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Frictionless Bead Sliding Down A Parabola

  1. Nov 21, 2008 #1
    1. The problem statement, all variables and given/known data
    It seems simple, doesn't it? A bead starts from (0,1/2) and simply slides down the parabola y=(1/2)(x-1)2 under gravity. The problem is to get its position rel. to origin in terms of time, ie.
    r = [x(t),y(t)]. Anyone into this?

    2. Relevant equations
    y' = x-1 = tan A where A is angle tangential veloc. (v) (and tang. accel.(a)) makes with x axis.
    a = -g sin A .

    3. The attempt at a solution
    For starters, can I say that sin A = -(x-1)/sqr(1+(x-1)^2) ?
  2. jcsd
  3. Nov 21, 2008 #2
    What quantities do you think are conserved? Can you use them?
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