Remark. This case generalizes to your real goal to have a system of more than two mass-points attached to each other by rods where the rods can rotate relative to each other. Basically, in a uniform force field like the gravitational force field near the surface of the earth the problem becomes equivalent to the geodesic motion of a point onto a torus of dimension equal to the number of rods.
Lagrangian approach. First, you place an inertial coordinate frame (say a frame attached to the ground) in the plane of the system. Denote it by $O\,\vec{e}_x\,\vec{e}_y$, where the coordinate vectors are of unit length and perpendicular to each other. Moreover, $O\,\vec{e}_x\,\vec{e}_y$ is oriented so that $\vec{e}_y$ is the vertical vector along which gravitational acceleration is $\vec{g} = - \, g \, \vec{e}_y$. The position vectors of the mass-points, originating from point $O$ and ending at the either of the mass points, are
$\vec{r}_M$ and $\vec{r}_m$. The Lagrangian is then
$$L\, =\, \frac{M}{2}\,\left|\frac{d\vec{r}_M}{dt}\right|^2
\, + \, \frac{m}{2}\,\left|\frac{d\vec{r}_m}{dt}\right|^2
\, - \, Mg\,\big(\vec{r}_M \cdot \vec{e}_y\big)
\, - \, mg\,\big(\vec{r}_m \cdot \vec{e}_y\big)$$
with holonomic constriant
$$\big|\vec{r}_M \, - \, \vec{r}_m\big|^2 \, = \, \frac{l^2}{4}$$
Perform the linear chainge of coordinates:
\begin{align}
&\vec{r}_G \, =\, \frac{M}{M+m} \, \vec{r}_M \, + \, \frac{m}{M+m} \, \vec{r}_m\\
&\vec{r}_l \, = \, \vec{r}_M \, - \, \vec{r}_m
\end{align}
with inverse
\begin{align}
&\vec{r}_M \, =\, \vec{r}_G \, + \, \frac{m}{M+m} \, \vec{r}_l\\
&\vec{r}_m \, = \, \vec{r}_G \, - \, \frac{M}{M+m} \, \vec{r}_l
\end{align}
In the new coordinates, the Lagrangian becomes
\begin{align}
L \, =& \, \frac{M}{2}\,\left|\frac{d\vec{r}_G}{dt} \, + \, \frac{m}{M+m} \, \frac{d\vec{r}_l}{dt}\right|^2
\, + \, \frac{m}{2}\,\left|\frac{d\vec{r}_G}{dt} \, - \, \frac{M}{M+m} \, \frac{d\vec{r}_l}{dt}\right|^2 \\
&\, - \,g\,\Big(\big(M\,\vec{r}_M \, + \, m \, \vec{r}_m \big)\cdot \vec{e}_y\Big) \\
\, =&\, \frac{M}{2} \left|\frac{d\vec{r}_G}{dt}\right|^2 \, + \, \frac{M\,m}{(M+m)}\left(\frac{d\vec{r}_G}{dt}\cdot\frac{d\vec{r}_l}{dt}\right) \, + \, \frac{M\,m^2}{2(M+m)^2}\, \left|\frac{d\vec{r}_l}{dt}\right|^2\\
&\, + \, \frac{m}{2} \left|\frac{d\vec{r}_G}{dt}\right|^2 \, - \, \frac{M\,m}{(M+m)}\left(\frac{d\vec{r}_G}{dt}\cdot\frac{d\vec{r}_l}{dt}\right) \, + \, \frac{M^2\,m}{2(M+m)^2} \left|\frac{d\vec{r}_l}{dt}\right|^2\\
&\, - \,(M+m)\,g\,\Big(\vec{r}_G \cdot \vec{e}_y\Big)
\end{align}
and after some regrouping and simplification the Lagrangian and the holonomic constraint become:
$$L \, = \, \frac{M\,m}{2(M+m)}\, \left|\frac{d\vec{r}_l}{dt}\right|^2
\, + \, \frac{(M+m)}{2}\, \left|\frac{d\vec{r}_G}{dt}\right|^2 \, - \, (M+m)\,g\,\Big(\vec{r}_G \cdot \vec{e}_y\Big) $$
$$\big|\vec{r}_l\big|^2 = \frac{l^2}{4}$$
Introduce the generalized coordinates:
$$\vec{r}_l \, = \, \frac{l}{2}\, \cos(q) \, \vec{e}_x \, + \, \frac{l}{2}\, \sin(q) \, \vec{e}_y$$
$$\vec{r}_G \, = \, x_G \, \vec{e}_x \, + \, y_G\, \vec{e}_y$$ which parametrixe the holonimc constrints (they tirivialize it). Then, the Lagrnagian simplifies to
$$L \, = \, \frac{M\,m\ l^2}{8(M+m)}\, \left(\frac{dq}{dt}\right)^2
\, + \, \frac{(M+m)}{2}\, \left(\frac{dx_G}{dt}\right)^2 \, + \, \frac{(M+m)}{2}\, \left(\frac{dy_G}{dt}\right)^2 \, - \, (M+m)\,g\,y_G $$
Consequently, the Euler-Lagrange equations are
\begin{align}
&\frac{d^2{x}_G}{dt^2} \, = \, 0 \\
&\frac{d^2{y}_G}{dt^2} \, = \, - \, g\,\\
& \frac{d^2q}{dt^2} \, = \, 0
\end{align}
You can solve them right away
\begin{align}
&x_G \, = \, x_G(0) \, + \, v_x\, t \\
&y_G \, = \, y_G(0) \, + \, v_y\, t \, - \, \frac{g}{2} \, t^2\,\\
&q \, = \, q(0) \, + \, \omega_0 \, t
\end{align}
where $x_G(0), \, y_G(0), \, q(0)$ are constants describing the initial configuration of the rod and $v_x, \, v_y, \, \omega_0$ are constants describing the initial velocities of the rod.
