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A falling object problem

  1. Feb 2, 2013 #1
    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.
     
  2. jcsd
  3. Feb 2, 2013 #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:
     
  4. Feb 2, 2013 #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 :)
     
  5. Feb 2, 2013 #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:
     
  6. Feb 2, 2013 #5
    I'm on it.
     
  7. Feb 2, 2013 #6
    I get elliptic integral !?
     
  8. Feb 2, 2013 #7

    tiny-tim

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    very likely!

    well, if you will start with a parabola! :rolleyes:
     
  9. Feb 2, 2013 #8
    Wow... Thanks a lot! :D this is great!
     
  10. Feb 2, 2013 #9
    If I take the whole speed, not just it's components I get a sin(t) :D
     
  11. Feb 2, 2013 #10

    tiny-tim

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    sorry, not following you :redface:

    you'll need to show the equations :smile:
     
  12. Feb 2, 2013 #11

    rollingstein

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    I was trying too. Do you get:

    dx/dt=sqrt(2g) sqrt((1-4*x^2)*(x1^2-x^2))
     
  13. Feb 2, 2013 #12
    Nope. I'll post the equations right away.
     
  14. Feb 2, 2013 #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).
     
  15. Feb 2, 2013 #14

    rollingstein

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    Somehow my expression for v_x has an additional (1-4x)^0.5 factor. Not sure what I'm doing wrong.
     
  16. Feb 2, 2013 #15

    rollingstein

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    This term (2g(x1^2-x^2))^0.5

    Is that your v or v_x?
     
  17. Feb 2, 2013 #16
    My equation is for v alone.
     
  18. Feb 2, 2013 #17
    I'm thinking how could i get it without using the conservation of energy and using forces.
     
  19. Feb 2, 2013 #18

    rollingstein

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    What did you do next?
     
  20. Feb 2, 2013 #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 )
     
  21. Feb 2, 2013 #20

    rollingstein

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    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! :)
     
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