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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
    Calculate the moment of inertia about the pivot point using the parallel axis theorem.
     
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