Relative Velocity & Acceleration — Use relative velocity...

SECTION 1

Historical Context & Motivation

The study of how points on a single rigid body move relative to one another has deep roots in classical mechanics, stretching back to the Enlightenment era when mathematicians and natural philosophers first grappled with the motion of extended objects rather than idealized particles. A fundamental difficulty arises because every point on a rotating body traces a different path in space, yet the body does not deform—its internal distances remain constant. The resolution of this paradox required the development of relative motion analysis, a framework that decomposes the velocity and acceleration of any point into contributions from translation, rotation, and—when applicable—changing rotation rate. This approach ultimately became one of the most powerful tools in mechanism analysis, robotics, and vehicle dynamics.

1736

Euler's Mechanica

Leonhard Euler published Mechanica, formulating the equations of motion for rigid bodies and introducing the concept of angular velocity as a vector quantity that governs relative velocities between points.

1834

Chasles' Theorem

Michel Chasles proved that the most general displacement of a rigid body can be decomposed into a translation plus a rotation about an axis, laying the geometric foundation for relative motion relations.

1876

Reuleaux's Kinematics of Machinery

Franz Reuleaux systematized the kinematic analysis of mechanisms, applying relative velocity and acceleration methods to linkages and gears, transforming rigid-body kinematics into an engineering discipline.

1897

Coriolis & Relative Acceleration

Building on Gaspard-Gustave de Coriolis's earlier work, engineers formalized the relative acceleration equation including normal and tangential components, enabling dynamic analysis of rotating machinery.

1960s

Computational Kinematics

The advent of digital computation allowed relative velocity and acceleration equations to be solved for complex multi-body systems, giving rise to modern CAD-based mechanism simulation software.

The central question these developments address is deceptively simple: given that point A on a rigid body has a known velocity and acceleration, how can we determine the velocity and acceleration of any other point B on the same body? Because the body is rigid, the answer depends entirely on the body's angular velocity ω and angular acceleration α, combined with the position vector from A to B. Mastery of these relations is essential for analyzing linkages, gear trains, robotic arms, and virtually every mechanical system with rotating components.

SECTION 2

Core Principles & Definitions

Before diving into the vector equations, it is essential to establish the physical and mathematical foundations that govern relative motion on a rigid body. The rigidity constraint—namely that the distance between any two points on the body never changes—imposes powerful restrictions on admissible velocity and acceleration fields. These restrictions manifest as elegant cross-product relationships that link the motion of every point to a common angular velocity and angular acceleration vector.

Rigid-Body Constraint

The distance |r_B/A| between any two points A and B on the body remains constant for all time. This eliminates stretching, compression, and shearing, reducing the body's internal degrees of freedom to pure rotation about any chosen base point.

Angular Velocity Vector ω

A single vector ω characterizes the instantaneous rate of rotation of the entire body. In planar (2-D) motion, ω = ω k̂ points perpendicular to the plane. Every relative velocity on the body is generated by this vector through the cross product ω × r_B/A.

Angular Acceleration Vector α

The time derivative of ω is the angular acceleration α = dω/dt. In planar motion, α = α k̂. It generates a tangential component of relative acceleration via α × r_B/A, complementing the centripetal term −ω²r_B/A.

Relative Position Vector r_B/A

Defined as r_B − r_A, this vector points from the base point A to the point of interest B. Its magnitude is fixed (rigidity), so its time derivative is purely due to rotation: dr_B/A/dt = ω × r_B/A.

Superposition of Translation & Rotation

General planar motion is the sum of a pure translation (carrying A) and a pure rotation about A. This decomposition is the conceptual heart of the relative velocity and relative acceleration equations and holds for any choice of base point.

✦ KEY TAKEAWAY

Think of a rigid body as a turntable with a person standing at the center (point A) and another at the edge (point B). The person at the edge has all the velocity of the center person plus the velocity due to spinning around the center. The relative velocity equation formalizes this: v_B = v_A + ω × r_B/A. The spin contributes a velocity perpendicular to the line joining A and B, and its magnitude scales with both the spin rate ω and the distance |r_B/A|.

SECTION 3

Visual Explanation — Relative Velocity on a Rigid Body

The velocity of point B is constructed by translating v_A (pink dashed) to point B and then adding the rotational contribution ω × r_B/A (cyan), which acts perpendicular to the line AB. The resultant blue vector is v_B.

