Crackling Mechanics from Springer.

Though it was tough to draw the arrows in the diagram due to lack of physics editor, i managed somehow and took the photo. Hope it is fine.

The centers of rotation ${ \Pi }_{ 1 }$ ( bar $CD$ ) and ${ \Pi }_{ 2 }$ (bar $BC$ ) are given by the intersections of the perpendiculars to

$v_D = r \omega$ ( $v _ D \bot ED$ ) and $v_C$ ( $v _ C \bot AC$ ) and to $v_C$ and $v_B$ , respectively. This leads in the layout diagram to the depicted representation.

With the chosen scale, we read from the figure

$v_C \cong 0.9 { v }_{ D }$ , $v_B \cong 1.0 { v }_{ D }$

and obtain with

${ v }_{ D } = r \omega = 0.6 m/s$

the velocities

${ v }_{ C } \cong 0.5 m/s$ , ${ v }_{ B } \cong 0.6 m/s$ .

So, $\large\ v_ A + v_B + v_C = \boxed{1.7 m/s}$

Call the angle between $AC$ and the vertical $\theta$ . Define the coordinates of $A$ as the origin. Some initial coordinate definitions (note that we define the $C$ coordinates in terms of $\theta$ because $AC$ rotates about $A$ ) :

$\large{A_x = 0 \\ A_y = 0 \\ E_x = b \\ E_y = -l \\ D_x = E_x + r \, cos \beta \\ D_y = E_y + r \, sin \beta \\ C_x = l \, sin \theta \\ C_y = -l \, cos \theta \\ B_x = 0 \\ B_y = B_y}$

Apply condition for length of CD:

$\large{(E_x + r \, cos \beta - l \, sin \theta)^2 + (E_y + r \, sin \beta + l \, cos \theta)^2 = a^2}$

The only unknown in the above equation is $\theta$ , so we solve this equation for $\theta$ . $\theta = 32.7564^\circ$

Apply length invariance principle to CD:

$\large{ \frac{d}{dt} (E_x + r \, cos \beta - l \, sin \theta)^2 + \frac{d}{dt} (E_y + r \, sin \beta + l \, cos \theta)^2 = \frac{d}{dt} a^2 \\ (E_x + r \, cos \beta - l \, sin \theta)(-r \, sin \beta \,\, \omega - l \, cos \theta \, \dot{\theta}) + (E_y + r \, sin \beta + l \, cos \theta)(r \, cos \beta \,\, \omega - l \, sin \theta \, \dot{\theta}) = 0 }$

Now that $\theta$ is known, the only unknown in the above equation is $\dot{\theta}$ , so we solve this equation for $\dot{\theta}$ . $\dot{\theta} = 1.7672 \, \text{rad/s}$ . Note that since the rotation is clockwise, I used $-4$ for $\omega$ .

Apply condition for length of BC:

$\large{(l \, sin \theta - 0)^2 + (-l \, cos \theta - B_y)^2 = l^2}$

Solve for $B_y$ . $B_y = -0.50459 \, \text{m}$

Apply length invariance principle to BC:

$\large{ \frac{d}{dt}(l \, sin \theta - 0)^2 + \frac{d}{dt} (-l \, cos \theta - B_y)^2 = \frac{d}{dt} l^2 \\ (l \, sin \theta)(l \, cos \theta \, \dot{\theta}) + (-l \, cos \theta - B_y)(l \, sin \theta \, \dot{\theta} - \dot{B_y}) = 0}$

The only unknown in the above equation is $\dot{B_y}$ , so we solve this equation for $\dot{B_y}$ . $\dot{B_y} = 0.5737 \, \text{m/s}$

Final sum of speeds:

$\large{v_B + v_C + v_D = | \dot{B_y} | + | l \, \dot{\theta} | + | r \, \omega |} \approx 1.704 \, \text{m/s}$