My solution is below
Given
\begin{align*} \vec{V}_{D} & =-8\hat{\imath }\\ \vec{a}_{D} & =-30\hat{\imath } \end{align*}
But also (assuming cord is not extensible)
\begin{align*} \vec{V}_{B} & =-8\hat{\imath }\\ \vec{a}_{B} & =-30\hat{\imath } \end{align*}
Since the point \(B\) is also on the large disk, its velocity can be used to find the angular velocity of the disk. The disk is spining in the clockwise direction. Using \(V_{B}=r\omega _{disk}\), where \(r=5\) inch, then \(\omega _{disk}=\frac{-8}{5}=\allowbreak -1.6\) rad/sec or \[ \vec{\omega }_{disk}=-1.6\hat{k}\]
Similarly \(a_{B}=r\alpha _{disk}\) in the clockwise direction, hence \(\alpha _{disk}=\frac{a_{B}}{r}=\frac{-30}{5}=-6\) rad/sec\(^{2}\)
\[ \vec{\alpha }_{disk}=-6\hat{k}\]
Now
\[ \vec{a}_{A}=\vec{a}_{B}+\vec{\alpha }_{AB}\times \vec{r}_{A/B}-\omega _{AB}^{2}\vec{r}_{A/B}\]
Where \(\vec{r}_{A/B}=\left ( r_{2}-r_{1}\right ) \hat{\jmath }=\left ( 5-3\right ) \hat{\jmath }=2\hat{\jmath }\) and the above becomes
\begin{align*} \vec{a}_{A} & =-30\hat{\imath }+\left ( -6\hat{k}\times 2\hat{\jmath }\right ) -\left ( -1.6\right ) ^{2}\left ( 2\hat{\jmath }\right ) \\ & =-30\hat{\imath }+\left ( 12\hat{\imath }\right ) -5.12\hat{\jmath }\\ & =-18\hat{\imath }-5.12\hat{\jmath } \end{align*}
Now
\[ \vec{a}_{C}=\vec{a}_{O}+\vec{\alpha }_{OC}\times \vec{r}_{C/O}-\omega _{OC}^{2}\vec{r}_{C/O}\]
Where \(O\) is the center of the disk. Since disk is not sliding, then \(\vec{a}_{O}=0\) and \(\vec{r}_{C/O}=5\hat{\imath }\). The above becomes
\begin{align*} \vec{a}_{C} & =-6\hat{k}\times 5\hat{\imath }-\left ( -1.6\right ) ^{2}5\hat{\imath }\\ & =-30\hat{\jmath }-12.8\hat{\imath } \end{align*}