The diagram above illustrates the fundamental superposition principle for rigid-body velocity. Point A serves as the base point whose velocity is known. To find the velocity of point B, we first carry v_A unchanged to the location of B (the translational part), then add the velocity that B has relative to A due to the body's rotation. This relative velocity ω × r_B/A is always perpendicular to the line segment from A to B, and its magnitude is ω multiplied by the distance |r_B/A|. In the planar case, the direction of this perpendicular velocity is determined by the sign of ω: counterclockwise rotation (positive ω) produces a velocity rotated 90° counterclockwise from r_B/A.

SECTION 4

Mathematical Framework

The relative motion equations for rigid bodies emerge directly from differentiating the position constraint. Consider two points A and B fixed on the same rigid body. The position of B is r_B = r_A + r_B/A. Because the body is rigid, the magnitude of r_B/A is constant, but its direction changes as the body rotates. Taking successive time derivatives yields the velocity and acceleration relations.

RELATIVE VELOCITY EQUATION

v_B = v_A + ω × r_{B/A}

v_B = velocity of point B; v_A = velocity of base point A; ω = angular velocity of the body; r_B/A = position vector from A to B. The cross product ω × r_B/A is perpendicular to r_B/A with magnitude ω|r_B/A|.

Differentiating the velocity equation with respect to time yields the relative acceleration equation. The key subtlety is that the time derivative of ω × r_B/A produces two terms: one from the change in ω (the tangential component) and one from the change in direction of r_B/A (the normal/centripetal component).

RELATIVE ACCELERATION EQUATION

a_B = a_A + α × r_{B/A} + ω × (ω × r_{B/A})

a_B = acceleration of point B; a_A = acceleration of base point A; α = angular acceleration (dω/dt); α × r_B/A = tangential acceleration component (⊥ to r_B/A, magnitude α|r_B/A|); ω × (ω × r_B/A) = centripetal (normal) acceleration component directed from B toward A, magnitude ω²|r_B/A|.

PLANAR SCALAR COMPONENTS (2-D)

v_B = v_A + ω k̂ × r_{B/A} → (v_Bx, v_By) = (v_Ax − ω·r_{B/A,y}, v_Ay + ω·r_{B/A,x})

In planar problems, ω = ω k̂ is perpendicular to the xy-plane. The cross product reduces to: ω k̂ × (r_x î + r_y ĵ) = −ω r_y î + ω r_x ĵ. Similarly, the centripetal term becomes −ω²(r_x î + r_y ĵ), directed from B toward A.

📐 Derivation Insight

The double cross product ω × (ω × r_B/A) can be expanded using the BAC-CAB identity: A × (B × C) = B(A·C) − C(A·B). For planar motion with ω = ω k̂, this simplifies to −ω² r_B/A, confirming that the centripetal acceleration has magnitude ω²|r_B/A| and points from B toward A.

SECTION 5

Detailed Breakdown — Acceleration Components

The relative acceleration equation contains three distinct physical contributions, and correctly identifying each is the most common source of difficulty for students. The tangential component arises from changing angular speed and acts perpendicular to the position vector r_B/A. The normal (centripetal) component arises from the change in direction of the velocity due to rotation and points radially inward from B toward A. Together with the base-point acceleration a_A, these three vectors sum to give the total acceleration of B.

At point B, three acceleration vectors are superimposed: the translated base-point acceleration a_A (violet dashed), the tangential component α × r_B/A (amber, perpendicular to the connecting line), and the centripetal component −ω²r_B/A (red, directed from B toward A). Their vector sum produces the total acceleration a_B (blue).

Summary of acceleration components in the relative acceleration equation

Component	Expression	Direction	Magnitude
Base-point acceleration	a_A	Known from problem data	\|a_A\|
Tangential (α × r_B/A)	α × r_B/A	⊥ to r_B/A (sense from α)	α · \|r_B/A\|
Normal/centripetal	ω × (ω × r_B/A) = −ω²r_B/A	From B toward A (along −r_B/A)	ω² · \|r_B/A\|

SECTION 6

Worked Example — Crank-Slider Mechanism

Consider a planar crank-slider mechanism. The crank OA has length 0.3 m and rotates counterclockwise at ω_OA = 10 rad/s with angular acceleration α_OA = 0 (constant speed). Pin O is fixed. At the instant shown, crank OA is at θ = 60° above the horizontal. The connecting rod AB has length 0.6 m. The slider at B moves horizontally. Find the velocity of slider B and the angular velocity of rod AB.

