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Simple cycloid time calculation problem

  1. May 20, 2007 #1
    1. The problem statement, all variables and given/known data

    Consider a wire bent into the shape of the cycloid

    [tex]x = a(\theta - \sin\theta)[/tex]
    [tex]y = a(\cos\theta -1)[/tex]

    If a bead is released at the origin and slides down the wire without friction, show that [itex]\pi\sqrt{a/g}[/tex] is the time it takes to reach the point [itex](\pi a, -2a)[/tex] at the bottom.

    2. Relevant equations

    (See below)

    3. The attempt at a solution

    Energy conservation gives

    [tex]\frac{1}{2}mv^{2} = mg(2a)[/tex]
    [tex]v^{2} = 4ga[/tex]

    For the point at the bottom, [itex]\theta = \pi[/itex]. So, the arc length is

    [tex]s = \int_{0}^{\theta}\sqrt{\left(\frac{dx}{d\theta}\right)^{2} + \left(\frac{dy}{d\theta}\right)^{2}}d\theta[/tex]

    [tex]v = \frac{ds}{dt}[/tex]

    How do I get rid of the [tex]d\theta/dt[/tex]? I know I'm missing something here...:rolleyes:
  2. jcsd
  3. May 20, 2007 #2
    Okay I got it. :tongue2:
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