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What torque will bring the balls to a halt in 5.6s?

  1. Oct 14, 2007 #1
    1. The problem statement, all variables and given/known data

    A 1.2kg ball and a 2.2kg ball are connected by a 1.0m long rigid, massless rod. The rod is rotating clockwise about its center of mass at 26rpm. What torque will bring the balls to a halt in 5.6s?

    2. Relevant equations

    T = I(alpha)
    Wf = Wi + alpha (delta t)
    I = m1(r1^2) + m2(r2^2)

    3. The attempt at a solution

    I tried finding I = (1.2kg)(.5m)^2 + (2.2kg)(.5m)^2 = .85. Then I tried using Wf = Wi + alpha(delta t) to find alpha, which I got to be -.829. Then I tried using T = I(alpha) and I got -7.047, but it was wrong.
     
  2. jcsd
  3. Oct 14, 2007 #2

    hage567

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    You need to find the centre of mass of the system. Since the balls are different masses, the axis of rotation will not be in the middle of the rod. Once you have the position of the centre of mass, you can find I.
     
  4. Oct 14, 2007 #3
    The center of mass = (x1m1 + x2m2)/(m1 + m2), but I'm not sure what the positions would be.
     
  5. Oct 14, 2007 #4

    hage567

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    Think of it like trying to balance this system on a fulcrum. You need to figure out where to set it so the system doesn't rotate. So the sum of the torques (rxF) will be zero. You can express x1 in terms of x2 since you've been given the total length of the rod. That lets you reduce your equation to one unknown variable (x2), that you can solve for and therefore determine the positions of both masses.
     
  6. Oct 15, 2007 #5
    So it would be the Xc = (1-x2)m1 + x2m2/(m1 + m2). I'm still not sure how I would solve for x2. I mean, if I set that equation to zero I get 1.201, but that can't be right because it's bigger than the whole length.
     
  7. Oct 15, 2007 #6

    hage567

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    For equilibrium: m1(1-x2) - m2x2 = 0 (summing the torques) The negative sign is what will make it work. (since one of the masses lies in the negative direction of your reference point)

    I hope I'm making some sense, I'm short on time.
     
  8. Oct 15, 2007 #7
    So x2 would be -1.2?
     
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