Three-Dimensional Kinetics of Rigid Bodies

Learning Objectives

  • Understand the general equations of motion for a rigid body in 3D.
  • Relate angular momentum and the inertia tensor.
  • Apply Euler's equations of motion to solve 3D rigid body problems.
  • Understand Eulerian angles and their application in 3D rotation.
  • Analyze torque-free motion and gyroscopic motion.
  • Calculate the kinetic energy of a rigid body in 3D motion.

The kinetics of rigid bodies in three dimensions relates the forces and moments acting on a body to its resulting translational and rotational motion. This is governed by Newton's Second Law and the angular momentum principle. Unlike planar kinetics, the angular momentum vector in 3D is generally not parallel to the angular velocity vector.

Eulerian Angles and 3D Rotation

Describing the orientation of a rigid body in 3D space requires a robust coordinate system. Eulerian angles are one of the standard ways to represent this orientation.

Eulerian Angles

A set of three independent angles used to specify the orientation of a rigid body with respect to a fixed coordinate system. They represent a sequence of three elemental rotations about the axes of a coordinate system.

Understanding Eulerian Angles

The standard Eulerian angle sequence involves:

  • Precession (ϕ\phi): A rotation about the vertical fixed ZZ-axis.
  • Nutation (θ\theta): A rotation about the moving x′x'-axis (line of nodes).
  • Spin (ψ\psi): A rotation about the body's zz-axis (symmetry axis).

These angles allow us to express the angular velocity vector ω\mathbf{\omega} in terms of its components along the body-fixed axes, which is crucial for solving Euler's equations of motion.

Equations of Motion

The general equations of motion for a rigid body in 3D are expressed in vector form. The translational motion is governed by the sum of external forces, and the rotational motion is governed by the sum of external moments.

General Equations of Motion

The general equations of motion relate the forces and moments to translational and rotational changes. The translational equation describes the motion of the center of mass, while the rotational equation describes the change in angular momentum.

Translational and Rotational Equations of Motion

Governs the translational and rotational motion of a rigid body.

∑F=maG\sum \mathbf{F} = m \mathbf{a}_G∑MG=H˙G\sum \mathbf{M}_G = \dot{\mathbf{H}}_G

Variables

SymbolDescriptionUnit
F\mathbf{F}External force vectorN
mmMass of the rigid bodykg
aG\mathbf{a}_GAcceleration vector of the mass centerm/s2m/s^2
MG\mathbf{M}_GExternal moment vector about the mass centerN⋅mN\cdot m
HG\mathbf{H}_GAngular momentum vector about the mass centerkg⋅m2/skg\cdot m^2/s

Angular Momentum and the Inertia Tensor

In 3D, the angular momentum HG\mathbf{H}_G depends on both the angular velocity ω\mathbf{\omega} and the mass distribution of the body, which is described by the inertia tensor I\mathbf{I}.

Inertia Tensor

A symmetric 3×33 \times 3 matrix that describes a rigid body's mass distribution and its resistance to rotational acceleration about various axes. It contains moments of inertia on the diagonal and products of inertia on the off-diagonals.

Principal Axes

A set of mutually orthogonal axes originating from a specific point (usually the center of mass) for which all products of inertia of a given body are zero, leaving only the principal moments of inertia.

Angular Momentum and Inertia

The angular momentum vector HG\mathbf{H}_G is related to the angular velocity ω\mathbf{\omega} through the inertia tensor IG\mathbf{I}_G. When using the principal axes, the products of inertia (Ixy,Ixz,IyzI_{xy}, I_{xz}, I_{yz}) become zero, significantly simplifying calculations.

Angular Momentum and Inertia Tensor

Relates angular momentum, inertia tensor, and angular velocity in 3D.

HG=IGω\mathbf{H}_G = \mathbf{I}_G \mathbf{\omega}IG=[Ixx−Ixy−Ixz−IyxIyy−Iyz−Izx−IzyIzz]\mathbf{I}_G = \begin{bmatrix} I_{xx} & -I_{xy} & -I_{xz} \\ -I_{yx} & I_{yy} & -I_{yz} \\ -I_{zx} & -I_{zy} & I_{zz} \end{bmatrix}

Variables

SymbolDescriptionUnit
HG\mathbf{H}_GAngular momentum vectorkg⋅m2/skg\cdot m^2/s
IG\mathbf{I}_GInertia tensor matrixkg⋅m2kg\cdot m^2
ω\mathbf{\omega}Angular velocity vectorrad/s
Ixx,Iyy,IzzI_{xx}, I_{yy}, I_{zz}Moments of inertia about the x, y, and z axeskg⋅m2kg\cdot m^2
Ixy,Ixz,IyzI_{xy}, I_{xz}, I_{yz}Products of inertiakg⋅m2kg\cdot m^2

Principal Axes Simplification

For any rigid body, there exists a set of mutually orthogonal principal axes for which the products of inertia are zero. When these axes are used, the inertia tensor becomes diagonal, simplifying the angular momentum equation.

