Home / Courses / Dynamics / Rigid-Body Energy and Momentum

Dynamics · Chapter 9 of 10 · Advanced

Rigid-Body Energy and Momentum

The energy and momentum methods extend to rigid bodies by adding rotation: kinetic energy gains a ½Iω² term, and angular momentum becomes Iω.

Readiness check

This chapter combines the methods of Chapters 4 and 5 with rotation. Tick only what you can do closed-notes.

Apply the work-energy theorem to a particle.
Use conservation of momentum.
Compute ½mv² and recall ½Iω².
Look up a moment of inertia.
Relate v_G = ωr in rolling.

0 or 1 weak itemsContinue with this chapter.

2 weak itemsReview work-energy (Chapter 4) and impulse-momentum (Chapter 5).

3 or more weak itemsRevisit rigid-body kinetics in Chapter 8 first.

The core idea

A rigid body stores kinetic energy in both motion of its center and its spin, and carries both linear and angular momentum; the same energy and momentum principles apply, now with rotational terms.

T = ½mv_G² + ½I_Gω²H_G = I_GωΣM_G dt = Δ(I_Gω)

Everything from Chapters 4 and 5 carries over once rotation is included. Kinetic energy adds a ½I_Gω² term; a couple does work U = ∫M dθ; angular momentum I_Gω plays the role of mv. When no external moment acts, angular momentum is conserved even as the moment of inertia changes.

The skill works when: you include both translational and rotational energy, and recognise when angular momentum is conserved.

The skill breaks down when: the ½I_Gω² term is dropped, or kinetic energy is assumed conserved while angular momentum changes I.

The concept. A rolling or spinning body holds energy in two forms: the translation of its center and its rotation. Both must be counted, and both contribute to its momentum.

The skills, taught in order

This chapter rounds out the rigid-body toolkit with energy and momentum. Five skills add rotation to the methods you already know.

9.1 Kinetic energy of a rigid body

The total kinetic energy is T = ½mv_G² + ½I_Gω²: the translation of the mass center plus the rotation about it. For pure rotation about a fixed axis O, this collapses to T = ½I_Oω².

Motion	Kinetic energy
Translation only	½mv_G²
Fixed-axis rotation	½I_Oω²
General plane motion	½mv_G² + ½I_Gω²

9.2 Work of forces and couples

Forces do work as before; in addition, a couple of moment M does work U = ∫M dθ, or simply Mθ when M is constant. This is how a motor's torque or a torsional spring enters the energy balance.

9.3 The work-energy principle and conservation

U₁₋₂ = T₂ − T₁ still holds, with T now including rotation. If only conservative forces act, mechanical energy is conserved. Energy methods are ideal for rolling problems, where they bypass the friction force entirely.

9.4 Linear and angular momentum of a rigid body

The body carries linear momentum G = mv_G and angular momentum H_G = I_Gω. Their rates of change equal the resultant force and moment, so impulse changes momentum just as for a particle.

Linear	Angular
G = mv_G	H_G = I_Gω
ΣF dt = Δ(mv_G)	ΣM_G dt = Δ(I_Gω)
conserved if ΣF = 0	conserved if ΣM = 0

9.5 Conservation of angular momentum

With no external moment, I_Gω stays constant. If the body redistributes its mass (a skater pulling in their arms, a diver tucking), I falls and ω rises to compensate. Kinetic energy, by contrast, changes, because internal work is done.

Engineering connection: flywheel energy storage, rolling-resistance and ride analysis, rotor and turbine spin-up, and the spin control of satellites and machinery.

Worked example 1: a cylinder rolling down by energy

A uniform solid cylinder rolls without slipping from rest down a slope, dropping a height of 2 m. Use energy to find the speed of its center at the bottom, and compare with a block that slides without friction.

Figure 1. Gravitational potential energy splits between translation and rotation. Two thirds goes to the center's motion and one third to the spin, so the cylinder arrives slower than a sliding block.

ProblemFind the center speed at the bottom for the cylinder in Figure 1, and compare with a sliding block.
Given / findSolid cylinder (I_G = ½mr²), h = 2 m, rolls without slipping, from rest. Find v_G.
AssumptionsNo slipping (ω = v_G/r), no rolling resistance, energy conserved.
ModelEnergy conservation: gravitational potential energy becomes translational plus rotational kinetic energy.
Equationsmgh = ½mv_G² + ½I_Gω² I_G = ½mr², ω = v_G/r v_G = √(4gh/3)
SolveSubstituting, mgh = ½mv² + ¼mv² = ¾mv², so v_G = √(4gh/3) = √(4 × 9.81 × 2/3) = 5.11 m/s. A frictionless sliding block would reach √(2gh) = 6.26 m/s.
CheckThe mass cancels, so the result is size-independent. The split is two thirds translation, one third rotation, and the cylinder's 5.11 m/s matches what the force method of Chapter 8 gives (a_G = ⅔g sinθ), a satisfying cross-check.
ConclusionEnergy bypassed the friction force entirely, the advantage over the force method here. Because some energy must spin the body up, rolling objects always trail a frictionless slider.

Result. v_G = 5.11 m/s, versus 6.26 m/s for a sliding block; one third of the energy is rotational.

Worked example 2: a skater pulling in their arms

A skater spinning at 2 rev/s has a moment of inertia of 5 kg·m² with arms out. Pulling them in drops it to 2 kg·m². Find the new spin rate and the change in kinetic energy.

Figure 2. With no external moment, angular momentum I ω is conserved. Reducing the moment of inertia raises the spin rate, and the kinetic energy rises because the skater's muscles do work pulling the arms in.

