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Radius of centripetal force when on an angle.

  1. Oct 5, 2013 #1
    http://puu.sh/4IjIT.png [Broken]

    I don't quite understand why this is correct. My initial try was this... (the only difference is at the end, bolded)

    r = 0.154m
    period = 0.426s
    rev per second = 1/0.426s
    w = 2*pi*(1/0.426)

    mg = Fn*cosx
    Fn = mg/cosx
    Fc = Fn*sinx = mg/cosx
    = mg*tanx

    mg*tanx = mv^2/r
    mg*tanx = m(r*sinx)w^2
    g*tanx = (r*sinx)*w^2
    9.8*tan(x) = 0.154*sinx*[2*pi*(1/0.426)]^2

    EDIT: So this way does work, why did the way in the screenshot get the same answer?
    EDIT 2: Oh... I just did it wrong and coincidentally got a close enough answer to round to the right answer.
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Oct 5, 2013 #2

    Doc Al

    User Avatar

    Staff: Mentor

    You are correct. The radius of the circular path made by the bead is not the radius of the hoop.
     
  4. Oct 5, 2013 #3
    So why did I get the correct answer by using the full radius of the hoop as the path made by the bead?

    When I use my way I get 90 degrees, so it's definitely wrong.

    EDIT: Did the math wrong, my way does work.

    9.8*tan(x) = 0.154*sinx*[2*pi*(1/0.426)]^2
    9.8/cosx = 0.154*[2*pi*(1/0.426)]^2
    9.8 = cosx*(0.154*[2*pi*(1/0.426)]^2)
    arccos(9.8 / [0.154*[2*pi*(1/0.426)]^2]) = 73
     
    Last edited: Oct 5, 2013
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