Rotational Kinematics

A thin rod is held resting on a rough ground with its length inclined at an angle (theta) to the horizontal. The coefficeint of friction between rod and ground is (mu). Show that when the rod is let go it will start slipping on the ground if (mu) < 3sin(theta)cos(theta)/(1+3sin^2(theta)

Attempt-

considering the torque acting about the point of contact,
mgcos(theta) l/2=ml^2/3 (alpha)
from here,
tangential acceleration is 3gcos(theta)/4

Using conservation of energy,
mgl/2 (1-cos(theta))=0.5 ml^2/3 w^2
(taking reference level through the initial position of centre of mass)
from here i get w^2=3g(1-cos(theta))/l

centripetal acceleration of the centre of mass=mw^2 l/2
=3g(1-cos(theta))/2

now the frictional force balances the force due to centre of mass in x-direction
taking components of tangential and centripetal acceleration along x- direction I got one equation involving N-the normal force which is perpendicular to the ground. ( (mu)N=m(ax))

then taking components of accelerations along y-direction ,
mg-N=m(ay)

solving for N I did not get the right answer.
Instead if i proceed without considering centripetal acceleration of the centre of mass, I got the correct answer. How?