ProblemFind the new spin rate and the kinetic-energy change for the skater in Figure 2.
Given / findI₁ = 5 kg·m², ω₁ = 2 rev/s, I₂ = 2 kg·m². Find ω₂ and ΔT.
AssumptionsFrictionless pivot, no external moment, so angular momentum is conserved.
ModelConservation of angular momentum sets ω₂; then compute kinetic energy before and after.
EquationsI₁ω₁ = I₂ω₂ T = ½Iω²
Solveω₂ = I₁ω₁/I₂ = 5(2)/2 = 5 rev/s. In rad/s, ω₁ = 12.57 and ω₂ = 31.42. T₁ = ½(5)(12.57²) = 395 J; T₂ = ½(2)(31.42²) = 987 J. Kinetic energy rises by 592 J, a factor of 2.5.
CheckAngular momentum H = Iω = 5 × 12.57 = 62.8 kg·m²/s is the same before and after (2 × 31.42 = 62.8). The energy increase equals the work the skater's muscles do pulling the arms inward against the rotation.
ConclusionAngular momentum is conserved; kinetic energy is not. The same principle spins up a collapsing star, a tucking diver, and a turbine rotor as mass moves inward.

Result. ω₂ = 5 rev/s; kinetic energy rises from 395 J to 987 J (a factor of 2.5).

Misconceptions and diagnostics

Mistake	Symptom	Diagnostic question	Correction
Dropping rotational energy	Rolling speed comes out too high	"Did I include ½I_Gω²?"	Total kinetic energy has both translation and rotation.
Assuming KE conserved with changing I	Energy "conserved" as skater pulls in	"Is internal work being done?"	Angular momentum is conserved; KE changes by the internal work.
Wrong axis for I in energy	I_G used where I_O belongs	"Rotation about G or a fixed axis?"	Use ½I_Oω² for fixed-axis, ½mv_G² + ½I_Gω² for general motion.
Forgetting couple work	Motor torque omitted from energy	"Does a couple act through an angle?"	Add U = ∫M dθ for any applied couple.

Practice ladder

Level 1 · Direct skill

A flywheel with I = 1.8 kg·m² spins at 100 rad/s. Find its stored kinetic energy.

Show answer

T = ½Iω² = ½(1.8)(100²) = 9000 J = 9 kJ. This is the energy a flywheel can store and release, the basis of flywheel energy storage.

Level 2 · Mixed concept

Race a solid sphere (I = 2/5 mr²) against the Worked Example 1 cylinder down the same slope. Which is faster at the bottom?

Show answer

For the sphere, mgh = ½mv² + ½(2/5 mr²)(v/r)² = (7/10)mv², so v = √(10gh/7) = 5.29 m/s, faster than the cylinder's 5.11 m/s. Less rotational inertia means more energy left for translation; the sphere wins.

Level 3 · Independent problem

A 0.4 kg·m² turntable spins freely at 60 rad/s. A 0.6 kg·m² ring is dropped on coaxially and they couple. Find the common speed and the energy lost.

Show answer

Angular momentum: 0.4(60) = (1.0)ω, so ω = 24 rad/s. T₁ = ½(0.4)(60²) = 720 J; T₂ = ½(1.0)(24²) = 288 J. Lost = 432 J (60%), to friction during coupling, the rotational version of a perfectly plastic impact.

Level 4 · Transfer to real engineering

Pick a real spinning or rolling system (a yo-yo, a flywheel, a spinning office chair). Apply energy or angular-momentum conservation, and quantify a speed, stored energy, or spin change.

What good work looks like

Both energy terms included where relevant, the conserved quantity identified correctly, and a result cross-checked against the force method or a limiting case.

Working with AI, and proving it yourself

Use AI as an examiner, not a solver

"Check that I included both translational and rotational kinetic energy."

"Ask me whether energy or angular momentum is conserved in this scenario, and why."

"Solve the rolling problem." Building the full energy balance is the skill.

"What is the final spin?" Applying conservation yourself is the point.

Portfolio task

Analyse one real rotating system with energy or angular-momentum conservation, state which quantity is conserved and why, and quantify the change with a check against the force method.

Must include: the full kinetic-energy expression or H = Iω, the conserved quantity justified, and a cross-check.

Retrieval and spaced review

Closed notes. Answer out loud, then reveal.

1. Write the kinetic energy of a rigid body in general plane motion.

T = ½mv_G² + ½I_Gω².

2. How much work does a constant couple do?

U = Mθ (the integral ∫M dθ).

3. Give the linear and angular momentum of a rigid body.

G = mv_G and H_G = I_Gω.

4. When is angular momentum conserved?

When the net external moment is zero.

5. Why does a skater speed up pulling their arms in?

Angular momentum I ω is conserved, so reducing I raises ω; the energy increase comes from internal muscle work.

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

+1 dayRe-solve the rolling cylinder by energy from a blank page.

+3 daysOne angular-momentum conservation problem.

+7 daysCarry rotational energy into vibration, Chapter 10.

+30 daysCompare all three rigid-body methods on one problem.

Textbook mapping

Item	Mapping
Primary source	Meriam and Kraige, Engineering Mechanics: Dynamics (7th ed), Chapter 6, Sections B and C (Work-Energy, Impulse-Momentum)
Cross-reference	Hibbeler, Dynamics, Ch. 18 and 19 · Beer and Johnston, Ch. 17
Core topics	9.1 Kinetic energy · 9.2 Work of couples · 9.3 Work-energy and conservation · 9.4 Linear and angular momentum · 9.5 Conservation of angular momentum
Engineering connection	Flywheel storage, rolling and ride analysis, rotor spin-up, and spin control.
Read next	Chapter 10: Vibration and Time Response.