The tangential acceleration at the point where disk A and disk B meet is R_{A}\alpha _{A}. But this is also must be the same as R_{B}\alpha _{B} since the gears assumed not to slip against each others. Therefore \alpha _{B}=-\frac{R_{A}}{R_{B}}\alpha _{A} The minus sign, is because gear A moves anti-clockwise, but B moves clockwise, hence in negative direction. Therefore\begin{align*} \alpha _{B} & =-\frac{201}{114}\left ( 51\right ) \\ & =-89.921\,\text{ rad/sec}^{2} \end{align*}
Since C moves with B as one body, then \alpha _{B}=\alpha _{C} and then \alpha _{C}=-89.921\,\text{ rad/sec}^{2} Similarly\begin{align*} \alpha _{D} & =-\frac{R_{C}}{R_{D}}\alpha _{C}\\ & =-\frac{162}{133}\left ( -89.921\right ) \\ & =109.528\text{ rad/sec}^{2} \end{align*}
Since a_{p}=r\alpha _{s} then \alpha _{s}=-\frac{a_{p}}{r}=-\frac{1}{1.31}=-0.763\text{ rad/sec}^{2} Since v_{p}=r\omega _{s} then \omega _{s}=-\frac{v_{p}}{r}=-\frac{9.2}{1.31}=-7.023\text{ rad/sec}
\begin{align*} RC & =105.5\text{ mm}\\ RS & =26.4\text{ mm}\\ RW & =\frac{720}{2}=360\text{ mm}\\ \omega _{C} & =2\pi \end{align*}
Hence \begin{align*} \omega _{wheel} & =\frac{RC}{RS}\omega _{C}\\ & =\frac{105.5\text{ }}{26.4}2\pi \\ & =25.109\text{ rad/sec} \end{align*}
Or\begin{align*} \omega _{wheel} & =\frac{25.109}{0.104719775}\\ & =239.773\text{ RPM} \end{align*}
Hence
\begin{align*} V & =\omega _{wheel}\left ( RW\right ) \\ & =25.109\left ( 360\left ( 10^{-3}\right ) \right ) \\ & =9.039\,\text{ m/s} \end{align*}
\begin{align} \bar{v}_{B} & =\bar{v}_{A}+\bar{\omega }\times \bar{r}_{B/A}\nonumber \\ & =v_{A}\left ( \sin \phi \hat{\imath }+\cos \phi \hat{\jmath }\right ) +\bar{\omega }\times \bar{r}_{B/A} \tag{1} \end{align}
And \bar{\omega }=-2^{0}\hat{k}=-0.03491\hat{k} And \bar{r}_{B/A}=d\cos \theta \hat{\imath }+d\sin \theta \hat{\jmath } Hence (1) becomes\begin{align} \bar{v}_{B} & =v_{A}\left ( \sin \phi \hat{\imath }+\cos \phi \hat{\jmath }\right ) +\omega \hat{k}\times \left ( d\cos \theta \hat{\imath }+d\sin \theta \hat{\jmath }\right ) \nonumber \\ & =v_{A}\left ( \sin \phi \hat{\imath }+\cos \phi \hat{\jmath }\right ) +\begin{vmatrix} \hat{\imath } & \hat{\jmath } & \hat{k}\\ 0 & 0 & \omega \\ d\cos \theta & d\sin \theta & 0 \end{vmatrix} \nonumber \\ & =v_{A}\left ( \sin \phi \hat{\imath }+\cos \phi \hat{\jmath }\right ) +\left ( -\omega d\sin \theta \hat{\imath }-\hat{\jmath }\left ( -\omega d\cos \theta \right ) \right ) \nonumber \\ & =\hat{\imath }\left ( v_{A}\sin \phi -\omega d\sin \theta \right ) +\hat{\jmath }\left ( v_{A}\cos \phi +\omega d\cos \theta \right ) \tag{2} \end{align}
But \begin{align*} v_{A} & =30\left ( 1.852\right ) \left ( \frac{1000}{km}\right ) \left ( \frac{hr}{3600}\right ) \\ & =30\left ( 1.852\right ) \left ( \frac{1000}{3600}\right ) \\ & =15.433\,\text{ m/s} \end{align*}
