Physics for ME · Module 8 of 16
Circular Motion and Rotating Systems
Turning at constant speed is still accelerating: the velocity's direction changes. Mechanical engineers live among rotating parts; this module builds the intuition.
Readiness check
From Modules 2 to 4. Tick only what you can do closed-notes.
- Apply ΣF = ma along chosen axes.
- Convert rpm to rad/s and Hz (Math Module 2).
- Use s = rθ and its rates.
- Solve for an unknown normal or friction force.
- Recognize when velocity changes direction but not magnitude.
The core idea
Circular motion needs a center-pointing force. Remove it and the body flies off on the tangent.
a = v²/r = ω²rv = ωrCentripetal acceleration is not a new force: it is the acceleration that some real force (friction, tension, a normal force, gravity) must supply. The angular quantities θ, ω, α mirror the linear story of Module 3 exactly.
The skills, taught in order
Circular motion is Newton's second law on a curve. The one new idea is that a center-pointing acceleration is always present; the rest is identifying which real force supplies it.
8.1 Angular position, velocity, and acceleration
Describe rotation with angle θ (radians), angular velocity ω = dθ/dt (rad/s), and angular acceleration α = dω/dt (rad/s²). These mirror the linear position, velocity, and acceleration of Module 3 exactly, with the radian as the natural angle unit.
8.2 Linking linear and angular
A point at radius r on a rotating body moves a distance s = rθ, at speed v = ωr, with tangential acceleration at = αr. The radius is the conversion factor between the two descriptions.
| Quantity | Linear | Angular | Link |
|---|---|---|---|
| Position | s | θ | s = rθ |
| Velocity | v | ω | v = ωr |
| Tangential acceleration | at | α | at = αr |
8.3 Centripetal acceleration
Even at constant speed, a turning body accelerates because its velocity direction changes. That acceleration points to the centre and has magnitude ac = v²/r = ω²r. It is separate from any tangential (speeding-up) acceleration.
Micro-example. A point 0.3 m from a shaft turning at 50 rad/s feels ac = ω²r = 50² × 0.3 = 750 m/s², about 76 times gravity.
8.4 The force that turns
"Centripetal force" is not a new kind of force; it is the demand mv²/r that some real force must meet. The whole skill is asking which force points to the centre here, then setting it equal to mv²/r.
| Situation | What supplies the centre-pointing force |
|---|---|
| Car on a flat curve | friction (Worked Example 1) |
| Banked curve, no friction | the inward part of the normal force (Worked Example 2) |
| Ball on a string, rotor blade | tension |
| Top of a vertical loop | gravity, plus any track normal force |
| Orbiting satellite | gravity |
8.5 Banked curves and rotors
Tilting a surface aims part of its normal force toward the centre, so geometry can do the turning that friction would otherwise have to. For no-friction design, tan θ = v²/(gr). Worked Example 2 sizes exactly this.
8.6 rpm thinking
Machine speeds come in rev/min; convert with ω = rpm × 2π/60 before any formula. Then tip speed is v = ωr and the rim's acceleration is ω²r, often thousands of g. That ω² is why overspeed destroys rotors.
Micro-example. 3000 rpm = 3000 × 2π/60 = 314 rad/s, so a 0.2 m rotor has a tip speed of 63 m/s.
Engineering connection: vehicle dynamics, centrifuges, bearing loads, and turbine and fan tip speeds. Rotation gains its own inertia in Module 9.
Worked example 1: how fast can the car take the curve?
A flat highway curve has radius 80 m. Tire-road friction can supply at most μs = 0.7 of the normal force. Find the maximum speed, independent of the car's mass.
- ProblemFind the maximum cornering speed in Figure 1.
- Given / findr = 80 m, μs = 0.7, flat road. Find vmax.
- AssumptionsFlat (unbanked) curve, steady speed, friction at its limit at vmax.
- ModelVertical: N = mg. Horizontal: friction supplies the centripetal demand mv²/r, capped at μsN.
- Equationsmv²/r = μsmg
- SolveMass cancels: vmax = √(μsgr) = √(0.7 × 9.81 × 80) = √549 = 23.4 m/s ≈ 84 km/h.
- CheckAt that speed a = v²/r = 6.87 m/s² = 0.70g, exactly the friction ceiling. Wet road (μ ≈ 0.4) drops the limit to 17.7 m/s (64 km/h): the √μ scaling explains rain-speed advisories.
- ConclusionThe mass cancelling means a loaded truck and a small car share the same flat-curve speed limit; what differs is tire μ and bank angle. Highway engineers bank curves precisely to stop borrowing all of friction for turning.
