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 / 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?