And d=231, hence (2) becomes\begin{align*} \bar{v}_{B} & =\hat{\imath }\left ( \left ( 15.433\right ) \sin \left ( 34\left ( \frac{\pi }{180}\right ) \right ) -\left ( -0.0349\right ) \left ( 231\right ) \sin \left ( 20\left ( \frac{\pi }{180}\right ) \right ) \right ) \\ & +\hat{\jmath }\left ( \left ( 15.433\right ) \cos \left ( 34\left ( \frac{\pi }{180}\right ) \right ) +\left ( -0.0349\right ) \left ( 231\right ) \cos \left ( 20\left ( \frac{\pi }{180}\right ) \right ) \right ) \end{align*}
Or
\bar{v}_{B}=11.388\,\,\hat{\imath }+5.217\hat{\jmath }
\begin{align*} \omega _{p}\left ( 2R\right ) & =v_{L}\\ \omega _{p} & =\frac{v_{L}}{2R}\\ & =\frac{5}{2\left ( \frac{2.3}{12}\right ) }=13.043\text{ rad/sec} \end{align*}
And\begin{align*} v_{o} & =\omega _{p}R\\ & =13.043\,48\left ( \frac{2.3}{12}\right ) \\ & =2.5\text{ m/s} \end{align*}
Since \omega L\sin \theta =v_{B} Then\begin{align*} \omega & =\frac{v_{B}}{L\sin \theta }\\ & =\frac{5}{\left ( \frac{35}{12}\right ) \sin \left ( 77\left ( \frac{\pi }{180}\right ) \right ) }\\ & =1.759\,\text{ rad/sec} \end{align*}
And \omega L\cos \theta =v_{A} Then\begin{align*} v_{A} & =-\left ( 1.759\,\right ) \left ( \frac{35}{12}\right ) \cos \left ( 77\frac{\pi }{180}\right ) \\ & =-1.154\text{ m/s} \end{align*}
The minus sign since A moves down.
This can also be solved using vector method as follows \vec{v}_{A}=\vec{v}_{B}+\vec{\omega }_{AB}\times \vec{r}_{A/B} Where \vec{r}_{A/B}=-L\cos \theta \hat{\imath }+L\sin \theta \hat{\jmath } and \vec{v}_{B}=5\hat{\imath } is given. Hence the above becomes\begin{align} \vec{v}_{A} & =5\hat{\imath }+\omega _{AB}\hat{k}\times \left ( -L\cos \theta \hat{\imath }+L\sin \theta \hat{\jmath }\right ) \nonumber \\ & =5\hat{\imath }+\left ( -\omega _{AB}L\cos \theta \hat{\jmath }-\omega _{AB}L\sin \theta \hat{\imath }\right ) \nonumber \\ & =\hat{\imath }\left ( 5-\omega _{AB}L\sin \theta \right ) +\hat{\jmath }\left ( -\omega _{AB}L\cos \theta \right ) \tag{1} \end{align}
And now comes the main point. We argue that A can only move in vertical direction, hence the \hat{\imath } component above must be zero. Therefore 5-\omega _{AB}L\sin \theta =0 There is only one unknown in the above. SOlving for \omega _{AB} gives \omega _{AB}=1.759\text{ rad/sec} Now we go back to (1) and find \vec{v}_{A}\begin{align*} \vec{v}_{A} & =\hat{\imath }\left ( 0\right ) -\hat{\jmath }\left ( 1.759\left ( \frac{35}{12}\right ) \cos \left ( 77\frac{\pi }{180}\right ) \right ) \\ & =\hat{\imath }\left ( 0\right ) -\hat{\jmath }\left ( 1.154\right ) \end{align*}
Which is the same as method earlier. Notice we did not need to use R, the radius of the disk.