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Lots of Forces

  1. Dec 30, 2003 #1
    Consider a particle moving without friction on a rippled surface, given by the equation h(x) = d Cos[k x]. Gravity acts down in the negative h direction. If the particle starts at x=0 with a speed in the x direction, for what value of v will the particle stay on the surface at all times?

    The answer is if v<= Sqrt[g/(k^2 d)].

    I found a general formula that involves x but is too complicated to solve. There must be another way to do this without using a calculator. Thanks!

    Ying
     
  2. jcsd
  3. Dec 30, 2003 #2

    enigma

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    Look at the boundary conditions.

    The point where the the particle will leave the surface is the point with the largest curvature. You need to solve for the required speed at the crest of the wave, and it will hold for all other points on the wave.
     
  4. Dec 30, 2003 #3
    How do you know that the particle will leave the surface at the crest of the wave?
     
  5. Dec 30, 2003 #4

    enigma

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    At the crest, the centripetal acceleration needed to maintain contact is the highest.

    It can leave at another point, but it has to be going faster than the crest's max speed to do so.
     
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