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Rotational Kinematics-Bead Constrained in a Hoop Rotating

  1. Sep 27, 2008 #1
    1. The problem statement, all variables and given/known data
    http://docs.google.com/Doc?id=d277r7r_58d3chgqfj


    2. Relevant equations
    a=w^2*r


    3. The attempt at a solution

    a=w^2*r
    w^2*r*cos(theta)=-mg

    theta=arcos(-mg/(Rw^2))

    I'm pretty sure that's right, because the vertical component of the acceleration has to equal the vertical component due to gravity. However, I wasn't sure if acceleration was simply w^2*R or if I had to do something more complicated to find it. Is my reasoning/answer correct?
     
  2. jcsd
  3. Sep 27, 2008 #2
    That's pretty close. You can see it's not quite right, though, because your second equation in your attempt at a solution part is dimensionally inconsistant. You have a force equal to an acceleration.
     
  4. Sep 27, 2008 #3
    oooh. didn't catch that. thank you. so then i just add a m to the left side of the equation, which makes the m's cancel out, and the final answer is theta=arcos(-g/(Rw^2)) Is that right now?
     
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