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Time Period in Vertical Circular Motion

  1. Mar 25, 2010 #1
    1. The problem statement, all variables and given/known data

    A bucket is rotated in a vertical circle such that the water just does not fall. Radius of the circle is R. Gravitational acceleration: g. Find the Time period of one circular motion.

    http://img148.imageshack.us/img148/6583/bucket.jpg [Broken]

    3. The attempt at a solution

    I obtained an expression for w (omega) (i.e. d(theta)/dt) in terms of theta and integrated. I want to know whether this is right and if there is a better approach (possibly not involving integration).

    For the water to just not fall, the velocity at the lowest point will be [tex]\sqrt{5gR}[/tex]

    Now here is the energy conservation equation for a general point,

    [tex]1/2m*5gR = 1/2mv^{2} + mgR(1 - cos\theta)[/tex]

    Solving for v,

    [tex]v = \sqrt{(3 + 2 cos\theta) gR}[/tex]
    [tex]\omega = d\theta/dt = \sqrt{(3 + 2 cos\theta) g/R}[/tex]

    Now separating the variables and integrating,

    [tex]T (Time Period) = \int d\theta/(\sqrt{(3 + 2 cos\theta) g/R})[/tex] (using limits 0 to 2 pi)

    Here I'm having trouble integrating this thing, so I'm wondering whether there is a simpler method.
    Last edited by a moderator: May 4, 2017
  2. jcsd
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