AQA A-Level Physics: Engineering physics (7408)

This study guide teaches 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. Section 3.11 is A-level only and is one of the five options (sections 3.9 to 3.13). You study one option, assessed in Paper 3 Section B (35 marks); Paper 3 Section A assesses practical skills and data analysis. The specification says questions may use unfamiliar contexts, with the information you need given.

Next, use the revision notes and practice questions. See also the course hub, the printable checklist and the free diagnostics. Radians, ω and the gas laws come from Further mechanics and thermal physics.

What this topic covers

Specification (all A-level only)	You must be able to
3.11.1.1 Moment of inertia	Use I = mr², I = Σmr²; factors affecting I
3.11.1.2 Rotational kinetic energy	Use E_k = ½Iω²; explain flywheels
3.11.1.3 Rotational motion	Use θ, ω, α, graphs and uniform-α equations
3.11.1.4 Torque and angular acceleration	Use T = Fr and T = Iα
3.11.1.5 Angular momentum	Use Iω, its conservation, TΔt = Δ(Iω)
3.11.1.6 Work and power	Use W = Tθ, P = Tω; frictional torque
3.11.2.1 First law	Apply Q = ΔU + W
3.11.2.2 Non-flow processes	Treat isothermal, adiabatic, constant p and V changes
3.11.2.3 The p–V diagram	Find work from areas and loops
3.11.2.4 Engine cycles	Petrol and diesel cycles; power and efficiencies
3.11.2.5 Second Law and engines	Source, sink, efficiency limits, CHP
3.11.2.6 Reversed heat engines	Heat pumps, refrigerators, COP

3.11.1.1 Moment of inertia

Moment of inertia, I (kg m²), measures how hard it is to change an object’s rotation about an axis. It plays the part of mass.

point mass:  I = mr²        extended object:  I = Σmr²

r is the distance from the axis. I increases with total mass and with mass placed further from the axis (r is squared), and depends on where the axis is. A ring has more I than a disc of equal mass and radius. The specification says expressions for particular shapes will be given where necessary.

Worked example 1. A light rod spins about its centre with 0.40 kg at each end, 0.25 m from the axis, and 0.20 kg at 0.10 m.

I = 2 × 0.40 × 0.25² + 0.20 × 0.10² = 0.052 kg m²

3.11.1.2 Rotational kinetic energy and flywheels

E_k = ½Iω²

A flywheel stores rotational kinetic energy. Its capacity rises with I (more mass, placed near the rim) and with ω, since E_k ∝ ω². The maximum ω is limited by the strength of the material. Uses in the specification:

• Smoothing torque and speed: an engine’s torque comes in pulses; the flywheel takes in energy when torque is high and returns it when low.
• Storing energy in vehicles: braking energy goes into the flywheel rather than heat, and is returned when accelerating.
• Production machines: a small motor spins the flywheel up slowly, and the energy is released in a short burst, as in a press.

Worked example 2. A flywheel (I = 0.80 kg m²) speeds up from 120 to 200 rad s⁻¹.

ΔE_k = ½ × 0.80 × (200² − 120²) = 1.0 × 10⁴ J

3.11.1.3 Rotational motion

Angular displacement θ (rad); angular speed or velocity ω = Δθ/Δt (velocity includes the sense of rotation); angular acceleration α = Δω/Δt. On an ω–t graph the gradient is α and the area is θ. A straight line shows uniform α; a curve shows non-uniform α (use a tangent).

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

The analogy with linear motion: s, v, a, m and F become θ, ω, α, I and T. So F = ma becomes T = Iα, ½mv² becomes ½Iω², mv becomes Iω, and Fv becomes Tω.

Worked example 3. A disc speeds up uniformly from 15 to 45 rad s⁻¹ in 6.0 s.

α = (45 − 15)/6.0 = 5.0 rad s⁻²
θ = ½(15 + 45) × 6.0 = 180 rad = 28.6 revolutions

3.11.1.4 Torque and angular acceleration

Torque T = Fr (N m), with r the perpendicular distance from the axis to the force’s line of action. Newton’s second law becomes T = Iα, where T is the resultant torque.

Worked example 4. A 40 N pull on a rope round a drum of radius 0.15 m; I = 0.60 kg m²; frictional torque 1.2 N m; starts from rest.

resultant T = 40 × 0.15 − 1.2 = 4.8 N m
α = 4.8/0.60 = 8.0 rad s⁻²;   after 3.0 s, ω = 24 rad s⁻¹

3.11.1.5 Angular momentum

Angular momentum = Iω (N m s). Angular impulse TΔt = Δ(Iω) for constant T. With no external resultant torque, angular momentum is conserved. A skater or diver who pulls in their limbs reduces I, so ω rises; kinetic energy rises too, because they do work pulling inwards.

Worked example 5. A turntable (I = 0.050 kg m²) spins freely at 3.0 rad s⁻¹. A ring (I = 0.025 kg m²) is dropped centrally onto it.

0.050 × 3.0 = 0.075 × ω₂   →   ω₂ = 2.0 rad s⁻¹
E_k: 0.225 J before, 0.150 J after

Angular momentum is conserved but kinetic energy is not, as in an inelastic collision.

