A falling object problem

Binaryburst

I did a simulation with vx and it matched the sinusoidal, it seems. Any help from the experts?

mfb

Where do you consider the motion along the bowl? It will lead to accelerations both in vx and vy, to follow the shape.


I'll try to start from scratch - with energy conservation, but it is possible to avoid this word if necessary.

If the object is at rest at position ##x_0>0##, it has a total energy of ##gx_0^2##, setting its mass and a meter to 1.
At position x, ##0<x<x_0##, energy conservation leads to ##v^2=2g(x_0^2-x^2)##.
The derivative of the parabola there is 2x, therefore ##v_x=\frac{1}{2x}v_y##. It follows that ##v^2 = v_x^2 + v_y^2 = v_x^2 (1+4x^2)##. Combining both,
$$\frac{dx}{dt}= v_x = \sqrt{\frac{v^2}{1+4x^2}} = \sqrt{\frac{2g(x_0^2-x^2)}{1+4x^2}}$$
The sign is arbitrary and depends on the current part of the oscillation.
This leads to ugly elliptic integrals. It can be written as
$$\frac{dt}{dx}=\sqrt{\frac{1+4x^2}{2g(x_0^2-x^2)}}$$
While this gives elliptic integrals as well, it allows to determine t(x) via numerical integration.

g=10 and x₀=1 leads to a period of T=2.36s.
For large x₀ (>>1), most of the parabola is like a free fall, and the period should approach 4 times this free-fall time. As an example, x=sqrt(500) leads to T=40.06, whereas a free fall would give T=40s.
For small x₀ (<<1), the deflection of the bowl is negligible, and we get a harmonic oscillator. The numerator for dt/dx can be approximated as 1, and the period approaches ##T \approx \frac{2\pi}{\sqrt{2g}}##.

Binaryburst

Thank you! At last :D