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Plane motion and relative velocity

Short answer

Tranlation: all points have the same velocity (e.g., piston). Fixed-axis rotation: points move in circles (e.g., flywheel). General plane motion combines both (e.g., connecting rod). For rigid-body points A and B, v_B = v_A + ω × r_B/A, derived by differentiating r_B = r_A + r_B/A noting r_B/A rotates with angular velocity ω.

Differentiate between translation, fixed-axis rotation and general plane motion of a rigid body with one example each. Derive the relation vB = vA + ω × rB/A for two points A and B of a rigid body in plane motion.

Types of plane motion

TypeDescriptionExample
TranslationEvery line in the body stays parallel to its original direction; all points have the same velocity and acceleration. The path may be straight (rectilinear) or curved (curvilinear).Piston in its cylinder; a car body on a straight road
Fixed-axis rotationAll particles move in circles about a fixed axis; v=ωrv = \omega r, at=αra_t = \alpha r, an=ω2ra_n = \omega^2 r.Flywheel, crankshaft, door on hinges
General plane motionA combination of translation and rotation; every point stays in a plane parallel to a fixed plane.Connecting rod of an engine; wheel rolling on a road

General plane motion can always be treated as a translation of a chosen base point plus a rotation about that point.

Relative velocity relation

Let A and B be two points of a rigid body having angular velocity ω\omega (counter-clockwise positive, along k\mathbf{k}).

        y
        |      B
        |     /
        |    /  r_B/A
        |  A
        | / r_A
        |/
        O------------ x

From the geometry, rB=rA+rB/A\mathbf{r}_B = \mathbf{r}_A + \mathbf{r}_{B/A}. Differentiating with respect to time:

vB=vA+r˙B/A\mathbf{v}_B = \mathbf{v}_A + \dot{\mathbf{r}}_{B/A}

In a rigid body the distance AB is constant, so the vector rB/A\mathbf{r}_{B/A} can only rotate with angular velocity ω\omega, and r˙B/A=ω×rB/A\dot{\mathbf{r}}_{B/A} = \boldsymbol{\omega}\times\mathbf{r}_{B/A}. Hence

vB=vA+ω×rB/A\mathbf{v}_B = \mathbf{v}_A + \boldsymbol{\omega}\times\mathbf{r}_{B/A}

The relative velocity vB/A\mathbf{v}_{B/A} has magnitude ω (AB)\omega\,(AB) and is perpendicular to AB, in the sense of ω\omega. To solve problems, draw the velocity triangle vB=vA+vB/A\mathbf{v}_B = \mathbf{v}_A + \mathbf{v}_{B/A} using the known directions, or write the equation in components and solve.

✓ Answer checked by Claude Opus 5.5 (high effort)

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