Worked example 2: the banked curve
The same 80 m curve is to be banked so that a car at the design speed of 20 m/s rounds it using no friction at all. Find the banking angle, and compare it with the flat-curve case of Worked Example 1.
- ProblemFind the banking angle θ that lets a car round the 80 m curve at 20 m/s with no friction (Figure 2).
- Given / findr = 80 m, design speed v = 20 m/s, g = 9.81 m/s², friction not needed. Find θ.
- AssumptionsFrictionless design point, car treated as a particle, only the weight and the road's normal force act.
- ModelTilt the free-body diagram: N is perpendicular to the road, so its horizontal part is the centripetal force and its vertical part carries the weight.
- EquationsN sin θ = mv²/r N cos θ = mg tan θ = v²/(gr)
- Solvetan θ = 20² / (9.81 × 80) = 400/784.8 = 0.510, so θ = 27.0°. Mass cancels, so the angle is the same for every vehicle.
- CheckOn this same curve, Worked Example 1's flat road needed friction μ = v²/(gr) = 0.51 at 20 m/s; banking at 27° supplies that same centripetal force through the normal force, leaving friction to spare. Above the design speed friction must point down the bank, below it, up the bank.
- ConclusionBanking lets geometry, not grip, turn the vehicle, which is why racetracks and highway ramps are banked and why the design still holds in the rain. The same "tilt a force inward" idea sizes governor weights and centrifuge rotors.
Misconceptions and diagnostics
| Mistake | Symptom | Diagnostic question | Correction |
|---|---|---|---|
| Centrifugal force on the FBD | An outward arrow balancing the turn | "What physical contact creates this outward push?" | None does, in the ground frame. The net force points inward; the outward feel is inertia. |
| "Constant speed, so no acceleration" | a = 0 claimed on a curve | "Is the velocity vector constant?" | Direction change is acceleration: a = v²/r toward the center. |
| Centripetal treated as its own force | mv²/r added on top of friction | "Which real force supplies the center pull here?" | mv²/r is the demand; friction, tension, or a normal force is the supplier. One entry, not two. |
| rpm fed into v²/r | Accelerations absurdly small or large | "Is ω in rad/s?" | Convert: ω = rpm × 2π/60. Radians everywhere in s = rθ, v = ωr, a = ω²r. |
Practice ladder
A grinder disc of 125 mm diameter spins at 12 000 rpm. Find ω in rad/s and the rim (tip) speed.
Show answer
ω = 12 000 × 2π/60 = 1257 rad/s; v = ωr = 1257 × 0.0625 = 78.5 m/s. Rim speeds near 80 m/s are why disc ratings matter.
What centripetal acceleration does the grinder rim experience, in g's, and what force does that mean for a 1-gram chip of the disc edge?
Show answer
a = ω²r = 1257² × 0.0625 = 98 800 m/s² ≈ 10 000g. The 1 g chip needs about 99 N of pull to stay on: when bonding fails, fragments leave at 78 m/s on the tangent.
A 2 kg mass swings on a 1.2 m cord as a pendulum, passing the lowest point at 4 m/s. Find the cord tension there.
Show answer
At the bottom, T − mg = mv²/r: T = 2(9.81) + 2(16)/1.2 = 19.6 + 26.7 = 46.3 N, well over double the weight. Swinging machinery sees these peaks every cycle.
Pick a rotating machine you can observe (washing machine spin, bike wheel, ceiling fan, lathe chuck). Measure or look up its rpm and radius, compute tip speed and g-level at the rim, and comment on one design feature that exists because of those numbers.
What good work looks like
rpm to rad/s shown, tip speed and ω²r in g's computed, and a sensible link (balancing, guard rings, drum perforations, blade root thickness) to the magnitude found.
Working with AI, and proving it yourself
Use AI as an examiner, not a solver
Portfolio task
Write a one-page "Rotation Numbers Card" for three machines (one household, one vehicle, one industrial from a datasheet): rpm, ω, tip speed, rim g-level, and the centripetal supplier, with one safety implication each.
Retrieval and spaced review
Closed notes. Answer out loud, then reveal.
1. Why is uniform circular motion accelerated motion?
The velocity vector's direction changes continuously; a = v²/r points at the center.
2. Write the three angular-linear links.
s = rθ, v = ωr, at = αr, with angles in radians.
3. Name four real forces that can serve as centripetal suppliers.
Friction (cars), tension (cords), normal force (banked tracks, drums), gravity (orbits).
4. Why does the flat-curve speed limit not depend on mass?
Both the demand mv²/r and the supply μmg scale with m; it cancels, leaving v = √(μgr).
5. What happens to rim stress when a rotor's speed doubles?
Centripetal loading goes as ω²: four times. Overspeed is the classic rotor killer.