Planar Rigid-Body Motion — Analyze planar rigid-body motion: translation, rotation, and general plane motion - Statics and Dynamics

SECTION 1

Historical Context & Motivation

The study of how rigid bodies move through space—without deformation—has been central to engineering mechanics since the dawn of the scientific revolution. Ancient Greek philosophers such as Aristotle recognized that objects move differently depending on whether they slide or spin, but a rigorous mathematical framework for planar rigid-body motion only crystallized over centuries of work by mathematicians, physicists, and engineers. Understanding how machines, linkages, gears, and vehicles move in a plane requires classifying motion into translation, rotation about a fixed axis, and general plane motion—a taxonomy that remains the backbone of modern kinematics.

1687 • Newton's Principia

Isaac Newton published the Principia Mathematica, establishing the laws of motion for particles and laying the groundwork for rigid-body dynamics.

1765 • Euler's Rigid-Body Theory

Leonhard Euler formalized the equations governing rigid-body rotation, introducing angular momentum concepts and the moment equation M = Iα that engineers still use today.

1830s • Chasles' Theorem

Michel Chasles proved that every spatial displacement of a rigid body can be decomposed into a translation and a rotation, unifying the classification of planar motion types.

1876 • Reuleaux & Mechanism Science

Franz Reuleaux published The Kinematics of Machinery, systematically applying planar rigid-body kinematics to the analysis and synthesis of mechanisms.

20th c. • Modern Engineering Dynamics

With the rise of automotive, aerospace, and robotic engineering, planar rigid-body kinematics became a core pillar of undergraduate dynamics curricula worldwide, enabling the design of everything from four-bar linkages to satellite attitude-control systems.

The central question that planar rigid-body kinematics answers is deceptively simple: Given a rigid body confined to move in a plane, how do we describe the velocity and acceleration of every point on that body? Because a rigid body cannot deform, the distance between any two points remains constant, and this geometric constraint dramatically reduces the problem's complexity. The entire motion can always be decomposed into at most two independent components—a translation and a rotation—and mastering this decomposition is the gateway to solving mechanism, vehicle, and structural dynamics problems.

SECTION 2

Core Principles & Definitions

A rigid body is an idealized solid in which the distance between every pair of material points remains invariant throughout the motion. When we restrict the body's motion so that every particle travels in a plane parallel to a single fixed reference plane, we call the resulting movement planar motion. This simplification applies to a vast number of engineering problems—wheels rolling along roads, connecting rods in reciprocating engines, robotic arms constrained to a vertical plane, and many more. Within planar motion three mutually exclusive categories exist, each defined by the nature of the body's velocity field.

Translation

Every line segment in the body remains parallel to its original orientation at all times. All points share the same velocity and acceleration vectors, so the motion of any single point fully characterizes the body's motion.

Rotation About a Fixed Axis

All particles move in concentric circles about a single fixed point (the axis, seen as a point in the plane). The body's angular velocity ω and angular acceleration α completely define the kinematics.

General Plane Motion

A simultaneous combination of translation and rotation. No single fixed point exists; however, at any instant, an instantaneous center of zero velocity (IC) can be identified, about which the body appears to purely rotate.

Relative-Motion Equation

The velocity of any point B on the body equals the velocity of a chosen base point A plus the velocity of B relative to A due to the body's rotation: v_B = v_A + ω × r_B/A.

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KEY TAKEAWAY

Think of planar rigid-body motion like a smartphone sliding across a table. If you push it straight without twisting, that is pure translation. If you pin one corner down and spin it, that is fixed-axis rotation. If you flick it so it both slides and spins freely, that is general plane motion. In every case the phone doesn't bend—it moves as a rigid body—and you can always decompose the complex motion into a translation of one chosen point plus a rotation about that point.

SECTION 3

Visual Explanation — The Three Motion Types

Figure 1. The three categories of planar rigid-body motion. Translation (left): every point traces a congruent path and all velocity vectors are identical. Fixed-axis rotation (center): all points orbit a stationary pivot O. General plane motion (right): the body translates and rotates simultaneously; the relative-motion equation governs point velocities.

