Coriolis acceleration; rod AB and BD velocity

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 frame. The term $\vec a_c = 2\vec\omega\times\vec v_{rel}$ is the Coriolis acceleration. Its magnitude is $2\omega v_{rel}$ and its direction is that of $\vec v_{rel}$ rotated through $90^\circ$ in the sense of $\omega$. It exists only when the point slides on the rotating body (for example a collar sliding on a rotating arm, or a slider in a slotted rotating link); it is zero for two points fixed in the same rigid body, such as the crank pin and piston pin of a connecting rod.

Velocity analysis

The figure is not fully legible, so I assume: $AB$ is a vertical rod of length $0.25$ m pinned at the top end $A$ to a fixed support; $BD$ runs from $B$ down to $D$ with horizontal projection $0.6$ m and drop $0.2$ m; and $D$ is constrained to move vertically (given $v_D = 2.4$ m/s upward). Take $x$ to the right, $y$ up.

$$\vec r_{D/B} = (0.6\,\hat i - 0.2\,\hat j)\ \text{m}, \qquad \vec v_D = 2.4\,\hat j\ \text{m/s}$$

Rod $AB$ rotates about $A$, so $\vec v_B = \omega_{AB}\hat k\times(-0.25\hat j) = 0.25\,\omega_{AB}\,\hat i$ is horizontal.

For rod $BD$:

$$\vec v_D = \vec v_B + \omega_{BD}\hat k\times\vec r_{D/B} = v_{B}\hat i + \omega_{BD}(0.2\,\hat i + 0.6\,\hat j)$$ $$\begin{aligned} \hat j:&\quad 2.4 = 0.6\,\omega_{BD} \;\Rightarrow\; \omega_{BD} = 4.00\ \text{rad/s (ccw)}\\ \hat i:&\quad 0 = v_B + 0.2(4.00) \;\Rightarrow\; v_B = -0.80\ \text{m/s} \end{aligned}$$

(a) Angular velocity of AB

$$\omega_{AB} = \frac{v_B}{0.25} = \frac{-0.80}{0.25} = -3.20\ \text{rad/s}$$

so $|\omega_{AB}| = 3.20$ rad/s, clockwise (the sign shows that $B$ moves to the left).

(b) Velocity of the midpoint of BD

$$\vec v_M = \vec v_B + \omega_{BD}\hat k\times\left(\tfrac12\vec r_{D/B}\right) = (-0.80\,\hat i) + 4.00(0.1\,\hat i + 0.3\,\hat j) = (-0.40\,\hat i + 1.20\,\hat j)\ \text{m/s}$$ $$v_M = 1.26\ \text{m/s}$$

Answer: $\omega_{AB} = 3.20$ rad/s clockwise; $v_M = 1.26$ m/s.