Dynamics · Chapter 7 of 10 · Intermediate
Plane Kinematics of Rigid Bodies
A rigid body is more than a point: it can spin as it travels. Its motion splits cleanly into translation plus rotation, and one special point is always momentarily at rest.
Readiness check
This chapter extends particle kinematics to bodies that rotate. Tick only what you can do closed-notes.
- Use v = ds/dt and a = dv/dt.
- Find normal acceleration an = v²/ρ.
- Add vectors by components.
- Convert rev/min to rad/s.
- Take a simple cross product in the plane.
The core idea
The velocity of any point on a rigid body equals the velocity of a reference point plus a rotation term; pick the reference well and the problem nearly solves itself.
vB = vA + ω × rB/Av = ωr (fixed axis)an = ω²r, at = αrPlane motion is translation plus rotation. Two points on the same rigid body are linked: their relative velocity is purely rotational, ω × r, perpendicular to the line joining them. There is always one point with zero velocity, the instantaneous center, and treating the body as rotating about it makes velocities a matter of geometry.
The skills, taught in order
Rigid-body kinematics is the relative-velocity equation applied with a good choice of reference. Five skills build from fixed-axis rotation to relative acceleration.
7.1 Rotation about a fixed axis
When a body spins about a fixed axis, every point moves on a circle. The angular quantities ω and α play the roles of v and a, and a point at radius r has speed v = ωr, tangential acceleration at = αr, and normal acceleration an = ω²r.
| Translation | Rotation |
|---|---|
| velocity v = ds/dt | angular velocity ω = dθ/dt |
| acceleration a = dv/dt | angular acceleration α = dω/dt |
| v = v₀ + at | ω = ω₀ + αt |
| v² = v₀² + 2as | ω² = ω₀² + 2αθ |
7.2 Absolute motion analysis
Sometimes a geometric constraint relates a body's angle to a position directly (a piston driven by a crank, a spool unwinding). Write that relation, then differentiate with respect to time to get velocities and accelerations. It is exact but can be algebraically heavy.
7.3 Relative velocity
The general tool: vB = vA + ω × rB/A. The relative term ω × rB/A is perpendicular to the line AB with magnitude ωrB/A. Choosing A as a point whose velocity is known (often a pin or a contact) reduces the equation to two scalar component equations.
7.4 The instantaneous center of zero velocity
At every instant there is a point (possibly off the body) with zero velocity, the instantaneous center. Every other point moves as if the body rotated about it, so v = ω·d, where d is the distance to the center. Find it as the intersection of lines drawn perpendicular to two known velocity directions.
7.5 Relative acceleration
Accelerations add the same way but with two rotation terms: aB = aA + α × rB/A − ω²rB/A. The α term is tangential and the ω² term points from B back toward A (centripetal). The instantaneous center shortcut does not apply to acceleration, so use the full equation.
Engineering connection: linkages, gears, cams, rolling wheels, and robot arms, supplying the velocities and accelerations that Chapter 8 turns into forces and torques.
Worked example 1: a rolling wheel
A wheel of radius 0.35 m rolls without slipping, its center moving at 12 m/s. Find the angular velocity and the velocity of the top and contact points, using the instantaneous center.
- ProblemFind ω and the velocities of the top and contact points for the wheel in Figure 1.
- Given / findr = 0.35 m, center speed vc = 12 m/s, rolling without slipping. Find ω, vtop, vcontact.
- AssumptionsNo slipping, so the contact point is the instantaneous center with zero velocity.
- ModelTreat the wheel as rotating about the contact point; velocity of any point is ω times its distance from that center.
- Equationsω = vc/r v = ω·d (d from IC)
- Solveω = 12/0.35 = 34.3 rad/s. The contact point is the IC, so vcontact = 0. The top is 2r = 0.70 m from the IC, so vtop = 34.3 × 0.70 = 24 m/s, exactly twice the center speed.
- CheckThe center is at r from the IC, giving ω·r = 12 m/s, consistent with the input. The top moving at 2vc is the well-known rolling result, visible as the blur at the top of a moving wheel.
- ConclusionThe instantaneous center turns a combined translation-and-rotation into a pure rotation, so velocities follow from distances alone. Note the contact point is the IC only for velocity; it does have acceleration.
Worked example 2: a sliding ladder
A 5 m ladder leans against a wall at 50° to the floor. Its base A slides away from the wall at 2 m/s. Find the angular velocity of the ladder and the downward velocity of its top B.
- ProblemFind the ladder's angular velocity and the top's speed in Figure 2.
- Given / findL = 5 m, θ = 50°, vA = 2 m/s horizontal. Find ω and vB.
- AssumptionsRigid ladder, base on floor (horizontal velocity), top on wall (vertical velocity).