In the left panel, both points A and B undergo identical displacements; every line in the body remains parallel to its original orientation, confirming ω = 0. In the center panel, point A sweeps through an angle θ about the fixed point O; its velocity is always tangent to the circular arc and has magnitude v = ωr, where r is the distance from O. The right panel illustrates the most common scenario in mechanism analysis: the body simultaneously translates (point A shifts from its initial position to A′) and rotates (the orientation of the body changes). The relative-velocity equation v_B = v_A + ω × r_B/A is the workhorse for computing velocities of arbitrary points during general plane motion.

SECTION 4

Mathematical Framework

The kinematic analysis of planar rigid-body motion rests on a small set of equations that relate the position, velocity, and acceleration of any point on the body to those of a chosen reference (base) point. Because a rigid body in the plane has exactly three degrees of freedom—two translational coordinates (x, y) and one rotational coordinate (θ)—the governing equations are compact yet powerful. The following equations form the mathematical core of planar rigid-body kinematics and should become second nature for any practising dynamicist.

Angular Kinematics

ANGULAR VELOCITY & ACCELERATION

ω = dθ/dt α = dω/dt = d²θ/dt²

θ = angular position (rad), ω = angular velocity (rad/s), α = angular acceleration (rad/s²). For constant α: ω = ω₀ + αt, θ = θ₀ + ω₀t + ½αt², ω² = ω₀² + 2α(θ − θ₀).

Relative-Velocity Equation

RELATIVE VELOCITY

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 (k̂ component), r_B/A = position vector from A to B. The cross product ω × r_B/A yields a velocity perpendicular to r_B/A with magnitude ω|r_B/A|.

Relative-Acceleration Equation

RELATIVE ACCELERATION

a_B = a_A + α × r_{B/A} − ω² r_{B/A}

The term α × r_B/A is the tangential acceleration component (perpendicular to r_B/A), while −ω²r_B/A is the normal (centripetal) acceleration component directed from B toward A.

Instantaneous Center of Zero Velocity

INSTANTANEOUS CENTER (IC)

v_P = ω × r_{P/IC} ⟹ |v_P| = ω · d_{P}

At any instant during general plane motion, a unique point IC exists where the velocity is zero. Every other point P on the body has velocity perpendicular to the line from IC to P, with magnitude equal to ω times the distance d_P from IC to P. This transforms general plane motion into apparent pure rotation about the IC.

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Sign Convention

In planar problems the angular velocity vector ω is always along the k̂ axis (perpendicular to the plane of motion). Counterclockwise rotation is taken as positive by convention. When evaluating cross products in 2-D, ω × r_B/A = ω k̂ × (x î + y ĵ) = ω(−y î + x ĵ), which is a 90° counterclockwise rotation of r_B/A scaled by ω.

SECTION 5

Detailed Classification & the Instantaneous Center

A deeper look at each motion type reveals the kinematic constraints engineers exploit in practice. Rectilinear translation occurs when every point follows a straight-line path (e.g., a piston sliding in a cylinder), while curvilinear translation means all points trace identical curved paths (e.g., a Ferris-wheel gondola that does not rotate about its own pivot). In both cases ω = 0, so the relative-velocity equation collapses to v_B = v_A. The concept of the instantaneous center of zero velocity (IC) is especially powerful for general plane motion: at any given instant, the IC is the point about which the entire body appears to rotate, so the velocity of every point equals ω times its distance from the IC. The IC migrates as the motion proceeds, tracing a locus called the centrode.

Figure 2. A wheel rolling without slipping exemplifies general plane motion. The instantaneous center (IC) lies at the contact point with the ground, where the velocity is zero. The center C has velocity ωR directed horizontally, while the topmost point T has velocity 2ωR—twice as fast—because it is twice as far from the IC.

Motion Type	Degrees of Freedom	ω	IC Location
Rectilinear Translation	1 (one translational coordinate)	0	At infinity (all velocity vectors are parallel)
Curvilinear Translation	2 (x, y linked by path constraint)	0	At infinity
Fixed-Axis Rotation	1 (θ)	≠ 0	Permanently at the fixed pivot
General Plane Motion	3 (x, y, θ)	≠ 0	Changes location at each instant

Summary of planar motion types with their kinematic characteristics

SECTION 6

Worked Example — Connecting Rod in a Slider-Crank

Consider a slider-crank mechanism in which the crank OA of length 0.15 m rotates counterclockwise with a constant angular velocity ω_OA = 10 rad/s. The connecting rod AB has a length of 0.40 m. At the instant shown, the crank is at θ = 90° (OA points straight up) and the slider B moves horizontally. Determine the angular velocity of the connecting rod AB and the velocity of the slider B.