Relative Velocity — Crank-Slider

Step 1 — Velocity of ASince O is fixed, v_A = ω_OA × r_A/O. The position of A: r_A/O = 0.3(cos 60° î + sin 60° ĵ) = 0.15 î + 0.2598 ĵ m. With ω_OA = 10 k̂ rad/s: v_A = 10 k̂ × (0.15 î + 0.2598 ĵ) = 10(0.15 ĵ − 0.2598 î) = −2.598 î + 1.5 ĵ m/s.

v_A = −2.598 î + 1.5 ĵ m/s (|v_A| = 3.0 m/s)

Step 2 — Write the relative velocity equation for rod ABApply v_B = v_A + ω_AB × r_B/A. We need the geometry. If A is at (0.15, 0.2598) and B slides along the x-axis, we can find the position of B by geometry. The vertical coordinate of B is 0, so r_B,y = 0. Using |AB| = 0.6 m and knowing A's coordinates: (x_B − 0.15)² + (0.2598)² = 0.36, giving x_B = 0.15 + √(0.36 − 0.0675) = 0.15 + 0.5413 = 0.6913 m. Therefore r_B/A = (0.6913 − 0.15) î + (0 − 0.2598) ĵ = 0.5413 î − 0.2598 ĵ m.

r_B/A = 0.5413 î − 0.2598 ĵ m

Step 3 — Expand and solve for unknownsLet ω_AB = ω_AB k̂ (unknown). Also v_B = v_B î (horizontal slider, unknown magnitude). The equation v_B î = (−2.598 î + 1.5 ĵ) + ω_AB k̂ × (0.5413 î − 0.2598 ĵ). Computing the cross product: ω_AB(0.5413 ĵ + 0.2598 î). Collecting components — î: v_B = −2.598 + 0.2598 ω_AB; ĵ: 0 = 1.5 + 0.5413 ω_AB. From the ĵ equation: ω_AB = −1.5/0.5413 = −2.772 rad/s (clockwise). From the î equation: v_B = −2.598 + 0.2598(−2.772) = −2.598 − 0.720 = −3.318 m/s.

ω_AB = −2.77 rad/s (clockwise); v_B = −3.32 î m/s (leftward)

Step 4 — Interpret the resultsThe slider B moves to the left at 3.32 m/s, which is physically consistent with the crank OA sweeping counterclockwise at the given orientation. The connecting rod AB rotates clockwise at 2.77 rad/s. These kinematic quantities could now feed into a subsequent dynamic analysis (Newton–Euler equations) or into the relative acceleration equation to find the acceleration of B and the angular acceleration of AB.

SECTION 7

Comparison of Kinematic Methods

The relative velocity/acceleration method is not the only approach to planar kinematics. It is useful to compare it with alternative methods—the instantaneous center of zero velocity (IC) method and rotating reference frame (Coriolis) formulations—to understand when each is most appropriate and what trade-offs exist.

Comparison of planar kinematic analysis methods

Criterion	Relative Velocity/Acceleration	Instantaneous Center (IC)	Rotating Frame (Coriolis)
Velocity analysis	Systematic vector equations; handles any base-point choice	Elegant for single links; magnitudes by proportion	Required when points slide on moving bodies
Acceleration analysis	Full acceleration equation with tangential + centripetal terms	IC cannot be used directly for acceleration	Adds Coriolis term 2ω × v_rel for sliding contacts
Complexity	Two scalar equations per vector equation (planar)	Graphical/geometric; minimal algebra	Three additional terms; highest algebraic complexity
Best for	Multi-link mechanisms, computer implementation, dynamics	Quick velocity answers for single links, conceptual checks	Slider-crank with slots, cams, or any sliding joint
Limitation	Requires careful sign conventions; no shortcut	Cannot find accelerations; IC may be at infinity for pure translation	Easy to forget or misapply the Coriolis term

⚙ WHEN TO USE WHICH METHOD

Use the relative velocity/acceleration approach as your default tool—it works for every rigid-body configuration and naturally extends to acceleration analysis and dynamics. Turn to the instantaneous center for quick velocity checks or when a geometric/graphical approach is more intuitive. Switch to the rotating frame (Coriolis) formulation whenever a point slides along a moving body—this is the only method that correctly captures the Coriolis acceleration 2ω × v_rel.

