Stuck on relative angular motion and inertial forces OK,before I start explaining my problem, let's consider this system - You have a particle rotating about the z-axis of an inertial frame of reference, with angular velocity $\omega$ with a radius $r$, let's label all the quantities measured with respect to this frame of reference with a subscript $A$. Now let's consider frame of reference $B$, which is rotating about it's z axis with angular velocity $\omega _0$, and the z-axis coincides with that of $A$.
If I were to find out the net acceleration of the particle in $B$, it obviously will be $\omega _B ^2 \cdot r$
And $\omega _B = \omega _A - \omega _0 = \omega - \omega _0$
So, I could say that the net acceleration is $\left( \omega - \omega _0 \right)^2\cdot r$
But if I were to approach it using inertial forces,
$$ma = -F_B + F_A$$
$F_B$ is the centrifugal force caused by the rotation of B
$$F_B = m\cdot w_0^2r, \space F_A = m\cdot w_a^2r$$
$$\therefore a = r\cdot \left[ w^2 - w_0^2 \right]$$


So, yeah, what am I doing wrong? I can't get my head around this concept..
EDIT: Made a few mistakes in the last two equations
 A: Centrifugal force $F_0=mr\omega_0^2$ is not the only fictitious force in a rotating frame of reference. When the object is moving with non-zero velocity $v$ in the rotating frame, there is also a Coriolis force, which has magnitude $F_1=2m\omega_0 v$. This force is what you are missing.
In your example, $v=r(\omega-\omega_0)$, so the Coriolis force is
$F_1=2mr\omega_0(\omega-\omega_0)=2mr(\omega_0\omega-\omega_0^2)$.
If $\omega>\omega_0$ as in your example - ie the object is moving in the direction of rotation - then the Coriolis force is outward, away from the axis, the same as the centrifugal force. If $\omega<\omega_0$ the Coriolis force would be inward, towards the axis. 
The centripetal force measured in the rotating frame of reference is therefore
$F_B=F_A-F_0-F_1$
$=mr\omega^2-mr\omega_0^2-2mr(\omega_0\omega-\omega_0^2)$
$=mr(\omega^2-2\omega_0\omega_0+\omega_0^2)$
$=mr(\omega-\omega_0)^2$
which agrees with your formula for the net acceleration $r(\omega-\omega_0)^2$ measured in B's frame of reference.
A: Newton's Second Law only holds in inertial reference frames.  
If $A$ is an inertial frame, then $F_A=ma_A$.  Since the $B$ reference frame is rotating with respect to the $A$ frame, $B$ is a non-inertial frame; thus $F_B\ne m a_B$ for the quantities measured in the $B$ frame.
