We will solve this using 3 separate bodies. So there are three free body diagrams as shown below
In this diagram, it is assumed the horizontal bar only moves in the \(x\) direction and this is all for small angle \(\theta \). Now we apply Newton laws to each body.
For disk, we apply \(\tau =I_{o}\ddot{\theta }\) but using the point \(D\) on the figure to take moments around in order to get rid of the friction \(F\) and \(N\) terms. This gives (using counter clock wise as positive)\begin{align} \left ( kr\theta \right ) r-p_{x_{1}}r & =-I_{o}\ddot{\theta }\nonumber \\ kr^{2}\theta -p_{x_{1}}r & =-\left ( I_{cg}+mr^{2}\right ) \ddot{\theta }\nonumber \\ & =-\left ( \frac{1}{2}mr^{2}+mr^{2}\right ) \ddot{\theta }\nonumber \\ & =-\frac{3}{2}mr^{2}\ddot{\theta }\tag{1} \end{align}
We now move to the second body, which is the horizontal bar. \begin{align} \sum F_{x} & =m_{bar}\ddot{x}\nonumber \\ -p_{x_{1}}+p_{x_{2}} & =\frac{m}{4}r\ddot{\theta }\tag{2} \end{align}
From (2) we solve for \(p_{x_{1}}\) and plug it into (1)\[ p_{x_{1}}=p_{x_{2}}-\frac{m}{4}r\ddot{\theta }\] Hence (1) now becomes\begin{align} kr^{2}\theta -\left ( p_{x_{2}}-\frac{m}{4}r\ddot{\theta }\right ) r & =-\frac{3}{2}mr^{2}\ddot{\theta }\nonumber \\ kr^{2}\theta -p_{x_{2}}r & =-\left ( \frac{3}{2}mr^{2}+\frac{m}{4}r^{2}\right ) \ddot{\theta }\nonumber \\ & =-\frac{7}{4}mr^{2}\ddot{\theta }\tag{3} \end{align}
To find \(p_{x_{2}}\), we use the third body, the vertical bar. Taking moments about C.G. of bar using counter clock wise as positive gives\begin{align*} \tau & =-I_{cg}\ddot{\theta }\\ \left ( kr\theta \right ) r\cos \theta +\left ( cr\dot{\theta }\right ) r\cos \theta +p_{x_{2}}r\cos \theta +p_{y_{2}}r\sin \theta & =-\frac{1}{12}\left ( \frac{m}{4}\right ) \left ( 2r\right ) ^{2}\ddot{\theta }\\ & =-\frac{1}{12}mr^{2}\ddot{\theta } \end{align*}
For small angle the above becomes\begin{equation} kr^{2}\theta +cr^{2}\dot{\theta }+p_{x_{2}}r+p_{y_{2}}r\theta =-\frac{m}{12}r^{2}\ddot{\theta }\tag{4} \end{equation} \(p_{y_{2}}\) is now found from vertical balance of horizontal bar. Since it does not move vertically and assumed to only move horizontally, then\begin{align*} \sum F_{y} & =0\\ -p_{y_{1}}-p_{y_{2}}-\frac{m}{4}g & =0 \end{align*}
Due to symmetry, \(p_{y_{1}}=p_{y2}\) and the above becomes\begin{align*} -2p_{y_{2}} & =\frac{m}{4}g\\ p_{y_{2}} & =-\frac{m}{8}g \end{align*}
Plugging this value for \(p_{y_{2}}\) into (4) and solving for \(p_{x_{2}}\) gives\begin{align*} kr^{2}\theta +cr^{2}\dot{\theta }+p_{x_{2}}r-\frac{m}{8}gr\theta & =-\frac{m}{12}r^{2}\ddot{\theta }\\ p_{x2} & =\frac{1}{r}\left ( -\frac{m}{12}r^{2}\ddot{\theta }+\frac{m}{8}gr\theta -kr^{2}\theta -cr^{2}\dot{\theta }\right ) \end{align*}
Plugging the above into (3) gives the equation of motion for disk\begin{align*} kr^{2}\theta -\left ( -\frac{m}{12}r^{2}\ddot{\theta }+\frac{m}{8}gr\theta -kr^{2}\theta -cr^{2}\dot{\theta }\right ) & =-\frac{7}{4}mr^{2}\ddot{\theta }\\ kr^{2}\theta +\frac{m}{12}r^{2}\ddot{\theta }-\frac{m}{8}gr\theta +kr^{2}\theta +cr^{2}\dot{\theta } & =-\frac{7}{4}mr^{2}\ddot{\theta }\\ \theta \left ( 2kr^{2}-\frac{m}{8}gr\right ) +cr^{2}\dot{\theta } & =-\frac{7}{4}mr^{2}\ddot{\theta }-\frac{m}{12}r^{2}\ddot{\theta }\\ \frac{11}{6}mr^{2}\ddot{\theta }+cr^{2}\dot{\theta }+\theta \left ( 2kr^{2}-\frac{m}{8}gr\right ) & =0 \end{align*}
Or\[ \ddot{\theta }+\frac{6c}{11m}\dot{\theta }+\theta \left ( \frac{12}{11}\frac{k}{m}-\frac{3}{44}\frac{g}{r}\right ) =0 \] Writing the above in the standard form \(\ddot{\theta }+2\zeta \omega _{n}\dot{\theta }+\omega _{n}^{2}\theta =0\) we see that \[ \omega _{n}^{2}=\sqrt{\frac{12}{11}\frac{k}{m}-\frac{3}{44}\frac{g}{r}}\] And\begin{align*} 2\zeta \omega _{n} & =\frac{6c}{11m}\\ \zeta & =\frac{3c}{11m\omega _{n}}\\ & =\frac{3c}{11m\sqrt{\frac{12}{11}\frac{k}{m}-\frac{3}{44}\frac{g}{r}}}\\ & =\frac{3c}{\sqrt{132km-\frac{363}{44}\frac{gm^{2}}{r}}} \end{align*}