Multibody Dynamics · Module 6 of 10
Newton-Euler Equations of Motion
This is where forces finally meet motion. Newton handles the mass center, Euler handles the spin, and together they give the equations of motion for one rigid body.
Readiness check
Tick only what you can do closed-notes before starting.
- State Newton’s second law F = m a.
- Recall that net moment equals inertia times angular acceleration in the plane.
- Recall the inertia tensor from Module 5.
- Separate applied forces from constraint forces.
- Recall that principal-axis inertias are the diagonal terms.
The core idea
The Newton-Euler equations state that the net force equals mass times the acceleration of the mass center, and the net moment equals the rate of change of angular momentum. In the plane they reduce to F = m a and M = I α.
translation: ΣF = m a_Grotation (planar): ΣM_G = I_G αrotation (spatial): M = I α + ω × (I ω)For a single rigid body the equations of motion come in two parts. Newton’s law governs the mass center: the sum of all forces equals the mass times the mass-center acceleration. Euler’s law governs rotation: the sum of moments about the mass center equals the rate of change of angular momentum. In planar motion the rotational law is simply M = Iα, because the spin axis is fixed. In three dimensions the inertia tensor can turn with the body, so the angular-momentum rate carries an extra gyroscopic term, omega cross I-omega; this is why a spinning body free of torque still changes its component spin rates. The forces divide into applied forces (gravity, springs, actuators, contact) and constraint forces from the joints. In this module the constraint forces are known or absent; the next modules add the machinery to solve for them. Written per body and stacked, these equations are the backbone of every multibody formulation.
The skills, taught in order
Five skills produce the equations of motion for one body.
6.1 The translational equation
Sum the forces on the body and set them equal to m a_G. This locates the mass center in time and is identical to particle dynamics applied to the center of mass.
6.2 The planar rotational equation
Sum the moments about the mass center and set them equal to I_G α. One scalar equation governs planar spin, which covers most linkage and mechanism problems.
6.3 Euler’s equations in space
About principal axes, use I₁ω̇₁ = (I₂ − I₃)ω₂ω₃ and its cyclic partners. The cross terms are the gyroscopic coupling that planar analysis never shows.
6.4 Applied versus constraint forces
Tag each force as applied or as a joint reaction. Applied forces are known inputs; constraint forces are unknowns tied to the Jacobian, solved for in later modules.
6.5 Free-body per body
Draw a free-body diagram for every rigid body and write its Newton-Euler pair. Assembling these per-body equations is exactly how a multibody model is built.
| Motion | Governing equation | Unknown |
|---|---|---|
| Planar translation | ΣF = m a_G | a_G |
| Planar rotation | ΣM_G = I_G α | α |
| Spatial rotation | M = I α + ω × Iω | α (3 components) |
Two equations in the plane, six in space, per rigid body; joints then couple the bodies together.
Engineering connection: the gyroscopic term is what makes a spinning rotor or wheel resist tilting and precess sideways instead, an effect a planar model would miss entirely.
Worked example 1: planar Newton-Euler for one body
A planar rigid body has mass m = 10 kg and central moment of inertia I_G = 0.5 kg·m². A net force of 50 N acts through a line offset from the center, producing a net moment of 20 N·m about the center. Find the linear and angular accelerations.
- ProblemFind a_G and α for the body in Figure 1.
- Given / findm = 10 kg, I_G = 0.5 kg·m², ΣF = 50 N, ΣM_G = 20 N·m. Find a_G and α.
- AssumptionsRigid body, planar motion, forces and moment already netted about the mass center.
- ModelNewton: a_G = ΣF / m. Euler (planar): α = ΣM_G / I_G.
- Equationsa_G = ΣF / mα = ΣM_G / I_G
- Solvea_G = 50 / 10 = 5 m/s². α = 20 / 0.5 = 40 rad/s².
- CheckUnits confirm: N/kg gives m/s², and N·m per kg·m² gives rad/s². Translation and rotation solve independently in the plane.
- ConclusionThe body accelerates at 5 m/s² and spins up at 40 rad/s². The two Newton-Euler equations fully determine planar motion of one body.
