Three-Dimensional Kinematics of a Rigid Body
This chapter extends rigid-body kinematics into three dimensions. We analyze a body rotating about a fixed point, develop the tool for differentiating a vector observed from a rotating frame, and finally build a general relative-motion analysis using axes that both translate and rotate.
Chapter Objectives
- Analyze the kinematics of a body undergoing rotation about a fixed point and general spatial motion.
- Carry out a relative-motion analysis of a rigid body using translating and rotating reference axes.
- Select fixed, translating, or translating-rotating axes from the physical constraints.
- Calculate spatial point velocity and acceleration using auditable cross products.
- Verify rigid-distance, guide-direction, units, signs, and limiting behavior.
Physical meaning
Spatial kinematics describes geometry and rates only; mass and force do not enter. Cross products encode both magnitude and the right-hand-rule direction, while transport terms account for basis vectors that rotate during observation.
Problem-solving strategy
- Classify the point as fixed in a rigid body, on a generally moving body, or moving relative to rotating axes.
- Declare one right-handed fixed frame and label every moving frame and its angular velocity.
- Build position vectors from geometry, then write the complete symbolic velocity or acceleration relation before substitution.
- Evaluate cross products component by component and include Coriolis acceleration only when relative motion and frame rotation both exist.
- Check dimensions, right-hand-rule signs, rigid-distance constraints, and guide or contact admissibility.
Rotation About a Fixed Point
When a rigid body turns about a fixed point, every particle stays a constant distance r from that point. Each particle therefore travels on the surface of a sphere of radius r centered on the fixed point. Because that motion is built from successive rotations, we begin by examining how rotational displacements behave.
Rotation About a Fixed Point
A body rotates about a fixed point when one point of the body remains stationary in space. Every other particle then moves on the surface of a sphere centered at that fixed point, since its radial distance to the point never changes.
Euler's Theorem
Two rotations taken about different axes through a common point can always be replaced by one equivalent rotation about a single axis through that same point. When more than two rotations act, they may be combined in pairs and each pair reduced to one resultant.
Condition: Two component rotations act about different axes passing through a common point.
Conclusion: They are equivalent to a single resultant rotation about one axis through that point. Successive pairs of rotations can be reduced the same way.
Finite vs. Infinitesimal Rotations
The order of application matters for finite rotations. Applying a 90° rotation about one axis then another generally leaves the body in a different orientation than the reverse order does. Because finite rotations are not commutative, they cannot be treated as vectors.
If instead the rotations are made infinitesimally small, the order ceases to matter. Because the vector cross product obeys the distributive law, two infinitesimal rotations produce the same displacement regardless of sequence. Infinitesimal rotations therefore do qualify as vectors.
Infinitesimal Rotations as Vectors
Only infinitesimally small rotations are used to define angular motion in three dimensions. Such rotations commute under addition, so they can be combined vectorially in any order. By Euler's theorem the two components reduce to one resultant, dθ = dθ₁ + dθ₂.
Angular Velocity & Angular Acceleration
The angular velocity is the time rate of the rotation vector. Its direction defines the instantaneous axis of rotation, the line about which the body momentarily turns. This axis generally swings in space from instant to instant.
Instantaneous Axis & the Space / Body Cones
As the instantaneous axis of rotation sweeps through space, its locus traces a fixed space cone. Viewed from the rotating body, the same axis traces a body cone. The two cones touch along the instantaneous axis, and the body cone appears to roll on the space cone. Unlike for fixed-axis motion, α need not lie along the instantaneous axis, because it captures changes in both the magnitude and the direction of ω.
Velocity & Acceleration of a Point
Once ω is known, the velocity and acceleration of any point of the body follow from the same cross-product relations used for fixed-axis rotation, with r measured from the fixed point.
Watch the trap in 3-D: α is NOT necessarily parallel to ω. In the fixed-axis problems from earlier chapters they were collinear, so students assume it carries over. It doesn't — here α also accounts for the swinging of the axis direction. If you only differentiate the magnitude of ω, you'll lose a whole term. Always use Eq. 20.6 (next section) to get α.
Time Derivative of a Vector — Fixed vs. Translating-Rotating System
When ω is given by its components, finding α is often easier in a frame that itself rotates with one of those components. We need a rule that links the time derivative of any vector seen from a fixed frame to its derivative seen from a moving (translating and rotating) frame.
