waise ek bande ne answer de diya hai but i will post my own solution it was a 6 out of 10 problem normally found in resonance paper.......

a=centre of mass accel.

A=angular acc.

Mg-N=ML/2Acos@

f=ML/2Asin@

torque about the end point

MgL/2cos@=IA

A=3gcos@/2L

applying SIR ISAAC NEWTON'S 2nd. LAW

friction=MLsin@3gcos@/4

uN=3Mgsin@cos@/4

Mg-N=MLcos@3gsin@/4l

N=Mg-3MGcos^@/4

u(Mg-3Mgcos^@/4)=3Mgsin@cos@/4

u=3sin@cos@/(4-3cos^@)

u=3sin@cos@/(1+3sin^@)