Answer
$\omega_{D}=105rad/s$
Work Step by Step
We can determine the required angular velocity as follows:
The velocity of $C$ is given as
$\vec{v_C}=\vec{\omega_{BC}}\times \vec{r_{BC}}$
$\implies \vec{v_C}=15\hat k\times (0.25\hat j)=3.75\hat i$
and $\vec{r_{E/C}}=0.05\hat j$
Similarly, $\vec{v_E}=\vec{\omega_A}\times \vec{r_A}=30\hat k(0.3\hat j)=9\hat i$
We know that
$\vec{v_E}=\vec{v_C}+\vec{\omega_D}\times \vec{r_{E/C}}$
We plug in the known values to obtain:
$9\hat i=3.75\hat i+(-\omega_D)\hat k\times (0.05\hat j)$
$\implies 9\hat i=(3.75+0.5\omega_D)\hat i$
Comparing the $i$ components on both sides, we obtain:
$9=3.75+0.05\omega_D$
This simplifies to:
$\omega_{D}=105rad/s$
