Determining the equations of motion for a 2 DOF in 2 directions I'm having trouble understanding a 2DOF system that moves in 2 directions. The problem asks me to determine the EOM for a system shown in the attached image:

My first idea is to transform this system into a simpler one in which the two masses move in the same direction instead of the horizontal and vertical directions. I have determined that the displacement of $m_1$ is directly related to the rotation of the pulley due to the inextensible string.
Let $x$ be the displacement of $m_1$ and y be the displacement of $m_2$, therefore, $x=φr$ and $\dot x= \dot φr$.
However, I am confused as to how to relate the displacement of $m_2$ to the pulley. Additionally, I am having trouble understanding how to incorporate the inertial moment of the pulley into the EOM.
Any help would be appreciated, however, please do not give me a direct answer as I understand better if I am able to solve the problem with only hints and pointers.
 A: If the pulley has negligible moment of inertia or the string slides around it without friction then you can ignore the rotation of the pulley.
It will simplify the problem if you measure displacements and extensions of springs from the equilibrium position when mass 2 hangs freely. Neither mass is accelerating in the equilibrium position so the tension in the string and both springs is then $T_0=m_2 g$.
At some arbitrary time let the displacements of masses 1 and 2 from their equilibrium positions be $x$ and $y$. The extensions of springs 1 and 2 are then $x$ and $y-x$ so the tensions in them above the equilibrium value $T_0$ are $k_1 x$ and $k_2 (y-x)$. The accelerations of the masses are $\ddot x$ and $\ddot y$.
Apply Newton's 2nd Law of Motion to each mass :
$$m_2 \ddot y=-k_2 (y-x)$$
$$m_1 \ddot x=-k_1x+k_2(y-x)$$
The general solutions $x(t), y(t)$ of these coupled differential equations are superpositions of 2 normal modes in which $y=+ax$ and $y=-ax$ each with a different frequency. ($a$ is a constant.)