Slider-Crank Velocity Analysis

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Step 1 — Identify the Motion Type of Each Link

The crank OA undergoes fixed-axis rotation about O. The slider B undergoes rectilinear translation along the horizontal guide. The connecting rod AB undergoes general plane motion because it both translates and rotates.
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Step 2 — Compute the Velocity of Point A

Point A is on the crank, at distance r = 0.15 m from the fixed pivot O. Since OA is at θ = 90° (pointing up), point A is directly above O. The velocity of A is tangent to its circular path, directed to the left (for CCW rotation). v_A = ω_OA × r_OA = 10 × 0.15 = 1.5 m/s to the left.

v_A = 1.5 m/s (←)
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Step 3 — Establish Geometry at the Given Instant

With OA vertical, point A is at coordinates (0, 0.15) relative to O. Point B lies on the horizontal line through O (since the slider guide passes through O in this standard configuration). The connecting rod AB has length 0.40 m, so the horizontal distance OB = √(AB² − OA²) = √(0.40² − 0.15²) = √(0.16 − 0.0225) = √0.1375 ≈ 0.3708 m. The angle φ that AB makes with the horizontal satisfies sin φ = 0.15/0.40 = 0.375, giving φ ≈ 22.02°.

φ ≈ 22.02°, OB ≈ 0.371 m
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Step 4 — Apply the Relative-Velocity Equation to Rod AB

Applying v_B = v_A + ω_AB × r_B/A in vector form: v_A = −1.5 î m/s, and r_B/A points from A (0, 0.15) to B (0.371, 0), giving r_B/A = 0.371 î − 0.15 ĵ m. With ω_AB k̂ as the unknown angular velocity of rod AB, the cross product evaluates to ω_AB k̂ × (0.371 î − 0.15 ĵ) = ω_AB(0.15 î + 0.371 ĵ). The full vector equation is therefore v_B î = −1.5 î + ω_AB(0.15 î + 0.371 ĵ). Since v_B is horizontal (constrained by the guide), its ĵ component is zero. Equating ĵ components: 0 = 0.371 ω_AB, which gives ω_AB = 0. Substituting back into the î equation: v_B = −1.5 + (0)(0.15) = −1.5 m/s, so the slider moves to the left at 1.5 m/s.

At θ = 90°: ω_AB = 0, v_B = 1.5 m/s (←)
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Step 5 — Interpret the Result

At the instant when the crank is vertical (θ = 90°), the connecting rod AB happens to have zero angular velocity—it is instantaneously in pure translation. This is a special geometric configuration: the ĵ component of the relative-velocity cross product is proportional to the horizontal component of r_B/A (which is nonzero), and the constraint that v_B has no vertical component forces ω_AB = 0. Consequently, the slider velocity equals the crank-tip velocity at this instant because the entire rod translates as a unit. At most other crank angles, ω_AB ≠ 0 and the rod executes general plane motion.

SECTION 7

Strengths & Limitations of Planar Analysis Methods

Engineers have several methods available for analyzing planar rigid-body motion. The relative-velocity equation and the instantaneous-center method are the two most common approaches taught in dynamics courses, and each has strengths and trade-offs. In addition, the graphical velocity-polygon technique, though largely superseded by computational tools, still provides valuable geometric intuition.

Method	Strengths	Limitations
Relative-Velocity Equation	Systematic and algebraically rigorous; extends naturally to acceleration analysis; handles any planar geometry	Requires careful bookkeeping of vector directions; can involve solving simultaneous equations for complex mechanisms
Instantaneous Center (IC)	Often provides velocities in a single step; gives excellent physical insight into the motion; no vector decomposition needed	Only valid for velocity analysis at one instant; cannot be used directly for acceleration; IC may lie at infinity for pure translation
Velocity Polygon (Graphical)	Visual; helpful for building intuition; quick checks of analytical results	Low precision without CAD tools; not scalable to large mechanisms; largely replaced by software

Comparison of planar kinematic analysis methods

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CHOOSING A METHOD

Use the IC method when you need a quick velocity for one or two points—think of it as a shortcut that converts general plane motion into momentary rotation. Switch to the relative-velocity equation when you need a complete velocity field or must proceed to acceleration analysis, since accelerations cannot be found from the IC alone (the IC itself is generally accelerating).

