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Classical mechanics-acc. of a rod using inertia

  1. Nov 23, 2009 #1
    1. The problem statement, all variables and given/known data

    two people are holding the ends of a rod length l and mass M, show that if one person lets go the initial acceleration of the free end is 3g/2


    3. The attempt at a solution
    i worked out the moment of inertia about centre mass (cm) and got = Ml2/12

    because L=Iw
    dL/dt = I*dw/dt
    torque (T) = I*ang.acc. (a)
    so
    a = T/I = mglsin[90] / Ml2/12
    = 12g/l
    this is wrong so i used l=l/2 as this is the distance to the cm.
    so a = 24g/l also wrong

    what am i doing wrong?
     
  2. jcsd
  3. Nov 23, 2009 #2
    Are you sure? As you state, this is the moment of inertia about the center of the rod. But if two people are holding the ends and one lets go, it is not rotating about the center, is it?
     
  4. Nov 24, 2009 #3
    yeah i realised that now and took the mom.of inertia at the pivot and with torqu found the answer.

    the next part of the question however is show how the load supported by the one person falls from mg/2 to mg/4.

    i have thought about balancing out the downward forces and the upward forces

    F1=mg/2 ...............F2=mg/2
    ^...........................^
    |__________________|
    .................|
    .................V
    ..............F=mg

    but for the next part after one person has let go i tried saying that F2 will become a downward force = 3mg/2 but this doesnt seem right
     
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