A falling object problem

  • #1

Main Question or Discussion Point

I have a bowl in the form of a parabola (say y=x^2, motion is in 2d! On the graph of the parabola) and i place an object on the edges of the bowl and let if fall in the bowl. I have gravity, and no friction. My question is plain simple: what are it's equations of motion? I know it has to oscillate. I know calculus quite well, but that doesn't seem to help. Is there something I'm missing out? I would simply love to see the correct equations :D. Thank you very, very much in advance.
 

Answers and Replies

  • #2
tiny-tim
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Welcome to PF!

Hi Binaryburst! Welcome to PF! :smile:

Start by using conservation of energy, and that will give you v as a function of y.

Carry on from there. :wink:
 
  • #3
Thanks for the tip :D but i've already gotten the speed with respect to x or y. I was wondering how I could get it as a function of time :)
 
  • #4
tiny-tim
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ok, now use cos or sin to get the x or y component of the speed …

you now have x' as a function of x (or y' as a function of y) :wink:
 
  • #5
I'm on it.
 
  • #6
I get elliptic integral !?
 
  • #7
tiny-tim
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very likely!

well, if you will start with a parabola! :rolleyes:
 
  • #8
Wow... Thanks a lot! :D this is great!
 
  • #9
If I take the whole speed, not just it's components I get a sin(t) :D
 
  • #10
tiny-tim
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sorry, not following you :redface:

you'll need to show the equations :smile:
 
  • #11
rollingstein
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I get elliptic integral !?
I was trying too. Do you get:

dx/dt=sqrt(2g) sqrt((1-4*x^2)*(x1^2-x^2))
 
  • #12
Nope. I'll post the equations right away.
 
  • #13
I get this: (2g(x1^2-x^2))^0.5. Then if i integrate it with respect to t i get x=x0*sin(t(2g)^0.5).
 
  • #14
rollingstein
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I get this: (2g(x1^2-x^2))^0.5. Then if i integrate it with respect to t i get x=x0*sin(t(2g)^0.5).


Somehow my expression for v_x has an additional (1-4x)^0.5 factor. Not sure what I'm doing wrong.
 
  • #15
rollingstein
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I get this: (2g(x1^2-x^2))^0.5. Then if i integrate it with respect to t i get x=x0*sin(t(2g)^0.5).
This term (2g(x1^2-x^2))^0.5

Is that your v or v_x?
 
  • #16
My equation is for v alone.
 
  • #17
I'm thinking how could i get it without using the conservation of energy and using forces.
 
  • #18
rollingstein
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  • #19
I had v=f(x). Rewritten it as follows: v/f(x)=1 ; 1/f(x)*dx/dt=1. ; Integrate with respect to t
Int( 1/f(x)* dx/dt * dt ) = int( 1 dt )
 
  • #20
rollingstein
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I had v=f(x). Rewritten it as follows: v/f(x)=1 ; 1/f(x)*dx/dt=1. ; Integrate with respect to t
Int( 1/f(x)* dx/dt * dt ) = int( 1 dt )
That's where I think you are wrong.

You've put v=dx/dt.

Only v_x = dx/dt

But I could be talking nonsense! Be warned. But I'd love to see your opinion.

Your solution does look super tempting! :)
 
  • #21
Hmmm.. That's interesting I actually got the vx. Sorry for the blunder. I was too excited :) correcting the mistake.
 
  • #22
rollingstein
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Hmmm.. That's interesting I actually got the vx. Sorry for the blunder. I was too excited :)
Funny point is your solution seems to satisfy all the boundary conditions etc. I'm puzzled.

My solution integrates to something super messy. :(
 
  • #23
I am super puzzelled as well. I can't figure out what I did.
 
  • #24
rollingstein
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Another reason why your solution seems wrong to me.

take dx/dt and dy/dt

[itex]v_x^2 + v_y^2 = v^2 [/itex]

But yours don't seem to sum up to v. Try.
 
Last edited:
  • #25
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).
 

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