SECTION 8

Connection to 3-D Kinematics & Advanced Dynamics

Planar rigid-body kinematics is the two-dimensional special case of a much richer three-dimensional theory. In 3-D, a rigid body has six degrees of freedom (three translational and three rotational), and Euler's rotation theorem guarantees that any displacement can be expressed as a rotation about a single axis combined with a translation along that axis—a screw displacement. The planar instantaneous center generalizes to the instantaneous screw axis in three dimensions. When kinetics (forces and torques) are introduced, the kinematic equations derived here feed directly into Newton-Euler equations of motion, enabling engineers to predict forces in mechanisms, design control systems for robotic arms, and simulate vehicle dynamics.

Concept	Planar (2-D)	Spatial (3-D)
Degrees of freedom	3 (x, y, θ)	6 (x, y, z, φ, θ, ψ)
Angular velocity	Scalar ω (along k̂)	Vector ω = ωₓ î + ω_y ĵ + ω_z k̂
Instantaneous zero-velocity locus	A point (IC)	A line (instantaneous screw axis)
Relative velocity	$v_B$ = $v_A$ + ω × $r_{B/A}$	Same form, but ω is a 3-D vector
Common applications	Linkages, gears, rolling wheels	Gyroscopes, spacecraft, robotic manipulators

Planar vs. spatial rigid-body kinematics

Mastering the planar case thoroughly prepares you for these extensions because the relative-velocity and relative-acceleration equations have exactly the same vector form in three dimensions—only the angular velocity vector ω gains additional components. In courses on mechanism design, robotics, or vehicle dynamics, you will encounter these equations repeatedly, augmented by mass-acceleration terms that transform kinematics into kinetics.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

A Ferris wheel gondola is designed so that passengers always remain upright (the gondola does not rotate about its own pivot). Classify the motion of the gondola and explain why every point on the gondola has the same velocity at any given instant.

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The gondola undergoes curvilinear translation. Although each point traces a circular arc, the orientation of every line segment in the gondola remains constant (it does not rotate), so ω = 0 at all times. From the relative-velocity equation v_B = v_A + ω × r_B/A, setting ω = 0 yields v_B = v_A, confirming all points share the same velocity vector. This is the defining characteristic of translation, whether rectilinear or curvilinear.

Question 2: BASIC CALCULATION

A disk of radius 0.3 m rotates about its fixed center O with a constant angular acceleration α = 4 rad/s². If ω₀ = 0, find the angular velocity and the speed of a point on the rim after t = 3 s.

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Using the constant-angular-acceleration equation ω = ω₀ + αt = 0 + 4(3) = 12 rad/s. The speed of a rim point is v = ωr = 12 × 0.3 = 3.6 m/s. Additionally, the tangential acceleration is a_t = αr = 4 × 0.3 = 1.2 m/s² and the centripetal acceleration is a_n = ω²r = 144 × 0.3 = 43.2 m/s². The total acceleration magnitude is √(1.2² + 43.2²) ≈ 43.2 m/s², dominated by the centripetal component at this relatively high angular velocity.

Question 3: INTERMEDIATE

A bar AB of length 1.2 m slides so that end A moves along a vertical wall and end B moves along a horizontal floor. At the instant when A is 0.9 m above the floor, end A slides downward at 2 m/s. Determine the angular velocity of the bar and the velocity of end B.

