thank u very much shubham for rewriting the ques....now its absolutely clear....

draw a rod abtouch roller(circle) at b and ground at a...ab=L...

draw the free body diagram...N1 and N2 be normals at a and b perpendicular to ground and circle respectively....

eqns are as follows..

N1+N2cos(theta)=mg

uN1=N2sin(theta)

N2L=(mgL/2)cos(theta).....

where u is the coeff of friction..

from third get N2 and substitute in 2 to get N1...now put both of them in 1 to get u=sin(theta)cos(theta)/(2-cos^2(theta))....

.