Two Observers, Two Derivatives
Let the moving x, y, z axes rotate with angular velocity Ω relative to the fixed X, Y, Z axes. An observer riding the moving frame sees only the change in the components of a vector A, written (Ȧ)xyz. A fixed observer additionally sees the unit vectors rotate. The two views differ by the term Ω × A.
This is one of the most important results in the chapter: the fixed-frame derivative equals the rotating-frame derivative plus the correction Ω × A caused by the spin of the axes. If the axes don't rotate (Ω = 0), the two derivatives are identical.
This single equation is the engine for the entire rest of the chapter — every α, every aᴮ in sections 20.3 and 20.4 is just this transport rule applied over and over. Memorize it cold. My tip: pick your rotating frame Ω so that the vector you're differentiating looks constant from inside it. Then (Ȧ)ₓᵧ𝓏 = 0 and you're left with only the clean Ω × A term. That choice often leaves only the clean transport cross-product term.
General Motion
A body in general spatial motion has both translation and rotation. If point A has known motion, the motion of any other point B follows from a relative-motion analysis using translating axes attached to A. The relative term is exactly the fixed-point rotation of B as seen from A.
General Motion via a Translating Base Point
Place a non-rotating frame at base point A. The body's motion is then an instantaneous translation matching A plus a rotation about an instantaneous axis through A. The relative velocity and acceleration of B are the fixed-point results ω × rB/A and α × rB/A + ω × (ω × rB/A).
These mirror the planar relative-motion equations, but in 3-D the term α now carries directional change as well. For a member jointed by ball-and-sockets, a useful constraint comes from noting that the relative velocity is perpendicular to the rod's axis.
Relative-Motion Analysis Using Translating and Rotating Axes
The most general approach uses x, y, z axes that both translate and rotate relative to a fixed frame X, Y, Z. This handles points on separate members of a mechanism and points moving along curved paths. The moving frame has angular velocity Ω and angular acceleration Ω̇, both measured from the fixed frame.
Reading the Acceleration Terms
The acceleration of B collects five contributions: the base-point acceleration aA; the angular-acceleration term Ω̇ × rB/A; the centripetal term Ω × (Ω × rB/A); the Coriolis term 2Ω × (vB/A)xyz; and the relative acceleration measured inside the moving frame, (aB/A)xyz. The Coriolis term appears only when the point moves relative to a rotating frame.
Procedure for Analysis
Place the origins coincident and axes parallel where possible. When several angular-velocity components act, choose the moving frame so only one component is observed from it, letting the frame rotate with the rest. Use nested primed axes and the transport rule (Eq. 20.6) to differentiate rA, rB/A, Ω, and Ωxyz term by term, then substitute numbers last.
The Coriolis term is the #1 thing students drop. If a point slides along a member that is itself rotating, you have a non-zero (v_B/A)ₓᵧ𝓏 and a non-zero Ω — that guarantees a 2Ω×(v_B/A)ₓᵧ𝓏 term. Whenever you see "collar moving along the rod" plus "arm rotating," circle both before you start. Use an independent fixed-frame reconstruction when possible; agreement between frame choices is a powerful check.
Chapter Review
Rotation About a Fixed Point
Points move on a sphere about O. Because α reflects both magnitude and direction changes of ω, specify ω by components and differentiate with the transport rule Ȧ = (Ȧ)xyz + Ω × A. Then vP = ω × r and aP = α × r + ω × (ω × r).
General Motion
Relate point B to base point A with translating axes at A: vB = vA + ω × rB/A and aB = aA + α × rB/A + ω × (ω × rB/A).
Translating & Rotating Axes
The full relative-motion equations (20.13, 20.14) connect two points on a body, on connected bodies, or on separate paths. Account for direction changes in rA, rB/A, Ω, and Ωxyz by applying Eq. 20.6 to every time derivative — including the Coriolis term 2Ω × (vB/A)xyz.
New integrated applications
New integrated applications
Recognition strategy
The base point is fixed and the probe is fixed in one rigid body. Therefore the fixed-point cross-product relations apply directly; no translating-base or Coriolis terms belong.
Given information
- All vectors use one right-handed inertial basis with O fixed.
- The probe has no relative sliding; the unknowns are its instantaneous velocity and acceleration.
Model and interpretation
Treat the bracket as perfectly rigid. This is kinematics, so no force or mass data are needed. Use \(\mathbf v=\boldsymbol\omega\times\mathbf r\) and the complete tangential-plus-normal acceleration relation.