The rigid body interpretation. The special case you are modelling in your OP can be thought of as a solid body, basically a rigid rod, which moves in a plane and whose mass is concentrated at its ends. First, you place an inertial coordinate frame (say a frame attached to the ground). Denote it by $O\,\vec{e}_x\,\vec{e}_y\,\vec{e}_z$, where the coordinate vectors are of unit length and perpendicular to each other. Moreover, $O\,\vec{e}_x\,\vec{e}_y\,\vec{e}_z$ is oriented so that $\vec{e}_y$ is the vertical vector along which gravitational acceleration is $\vec{g} = - \, g \, \vec{e}_y$, the system moves in the coordinate plane $O\,\vec{e}_x\,\vec{e}_y$, just like on your picture, and $\vec{e}_z$ is pointing perpendicularly to the picture.
Denote by $\vec{r}_M$ the position vector, pointing from the origin $O$ of the coordinate frame to the mass-point $M$ and by $\vec{r}_m$ the position vector, pointing from the origin $O$ of the coordinate frame to the mass-point $m$. The condition that the two mass points are connected by a mass-less rod of length $\frac{l}{2}$, can be written as the quadratic equation.
$$\big|\vec{r}_M - \vec{r}_m\big|^2 = \frac{l^2}{4}$$ Let $G$ be the center of gravity of the rod, i.e. the center of gravity, which is
$$\vec{r}_G = \frac{M}{M+m} \, \vec{r}_M \, + \, \frac{m}{M+m} \, \vec{r}_m$$
The force that acts on the system of the two mass points on the rod is gravity and is applied to the center of mass $\vec{r}_G$ of the rod. The general equations of motion of a rigid body can be written as
\begin{align}
&\frac{d^2\vec{r}_G}{dt^2} \, = \, - \, g\, \vec{e}_y\\
& \frac{d}{dt}\,I \,\omega\, \vec{e}_z \, = \, \vec{T}
\end{align} where $\omega \, \vec{e}_z$ is the angular velocity of the rod and $$I = \frac{M\,m\,l^2}{4\,(M+m)}$$ is its moment of inertia along the $O\, \vec{e}_z$ axis with respect to the center of mass $G$ of the rod. The other moments of inertia are zero because this is a one dimensional rod. The vector $\vec{T}$ is the sum of the torques of all forces acting on the rod, calculated with respect to the center of mass $G$. However, there is only one force, the gravity force $-(M+m)\, g \, \vec{e}_y$, acting on the rod and it is applied to the center of gravity $G$. Since $$\vec{T} = \vec{GG} \times \big(-(M+m)\, g \, \vec{e}_y\big) = \vec{0}\times \big(-(M+m)\, g \, \vec{e}_y\big) = \vec{0}$$ i.e. the torque is zero, the equations of motion become
\begin{align}
&\frac{d^2\vec{r}_G}{dt^2} \, = \, - \, g\, \vec{e}_y\\
& \frac{d}{dt}\,I \,\omega\, \vec{e}_z \, = \, \vec{0}
\end{align}
As you can see, these equations are decoupled and can be written as
\begin{align}
&\frac{d^2\vec{r}_G}{dt^2} \, = \, - \, g\, \vec{e}_y\\
& \,I \, \frac{d\omega}{dt} \, = \, {0}
\end{align}
But, the angular velocity, in your notations, is simply $\omega = \frac{dq}{dt}$ and also
$$\vec{r_G} = x_G \, \vec{e}_x + y_G\, \vec{e}_y$$ so
\begin{align}
&\frac{d^2{x}_G}{dt^2} \, = \, 0 \\
&\frac{d^2{y}_G}{dt^2} \, = \, - \, g\,\\
& \,I \, \frac{d^2q}{dt^2} \, = \, {0}
\end{align}
You can solve them right away
\begin{align}
&x_G \, = \, x_G(0) \, + \, v_x\, t \\
&y_G \, = \, y_G(0) \, + \, v_y\, t \, - \, \frac{g}{2} \, t^2\,\\
&q \, = \, q(0) \, + \, \omega_0 \, t
\end{align}
where $x_G(0), \, y_G(0), \, q(0)$ are constants describing the initial configuration of the rod and $v_x, \, v_y, \, \omega_0$ are constants describing the initial velocities of the rod.
If I have time, I can get to these equations from Lagrangian point of view too. Basically you just have to change coordinates of your Lagrangian and the holonomic constrant $\big|\vec{r}_M - \vec{r}_m\big|^2 = \frac{l^2}{4}$ like this:
\begin{align}
&\vec{r}_G \, =\, \frac{M}{M+m} \, \vec{r}_M \, + \, \frac{m}{M+m} \, \vec{r}_m\\
&\vec{r}_l \, = \, \vec{r}_M \, - \, \vec{r}_m
\end{align}
Remark. This case does not generalize to your real goal to have a system of more than two mass-points attached to each other by rods where the rods can rotate relative to each other.