Angular Momentum with Principal Axes

Simplified angular momentum equation using principal axes where products of inertia are zero.

HG=Ixωxi+Iyωyj+Izωzk\mathbf{H}_G = I_x \omega_x \mathbf{i} + I_y \omega_y \mathbf{j} + I_z \omega_z \mathbf{k}

Variables

SymbolDescriptionUnit
HG\mathbf{H}_GAngular momentum vector about the mass centerkg⋅m2/skg\cdot m^2/s
Ix,Iy,IzI_x, I_y, I_zPrincipal moments of inertiakg⋅m2kg\cdot m^2
ωx,ωy,ωz\omega_x, \omega_y, \omega_zComponents of angular velocity along principal axesrad/s

Euler's Equations of Motion

When the reference frame is attached to the rigid body and aligned with its principal axes, the rotational equations of motion simplify to Euler's Equations.

Euler's Equations Concepts

Euler's equations expand the vector equation ∑MG=H˙G\sum \mathbf{M}_G = \dot{\mathbf{H}}_G into three scalar equations along the principal axes. These nonlinear differential equations relate the applied moments to the resulting angular acceleration and the gyroscopic effects caused by rotation.

Euler's Equations

Scalar equations of rotational motion aligned with the principal axes.

∑Mx=Ixω˙x−(Iy−Iz)ωyωz∑My=Iyω˙y−(Iz−Ix)ωzωx∑Mz=Izω˙z−(Ix−Iy)ωxωy\begin{aligned} \sum M_x &= I_x \dot{\omega}_x - (I_y - I_z) \omega_y \omega_z \\ \sum M_y &= I_y \dot{\omega}_y - (I_z - I_x) \omega_z \omega_x \\ \sum M_z &= I_z \dot{\omega}_z - (I_x - I_y) \omega_x \omega_y \end{aligned}

Variables

SymbolDescriptionUnit
∑Mx,∑My,∑Mz\sum M_x, \sum M_y, \sum M_zSum of moments about the principal axesN⋅mN\cdot m
Ix,Iy,IzI_x, I_y, I_zPrincipal moments of inertiakg⋅m2kg\cdot m^2
ωx,ωy,ωz\omega_x, \omega_y, \omega_zComponents of angular velocity along principal axesrad/s
ω˙x,ω˙y,ω˙z\dot{\omega}_x, \dot{\omega}_y, \dot{\omega}_zComponents of angular acceleration along principal axesrad/s2rad/s^2

Solving 3D Kinetics Problems using Euler's Equations

  1. Define a Coordinate System: Establish a coordinate system attached to the rigid body, with the origin at the center of mass or a fixed point.
  2. Determine Principal Axes and Moments of Inertia: Align the coordinate axes with the principal axes of the body to eliminate products of inertia, and calculate the principal moments of inertia (Ix,Iy,IzI_x, I_y, I_z).
  3. Express Angular Velocity and Acceleration: Write the angular velocity vector ω\mathbf{\omega} and the angular acceleration vector ω˙\dot{\mathbf{\omega}} in terms of components along the chosen body-fixed axes.
  4. Identify External Moments: Determine the components of all external moments acting on the body about the coordinate origin (∑Mx,∑My,∑Mz\sum M_x, \sum M_y, \sum M_z).
  5. Apply Euler's Equations: Substitute the moments, moments of inertia, angular velocities, and angular accelerations into Euler's equations.
  6. Solve the Equations: Solve the resulting system of differential equations for the unknown angular accelerations, velocities, or applied moments.

Torque-Free Motion

An important class of problems in 3D rigid body dynamics involves bodies that undergo rotation while the net external moment applied to their center of mass is zero (∑MG=0\sum \mathbf{M}_G = 0). Examples include spacecraft, satellites, and tossed objects (like a football in flight).

Torque-Free Motion

The rotational motion of a rigid body when the resultant external moment acting about its center of mass is zero. Under these conditions, the angular momentum of the body is conserved in both magnitude and direction.

Torque-Free Motion Principles

Because the net moment is zero (∑MG=0\sum \mathbf{M}_G = 0), two fundamental conservation laws apply to the body during torque-free motion: the conservation of angular momentum (HG\mathbf{H}_G) and the conservation of kinetic energy (TT).

