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Angular momentum of a uniform bar w/ two forces applied

  1. Nov 6, 2009 #1
    1. The uniform bar shown in the diagram has a length of 0.80 m. The bar begins to rotate from rest in the horizontal plane about the axis passing through its left end. What will be the magnitude of the angular momentum of the bar 6.0 s after the motion has begun? The forces acting on the bar are shown.

    170580A.jpg




    2. L = Iw, L = I(alpha)t



    3. I know that I = (1/3)mL^2 = 16m/75

    So, L = (16m/75)*6sec*alpha

    Am I going about this correctly so far? How do I bring the forces into the equation?
     
  2. jcsd
  3. Nov 6, 2009 #2
    What do you apply to a rigid, rotating body to increase/decrease it's angular momentum?
     
  4. Nov 7, 2009 #3
    ... torque. I still don't see what I'm doing. net torque = 4*.6*sin(90) = 2.4 = the derivative of L
     
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