# A falling object problem

rollingstein
Gold Member
Actually i got the total speed dependent only on the x axis so saying that v.total is dx/dt is correct because it's no longer the slope of the parabola but the slope of f(x).
I don't agree.

The only way v = dx/dt is if $\vec{v}$ was directed along the x axis.

i.e. v_y was zero.

But here the velocity does have a y component so I don't see how you are allowed to do what you did.

OTOH, your elegant solution form tempts me but still.....

PS. Maybe I am wrong. I'm no expert.

Omg! What a humongous mistake in my formulation of the conservation of energy!

rollingstein
Gold Member
Omg! What a humongous mistake in my formulation of the conservation of energy!
Conservation is fine. Your v is ok.

It's where you put v=dx/dt that the problems start.

rollingstein
Gold Member
Here's what I get :

$\frac{dx}{dt} = \sqrt{2g} \sqrt{\frac{x_1^2-x^2}{1+4x^2}}$

Sad part: If this is right the integral is ugly.

It is right. That's what I got too... That's just v*cos(theta). It results that v is equal to f(x).

rollingstein
Gold Member
It is right. That's what I got too... That's just v*cos(theta). It results that v is equal to f(x).
Now try writing x(t). :)

In the consevation of Energy formula you got the speed with respect to the height only which is x^2 so we actually got vx :~> when you replace y with x^2 and vy when you leave y unchanged. PS: i'm not an expert either. :D

rollingstein
Gold Member
In the consevation of Energy formula you got the speed with respect to the height only which is x^2 so we actually got vy :~>
Nope. I don't think so.

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

Last edited:
mfb
Mentor
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}}##.

Thank you! At last :D