Physics for ME · Module 9 of 16
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 Modules 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 (Module 6).
- State momentum conservation and when it holds (Module 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.
The skills, taught in order
Rotation reruns all of mechanics with new names: torque for force, moment of inertia for mass, angular velocity for velocity. Learn the dictionary and every linear result you already own has a rotational twin.
9.1 Torque
Torque is the turning effect of a force, τ = Fd, where d is the perpendicular distance from the axis to the line of action (the lever arm); equivalently τ = rF sin φ. It plays the role that force played in straight-line motion.
9.2 Moment of inertia
Moment of inertia I = Σmr² is rotational mass: it measures not just how much mass there is but how far it sits from the axis. Rim mass counts far more than hub mass, since r is squared.
| Shape (axis) | Moment of inertia |
|---|---|
| Thin hoop, central axis | MR² |
| Solid disc or cylinder, central axis | ½MR² |
| Solid sphere, diameter | 2MR²/5 |
| Thin rod, through centre | ML²/12 |
| Thin rod, through one end | ML²/3 |
Micro-example. A hoop and a disc of equal mass and radius differ by a factor of two in I (MR² versus ½MR²), so the hoop is twice as hard to spin up.
9.3 Rotational Newton's law
Net torque produces angular acceleration in proportion to the moment of inertia: τ = Iα. It is F = ma with rotational quantities, and it sizes every motor, brake, and spin-up time.
9.4 Rotational kinematics
For constant α, the angle, angular velocity, and angular acceleration obey the same three equations as Module 3, with θ, ω, α in place of s, v, a: ω = ω₀ + αt, and so on. Keep all angles in radians.
9.5 Rotational kinetic energy
A spinning body stores KE = ½Iω². Because I can be made large and ω larger still, a modest flywheel banks a surprising amount of energy, which it can release far faster than the motor that filled it.
9.6 Rolling
A rolling body both moves and spins, so its kinetic energy carries two terms: ½mv² + ½Iω², tied together by the no-slip condition v = ωr. This is why a hoop and a disc released together down a ramp arrive at different speeds.
9.7 Angular momentum and its conservation
Angular momentum is L = Iω. With no external torque it cannot change, so reducing I forces ω up. A skater pulling her arms in, a gyroscope holding its heading, and a satellite's attitude wheels all run on this one law.
Micro-example. Halve a spinning body's moment of inertia with no external torque and its angular speed doubles, since Iω stays fixed.
| Linear quantity | Rotational twin |
|---|---|
| Force F | Torque τ = Iα |
| Mass m | Moment of inertia I = Σmr² |
| Velocity v | Angular velocity ω |
| Momentum p = mv | Angular momentum L = Iω |
| Kinetic energy ½mv² | ½Iω² |
Engineering connection: machine design, shafts and drivetrains, flywheels, and gyroscopic effects. The mass moment of inertia previewed in Statics comes alive here.
Worked example 1: 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.
Worked example 2: engaging a clutch
The flywheel from Worked Example 1 (I₁ = 1.8 kg·m²) is spinning at 3000 rpm when a stationary load disc of I₂ = 1.2 kg·m² is suddenly clutched onto the same shaft. They lock and turn together. Find the common speed and the energy lost in the engagement.
- ProblemFind the common angular speed and the energy lost when the flywheel is clutched to the load disc in Figure 2.
- Given / findI₁ = 1.8 kg·m² at ω₁ = 314.2 rad/s; I₂ = 1.2 kg·m² at rest; they lock together. Find ωf and the energy lost.
- AssumptionsNegligible external torque during the brief engagement, so angular momentum is conserved; the clutch then forces one common speed.
- ModelThe rotational twin of a perfectly inelastic collision: angular momentum is conserved, rotational kinetic energy is not.
- EquationsI₁ω₁ = (I₁ + I₂)ωf ΔKE = ½I₁ω₁² − ½(I₁ + I₂)ωf²
- Solveωf = (1.8 × 314.2)/(1.8 + 1.2) = 565.5/3.0 = 188.5 rad/s (1800 rpm). KE before = 88.8 kJ; KE after = ½(3.0)(188.5²) = 53.3 kJ; energy lost = 35.5 kJ.
- CheckThe fraction lost is I₂/(I₁ + I₂) = 1.2/3.0 = 40%, the same clean result as the linear inelastic collision in Module 7. The common speed sits below ω₁, as it must.
- ConclusionCoupling a load onto a spinning shaft always dumps energy as heat, which is why a slipping clutch or a hard gear engagement gets hot. Matching inertias or ramping the engagement is how real drivetrains keep that 35 kJ from cooking the clutch.
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.