Coupled wheel and rod (analytical mechanics) 
I am struggling with formulating the equations of motion.
Consider a coordinate system with origin in $O$ ($y$ upwards and $x$ to the right), label the center of mass of rod $AB$ with $G$ then: 
$$ \begin{cases} x_A = ( R - r) \sin \theta \\ y_A = -(R - r) \cos \theta \end{cases} \quad \text{&} \quad \begin{cases} x_G = (R - r) \sin \theta + \frac l2 \sin \varphi \\ y_G = - (R - r) \cos \theta - \frac l2 \cos \varphi \end{cases} $$
The plan is that we want to use $L = T - V$ together with $$ \frac {\partial L}{\partial q_k} = \frac { d }{dt} \left( \frac {\partial L}{\partial \dot{q}_k } \right)  $$
where $T$ is the sum of the kinetic energy of the rod and wheel (given individually by $\frac 12 m v^2 + \frac 12 I \omega^2$) and $V$ is the potential energy. We can differentiate the position equations in order to obatin $v_G^2$ and $v_A^2$ respectively. Ultimately I have arrived at
$$ \begin{cases} T_G = \frac 12 m \bigg( (R-r)^2 \dot{\theta}^2 + \frac {l^2}4 \dot{\varphi}^2 + (R - r) \dot{\theta} l \dot{\varphi} \cos (\theta - \varphi) \bigg) \\ T_A = \frac {\dot{\theta}^2}4 \bigg( 2m (R - r)^2 + mr^2 \bigg) \end{cases}$$
and
$$ \begin{cases} V_A = -mg (R - r)\cos \theta + \text{const.} \\ V_G = -mg \left( (R - r)\cos \theta + \frac l2 \cos \varphi  \right)    \end{cases}  $$
I worry though that there must first be some relationship established between the angular velocities of the two bodies but I am unable to pinpoint it. Also I am not entirely confident of my hitherto progress, is it correct thus far?

nevermind, solved the problem eventually
 A: The angular velocity for the disk is given by $ \dot{x}_G^2 + \dot{y}_G^2 = v_G^{\, 2} = \omega_{disk}^2 r^2 $. Insert $\omega_{\text{disk}}^2$ into $$ T_A = \frac 12 m v_G^2 + \frac 12 I_A \omega_{\text{disk}}^2 $$
We can then add up $T_A$ and $T_G$ to form $T$, likewise for $V = V_A + V_G$. The Lagrange function is then given be $L = T - V$ and we need to study $$ \begin{gather*} \frac {\partial L}{\partial \theta} = \frac {d}{dt} \left( \frac {\partial L}{\partial \dot{\theta}} \right) \end{gather*} \\ \frac {\partial L}{\partial \varphi} = \frac d{dt} \left( \frac { \partial L }{\partial \dot{\varphi}} \right) $$
These equations will give rise to two differential which upon linearization can be written as $$ \begin{gather*}  5(R -r) \ddot{\theta} + l \ddot{ \varphi } +  4g\theta = 0  \\ 3(R-r)\ddot{\theta} + 2l\ddot{ \varphi } + 3g \varphi  = 0\end{gather*} $$
Let $\theta = A \cos {\omega t + \alpha} $ and $\varphi = B \cos {\omega t + \alpha} $, and insert into the differential equations. A linear system will arise with $A$ and $B$ as unknowns. Upon computing the determinant one can determine the frequencies $\omega$.