Worked example 2: Euler’s equation for a torque-free body
A rigid body has principal moments of inertia I = (2, 3, 4) kg·m². At an instant its angular velocity is ω = (1, 2, 3) rad/s with no applied torque. Find the angular acceleration about the first axis.
- ProblemFind ω̇₁ for the torque-free body in Figure 2.
- Given / findI₁ = 2, I₂ = 3, I₃ = 4 kg·m²; ω = (1, 2, 3) rad/s; torque T = 0. Find ω̇₁.
- AssumptionsRotation about the mass center, principal axes, no external torque.
- ModelEuler’s first equation: I₁ω̇₁ = (I₂ − I₃)ω₂ω₃ + T₁, with T₁ = 0.
- Equationsω̇₁ = (I₂ − I₃) ω₂ ω₃ / I₁
- Solveω̇₁ = (3 − 4)(2)(3) / 2 = (−1)(6) / 2 = −3 rad/s².
- CheckNonzero angular acceleration with zero torque is the gyroscopic effect: unequal principal inertias couple the axes. A planar model (single inertia) would predict zero.
- ConclusionEven torque-free, the first spin rate changes at −3 rad/s². This coupling is the signature of true three-dimensional rigid-body dynamics.
Misconceptions and diagnostics
| Mistake | Symptom | Diagnostic question | Correction |
|---|---|---|---|
| Summing moments about the wrong point | Extra terms or missing ones in the rotational law | Am I taking moments about the mass center? | Use the mass center, or add the transport terms if not. |
| Dropping the gyroscopic term | Three-dimensional spins come out wrong | Is the motion truly planar? | Keep ω × Iω for spatial rotation. |
| Mixing applied and constraint forces | Reactions treated as known inputs | Is this force applied or a joint reaction? | Tag reactions as unknowns for the constrained solve. |
| Using a single inertia in space | Angular momentum assumed parallel to spin | Are these principal axes with equal inertias? | Use the full tensor unless inertias are equal. |
Practice ladder
A 4 kg body feels a net force of 24 N. Find its mass-center acceleration.
Show answer
a = F/m = 24/4 = 6 m/s².
A flywheel with I = 0.25 kg·m² feels a net moment of 10 N·m. Find its angular acceleration.
Show answer
α = M/I = 10/0.25 = 40 rad/s².
For I = (2, 3, 4) and ω = (1, 2, 3), find ω̇₂ with no torque.
Show answer
ω̇₂ = (I₃ − I₁)ω₃ω₁/I₂ = (4−2)(3)(1)/3 = 2 rad/s².
For one body in a mechanism you know, write its Newton-Euler equations and identify which forces are applied and which are joint reactions.
What good work looks like
A good answer gives the translational and rotational equations, lists gravity and actuator forces as applied, and marks pin or slider reactions as unknown constraint forces.
Working with AI, and proving it yourself
Ask AI to check your free body, not to write the equations
Portfolio task
Draw the free-body diagram of one body in your model and write its Newton-Euler equations, separating applied and constraint forces.
Retrieval and spaced review
Closed notes. Answer out loud, then reveal.
1. State the translational equation of motion.
Net force equals mass times mass-center acceleration.
2. State the planar rotational equation.
Net moment about the center equals I_G times angular acceleration.
3. What extra term appears in space?
The gyroscopic term ω × Iω.
4. How do applied and constraint forces differ?
Applied forces are known inputs; constraint forces are joint reactions to solve for.
5. Why can a torque-free body change its spin rates?
Unequal principal inertias couple the axes through Euler’s equations.
Textbook mapping
The Newton-Euler equations are the core of Wittenburg’s rigid-body dynamics. Use these references to read further.
| Topic in this module | Where to read more |
|---|---|
| Law of angular momentum and Euler’s equations | Wittenburg, Dynamics of Multibody Systems, ch. 3 |
| Classical spinning-body problems | Wittenburg, Dynamics of Multibody Systems, ch. 4 |
| Newton-Euler formulation per body | Nikravesh, Computer-Aided Analysis of Mechanical Systems |
Chapter references are to Wittenburg, Dynamics of Multibody Systems (Springer).