Recalling #theta# as the angle between the #x# axis and the rod, (this new definition is more according to the positive angle orientation) , and considering #L# as the rod length, the rod's center of mass is given by
#(X,Y)=(x_A+L/2cos(theta),L/2 sin(theta))#
the horizontal sum of intervening forces is given by
#mu N "sign"(dot x_A)=m ddot X#
the vertical sum gives
#N-mg=m ddotY#
Considering the origin as the moment reference point we have
#-(Y m ddot X+X m ddot Y)+x_A N-X m g= J ddot theta#
Here #J = mL^2/3# is the inertia moment.
Now solving
#{(mu N "sign"(dot x_A)=m ddot X),(N-mg=m ddotY),(-(Y m ddot X+X m ddot Y)+x_A N-X m g= J ddot theta):}#
for #ddot theta, ddot x_a, N# we obtain
#ddot theta = (L m(cos(theta)+mu"sign"(dot x_A)sin(theta))f_1(theta,dot theta))/f_2(theta,dot x_A)#
#N=-(2Jm f_1(theta,dot theta))/f_2(theta,dot x_A)#
#ddot x_A=f_3(theta,dot theta,dot x_A)/(2f_2(theta,dot x_A))#
with
#f_1(theta,dot theta)=Lsin(theta)dot theta^2-2g#
#f_2(theta,dot x_A)=mL^2(cos^2(theta)+mu cos(theta)sin(theta)"sign"(dot x_A)+4J#
#f_3(theta,dot theta,dot x_A)=(g mu (8 J - L^2 m + L^2 m Cos(2theta) "sign"(dot x_A)-
g L^2 m Sin(2theta)+
L ((4 J + L^2 m) Cos(theta) + (L^2 m-4J)mu "sign"(dot x_A) Sin(theta)) dot theta^2)#