SECTION 8

Connection to Advanced Theory

The relative motion equations for rigid bodies in planar kinematics are a special case of far more general frameworks. Understanding where they sit in the broader landscape of dynamics helps you appreciate both their power and their limitations. In particular, these relations form the kinematic foundation upon which the Newton–Euler equations of rigid-body dynamics are built. Once you know the accelerations of key points (especially the center of mass), you can apply ΣF = ma_G and ΣM_G = I_Gα to find forces, torques, and reactions.

Planar kinematics as a gateway to advanced dynamics

This Course: Planar Rigid-Body Kinematics	Advanced Extension
ω = ω k̂ (single scalar)	3-D: ω is a general vector; Euler angles or quaternions parameterize orientation
v_B = v_A + ω × r_{B/A}	Same equation holds in 3-D; cross product yields three scalar equations
Two points on same rigid body	Multi-body systems: add joint constraints, use Denavit–Hartenberg parameters in robotics
Constant or given ω and α	Dynamic coupling: ω and α found simultaneously with forces via Newton–Euler or Lagrangian mechanics
Analytical (closed-form) solutions	Numerical multi-body dynamics solvers (e.g., Adams, Simscape) for complex mechanisms

In your later coursework—particularly in machine design, robotics, and vehicle dynamics—you will extend these relative motion equations to three dimensions using rotation matrices, apply them to open and closed kinematic chains, and couple them with the equations of motion to solve for unknown forces and moments. The conceptual core, however, remains identical: the velocity and acceleration of any point on a rigid body can always be expressed in terms of a base point's motion plus contributions from angular velocity and angular acceleration.

SECTION 9

Practice Problems

PROBLEM 1 — CONCEPTUAL

Explain physically why the relative velocity ω × r_B/A must always be perpendicular to the line segment from A to B on a rigid body. What would it mean if there were a component along the line AB?

PROBLEM 2 — BASIC CALCULATION

A rigid bar AB of length 0.5 m rotates about fixed pin A at ω = 6 rad/s counterclockwise with α = 0. At the instant shown, the bar is horizontal (B is directly to the right of A). Find the velocity and acceleration of point B.

PROBLEM 3 — INTERMEDIATE

A wheel of radius 0.4 m rolls without slipping on a flat surface. The center C of the wheel has velocity v_C = 2.0 î m/s and acceleration a_C = 1.0 î m/s². Find the velocity and acceleration of the topmost point T on the wheel.

PROBLEM 4 — APPLIED

In a four-bar linkage, the input crank (link 2) has length 100 mm and rotates at ω₂ = 20 rad/s CCW. The coupler (link 3) has length 300 mm, and the output rocker (link 4) has length 200 mm. At the instant when the crank is horizontal (pointing right) and the coupler makes 30° below horizontal, use the relative velocity equation to determine ω₃ and ω₄. Assume ground link length is 350 mm.

PROBLEM 5 — CRITICAL THINKING

Prove that the relative acceleration equation a_B = a_A + α × r_B/A + ω × (ω × r_B/A) is independent of the choice of base point. That is, show that using point B as the base and A as the target yields a consistent result for a_A.

SUMMARY

Lesson Summary

The motion of any point on a rigid body can be determined from the motion of a known base point using the relative velocity equation v_B = v_A + ω × r_B/A and the relative acceleration equation a_B = a_A + α × r_B/A + ω × (ω × r_B/A). The angular velocity ω and angular acceleration α are properties of the entire body; combined with the position vector r_{B/A}, they generate a tangential component (perpendicular to the line joining the two points) and a centripetal (normal) component (directed from the point of interest toward the base point).

In planar analysis, each vector equation yields two scalar equations—one per coordinate—providing a systematic means to solve for unknown angular velocities and angular accelerations in linkages, gear trains, and other mechanisms. The method is base-point independent, extends naturally to three dimensions, and serves as the kinematic prerequisite for rigid-body kinetics (Newton–Euler equations). For problems involving sliding contacts on moving bodies, augment the equations with a Coriolis term 2ω × v_rel using the rotating reference-frame formulation.