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Multibody Dynamics · Module 5 of 10

Mass, Inertia, and Momentum

Forces cause acceleration in proportion to inertia. For rotation that inertia is a tensor, and the parallel-axis theorem lets you move it wherever the joint sits.

Readiness check

Tick only what you can do closed-notes before starting.

Recall the moment of inertia of a rod and a disk.
State the parallel-axis theorem in words.
Compute translational kinetic energy ½ m v².
Recall that angular momentum is inertia times angular velocity.
Distinguish mass (a scalar) from the inertia tensor (a matrix).

0 or 1 weak itemsContinue with this module.

2 weak itemsRevisit moments of inertia in Statics, moments of inertia.

3 or more weak itemsReview rotational motion in Dynamics, Module 8.

The core idea

Inertia measures resistance to acceleration: mass for translation, the inertia tensor for rotation. These set a body’s momentum, angular momentum, and kinetic energy, the quantities the equations of motion balance.

momentum: p = m vangular momentum: H = I ωparallel axis: I = I_G + m d²

Newton’s law says force equals mass times acceleration, so mass is the translational inertia. Rotation is richer: how hard a body is to spin depends on the axis, so the rotational inertia is a three-by-three inertia tensor, with moments of inertia on the diagonal and products of inertia off it. About the principal axes the products vanish and the tensor is diagonal, which simplifies the dynamics. The inertia tensor is usually known about the center of mass; the parallel-axis theorem shifts it to any other point by adding m times the squared offset, which is exactly what you need when a body rotates about a joint rather than its own center. From mass and inertia follow the momentum p = m v, the angular momentum H = Iω, and the kinetic energy, a translational half m v-squared plus a rotational half omega-dot-I-omega. These conserved and balanced quantities are the currency of every method in the rest of the course.

The skill works when: you track mass and the inertia tensor separately and move inertia to the right point with the parallel-axis theorem.

The skill breaks down when: you use a center-of-mass inertia for a body that actually rotates about a joint, understating the true resistance to spin.

The concept. A rigid body’s inertia is mass for translation and the inertia tensor for rotation. Together they set its momentum, angular momentum, and kinetic energy.

The skills, taught in order

Five skills quantify how a body stores motion.

5.1 The inertia tensor

Assemble the moments and products of inertia into a symmetric tensor. It relates angular velocity to angular momentum and is the rotational analogue of mass.

5.2 Principal axes

Find the axes where the products of inertia vanish and the tensor is diagonal. Along principal axes, angular momentum lines up with angular velocity and Euler’s equations take their simplest form.

5.3 The parallel-axis theorem

Shift a center-of-mass inertia to a parallel axis a distance d away by adding m d². This is essential when a body pivots about a joint offset from its center.

5.4 Momentum and angular momentum

Compute p = m v for the mass center and H = Iω about the reference point. These are what impulses change and what conservation laws track.

5.5 Kinetic energy

Split kinetic energy into translational half m v-squared and rotational half omega-dot-I-omega. Energy checks built on this split are one of the best validations of a simulation.

Body (about center)	Moment of inertia
Slender rod, length L	m L² / 12
Slender rod, about end	m L² / 3
Solid disk, radius R	m R² / 2
Solid sphere, radius R	2 m R² / 5

Standard inertias; the parallel-axis theorem moves any of them to the axis your joint actually uses.

Engineering connection: a motor sizing a robot joint must overcome the link inertia about that joint, not about the link center, so the parallel-axis shift changes the torque you specify.

Worked example 1: parallel-axis theorem for a rod

A uniform slender rod has mass m = 2 kg and length L = 1 m. Find its moment of inertia about the center and about one end.

Figure 1. The parallel-axis theorem: shifting the spin axis from the center to the end of a rod adds m(L/2)², raising the moment of inertia from mL²/12 to mL²/3.

