AQA A-Level Physics: Engineering physics (7408) – Revision Notes

These revision notes condense section 3.11 Engineering physics of the AQA AS and A-level Physics specification (7407/7408, version 1.4, AS and A-level exams June 2016 onwards): 3.11.1 Rotational dynamics and 3.11.2 Thermodynamics and engines. The section is A-level only. It is one of the five options (sections 3.9 to 3.13); you study one, and it is assessed in Paper 3 Section B (35 marks). Paper 3 Section A assesses practical skills and data analysis.

For full explanations and worked examples, use the study guide. Then test yourself with the practice questions. See also the course hub, the printable checklist and the free diagnostics. Radians, ω and the ideal gas equation come from Further mechanics and thermal physics.

Key equations

Quantity	Equation	Unit
Moment of inertia	I = mr² (point); I = Σmr²	kg m²
Rotational kinetic energy	E_k = ½Iω²	J
Angular speed	ω = Δθ/Δt	rad s⁻¹
Angular acceleration	α = Δω/Δt	rad s⁻²
Torque	T = Fr; T = Iα	N m
Angular momentum	L = Iω	N m s
Angular impulse	TΔt = Δ(Iω)	N m s
Work, power	W = Tθ; P = Tω	J, W
First law	Q = ΔU + W	J
Ideal gas	pV = nRT
Isothermal	pV = constant
Adiabatic	pV^γ = constant
Constant pressure	W = pΔV	J
Input power	calorific value × fuel flow rate	W
Indicated power	loop area × cycles per second × cylinders	W
Friction power	indicated − brake	W
Engine efficiency	W/Q_H = (Q_H − Q_C)/Q_H
Maximum efficiency	(T_H − T_C)/T_H
Refrigerator	COP = Q_C/W, max T_C/(T_H − T_C)
Heat pump	COP = Q_H/W, max T_H/(T_H − T_C)

Uniform angular acceleration:

ω₂ = ω₁ + αt          θ = ½(ω₁ + ω₂)t
θ = ω₁t + ½αt²        ω₂² = ω₁² + 2αθ

3.11.1 Rotational dynamics

Moment of inertia (3.11.1.1). Depends on the mass, how far the mass is from the axis (r²), and where the axis is. Mass at the rim counts most. The specification says expressions for particular shapes will be given where necessary.

Flywheels (3.11.1.2). Store energy as ½Iω². Capacity rises with I (heavy rim) and ω (limited by the material’s strength). Uses: smoothing torque and speed in engines; storing braking energy in vehicles; releasing stored energy quickly in production machines such as presses.

Graphs (3.11.1.3). ω–t: gradient = α, area = θ. Straight line = uniform α; curve = non-uniform α (use a tangent). θ–t: gradient = ω.

Torque (3.11.1.4). Use the resultant torque in T = Iα: applied torque minus frictional torque.

Angular momentum (3.11.1.5). Conserved when no external resultant torque acts. Pulling mass in reduces I and raises ω; kinetic energy rises because work is done pulling inwards. Joining two rotating bodies: L is conserved, E_k is not.

Work and power (3.11.1.6). At constant speed, driving torque = frictional torque. Power lost to friction = T_friction × ω.

Method: rotational dynamics problem

1. Convert rpm or rev s⁻¹ to rad s⁻¹ (multiply rev s⁻¹ by 2π).
2. Find the resultant torque.
3. α = T/I.
4. Pick the uniform-α equation that contains your known and wanted quantities.
5. Check with energy: Tθ (work by resultant torque) = Δ(½Iω²).

3.11.2 Thermodynamics and engines

First law (3.11.2.1). Q = energy to the gas by heating; ΔU = increase in internal energy; W = work done by the gas. Compression: W negative. Losing heat: Q negative.

Non-flow processes (3.11.2.2).

Process	What is fixed	Result
Isothermal	T	ΔU = 0, Q = W
Adiabatic	no heat flow	Q = 0, W = −ΔU
Constant pressure	p	W = pΔV
Constant volume	V	W = 0, Q = ΔU

Adiabatic compression heats a gas; adiabatic expansion cools it. γ is a constant for the gas.

Worked reminder: halving the volume adiabatically with γ = 1.4 multiplies p by 2^1.4 = 2.64 and T by 2^0.4 = 1.32. Isothermally, p would only double.

p–V diagrams (3.11.2.3). Work = area under the line. Cycle: work per cycle = loop area; ΔU = 0 per cycle. Only W = pΔV needs a formula; other areas are estimated (count squares).

Engine cycles (3.11.2.4).

