Relative motion analysis

The idea

For a body in general plane motion (like a connecting rod), the motion of any point B equals the motion of another point A plus the motion caused by the body rotating about A. You don’t need a position equation; you work directly with vectors at the instant you care about.

The equations

${\mathbf{v}}_B={\mathbf{v}}_A+\mathbf{\omega }\times {\mathbf{r}}_{B/A}$ ${\mathbf{a}}_B={\mathbf{a}}_A+\mathbf{\alpha }\times {\mathbf{r}}_{B/A}-{\omega }^2{\mathbf{r}}_{B/A}$

• ${\mathbf{r}}_{B/A}$ is the vector from A to B.
• For flat motion, $\mathbf{\omega }=\omega \mathbf{k}$ and $\mathbf{\alpha }=\alpha \mathbf{k}$ (anticlockwise positive).
• Handy: $\mathbf{k}\times \mathbf{i}=\mathbf{j}$ and $\mathbf{k}\times \mathbf{j}=-\mathbf{i}$.
• The $-{\omega }^2{\mathbf{r}}_{B/A}$ term is the normal (centripetal) part; $\mathbf{\alpha }\times {\mathbf{r}}_{B/A}$ is the tangential part.

The method

1. Pick A as a point whose motion you know (e.g. a pin on a crank, or a slider).
2. Write ${\mathbf{r}}_{B/A}$ in components.
3. Write what you know about ${\mathbf{v}}_B$‘s direction (e.g. a slider can only move along its guide).
4. Expand, then match the $\mathbf{i}$ and $\mathbf{j}$ components: two equations, two unknowns.
5. Repeat for acceleration, using the ω you just found.

Worked example

A 0.5 m rod AB has end A sliding along the floor to the right at 2 m/s (constant) and end B sliding down a vertical wall. The rod is at 60° to the floor. Find the rod’s ω and α, and B’s velocity and acceleration.

Answer

${\mathbf{r}}_{B/A}=-0.5\cos 60°\mathbf{i}+0.5\sin 60°\mathbf{j}=-0.25\mathbf{i}+0.433\mathbf{j}$ Velocity: $v_B\mathbf{j}=2\mathbf{i}+\omega \mathbf{k}\times (-0.25\mathbf{i}+0.433\mathbf{j})=2\mathbf{i}-0.433\omega \mathbf{i}-0.25\omega \mathbf{j}$ $\mathbf{i}$: $0=2-0.433\omega$ → $\omega =4.62$ rad/s (anticlockwise) $\mathbf{j}$: $v_B=-0.25\times 4.62=-1.15$ m/s (moving down) Acceleration (${\mathbf{a}}_A=0$, ${\omega }^2=21.3$): $a_B\mathbf{j}=\alpha \mathbf{k}\times {\mathbf{r}}_{B/A}-21.3(-0.25\mathbf{i}+0.433\mathbf{j})$ $\mathbf{i}$: $0=-0.433\alpha +5.33$ → $\alpha =12.3$ rad/s² $\mathbf{j}$: $a_B=-0.25\times 12.3-21.3\times 0.433=-12.3$ m/s² Check with Implicit differentiation: $x^2+y^2=0.25$ gives $\dot{y}=-x\dot{x}/y=-0.25\times 2/0.433=-1.15$ m/s ✓

Common mistakes

• Using ${\mathbf{r}}_{A/B}$ instead of ${\mathbf{r}}_{B/A}$, which flips every sign.
• Getting the cross products wrong: $\mathbf{k}\times \mathbf{j}=-\mathbf{i}$.
• Forgetting the $-{\omega }^2\mathbf{r}$ term in the acceleration.

Check yourself

1. In the example, why is B's acceleration non-zero even though A moves at constant speed?

The rod’s angle keeps changing, so ω changes (α ≠ 0) and there’s a centripetal term. B’s motion is constrained to the wall, so it speeds up as it falls.

Next

Put it all together in the Slider-crank worked example: a test-style question with a speeding-up crank and a horizontal slider.