ProblemFind I_center and I_end for the rod in Figure 1.
Given / findm = 2 kg, L = 1 m. Find the moment of inertia about the center and about the end.
AssumptionsUniform slender rod, so its cross-section inertia is negligible.
ModelI_center = mL²/12; I_end = I_center + m(L/2)² by the parallel-axis theorem.
EquationsI_c = mL²/12I_end = I_c + m(L/2)²
SolveI_c = 2(1)/12 = 0.167 kg·m². I_end = 0.167 + 2(0.5)² = 0.167 + 0.5 = 0.667 kg·m².
CheckThe end value equals mL²/3 = 2(1)/3 = 0.667, the known result, confirming the parallel-axis shift.
ConclusionMoving the axis from center to end raises the inertia four-fold. A rod pinned at its end resists angular acceleration much more than one spun about its middle.

Result. I_center = 0.167, I_end = 0.667 kg·m².

Worked example 2: angular momentum and energy of a disk

A solid disk of mass m = 5 kg and radius R = 0.2 m spins at ω = 20 rad/s about its central axis. Find its moment of inertia, angular momentum, and rotational kinetic energy.

Figure 2. A spinning disk: its moment of inertia sets both angular momentum (Iω) and rotational kinetic energy (half Iω²) from the same inertia value.

ProblemFind I, H, and rotational KE for the disk in Figure 2.
Given / findm = 5 kg, R = 0.2 m, ω = 20 rad/s. Find I, H = Iω, and KE = ½ Iω².
AssumptionsUniform solid disk spinning about its central symmetry axis, a principal axis.
ModelI = ½ m R²; H = Iω; KE = ½ Iω².
EquationsI = ½ m R²H = IωKE = ½ Iω²
SolveI = ½(5)(0.2)² = 0.1 kg·m². H = 0.1(20) = 2.0 kg·m²/s. KE = ½(0.1)(20)² = 20 J.
CheckKE = ½ H ω = ½(2.0)(20) = 20 J, a consistent cross-check between momentum and energy.
ConclusionThe same inertia value sets both the angular momentum and the energy. Doubling the spin doubles H but quadruples the kinetic energy.

Result. I = 0.1 kg·m², H = 2.0 kg·m²/s, KE = 20 J.

Worked example 3: the inertia tensor and principal axes

A rigid body has the inertia tensor I = [[3, −1, 0], [−1, 3, 0], [0, 0, 5]] kg·m² about its mass center. Find the principal moments of inertia, and show that a spin ω = (1, 0, 0) rad/s gives an angular momentum not parallel to ω.

Figure 3. A full inertia tensor with products of inertia. Spinning about the x-axis, which is not a principal axis, the angular momentum H leans 18.4° off the spin ω; the principal moments are 2, 4, and 5 kg·m².

ProblemFind the principal moments and the angle between H and ω for the tensor in Figure 3.
Given / findThe inertia tensor above; ω = (1, 0, 0) rad/s. Find the principal moments (eigenvalues), H = Iω, and the angle between them.
AssumptionsRigid body, inertia taken about the mass center; the nonzero off-diagonal terms are genuine products of inertia.
ModelPrincipal moments are the eigenvalues of I. Angular momentum H = Iω; it is parallel to ω only when ω lies along a principal axis.
Equationsdet(I − λ 1) = 0H = I ω
SolveThe z-axis is already principal (I_zz = 5). The x-y block [[3, −1], [−1, 3]] has eigenvalues 3 − 1 = 2 and 3 + 1 = 4, so the principal moments are 2, 4, 5 kg·m². For ω = (1, 0, 0), H = (3, −1, 0) kg·m²/s, with |H| = √10 = 3.16 and an angle arccos(3/√10) = 18.4° from ω.
CheckThe trace is invariant: 3 + 3 + 5 = 2 + 4 + 5 = 11. H aligns with ω only along a principal axis, and the x-axis is not one here, so the 18.4° tilt is expected.
ConclusionThe principal moments are 2, 4, and 5 kg·m². Because the spin axis is not principal, H leans 18.4° off ω, the very coupling Euler’s equations carry. The rotational energy is ½ ω · H = 1.5 J.

