A falling object problem

[rollingstein]

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.

 

[Binaryburst]

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

 

[rollingstein]

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]

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.

 

[Binaryburst]

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]

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). :)

 

[Binaryburst]

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]

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.

 

[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