# Intrinsic relations between two solutions

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.

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.