Result. principal moments 2, 4, 5 kg·m²; H = (3, −1, 0), 18.4° off ω.

Misconceptions and diagnostics

Mistake	Symptom	Diagnostic question	Correction
Confusing mass with inertia tensor	Using one scalar for rotation in space	Is rotation about one axis or general?	Rotation needs the full inertia tensor, not a single number.
Skipping the parallel-axis shift	Torque estimates too low for a pivoted link	Does the body spin about its center or a joint?	Shift inertia to the actual axis with m d².
Ignoring products of inertia	Angular momentum not aligned with spin, but assumed to be	Are these principal axes?	Off principal axes, H and ω differ in direction.
Forgetting the rotational energy term	Energy checks fail for spinning bodies	Did I add half omega-dot-I-omega?	Kinetic energy has translational and rotational parts.

Practice ladder

Level 1 · Direct skill

A disk has m = 4 kg and R = 0.5 m. Find its central moment of inertia.

Show answer

I = ½ m R² = ½(4)(0.25) = 0.5 kg·m².

Level 2 · Mixed concept

A rod (m = 3 kg, L = 2 m) rotates about one end. Find its moment of inertia.

Show answer

I_end = mL²/3 = 3(4)/3 = 4 kg·m².

Level 3 · Independent problem

The disk from Level 1 spins at 10 rad/s. Find its angular momentum and rotational kinetic energy.

Show answer

H = Iω = 0.5(10) = 5 kg·m²/s; KE = ½ Iω² = ½(0.5)(100) = 25 J.

Transfer task | Real engineering

For a link in a mechanism you know, find its inertia about the joint it pivots on, starting from the center-of-mass value.

What good work looks like

A good answer states the center-of-mass inertia, measures the offset d to the joint, and applies I = I_G + m d² with correct units.

Working with AI, and proving it yourself

Ask AI to check an inertia value, not to skip the shift

"Confirm the end inertia of this rod using the parallel-axis theorem."

"Check my angular momentum and kinetic energy for this spinning disk."

"Give me the inertia." Setting up the axis and shift is the skill.

"How much energy does it have?" Splitting translational and rotational parts is the point.

Portfolio task

For one body in your model, compute the inertia about its joint, then its angular momentum and kinetic energy at a chosen spin rate.

Must include: a center-of-mass inertia, a parallel-axis shift to the joint, and consistent H and KE values.

Retrieval and spaced review

Closed notes. Answer out loud, then reveal.

1. What is the rotational analogue of mass?

The inertia tensor.

2. State the parallel-axis theorem.

I = I_G + m d², shifting inertia a distance d from the center.

3. Write angular momentum for a rigid body.

H = I ω about the reference point.

4. What are principal axes?

Axes where products of inertia vanish and the tensor is diagonal.

5. Write the kinetic energy of a rigid body.

Half m v-squared plus half omega-dot-I-omega.

TodayFinish this quiz and Levels 1 and 2 of the ladder.

+1 dayRe-derive the rod end inertia from a blank page.

+3 daysCompute inertia about a joint for three bodies.

+7 daysMove on to Newton-Euler equations in Module 6.

+30 daysReuse the parallel-axis shift whenever a body pivots off its center.

Textbook mapping

Mass, inertia, and momentum are Wittenburg’s dynamics fundamentals. Use these references to read further.

Topic in this module	Where to read more
Inertia tensor and angular momentum	Wittenburg, Dynamics of Multibody Systems, ch. 3
Kinetic energy of rigid bodies	Wittenburg, Dynamics of Multibody Systems, ch. 3
Parallel-axis theorem and standard inertias	Statics and dynamics texts on moments of inertia

Chapter references are to Wittenburg, Dynamics of Multibody Systems (Springer).