Axisymmetric Bodies: For bodies with an axis of symmetry (e.g., a cylinder where Ix=Iy=II_x = I_y = I), the motion simplifies significantly. The body will rotate about its axis of symmetry at a constant rate ωz\omega_z, while simultaneously undergoing regular precession about the fixed angular momentum vector HG\mathbf{H}_G.

Torque-Free Euler's Equations

Euler's equations when the net external moment is zero.

0=Ixω˙x−(Iy−Iz)ωyωz0 = I_x \dot{\omega}_x - (I_y - I_z) \omega_y \omega_z0=Iyω˙y−(Iz−Ix)ωzωx0 = I_y \dot{\omega}_y - (I_z - I_x) \omega_z \omega_x0=Izω˙z−(Ix−Iy)ωxωy0 = I_z \dot{\omega}_z - (I_x - I_y) \omega_x \omega_y

Variables

SymbolDescriptionUnit
Ix,Iy,IzI_x, I_y, I_zPrincipal moments of inertiakg⋅m2kg\cdot m^2
ωx,ωy,ωz\omega_x, \omega_y, \omega_zComponents of angular velocity along principal axesrad/s
ω˙x,ω˙y,ω˙z\dot{\omega}_x, \dot{\omega}_y, \dot{\omega}_zComponents of angular acceleration along principal axesrad/s2rad/s^2

Torque-Free Conservation Laws

Angular momentum and kinetic energy remain constant in torque-free motion.

HG=constant\mathbf{H}_G = \text{constant}T=12ω⋅HG=constantT = \frac{1}{2} \mathbf{\omega} \cdot \mathbf{H}_G = \text{constant}

Variables

SymbolDescriptionUnit
HG\mathbf{H}_GConstant angular momentum vectorkg⋅m2/skg\cdot m^2/s
TTConstant rotational kinetic energyJ
ω\mathbf{\omega}Angular velocity vectorrad/s

Kinetic Energy in 3D

The kinetic energy (TT) of a rigid body undergoing general 3D motion is the sum of its translational and rotational kinetic energy. Unlike 2D where rotational energy is simply 12Izω2\frac{1}{2}I_z\omega^2, in 3D, we must use the dot product of angular velocity and angular momentum vectors.

3D Kinetic Energy Formula

Calculates the total kinetic energy for a rigid body in 3D motion.

T=12mvG⋅vG+12ω⋅HGT = \frac{1}{2}m\mathbf{v}_G \cdot \mathbf{v}_G + \frac{1}{2}\mathbf{\omega} \cdot \mathbf{H}_G

If the angular velocity vector is expressed in terms of components along the principal axes, this expands to:

T=12mvG2+12(Ixωx2+Iyωy2+Izωz2)T = \frac{1}{2}mv_G^2 + \frac{1}{2}(I_x \omega_x^2 + I_y \omega_y^2 + I_z \omega_z^2)

Variables

SymbolDescriptionUnit
TTTotal kinetic energyJ
mmMass of the bodykg
vG\mathbf{v}_GVelocity vector of the center of massm/s
vGv_GMagnitude of the center of mass velocitym/s
ω\mathbf{\omega}Angular velocity vectorrad/s
HG\mathbf{H}_GAngular momentum vectorkg⋅m2/skg\cdot m^2/s
Ix,Iy,IzI_x, I_y, I_zPrincipal moments of inertiakg⋅m2kg\cdot m^2
ωx,ωy,ωz\omega_x, \omega_y, \omega_zComponents of angular velocity along principal axesrad/s

Gyroscopic Motion

Gyroscopic motion involves the rotation of a symmetric body at high speed about its axis of symmetry, while being subjected to a torque that causes it to rotate about a mutually perpendicular axis.

Precession

The change in the orientation of the rotational axis of a spinning body. It occurs when a torque is applied perpendicular to the axis of spin.

Gyroscopes

A gyroscope is a device containing a rapidly spinning rotor that tends to maintain its orientation in space due to conservation of angular momentum. When a torque is applied perpendicular to the spin axis, the gyroscope exhibits precession—a rotation about an axis mutually perpendicular to the spin and torque axes.

Steady Precession of a Gyroscope

Relates the applied moment to the precession angular velocity and spin angular momentum.

M=Ω×H\mathbf{M} = \mathbf{\Omega} \times \mathbf{H}

Variables

SymbolDescriptionUnit
M\mathbf{M}Applied moment vectorN⋅mN\cdot m
Ω\mathbf{\Omega}Precession angular velocity vectorrad/s
H\mathbf{H}Spin angular momentum vectorkg⋅m2/skg\cdot m^2/s

Interactive 3D Kinetics Lab

Change the body dimensions, force direction, eccentric load point, pure couple, and initial angular rates. Play or pause the teaching loop, inspect the frame vectors and moment ring, switch to the section cutaway, and run the guided camera sequence to compare translation with rotation.

