This mini-lesson covers the AQA 3.11 Engineering physics option: rotational dynamics (moment of inertia, torque, angular momentum, rotational kinetic energy, flywheels) and thermodynamics (the first law, p–V diagrams, engine cycles, indicated power, efficiency and the second law).
Optional topic — examined in Paper 3 Section B.
Optional topic — AQA sections 3.9–3.13 are options: you study exactly ONE of the five (Astrophysics, Medical physics, Engineering physics, Turning points in physics, Electronics). It is examined in Paper 3 Section B. Check with your teacher which option your school teaches before revising this one.
Work through each screen, answer the questions as you go (many are full A-level calculations) and collect ⭐ stars. Press Start when you're ready.
3.11.1 · rotational dynamics
Moment of inertia and angular kinematics
Rotational mechanics mirrors linear mechanics term for term. Moment of inertia I is the rotational equivalent of mass: it measures how hard it is to change a body’s rotation.
I = Σmr²uniform disc or flywheel about its axis: I = ½MR² · unit: kg m²
ω2 = ω1 + αt · θ = ω1t + ½αt² · ω2² = ω1² + 2αθthe angular suvat equations — identical in form to the linear ones
The r² matters enormously: mass placed far from the axis contributes far more inertia. A hoop (I = MR²) is twice as hard to spin up as a disc (I = ½MR²) of the same mass and radius.
Angular displacement θ is in radians, ω in rad s−1, α in rad s−2.
Worked example
A uniform disc: M = 4.0 kg, R = 0.30 m. I = ½MR² = 0.5 × 4.0 × 0.30² = 0.5 × 4.0 × 0.09 = 0.18 kg m².
Calculate
Your turn — calculation 1
1A uniform disc has mass 4.0 kg and radius 0.30 m. Calculate its moment of inertia about its central axis. (I = ½MR²)
kg m²
Hint: I = ½ × 4.0 × 0.30² = 0.5 × 4.0 × 0.09.
3.11.1 · torque & energy
Torque, work, power and rotational kinetic energy
T = Iαthe rotational form of Newton’s second law — torque T in N m
W = Tθ · P = Tω · Ek = ½Iω²compare: W = Fs, P = Fv, Ek = ½mv²
Flywheels store energy as rotational kinetic energy. A good flywheel has a large I (heavy rim, large radius) and spins fast (Ek goes as ω²).
Uses: smoothing the output of a reciprocating engine, regenerative braking in vehicles and trains, and short-term energy storage for a printing press or power grid.
Limits: the rim experiences enormous stress at high ω and can burst; friction and air drag mean energy leaks away, so real flywheels run in a vacuum on magnetic bearings.
Worked example
The same disc (I = 0.18 kg m²) is given an angular acceleration of 20 rad s−2:
T = Iα = 0.18 × 20 = 3.6 N m.
Spinning at ω = 30 rad s−1: Ek = ½Iω² = 0.5 × 0.18 × 900 = 81 J.
Calculate
Your turn — calculation 2
2A torque is applied to the disc (I = 0.18 kg m²), giving it an angular acceleration of 20 rad s−2. Calculate the torque.
N m
Hint: T = Iα = 0.18 × 20.
Calculate
Your turn — calculation 3
3The same disc (I = 0.18 kg m²) spins at 30 rad s−1. Calculate its rotational kinetic energy.
J
Hint: Ek = ½Iω² = 0.5 × 0.18 × 30².
Quick check
Think it through
?Two flywheels have the same mass and radius: one is a uniform disc (I = ½MR²), the other a hoop with all its mass at the rim (I = MR²). Which stores more kinetic energy at the same angular speed, and why?
3.11.1 · angular momentum
Angular momentum and its conservation
L = Iω · TΔt = Δ(Iω)angular momentum in kg m² s−1 · angular impulse = change in angular momentum
Conservation of angular momentum: if there is no external torque, Iω stays constant.
A spinning skater pulls her arms in → I falls → ω rises. Her kinetic energy actually increases, and the extra energy comes from the work she does pulling her arms inwards against the rotation.
The same law explains why a collapsing star spins up into a rapidly rotating pulsar, and why a cat can twist in mid-air.
Worked example
Skater: I1 = 6.0 kg m² at ω1 = 2.0 rad s−1 → L = 12 kg m² s−1.
Arms in: I2 = 2.0 kg m². ω2 = L ÷ I2 = 12 ÷ 2.0 = 6.0 rad s−1 — three times faster.
Calculate
Your turn — calculation 4
4A skater spins with moment of inertia 6.0 kg m² at 2.0 rad s−1. She pulls her arms in, reducing her moment of inertia to 2.0 kg m². Calculate her new angular speed.
rad s⁻¹
Hint: No external torque → I1ω1 = I2ω2. So 6.0 × 2.0 = 2.0 × ω2.
Quick check
Think it through
?The skater above spins faster after pulling her arms in. What has happened to her rotational kinetic energy?
3.11.2 · first law
The first law of thermodynamics and p–V diagrams
Q = ΔU + WQ = heat energy supplied to the gas · ΔU = increase in internal energy · W = work done by the gas
Work done by an expanding gas at constant pressure: W = pΔV. On a p–V diagram, the work done is the area under the curve.
Isothermal: T constant, so ΔU = 0 and Q = W. Follows pV = constant — needs to be slow, with good thermal contact.
Adiabatic: Q = 0 (no heat flows), so W = −ΔU — an expanding gas cools. Follows pVγ = constant, and happens fast (no time for heat flow).
Isobaric: constant pressure — a horizontal line on the p–V diagram; W = pΔV.
Isovolumetric: constant volume — a vertical line; no work is done, so Q = ΔU.
Worked example
A gas absorbs 500 J of heat and does 200 J of work on its surroundings.
ΔU = Q − W = 500 − 200 = +300 J — its internal energy (and so its temperature) rises.
Calculate
Your turn — calculation 5
5A gas absorbs 500 J of thermal energy and does 200 J of work on its surroundings. Calculate the increase in its internal energy.
J
Hint: Q = ΔU + W → ΔU = Q − W = 500 − 200.
Sort it
Isothermal, adiabatic or isobaric?
Tap a statement, then tap the process it describes.
🌡️ Isothermal
🚫 Adiabatic
📏 Isobaric
3.11.2 · engine cycles
Engine cycles, indicated power and efficiency
A heat engine takes in heat QH from a hot reservoir, does useful work W, and dumps waste heat QC into a cold reservoir.
W = QH − QC · efficiency = W ÷ QHthe net work per cycle is the area enclosed by the p–V loop
maximum theoretical efficiency = 1 − (TC ÷ TH)the Carnot limit — temperatures in kelvin. No engine can beat it.
Four-stroke petrol engine: induction → compression → ignition/power → exhaust. The indicator diagram is the real p–V loop.
Indicated power = (area of the p–V loop) × (number of cycles per second) × (number of cylinders).
Output (brake) power = Tω measured at the crankshaft. Friction power = indicated power − brake power.
Diesel engines compress air alone to a very high ratio, so it becomes hot enough to ignite the injected fuel — higher compression means higher efficiency.
Second law: heat cannot flow spontaneously from a colder to a hotter body, and no heat engine can be 100% efficient — some heat must always be rejected to the cold reservoir.
Worked example
An engine works between TH = 600 K and TC = 300 K.
Maximum efficiency = 1 − (300 ÷ 600) = 1 − 0.50 = 0.50 = 50%. A real engine will do considerably worse.
Calculate
Your turn — calculation 6
6A heat engine operates between a hot reservoir at 600 K and a cold reservoir at 300 K. Calculate its maximum theoretical efficiency, as a percentage.