General plane motion; angular acceleration of arm DE

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.

Given data and assumed geometry

Place $A$ at the origin with $AB$ horizontal (40 mm), $BD$ horizontal (50 mm) and arm $DE$ pivoted at $E$, which is 50 mm to the right of $D$ and 30 mm above it (the figure's dimensions 40, 50, 50 mm and 30 mm):

$$\vec r_{B/A} = 0.040\hat i,\quad \vec r_{D/B} = 0.050\hat i,\quad \vec r_{D/E} = (-0.050\,\hat i - 0.030\,\hat j)\ \text{m}$$

$\omega_{AB} = 15$ rad/s ccw, $\alpha_{AB}=0$. A and E are fixed.

Velocities

$$\vec v_B = 15\hat k\times0.04\hat i = 0.6\,\hat j\ \text{m/s}$$ $$\vec v_D = \vec v_B + \omega_{BD}\hat k\times\vec r_{D/B} = \omega_{DE}\hat k\times\vec r_{D/E}$$ $$\begin{aligned} \hat i:&\quad 0 = 0.03\,\omega_{DE} \;\Rightarrow\; \omega_{DE} = 0\ \text{rad/s}\\ \hat j:&\quad 0.6 + 0.05\,\omega_{BD} = -0.05\,\omega_{DE} \;\Rightarrow\; \omega_{BD} = -12.00\ \text{rad/s} \end{aligned}$$

Accelerations of B and D

$$\vec a_B = -\omega_{AB}^2\vec r_{B/A} = -225(0.04)\hat i = -9\,\hat i\ \text{m/s}^2$$

From rod $BD$: $\vec a_D = \vec a_B + \alpha_{BD}\hat k\times\vec r_{D/B} - \omega_{BD}^2\vec r_{D/B} = (-9 - 144(0.05))\hat i + 0.05\,\alpha_{BD}\hat j = -16.2\,\hat i + 0.05\,\alpha_{BD}\hat j$

From arm $DE$ (since $\omega_{DE}=0$): $\vec a_D = \alpha_{DE}\hat k\times\vec r_{D/E} = \alpha_{DE}(0.03\,\hat i - 0.05\,\hat j)$

Equating components:

$$\begin{aligned} \hat i:&\quad -16.2 = 0.03\,\alpha_{DE} \;\Rightarrow\; \alpha_{DE} = -540.0\ \text{rad/s}^2\\ \hat j:&\quad 0.05\,\alpha_{BD} = -0.05\,\alpha_{DE} \;\Rightarrow\; \alpha_{BD} = 540.0\ \text{rad/s}^2 \end{aligned}$$ $$\vec a_D = (-16.2\,\hat i + 27.0\,\hat j)\ \text{m/s}^2, \qquad a_D = 31.5\ \text{m/s}^2$$

Answer: $\alpha_{DE} = 540.0$ rad/s² clockwise; $a_D = 31.5$ m/s² (components $-16.2\,\hat i + 27.0\,\hat j$). The result depends on the assumed layout; the method is the same for the exact figure.