oh yeah ! sorry for that error !! ( i wrote energy as Iw lol )
let normal and friction due to horizontal surface be denoted by 1 and due to vertical surface be denoted by 2
mg = N1+f2
N2=f1
f1=μN1, f2=μN2=μf1
so μmg = f1(1+μ2)
and (f1+f2)Rθ = 1/2 I w2
then no of rotations = theta /2pi
u will get ans = Rw2(1+μ28pigμ(1+μ)