- #1

indie452

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two people are holding the ends of a rod length l and mass M, show that if one person let's 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?