average of any quantity Q over time t= (1/t)∫Qdt
here et at any instant frictional torque be Q
then
dL/dt=Q
or
Qdt=dL
∫Qdt=Lf-L0
finally there will be no slipping and hence same angular velocities
conserving angular momentum
I2w + 2Iw = (I+2I)w'
gives w'=(4w/3)
thus Lf(of let A)-L0 = 2I(4w/3)-2I(w)=2Iw/3
thus
∫Qdt=2Iw/3 hence (1/t)∫Qdt =2Iw/3t
loss of kinetic energy =
I(4w/3)2/2 + 2I(4w/3)2/2 - I(2w)2/2-2I(w)2/2
= -Iw2/3 thus loss = Iw2/3