgh=\frac{mv_A^2}{2}$$ Hence $$v_A^2=2gh$$ Since the mass experiences a centripedal force pointing towards the centre of curvature at point $A$, we have ( N is the contact force ): $$N-mg=\frac{mv_A^2}{R_A}$$ Therefore $$N=mg(1+\frac{2h}{R_A})$$

Solution #2: The displacement vector of the particle is $$\vec{s}(t)=x(t)\vec{i}+\frac{1}{2R_A}x^2(t)\vec{k}$$ (where ${x(t)}$ is the horizontal displacement) Therefore the acceleration is $$\vec{a}(t)=\frac{d^2\vec{s}}{dt^2}=\ddot{x}(t)\vec{i}+\frac{1}{R_A}\Big(\dot{x}^2(t)+x(t)\ddot{x}(t)\Big)\vec{k}$$ At point $A$, $\ddot{x}(t)=0$, Therefore $$\vec{a}=\frac{1}{R_A}\dot{x}^2(t)=\frac{v_A^2}{R_A}$$ Hence both of my solutions give the same value of acceleration and hence the same results. My questions are: Is there a rigorous explanation to the validity of my first solution? Is it just a coincidence that the acceleration at $A$ is identical to the centripetal force related to the curvature circle? Are there any "intrinsic links" between the two solutions? Thanks in advance.

Answer 1

At A the tangential acceleration is zero. Thus the only nonzero component of acceleration is the perpendicular one. This is acceleration that curves the trajectory to produce observed curvature of the path. The centripetal force related to the curvature is force that makes objects trajectory curve with given curvature. They are the same quantity up to a mass scale factor and philosophical interpretation.

Answer 2

There is an intrinsic connection of sorts. If one studies curved surfaces embedded in 3D space (or differential geometry in general), one can see that the equation of motion of a free particle, i.e., its acceleration, is determined by the curvature of the surface. In the case of classical mechanics like the case you have noted, recall that the normal force is merely an ad-hoc force that we add into the equation of motion such that the particle's motion is constrained onto the curved surface. In principle of course, one might have that the normal force is "too weak" in which case the particle must "fall through" or break the curved surface under the net force of gravity. However, when we assume that the particle's motion is in fact constrained, then its equation of motion is determined precisely by the curvature of the surface, or Newton's laws after including a normal force constraint, which amount to the same thing.

After the above connection is noted, the connection between centripetal acceleration and the computed acceleration is merely that when describing such a 1D curve the curvature at some given point is locally the same as the curvature of a circle with the same radius of curvature.

Answer 3

In your first solution, you have assumed that the centripetal acceleration is $v^2$/R (based on what you know about circular motion). In the second, you have shown that, in this situation, the vertical acceleration is $v^2$/R at point A.