Verification
Units: Angular rate times position gives \(\mathrm{m/s}\), and each acceleration cross product gives \(\mathrm{m/s^2}\); all components are dimensionally consistent.
Signs and direction: The dot product \(\mathbf r\cdot\mathbf v=0\), confirming velocity is tangent to the sphere and the right-hand-rule signs are self-consistent.
Plausibility: Differentiating \(\mathbf r\cdot\mathbf r\) twice requires \(\mathbf r\cdot\mathbf a=-\mathbf v\cdot\mathbf v=-2.85\ \mathrm{m^2/s^2}\); both sides agree.
Engineering interpretation
The velocity is purely tangent to the probe’s spherical path about O, while acceleration includes both turning and changing-rate effects. Clearance checks should use the full vector trajectory, not component magnitudes alone.
Recognition strategy
A rotating carrier changes the direction of a rotor spin component even when its body-frame direction is fixed. The transport theorem is needed before fixed-point point-motion equations can be used.
Given information
- Carrier angular velocity is \(\boldsymbol\Omega=2\mathbf k\ \mathrm{rad/s}\), with zero carrier angular acceleration.
- The total angular velocity is \(6\mathbf i+2\mathbf k\ \mathrm{rad/s}\), and \(\mathbf r=0.300\mathbf i\ \mathrm m\).
Model and interpretation
Use fixed right-handed axes coincident with the pedestal axes at this instant. The point is fixed in the rotor and the axis intersection is stationary, so there is no translational or relative-sliding term.
Verification
Units: The transport term has \(\mathrm{rad/s^2}\), velocity has \(\mathrm{m/s}\), and both acceleration terms reduce correctly to \(\mathrm{m/s^2}\).
Signs and direction: Positive \(\mathbf j\) velocity follows from \(\mathbf k\times\mathbf i=\mathbf j\); the equal opposite k acceleration components cancel rather than being discarded.
Plausibility: Since the selected point lies on the rotor spin axis, spin alone gives no point speed; the computed \(0.600\ \mathrm{m/s}\) equals pedestal rate times radius.
Engineering interpretation
The large transverse angular acceleration exists even though pedestal speed is constant, because the carrier continually redirects rotor angular momentum geometry. Shaft-alignment and sensor models must retain this transport effect.
Recognition strategy
The point slides relative to a frame that also rotates, so the full rotating-frame relations are mandatory. In particular, nonzero outward speed and nonzero frame rotation guarantee a Coriolis term.
Given information
- Take \(+\mathbf i\) outward, \(+\mathbf j\) tangential in the rotation sense, and \(+\mathbf k\) upward.
- The origin is fixed; relative velocity is \(1.20\mathbf i\ \mathrm{m/s}\) and relative acceleration is \(-0.400\mathbf i\ \mathrm{m/s^2}\).
Model and interpretation
Model the carriage as a point constrained to a rigid straight guide. This is kinematics only. Evaluate base, Euler, centripetal, Coriolis, and relative terms separately so guide travel and direction can be audited.
Verification
Units: Every acceleration contribution contains angular rate squared times length, angular acceleration times length, or angular rate times speed, all yielding \(\mathrm{m/s^2}\).
Signs and direction: Outward relative motion with positive k rotation produces positive j Coriolis acceleration by the right-hand rule; inward motion would reverse only that term.
Plausibility: With relative speed set to zero, the Coriolis term vanishes and the result reduces to ordinary rotating-arm acceleration, confirming the expected limiting behavior.
Engineering interpretation
The tangential acceleration is dominated by Coriolis motion and will influence guide reaction and actuator loading in a later kinetics analysis. Engineers must also verify remaining guide travel and end-stop clearance.
Exam strategy for this chapter: it's all bookkeeping. Write the symbolic equation first, list every vector with its frame label, decide your Ω before plugging numbers, and keep i-j-k columns aligned. 90% of lost marks here are sign errors and dropped terms, not concept failures. Skip the algebra shortcuts until you've done five clean problems the long way.
Chapter recap
- Add angular velocities as vectors and use the transport theorem whenever the observer’s basis rotates.
- Use fixed-point or translating-base rigid-body relations only when the selected points share one rigid body.
- Use the full rotating-frame equation for sliding points and identify Coriolis acceleration before calculating.
- Before moving on, be able to declare frames, expand cross products, verify constant-distance constraints, and check guide or contact admissibility.