Link CD 6 rad/s, point E and link BC

Coriolis acceleration is the extra acceleration a point has when it moves relative to a rotating (moving) frame or body. If a point $P$ moves with relative velocity $\vec v_{rel}$ on a body rotating with angular velocity $\vec\omega$, then

$$\vec a_P = \vec a_{P'} + \vec a_{rel} + 2\vec\omega\times\vec v_{rel}$$

where $\vec a_{P'}$ is the acceleration of the coincident point of the rotating body, $\vec a_{rel}$ is the acceleration of $P$ relative to the body, and $2\vec\omega\times\vec v_{rel}$ is the Coriolis acceleration (magnitude $2\omega v_{rel}$, perpendicular to $\vec v_{rel}$). The velocity problem below involves only points of rigid links joined by pins, so no Coriolis term appears.

Assumed layout (from the figure): $D$ and $A$ are fixed pins at the same level; $CD$ is vertical (0.6 m) with $C$ at the top; $BC$ is horizontal (0.6 m) with $B$ to the left of $C$ and $E$ its midpoint; $AB$ rises at $30^\circ$ from $A$ to $B$, so $AB = 0.6/\sin30^\circ = 1.2$ m. Take $\omega_{CD}=6$ rad/s counter-clockwise, $x$ to the right and $y$ up (reverse all senses if $CD$ turns the other way).

Step 1: velocity of C

$$\vec v_C = \omega_{CD}\hat k\times\vec r_{C/D} = 6\hat k\times0.6\hat j = -3.6\,\hat i\ \text{m/s}$$

Step 2: direction of v_B

Link $AB$ rotates about $A$, so $\vec v_B \perp AB$. With $\vec v_B = (v_{Bx}, v_{By})$ and $AB$ along $(\cos30^\circ, \sin30^\circ)$:

$$v_{Bx}\cos30^\circ + v_{By}\sin30^\circ = 0$$

Link $BC$ is horizontal, so $\omega_{BC}\hat k\times\vec r_{B/C}$ is vertical and $v_{Bx} = v_{Cx} = -3.6$ m/s. Hence

$$v_{By} = \frac{3.6\cos30^\circ}{\sin30^\circ} = 6.24\ \text{m/s}, \qquad v_B = \sqrt{3.6^2 + 6.24^2} = 7.20\ \text{m/s}$$

Step 3: link BC

$$\vec v_B = \vec v_C + \omega_{BC}\hat k\times(-0.6\hat i) \;\Rightarrow\; v_{By} = -0.6\,\omega_{BC} \;\Rightarrow\; \omega_{BC} = -10.39\ \text{rad/s}\ \text{(clockwise)}$$

Step 4: point E

$$\vec v_E = \vec v_C + \omega_{BC}\hat k\times(-0.3\hat i) = -3.6\hat i + (-10.39)(-0.3)\hat j = (-3.60\,\hat i + 3.12\,\hat j)\ \text{m/s}$$

$v_E = 4.76$ m/s, directed 40.9° above the horizontal, towards the left.

Answer: $v_E = 4.76$ m/s (40.9° above horizontal, towards the left); $\omega_{BC} = 10.39$ rad/s clockwise.