Physics for ME · Chapter 9 of 16 · Intermediate
Torque, Angular Momentum, and Rigid-Body Rotation
Shafts, motors, gears, gyroscopes, and flywheels: rotation has its own mass (I), its own F = ma, and its own momentum. Same physics, new bookkeeping.
Readiness check
From Chapters 6 to 8 and Statics Module 4. Tick only what you can do closed-notes.
- Compute a torque as force times perpendicular arm.
- Convert rpm to rad/s instantly.
- Use the kinematics equations with θ, ω, α in place of s, v, a.
- Run an energy audit (Chapter 6).
- State momentum conservation and when it holds (Chapter 7).
The core idea
Every linear law has a rotational twin: τ replaces F, I replaces m, ω replaces v.
τ = IαKE = ½Iω²L = IωThe moment of inertia I = Σmr² says where the mass sits, not just how much: rim mass counts far more than hub mass. Angular momentum L is conserved without external torque, which is why spinning skaters speed up and gyroscopes hold direction.
What this chapter covers
- 9.1 Torque: the turning force, from Statics, now causing α.
- 9.2 Moment of inertia: Σmr² and the standard shapes (disc ½mr², hoop mr², rod).
- 9.3 Rotational Newton's law: τ = Iα.
- 9.4 Rotational kinematics: θ, ω, α with the same three equations.
- 9.5 Rotational kinetic energy: ½Iω² and flywheel storage.
- 9.6 Rolling: translation plus rotation, linked by v = ωr.
- 9.7 Angular momentum and its conservation: L = Iω; skaters, gyros, satellites.
Engineering connection: machine design, shafts and drivetrains, flywheels, gyroscopic effects; Statics Module 10's mass moment preview comes alive here.
Worked example: spinning up a flywheel
A solid steel flywheel (m = 40 kg, r = 0.30 m) is driven by a motor delivering a steady 12 N·m. Find the angular acceleration, the time to reach 3000 rpm, and the energy then stored.
- ProblemFind α, the spin-up time to 3000 rpm, and the stored energy for the flywheel in Figure 1.
- Given / findm = 40 kg, r = 0.30 m, τ = 12 N·m, target 3000 rpm.
- AssumptionsSolid uniform disc, frictionless bearings, constant torque, rigid wheel.
- ModelI = ½mr² for a disc; τ = Iα; rotational kinematics from rest; KE = ½Iω².
- EquationsI = ½mr² α = τ/I t = ω/α KE = ½Iω²
- SolveI = 0.5 × 40 × 0.09 = 1.8 kg·m². α = 12/1.8 = 6.67 rad/s². Target ω = 3000 × 2π/60 = 314.2 rad/s, so t = 314.2/6.67 = 47.1 s. KE = 0.5 × 1.8 × 314.2² = 88.8 kJ.
- CheckEnergy route: work = τθ; θ = ½αt² = ½ × 6.67 × 47.1² = 7398 rad; τθ = 12 × 7398 = 88.8 kJ. The two books agree. Units: kg·m² × (rad/s)² = J.
- Conclusion88.8 kJ is the braking energy of a small car at 38 km/h, stored in a 40 kg disc: that is why flywheels buffer presses and hybrid drivetrains. Spin-up time scales with I, which is why the same metal moved to the rim (a hoop, I = mr²) would take twice as long.
Misconceptions and diagnostics
| Mistake | Symptom | Diagnostic question | Correction |
|---|---|---|---|
| Mass used where I belongs | τ = mα written; spin-up times absurd | "Does my inertia know where the mass sits?" | Rotation uses I = Σmr², shape-dependent. Look up or derive the form. |
| rpm inside the formulas | Energies off by (2π/60)² | "Is ω in rad/s?" | Convert first. All the clean laws speak radians. |
| Rolling treated as pure rotation or pure sliding | Rolling KE missing one of its halves | "Does the body translate and spin?" | Rolling KE = ½mv² + ½Iω², with v = ωr tying them. |
| Angular momentum expected to fade on its own | "It just slows down" with no torque named | "What external torque acts?" | Without one, L = Iω persists; pull mass inward and ω must rise (skater effect). |
Practice ladder
Find I for a 2 kg, 0.4 m rod about its end (I = ⅓mL²), and the torque needed for α = 5 rad/s².
Show answer
I = ⅓ × 2 × 0.16 = 0.1067 kg·m²; τ = 0.533 N·m.
The worked-example flywheel must dump its 88.8 kJ into a press stroke lasting 0.5 s. What average power does it deliver, and how far does ω fall if the stroke takes 20 kJ?
Show answer
P = 88 800/0.5 = 178 kW available at full discharge. For a 20 kJ stroke: ½I(ω₁² − ω₂²) = 20 000, so ω₂² = 314.2² − 22 222 = 76 500, ω₂ = 276.6 rad/s (2641 rpm). The wheel sags only 12% while delivering a punch the motor never could: the flywheel's whole job.
A skater spins at 2 rev/s with I = 4 kg·m², then pulls her arms in to I = 1.6 kg·m². Find the new spin rate and the kinetic-energy change, and explain where the extra energy came from.
Show answer
L conserved: ω₂ = 4 × 2/1.6 = 5 rev/s. KE ratio = I₁ω₁²/I₂ω₂²: KE rises by the factor 2.5 (from ½Lω). The skater's muscles did work pulling mass inward against the spin: conservation of L, not of KE.
Estimate the moment of inertia of a real wheel you can spin (bike wheel, office chair, shop grinder) by timing its spin-down under a known small friction torque, or by the falling-mass method. Compare with a calculated hoop or disc estimate.
What good work looks like
The method stated with its equation (τ = IΔω/Δt or energy), measured numbers, the calculated geometric estimate, and a percent gap with one honest explanation.
Working with AI, and proving it yourself
Use AI as an examiner, not a solver
Portfolio task
Write a one-page flywheel mini-design: choose a target energy (state why), pick disc dimensions and speed within a rim-speed limit of 100 m/s, compute I, KE, and spin-up time for a chosen motor torque, and state one safety consideration.
Retrieval and spaced review
Closed notes. Answer out loud, then reveal.
1. Write the rotational twins of F = ma, p = mv, and KE = ½mv².
τ = Iα, L = Iω, KE = ½Iω².
2. Define the moment of inertia and give disc and hoop values.
I = Σmr², the placement-weighted mass. Solid disc ½mr²; hoop mr² about their axes.
3. What links translation and rotation in rolling?
v = ωr at the contact (no slip); kinetic energy carries both terms.
4. When is angular momentum conserved, and one engineering consequence?
When net external torque is zero. Consequences: gyroscopic stability, skater spin-up, satellite attitude control.
5. Why do flywheels favor rim mass?
I grows as r²: metal at the rim stores far more energy per kilogram at a given ω.
Textbook mapping
| Item | Mapping |
|---|---|
| Main source | OpenStax University Physics Vol. 1, Fixed-Axis Rotation and Angular Momentum |
| Benchmark / reference | MIT 8.01 (torque and angular momentum as core concepts) · Young and Freedman |
| Core topics | 9.1 Torque · 9.2 Moment of inertia · 9.3 τ = Iα · 9.4 Rotational kinematics · 9.5 Rotational energy · 9.6 Rolling · 9.7 Angular momentum |
| Engineering connection | Machine design, shafts, drivetrains, gyroscopes, flywheels; the dynamics half of Statics Module 10's inertia story. |
| Read next | Chapter 10: Oscillations, Mechanical Waves, and Resonance. |