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Time calculation with non uniform acceleration

  1. Aug 23, 2009 #1
    How long does it take a skateboarder to roll from the top of a half pipe of radius 2 meters to the bottom? Ignore friction, and assume the half pipe is in the shape of a semi circle. I tried to use integration to solve this but failed miserably, would someone please show me how to do this? Show full solution please.
     
  2. jcsd
  3. Aug 23, 2009 #2

    rcgldr

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  4. Aug 23, 2009 #3

    tiny-tim

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    Welcome to PF!

    Hi woodie37! Welcome to PF! :smile:

    Use conservation of energy to find angular speed as a function of angle, and then integrate. :wink:

    Show us what you get. :smile:
     
  5. Aug 23, 2009 #4
    We love to help students like you, but please show us what you've tried so we can help :).

    You can use either integration or conservation of energy as tiny-tim has said.

    Using integration is somehow more difficult, and has a longer path, and will lead eventually to the law of conservation of energy.

    But using the law of conservation of energy is easier, and will give you a very direct answer.

    Good luck :)
     
  6. Aug 23, 2009 #5
    But will he be able to "work" with time in all this calculations??

    Would't it be the same than calculating the pendulum equation?

    It happened to me that I tried for many months when i was younger (about 16 y.o.) but y could only find w(&), that is to say omega (theta) what is angular speed in function of angle.

    But I always wondered if it was possible to aply power relationships to the energy conservation in order to calculate such times. Then, is it possible?

    Thanks in advance and sorry my english.
     
  7. Aug 23, 2009 #6
    That's what I've been saying, pal. He can either use the chain rule with circular motion laws, and derive the energy conservation laws, or he can directly apply the energy conservation laws, which I can do in 2 lines!!!
     
  8. Aug 23, 2009 #7
    I don't think he's gonna show up again, by the way. I think he expected us to solve problems for him, that's very bad. Anyway if you want to discuss the answer you're welcome too ;). Show us something and let's talk ;)
     
  9. Aug 24, 2009 #8

    tiny-tim

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    Hi TheDestroyer! :smile:
    Conservation of energy saves energy! :wink: :biggrin:
     
  10. Aug 24, 2009 #9
    hehehehehe :D, that's right ;). I really used to do it using chain rules and integrals, and noticed how stupid I had been been when I discovered the energy easy way :P
     
  11. Aug 24, 2009 #10
    but chain rule way is more "mathematically" beautyful :tongue2:
     
  12. Aug 24, 2009 #11
    Yeah, it shows how consistent physics and mathematics are :D, I really was surprised when discovered how whatever you do, if you do it right, it gives you the same answer :D
     
  13. Aug 24, 2009 #12
    Yehh, that's the beauty of natural science even in this very early and simple level.

    Good science Destroyer ;)
     
  14. Sep 23, 2009 #13
    Ok sorry about the delay but I haven't been on the computer for awhile, just checked my email and read that someone gave a reply.

    My solution is as follows

    starting from the top of the half circle, angle x is equal to 0 while after passing a half circle, angle x is 180

    Btw I did use conservation of energy to figure out the velocity (not angular) as a function of x

    -U=K where U is the potential energy and K is the kinetic
    so 2grsinx=v^2 v=[tex]\sqrt{2grsinx}[/tex]
    The distance, d = [tex]\pi[/tex]r where r is the radius
    and t = d/v = d(2grsinx)^(-1/2)

    Then I integrated...

    t=[tex]\int[/tex]d(2grsinx)^(-1/2)dx

    Then the thing became difficult...someone help me out plz
    and could someone, while explaining the physics, also give me a quick tutor of how to use the latex reference to insert symbols? cuz i dont understand programming
     
  15. Sep 24, 2009 #14

    tiny-tim

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    Hi woodie37! :smile:

    (have a theta: θ and an omega: ω and a pi: π and a square-root: √ and try using the X2 tag just above the Reply box :wink:)
    Nooo :redface: … use v = r(dx/dt)

    (or preferably write θ instead of x, and ω instead of dx/dt, to give you v = √(2grsinθ) = ωr)

    [noparse]For LaTeX on PF, just type [tex] before and [/tex] after, [/noparse]and see http://www.physics.udel.edu/~dubois/lshort2e/node61.html#SECTION008100000000000000000" [Broken] for symbols. :smile:
     
    Last edited by a moderator: May 4, 2017
  16. Sep 24, 2009 #15
    ok thx and if i get more problems i'll ask again =D
     
  17. Sep 24, 2009 #16
    Looking at the bastard of an integral I ended up with, along with the wiki page, is there any way to solve this without resorting to stuff like Elliptic functions?

    Other than giving us a function for the angular velocity as a function of sine of theta, how does conservation of energy really help here?

    Solving for velocities, accelerations and angular positions is straight-forward, but time mucks everything up. :\ I always did suspect there was a reason we used the small angle approximation.
     
  18. Sep 27, 2009 #17
    Sorry to double-post, but this is an interesting question.
    I've looked for a solution for the last couple of days, but the only things I've found are with the Elliptic functions and potential wells and whatnot, and that seems a bit out of the scope of this question. Further insight would be greatly appreciated.
     
  19. Sep 28, 2009 #18

    Doc Al

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    If anyone thinks they can solve for the time using conservation of energy, I'd love to see it. The only way I know involves elliptical integrals (as RoyalCat found).

    If I'm missing something simple, I await enlightenment. (It wouldn't be the first time I missed the obvious!)
     
  20. Sep 30, 2009 #19
    Can someone post a solution here involving elipticals? I'd like to see a solution plz, any would be fine, be it elliptical or be it other. =D
     
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