The figure from below shows a system featuring two coupled pulleys have $R_1=1.2\,m$ and $R_2=0.4\,m$, The mass of $m_1=20\,kg$ and $m_2=30\,kg$ and the total moment of inertia of the two pulleys is $26.4\,kg \cdot m^2$ . Find the angular acceleration in $\frac{rad}{s^2}$ of the system of pulleys. The acceleration due to gravity is $g=10\,\frac{m}{s^2}$.

The alternatives are as follows:

$\begin{array}{ll}

1.&10\,\frac{rad}{s^2}\\

2.&8\,\frac{rad}{s^2}\\

3.&6\,\frac{rad}{s^2}\\

4.&4\,\frac{rad}{s^2}\\

5.&2\,\frac{rad}{s^2}\\

\end{array}$

What I've attempted to do is displayed in the sketch from the figure from below:

After identifying the forces acting on the pulleys, I could spot that the best approach would be to find the torque acting on the wheels and from then finding the angular acceleration as follows:

$\tau=I\alpha$

In a similar fashion to that of the Atwood's machine, what I did was the following, which is to find the values of each tension acting on the wheels.

For the block which mass is the heavier of the two:

$T-mg=m(-a)$

$T=300-30a$

For the block which has a less heavy mass:

$T-mg=ma$

$T=mg+ma$

$T=200+20a$

Therefore:

$-(1.2)\left(200+\frac{20\alpha}{1.2}\right)-(0.4)\left(300-\frac{30\alpha}{0.4}\right)=26.4\alpha$

Solving this equation results into:

$\alpha\approx -21.95$

But this answer doesn't get any close to what the alternatives do say. Supposedly, this answer results into $2\frac{rad}{s^2}$, but I'm confused because each time I go back in my steps I end up getting the same answer. Can somebody help me with this please?