Consider a sliding cursor C that is fixed to the disk which has a constant angular velocity counterclockwise at 4 rad/s. Determine the angular velocity and angular acceleration of the AB bar at the instant shown. [0.667 rad/s (counterclockwise); 3.84 rad/s2 (Clockwise)]
to solve this exercise I draw the two axes
I've done
$$\begin{align} v_C &= v_O + \Omega_{OC} \times r_{OC} \\ &= 0 + 4k * (-0.06\sin(30)i+0.06\cos(30)j) \\ &= -0.12j-0.2078i \\ v_C &= v_A + \Omega_{AC}r_{AC} + v_{rel} \\ &= 0 + \Omega_{AC}k\cdot 0.180\cos(60)i + 0.18\sin(60)j) + v_{rel}\cos(60)i + v_{rel}\sin(60)j \\ &= (-0.156\Omega_{AC} + 0.5v_{rel})i + (0.09\Omega_{AC} + 0.866v_{rel})j \end{align}$$
and therefore i got $v_{rel} = -0.2079\text{ m/s}$ and $\Omega_{AC}=0.667\text{ rad/s}$ which are the correct results
$$\begin{align} a_C &= a_0 + \alpha_{OC}\times r_{OC} + \Omega_{OC}\times (\Omega_{OC}\times r_{OC}) \\ &= 0 + 0 + 4k(4k(-0.06\sin(30)i + 0.06\cos(30)j)) \\ &= 0.48i-0.8312j \\ a_C &= a_A + \alpha_{AC}\times r_{AC} + \Omega_{AC}\times (\Omega_{AC}\times r_{AC}) + a_{coriolis} + a_{rel} \\ &= 0 + 0 + \alpha_{AC}k(0.18\cos(60)i + 0.18\sin(60)j) + (-0.44i - 0.069j) +(-0.139j + 0.24i) + a_{rel}(\cos(60)i + \sin(60j)) \\ &= (-0.156\alpha_{AC} - 0.2 + 0.5a_{rel})i + (0.99\alpha_{AC} - 0.208 + 0.866a_{rel})j \end{align}$$
and therefore i got $a_{rel}=0.56\text{ m/s}^2$ and $\alpha_{AC} = -6.156\text{ rad/s}^2$, which is the wrong value for $α_{AC}$.
is it possible to solve the exercise using the 2 axes that I draw? because I can't understand where my resolution is wrong