- ModelRelative velocity: vB = vA + ω × rB/A, with B constrained to move vertically. (The instantaneous center gives the same answer.)
- EquationsvB = vA + ω × rB/A ω = vA/(L sinθ), vB = ω(L cosθ)
- SolveThe horizontal component of vB must be zero (it stays on the wall), which gives ω = vA/(L sinθ) = 2/(5 sin50°) = 2/3.83 = 0.52 rad/s. Then vB = ω(L cosθ) = 0.52 × 3.21 = 1.68 m/s downward.
- CheckThe instantaneous center sits where the two perpendiculars meet, at horizontal distance L cosθ from B and L sinθ from A; using v = ω·d reproduces both speeds. As the ladder flattens (θ → 0), ω grows without bound, the reason a slipping ladder whips down at the end.
- ConclusionOne rigid body links the two slider velocities through a single ω. Either the relative-velocity equation or the instantaneous center solves it; they are two views of the same geometry.
Misconceptions and diagnostics
| Mistake | Symptom | Diagnostic question | Correction |
|---|---|---|---|
| Relative term not perpendicular | ω × r drawn along AB | "Is the rotation term perpendicular to AB?" | ω × rB/A is always perpendicular to the joining line. |
| IC used for acceleration | Acceleration found as ω²·d wrongly | "Am I doing velocity or acceleration?" | The IC shortcut is for velocity only; use the full relative-acceleration equation. |
| Treating IC as fixed | Same center assumed at all instants | "Has the geometry changed?" | The IC moves over time; relocate it each instant. |
| Forgetting the ω² term | Centripetal acceleration omitted | "Did I include −ω²r?" | Relative acceleration has both α × r and −ω²r terms. |
Practice ladder
A disk spins up from rest at α = 5 rad/s². Find its angular velocity and the speed of a point 0.2 m out after 4 s.
Show answer
ω = αt = 5 × 4 = 20 rad/s; v = ωr = 20 × 0.2 = 4 m/s. The rotational formulas mirror the rectilinear ones exactly.
For the Worked Example 1 wheel, find the velocity of the point at the front of the wheel (same height as the center).
Show answer
That point is at distance √(r² + r²) = r√2 from the IC, so v = ω r√2 = 34.3 × 0.35 × 1.414 = 17.0 m/s, directed up and forward at 45°. Every point's speed is just ω times its distance from the contact point.
A 0.6 m connecting rod has one end on a crank pin moving at 3 m/s and the other on a piston constrained to move horizontally. At the instant the rod is horizontal and the pin velocity is vertical, find the rod's angular velocity.
Show answer
With the rod horizontal and the pin velocity vertical, the relative term must supply the piston's horizontal motion. ω = vpin/L = 3/0.6 = 5 rad/s. The exact value depends on the configuration; setting up vB = vA + ω × r is the reliable route.
Observe a real linkage or rolling body (a bicycle wheel, a door hinge, a folding mechanism). Identify the instantaneous center and estimate an angular velocity from a measured point velocity.
What good work looks like
The instantaneous center located from two velocity directions, ω found from v = ω·d, and a second point's velocity predicted and sanity-checked.
Working with AI, and proving it yourself
Use AI as an examiner, not a solver
Portfolio task
Analyse one real mechanism: locate its instantaneous center, find ω from a known velocity, and predict another point's velocity, confirming with the relative-velocity equation.
Retrieval and spaced review
Closed notes. Answer out loud, then reveal.
1. Write the relative-velocity equation.
vB = vA + ω × rB/A, with the last term perpendicular to AB.
2. For a point at radius r on a fixed axis, give v, at, an.
v = ωr, at = αr, an = ω²r.
3. What is the instantaneous center?
The point (possibly off the body) with zero velocity at that instant; the body momentarily rotates about it.
4. How do you locate the instantaneous center?
Intersect the perpendiculars to two known velocity directions.
5. Why can't the IC be used for acceleration?
It has zero velocity but generally nonzero acceleration; use the full relative-acceleration equation.
Textbook mapping
| Item | Mapping |
|---|---|
| Primary source | Meriam and Kraige, Engineering Mechanics: Dynamics (7th ed), Chapter 5 (Plane Kinematics of Rigid Bodies) |
| Cross-reference | Hibbeler, Dynamics, Ch. 16 · Beer and Johnston, Ch. 15 |
| Core topics | 7.1 Fixed-axis rotation · 7.2 Absolute motion · 7.3 Relative velocity · 7.4 Instantaneous center · 7.5 Relative acceleration |
| Engineering connection | Linkages, gears, cams, rolling wheels, and robot arms. |
| Read next | Chapter 8: Plane Kinetics of Rigid Bodies. |