My solution is below
Notice that the point \(E\) is not on the bar \(CD\). It is the point where the disks meet at this instance shown.\begin{align} \bar{V}_{D} & =\bar{V}_{C}+\bar{\omega }_{CD}\times \bar{r}_{D/C}\nonumber \\ & =0+\omega _{CD}\hat{k}\times \left ( r_{A}+r_{B}\right ) \hat{\imath }\nonumber \\ & =\omega _{CD}\left ( r_{A}+r_{B}\right ) \hat{\jmath }\tag{1} \end{align}
But we also see that \(\bar{V}_{D}\) can be written as\begin{align} \bar{V}_{D} & =\bar{V}_{E}+\bar{\omega }_{disk}\times \bar{r}_{D/E}\nonumber \\ & =0+\omega _{disk}\hat{k}\times r_{B}\hat{\imath }\nonumber \\ & =\omega _{disk}r_{B}\hat{\jmath }\tag{2} \end{align}
Where in the above we used the fact that \(\bar{V}_{E}=\bar{V}_{C}=0\) at the instance shown. Equating (1) and (2)\begin{align} \omega _{CD}\left ( r_{A}+r_{B}\right ) & =\omega _{disk}r_{B}\nonumber \\ \omega _{disk} & =\omega _{CD}\frac{r_{A}+r_{B}}{r_{B}}\tag{3} \end{align}
\begin{align*} \bar{V}_{F} & =V_{D}+\bar{\omega }_{disk}\times \bar{r}_{F/D}\\ & =\omega _{CD}\left ( r_{A}+r_{B}\right ) \hat{\jmath }+\omega _{disk}\hat{k}\times r_{B}\hat{\jmath } \end{align*}
Hence\begin{align*} \bar{V}_{F} & =-\omega _{disk}r_{B}\hat{\imath }+\omega _{CD}\left ( r_{A}+r_{B}\right ) \hat{\jmath }\\ & =-\omega _{CD}\frac{r_{A}+r_{B}}{r_{B}}r_{B}\hat{\imath }+\omega _{CD}\left ( r_{A}+r_{B}\right ) \hat{\jmath }\\ & =\omega _{CD}\left ( r_{A}+r_{B}\right ) \hat{\imath }+\omega _{CD}\left ( r_{A}+r_{B}\right ) \hat{\jmath } \end{align*}
\begin{align} \bar{V}_{D} & =\bar{V}_{B}+\bar{V}_{D/B}\nonumber \\ & =\bar{V}_{B}+\bar{\omega }_{BD}\times \bar{r}_{D/B} \tag{1} \end{align}
But \begin{align} \bar{V}_{B} & =\bar{V}_{A}+\bar{\omega }_{AB}\times \bar{r}_{B/A}\nonumber \\ & =0-\omega _{AB}\hat{k}\times \left ( L_{1}\cos \theta \hat{\imath }+L_{1}\sin \theta \hat{\jmath }\right ) \nonumber \\ & =-\omega _{AB}L_{1}\cos \theta \hat{\jmath }+\omega _{AB}L_{1}\sin \theta \hat{\imath } \tag{2} \end{align}
Where \[ \omega _{AB}=2000\left ( \frac{2\pi }{60}\right ) =\frac{200}{3}\pi =209.4395\text{ rad/sec}\] The angle \(\beta \) can be found as follows\begin{align*} \frac{\sin \theta }{L_{2}} & =\frac{\sin \beta }{L1}\\ \sin \beta & =\frac{L_{1}}{L_{2}}\sin \theta =\frac{3}{8}\sin \left ( 40\left ( \frac{\pi }{180}\right ) \right ) =0.241\text{ radians}\\ & =13.808^{0} \end{align*}
Now we know everything to evaluate (1). Therefore\begin{align} \bar{V}_{D} & =\bar{V}_{B}+\bar{\omega }_{BD}\times r_{D/B}\nonumber \\ & =\left ( -\omega _{AB}L_{1}\cos \theta \hat{\jmath }+\omega _{AB}L_{1}\sin \theta \hat{\imath }\right ) +\omega _{BD}\hat{k}\times \left ( L_{2}\cos \beta \hat{\imath }-L_{2}\sin \beta \hat{\jmath }\right ) \nonumber \\ & =\left ( -\omega _{AB}L_{1}\cos \theta \hat{\jmath }+\omega _{AB}L_{1}\sin \theta \hat{\imath }\right ) +\omega _{BD}L_{2}\cos \beta \hat{\jmath }+\omega _{BD}L_{2}\sin \beta \hat{\imath }\nonumber \\ & =\hat{\imath }\left ( \omega _{AB}L_{1}\sin \theta +\omega _{BD}L_{2}\sin \beta \right ) +\hat{\jmath }\left ( -\omega _{AB}L_{1}\cos \theta +\omega _{BD}L_{2}\cos \beta \right ) \tag{3} \end{align}
But the \(y\) componenent of \(\bar{V}_{D}=0\) since \(D\) can only move in \(x\) direction. Therefore from the above\begin{align} -\omega _{AB}L_{1}\cos \theta +\omega _{BD}L_{2}\cos \beta & =0\nonumber \\ \omega _{BD} & =\omega _{AB}\frac{L_{1}\cos \theta }{L_{2}\cos \beta } \tag{4} \end{align}
Substituting (4) into the \(x\) component of (3) gives the answer we want\begin{align*} \bar{V}_{D} & =\hat{\imath }\left ( \omega _{AB}L_{1}\sin \theta +\left ( \frac{L_{1}\cos \theta }{L_{2}\cos \beta }\omega _{AB}\right ) L_{2}\sin \beta \right ) \\ & =\hat{\imath }\left ( L_{1}\sin \theta +\left ( \frac{L_{1}\cos \theta }{L_{2}\cos \beta }\right ) L_{2}\sin \beta \right ) \omega _{AB}\\ & =\hat{\imath }\left ( 3\sin \left ( 40\left ( \frac{\pi }{180}\right ) \right ) +\frac{3\cos \left ( 40\left ( \frac{\pi }{180}\right ) \right ) \sin \left ( 13.808\left ( \frac{\pi }{180}\right ) \right ) }{\cos \left ( 13.808\left ( \frac{\pi }{180}\right ) \right ) }\right ) 209.4395\\ & =522.170\hat{\imath }\text{ inch/sec}\\ & =43.514\hat{\imath }\text{ ft/sec} \end{align*}
And \(\omega _{BD}=\omega _{AB}\frac{L_{1}\cos \theta }{L_{2}\cos \beta }=209.4395\frac{3\cos \left ( 40\left ( \frac{\pi }{180}\right ) \right ) }{8\cos \left ( 13.808\left ( \frac{\pi }{180}\right ) \right ) }=61.955\) rad/sec