Three-dimensional kinetics: force → moment → motion

Concept and model scope

A principal-axis rigid-body lab for seeing how an eccentric force and a pure couple create simultaneous center-of-mass acceleration, angular acceleration, and angular momentum.

This lab uses one canonical rectangular-body state for both the numerical results and the scene. The force is applied at PP, the mass center is GG, and the red/green/blue triad identifies the body principal axes.

∑F=maG\sum \mathbf{F}=m\mathbf{a}_G gives translation.
MG=Iα−ω×(Iω)\mathbf{M}_G=\mathbf{I}\boldsymbol{\alpha}-\boldsymbol{\omega}\times(\mathbf{I}\boldsymbol{\omega}) gives rotation.
Dashed path: scaled center-of-mass translation.
Cutaway: translucent shell exposes the internal mass model.

All values are SI. Vector lengths are normalized for visual readability and do not change the computed values.

Learning objective. Use the scene to predict which part of the response changes when the load direction, load offset, body inertia, or initial angular rates change.

∑F=maG\sum \mathbf{F}=m\mathbf{a}_G+MG=Iα−ω×(Iω)\mathbf{M}_G=\mathbf{I}\boldsymbol{\alpha}-\boldsymbol{\omega}\times(\mathbf{I}\boldsymbol{\omega})

Controls

One shared frame · labels open focused definitions · title opens shared model context.

120 kg
2.2 m
1.2 m
0.9 m
900 N
+35°
+20°
0.80 ×
+70 N·m
+0.8 rad/s
+1.4 rad/s
+5.5 rad/s
0.50×
Motion & overlays
t = 0.00 s · loop 2.40 s

Engineering visual

Rigid body, reference frames, and load path

At t = 0.00 seconds, the rigid body has center-of-mass acceleration (5.77, 2.57, 4.04) metres per second squared, total moment (272.95, -541.30, -1.91) newton metres, angular acceleration (14.29, -6.24, -0.83) radians per second squared, and angular momentum (18.00, 87.92, 310.75) kilogram metres squared per second. The load point P is offset from G by (0.74, 0.33, -0.36) metres in the body frame.

Translation path display scale: 1:50 · force/moment arrows are normalized for legibility · camera presets preserve the engineering frame.

Direct learning results

Computed state

SI · body axes
∣aG∣|\mathbf{a}_G|
7.50 m/s²
∣MG∣|\mathbf{M}_G|
606.23 N·m
∣α∣|\boldsymbol{\alpha}|
15.61 rad/s²
∣HG∣|\mathbf{H}_G|
323.45 kg·m²/s
Rotational energy
923.3 J
Total energy at t
923.3 J
Coupled translation + rotation. The gyroscopic term is 163.62 N·m; changing a second angular-rate component changes the response.

The offset force contributes 669.5 N·m to M_G before the pure couple is added.

aG = (5.77, 2.57, 4.04) m/s² · MG = (272.95, -541.30, -1.91) N·m

Governing relationship

Translation
∑F=maG\sum \mathbf{F}=m\mathbf{a}_G
Rotation
MG=Iα−ω×(Iω)\mathbf{M}_G=\mathbf{I}\boldsymbol{\alpha}-\boldsymbol{\omega}\times(\mathbf{I}\boldsymbol{\omega})

Read the response

Move PP toward GG to remove the force-induced moment without changing aG\mathbf{a}_G. Then vary two initial rates to make the cross-axis gyroscopic terms visible in α\boldsymbol{\alpha}.

Key Takeaways
  • Equations of Motion: In 3D, the equations are ∑F=maG\sum \mathbf{F} = m \mathbf{a}_G and ∑MG=H˙G\sum \mathbf{M}_G = \dot{\mathbf{H}}_G.
  • Inertia Tensor: Describes the mass distribution in 3D and is used to relate angular velocity to angular momentum.
  • Principal Axes: A specialized coordinate frame where products of inertia are zero, significantly simplifying the angular momentum and kinetic energy equations.
  • Euler's Equations: Provide the rotational equations of motion aligned with the principal axes.
  • Eulerian Angles: Precession, nutation, and spin define the 3D orientation of a rigid body.
  • Torque-Free Motion: Occurs when the net external moment is zero, resulting in constant angular momentum and constant kinetic energy. Axisymmetric bodies undergo steady precession in this state.
  • Gyroscopic Motion: Characterized by precession, where an applied torque causes rotation about a perpendicular axis.