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

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.

Step 5: angular velocity of AB

$$\omega_{AB} = \frac{v_B}{AB} = \frac{7.20}{1.2} = 6.00\ \text{rad/s}\ \text{(counter-clockwise)}$$

Answer: $v_E = 4.76$ m/s; $\omega_{AB} = 6.00$ rad/s counter-clockwise.

General plane motion

General plane motion (GPM) is the motion of a rigid body in which every particle stays in a plane parallel to a fixed plane, and the motion is neither pure translation nor pure rotation about a fixed axis. It is the sum of (i) a translation of a chosen reference point and (ii) a rotation of the body about that point.

For two points $A$ and $B$ of the same rigid body:

$$\vec v_B = \vec v_A + \vec\omega \times \vec r_{B/A}, \qquad \vec a_B = \vec a_A + \vec\alpha \times \vec r_{B/A} - \omega^2 \vec r_{B/A}$$

Here $v_{B/A} = \omega\, r_{B/A}$ is perpendicular to $AB$, and $a_{B/A}$ has a tangential part $\alpha r$ and a normal part $\omega^2 r$ towards $A$.

Examples

• A wheel rolling on the road: the centre translates while the wheel rotates about the centre.
• The connecting rod of an engine: end $A$ (crank pin) moves on a circle, end $B$ (piston) moves in a straight line.
• A ladder sliding down a wall.