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Position geometry: with A at height y_A = 0.9 m, the horizontal distance OB = √(1.2² − 0.9²) = √(1.44 − 0.81) = √0.63 ≈ 0.7937 m. The angle the bar makes with the horizontal is φ = arcsin(0.9/1.2) = 48.59°. Applying the relative-velocity equation v_B = v_A + ω × r_B/A: v_A = −2 ĵ, r_B/A = 0.7937 î − 0.9 ĵ. The cross product ω k̂ × r_B/A = ω(0.9 î + 0.7937 ĵ). From the y-component: 0 = −2 + 0.7937ω → ω = 2/0.7937 ≈ 2.52 rad/s (CW). From the x-component: v_B = 0 + 0.9(2.52) = 2.268 m/s to the right. Alternatively, the IC for this configuration lies at the intersection of lines perpendicular to v_A (horizontal through A) and v_B (vertical through B), confirming the same result.

Question 4: APPLIED

A 0.6 m radius wheel rolls without slipping on a flat road. The wheel's center has a velocity of 4 m/s to the right. Determine (a) the angular velocity of the wheel, (b) the velocity of the topmost point, and (c) the velocity of a point located at the same height as the center but on the leading edge (front) of the wheel.

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(a) For rolling without slipping the IC is at the contact point. v_C = ωR → ω = 4/0.6 = 6.667 rad/s (CW, since the wheel rolls to the right). (b) The top point is 2R = 1.2 m from the IC, so v_top = ω × 2R = 6.667 × 1.2 = 8.0 m/s to the right. (c) The leading-edge point P at center height is located at position (R, R) relative to the IC (R forward, R up). Its distance from the IC is d = R√2 = 0.6√2 ≈ 0.8485 m. The velocity magnitude is v_P = ωd = 6.667 × 0.8485 ≈ 5.66 m/s. The direction is perpendicular to the line from IC to P, which makes 45° with the horizontal, so v_P has equal horizontal and vertical components: v_Px = 4.0 m/s (→) and v_Py = 4.0 m/s (↑).

Question 5: CRITICAL THINKING

Consider a rigid bar PQ undergoing general plane motion. At a given instant, the velocity of P is 3 m/s directed at 30° above the horizontal, and the velocity of Q is 3 m/s directed at 30° below the horizontal. The bar is 2 m long and currently horizontal. (a) Locate the instantaneous center. (b) Determine the angular velocity of the bar. (c) Discuss what type of motion the bar would exhibit if both velocities were equal in magnitude and direction, and explain physically why the IC moves to infinity in that case.

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(a) The IC lies at the intersection of lines perpendicular to v_P and v_Q. The velocity at P is directed 30° above horizontal, so the perpendicular line through P is directed 30° from vertical (or 60° from horizontal). Similarly, the perpendicular through Q makes 60° with the horizontal on the opposite side. By symmetry (the bar is horizontal and the velocity magnitudes are equal), the IC lies directly above the midpoint M of PQ. The perpendicular from P intersects the perpendicular from Q at height h above the bar. Using geometry: tan 60° = h / 1.0, so h = √3 ≈ 1.732 m above the bar's midpoint. (b) The distance from the IC to P is d_P = 1.0/cos 60° = 2.0 m. Therefore ω = v_P/d_P = 3/2 = 1.5 rad/s. (c) If both velocities were equal in magnitude and direction, the bar would be in pure translation (ω = 0). In this case the perpendicular lines from P and Q to their velocity vectors would be parallel—they never intersect, meaning the IC is at infinity. Physically, an IC at infinity corresponds to zero angular velocity: all points have the same velocity, which is the definition of translation. The IC concept elegantly unifies translation and rotation as limiting cases of the same framework.

SUMMARY

Lesson Summary

Summary

Planar rigid-body motion encompasses three distinct categories: translation (all points share the same velocity, ω = 0), fixed-axis rotation (all points orbit a stationary pivot, v = ωr), and general plane motion (simultaneous translation and rotation). The relative-velocity equation v_B = v_A + ω × r_B/A is the fundamental tool for computing velocities of any point on a rigid body, decomposing the motion into the translation of a base point plus the rotation about that base point.

The instantaneous center of zero velocity (IC) provides a powerful shortcut for velocity analysis: at every instant during general plane motion, one unique point has zero velocity, and all other points appear to rotate about it. The relative-acceleration equation extends the analysis to include tangential and centripetal acceleration components. Mastery of these planar equations prepares engineers for three-dimensional rigid-body kinematics, mechanism synthesis, and the kinetic analyses that govern the design of machines, vehicles, and robotic systems.