• Petrol (four-stroke): induction, compression, power, exhaust. Theoretical: two adiabatics, heat in and out at constant volume. Spark ignition.
• Diesel: air only is compressed; fuel ignites in the hot air. Theoretical: heat in at constant pressure, out at constant volume.
• Real indicator diagram: rounded corners, lower peak, smaller area, small negative pumping loop.
• Four-stroke: one cycle per two crankshaft revolutions per cylinder.

Efficiencies.

overall    = brake / input
thermal    = indicated / input
mechanical = brake / indicated
overall    = thermal × mechanical

Method: engine data problem

1. Convert rpm to rev s⁻¹, then halve it for cycles per second (four-stroke).
2. Indicated power = loop area × cycles per second × cylinders.
3. Brake power = Tω, with ω = 2π × rev s⁻¹.
4. Input power = calorific value × fuel flow rate.
5. Check that overall = thermal × mechanical and that every efficiency is below 1.

Second Law (3.11.2.5). A heat engine needs a source and a sink; it cannot turn all of Q_H into W. Real engines fall short of (T_H − T_C)/T_H because of friction, heat losses and non-ideal processes. CHP schemes use Q_C for heating.

Reversed heat engines (3.11.2.6). Work moves energy from cold to hot: Q_H = Q_C + W. A fridge’s useful output is Q_C; a heat pump’s is Q_H. COP can be greater than 1. For the same temperatures, COP_hp = COP_ref + 1.

Must-know distinctions

• Angular speed vs angular velocity: angular velocity includes the sense of rotation.
• Torque vs resultant torque: T = Iα needs the resultant.
• Indicated vs brake power: indicated comes from the p–V loop (work done by the gas); brake is measured at the output shaft (Tω).
• Thermal vs mechanical efficiency: thermal is how well fuel energy becomes gas work; mechanical is how much gas work reaches the shaft.
• Isothermal vs adiabatic: isothermal is slow with heat flow, T fixed; adiabatic has no heat flow, T changes.
• Refrigerator vs heat pump: same device principle; different useful energy (Q_C vs Q_H).

Quick self-test

1. Find I for a 3.0 kg point mass 0.20 m from an axis.
2. A flywheel (I = 2.5 kg m²) spins at 40 rad s⁻¹. Find its kinetic energy.
3. Convert 1200 rpm to rad s⁻¹.
4. A wheel starts from rest with α = 4.0 rad s⁻². Find the angle turned in 5.0 s.
5. A resultant torque of 12 N m acts on a body with I = 0.30 kg m². Find α.
6. A diver’s I drops from 3.2 kg m² to 0.80 kg m². Her ω was 2.0 rad s⁻¹. Find her new ω.
7. Find the power delivered by a torque of 50 N m at 60 rad s⁻¹.
8. 250 J of work is done on a gas while it loses 70 J by heating. Find ΔU.
9. A gas at 2.0 × 10⁵ Pa is compressed isothermally from 3.0 L to 1.2 L. Find the new pressure.
10. Find the maximum efficiency of an engine working between 600 K and 300 K.
11. Find the maximum COP of a refrigerator with T_C = 255 K and T_H = 300 K.
12. Indicated power 30 kW; brake power 24 kW. Find the mechanical efficiency and friction power.

Answers

1. I = 3.0 × 0.20² = 0.12 kg m²
2. E_k = ½ × 2.5 × 40² = 2000 J
3. 1200/60 × 2π = 126 rad s⁻¹
4. θ = ½ × 4.0 × 5.0² = 50 rad
5. α = 12/0.30 = 40 rad s⁻²
6. ω = 3.2 × 2.0/0.80 = 8.0 rad s⁻¹
7. P = 50 × 60 = 3000 W
8. Q = −70 J, W = −250 J; ΔU = Q − W = −70 + 250 = +180 J
9. p = 2.0 × 10⁵ × 3.0/1.2 = 5.0 × 10⁵ Pa
10. (600 − 300)/600 = 0.50
11. 255/45 = 5.7
12. 24/30 = 0.80; friction power = 6 kW

Where marks are usually lost

• Leaving ω in rpm or rev s⁻¹ inside E_k = ½Iω², P = Tω or L = Iω.
• Using the motor torque in T = Iα and ignoring frictional torque.
• Claiming kinetic energy is conserved when two rotating bodies couple.
• Writing ΔU = Q + W with AQA’s definition of W (work done by the gas).
• Treating an adiabatic change as “no temperature change”.
• Using T in °C in pV = nRT, efficiency or COP formulas.
• Forgetting the factor of ½ in cycles per second for a four-stroke engine.
• Quoting an efficiency greater than 1, or a COP formula upside down.
• Describing a diesel engine with a spark plug.

Official syllabus

AQA AS and A-level Physics (7407/7408) specification, version 1.4, July 2026, AS and A-level exams June 2016 onwards, published by AQA. Section 3.11 Engineering physics (A-level only option).