A falling object problem

  • #26
rollingstein
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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 [itex]\vec{v}[/itex] 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.
 
  • #27
Omg! What a humongous mistake in my formulation of the conservation of energy!
 
  • #28
rollingstein
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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.
 
  • #29
rollingstein
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Here's what I get :

[itex] \frac{dx}{dt} = \sqrt{2g} \sqrt{\frac{x_1^2-x^2}{1+4x^2}} [/itex]

Sad part: If this is right the integral is ugly.
 
  • #30
It is right. That's what I got too... That's just v*cos(theta). It results that v is equal to f(x).
 
  • #31
rollingstein
Gold Member
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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). :)
 
  • #32
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
 
  • #33
rollingstein
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646
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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.
 
  • #34
I did a simulation with vx and it matched the sinusoidal, it seems. Any help from the experts?
 
Last edited:
  • #35
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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}}##.
 
  • #36
Thank you! At last :D
 

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