* Taking Torque about the bottom most point
$$ 2kxr + \mathcal{N}x = 2Mr^2\alpha $$
$$ mg – \mathcal{N} = m\alpha x $$
Substituting $\mathcal{N}$:
$$ 2kxr + m(g – \alpha x)x = 2Mr^2\alpha $$
$$ 2kxr + mgx = \alpha(2Mr^2 + mx^2) $$
$$ \alpha = \frac{(2kr + mg)x}{(2Mr^2 + mx^2)} $$
$$ a_c = \alpha r = \left( \frac{2kr + mg}{2Mr^2 + mx^2} \right) xr $$