3.11.1.6 Work and power

W = Tθ and P = Tω. In machines, some driving torque works against frictional torque; at steady speed the two are equal.

Worked example 6. A motor delivers 2.4 kW at 1500 rpm: ω = 1500 × 2π/60 = 157 rad s⁻¹, so T = 2400/157 = 15 N m.

3.11.2.1 First law of thermodynamics

Q = ΔU + W, where Q is energy transferred to the system by heating, ΔU is the increase in internal energy and W is work done by the system. For an ideal gas, ΔU = 0 when T is unchanged.

Worked example 7. A gas receives 500 J by heating and does 180 J of work: ΔU = 500 − 180 = 320 J.

3.11.2.2 Non-flow processes

A fixed mass of gas stays in its container, and pV = nRT throughout.

Process	Condition	First law
Isothermal	pV = constant	ΔU = 0, Q = W
Adiabatic	pV^γ = constant, Q = 0	W = −ΔU
Constant pressure	W = pΔV	Q = ΔU + pΔV
Constant volume	W = 0	Q = ΔU

γ is a constant for the gas. Isothermal changes are slow; adiabatic ones are fast or insulated. Adiabatic compression makes W negative, so ΔU and T rise.

Worked example 8. Air (γ = 1.4) at 1.0 × 10⁵ Pa and 300 K is compressed adiabatically from 4.0 × 10⁻⁴ m³ to 0.50 × 10⁻⁴ m³.

p₂ = 1.0 × 10⁵ × 8^1.4 = 1.8 × 10⁶ Pa
T₂ = T₁p₂V₂/(p₁V₁) = 300 × 8^0.4 = 689 K

3.11.2.3 The p–V diagram

Isothermals are curves; adiabatics are steeper curves; constant-pressure lines are horizontal; constant-volume lines vertical. Work done = area under the graph (estimate curved areas by counting squares). In a cycle, ΔU = 0 and work done per cycle = area of the loop; a clockwise loop is an engine. The specification says expressions for work done are not required except W = pΔV.

3.11.2.4 Engine cycles

Four-stroke petrol engine: induction, compression, power (a spark ignites the fuel–air mixture), exhaust. Theoretical cycle: two adiabatics joined by two constant-volume lines (heat in, heat out).

Diesel engine: air alone is compressed until hot enough to ignite injected fuel. Theoretical cycle: adiabatic compression, heat in at constant pressure, adiabatic expansion, heat out at constant volume.

Indicator diagrams are measured p–V loops. Compared with theory they have rounded corners, lower peak pressure and smaller area: burning takes time, valves do not move instantly, energy leaks through cylinder walls and the gas is not ideal. Induction and exhaust add a small loop of work done on the gas. Engine construction details are not required.

input power      = calorific value × fuel flow rate
indicated power  = loop area × cycles per second × number of cylinders
brake power      = Tω;   friction power = indicated − brake
overall = brake/input;  thermal = indicated/input;  mechanical = brake/indicated

A four-stroke cylinder completes one cycle every two crankshaft revolutions.

Worked example 9. Four cylinders, four-stroke, 3000 rpm; loop area 380 J; brake torque 100 N m; fuel 4.6 × 10⁷ J kg⁻¹ at 3.0 × 10⁻³ kg s⁻¹.

cycles per second = 50/2 = 25;  indicated = 380 × 25 × 4 = 38 kW
brake = 100 × 2π × 50 = 31.4 kW;  friction = 6.6 kW;  input = 138 kW
overall 0.23;  thermal 0.28;  mechanical 0.83  (overall = thermal × mechanical)

3.11.2.5 Second Law and engines

An engine cannot turn all the energy it takes in by heating into work, so it cannot work by the First Law alone. The Second Law: a heat engine must operate between a hot source and a cold sink, taking in Q_H, doing W and rejecting Q_C.

efficiency = W/Q_H = (Q_H − Q_C)/Q_H;   maximum = (T_H − T_C)/T_H  (kelvin)

Practical engines do worse because of friction, heat losses and non-ideal processes. Combined heat and power schemes use Q_C to heat buildings.

Worked example 10. Source 800 K, sink 300 K: maximum efficiency 0.625. A real engine taking in 2000 J and doing 600 J per cycle has efficiency 0.30 and rejects 1400 J.

3.11.2.6 Reversed heat engines

Work W moves Q_C out of a cold space and delivers Q_H = Q_C + W to a hot one. A refrigerator keeps the cold space cold; a heat pump warms the hot space.

COP_ref = Q_C/W = Q_C/(Q_H − Q_C);  maximum T_C/(T_H − T_C)
COP_hp  = Q_H/W = Q_H/(Q_H − Q_C);  maximum T_H/(T_H − T_C)

COP can exceed 1. Practical device cycles are not required.

Worked example 11. Between 278 K and 308 K, maximum COP_hp = 308/30 = 10.3. A heat pump with COP 3.5 delivering 6.0 kW uses 1.7 kW and extracts 4.3 kW from the cold side.

Common errors

• Leaving ω in rpm, or using applied rather than resultant torque.
• Assuming kinetic energy is conserved when angular momentum is.
• Sign errors in Q = ΔU + W, or °C in temperature formulas.
• One cycle per revolution in a four-stroke engine.

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).