3.11 Engineering physics (A-level only) — AQA 7408 spec coverage

3.11.1.1

Concept of moment of inertia

Moment of inertia is the rotational analogue of mass: it measures resistance to angular acceleration about a specified axis and has unit $\text{kg m}^2$ .
For a point mass, $I=mr^2$ ; for separated point masses, $I=\sum mr^2$ . Expressions for continuous extended objects are supplied when needed.
Moving the same mass farther from the axis increases $I$ strongly because the perpendicular distance is squared; both the mass distribution and rotation axis must therefore be stated.
A common error is to use an object's diameter for $r$ . Use the perpendicular distance from each mass to the stated axis and square it before summing.

Tier 1 · Easy

2 marks

ORIGINAL

Two small balancing masses rotate about the same shaft. A $0.30\,\text{kg}$ mass is $0.22\,\text{m}$ from the shaft and a $0.45\,\text{kg}$ mass is $0.12\,\text{m}$ from it. Calculate their combined moment of inertia.

Tier 2 · Standard

3 marks

ORIGINAL

A training flywheel has moment of inertia $0.080\,\text{kg m}^2$ before four identical $0.25\,\text{kg}$ sliders are attached. Determine the total moment of inertia when every slider is fixed $0.30\,\text{m}$ from the axis.

Tier 3 · Hard

5 marks

ORIGINAL

A prototype rotor consists of a uniform disc of mass $6.0\,\text{kg}$ and radius $0.25\,\text{m}$ , a shaft with moment of inertia $0.028\,\text{kg m}^2$ , and three identical point masses fixed $0.35\,\text{m}$ from the axis. The total moment of inertia is $0.456\,\text{kg m}^2$ . The disc expression is $I=\frac12MR^2$ . Determine the mass of each point mass.

3.11.1.2

Rotational kinetic energy

The rotational kinetic energy of a rigid object is $E_k=\frac12I\omega^2$ , directly analogous to $E_k=\frac12mv^2$ for translation.
A flywheel stores more energy when its moment of inertia or angular speed is larger; doubling $\omega$ makes the stored energy four times as large.
Flywheels smooth variations in torque and speed and can store energy in vehicles or production machinery, releasing it when demand rises.
Use $I$ in $\text{kg m}^2$ and $\omega$ in $\text{rad s}^{-1}$ . A common error is to substitute revolutions per second without multiplying by $2\pi$ .

Tier 1 · Easy

2 marks

ORIGINAL

A small flywheel has moment of inertia $0.85\,\text{kg m}^2$ and angular speed $14\,\text{rad s}^{-1}$ . Calculate its rotational kinetic energy.

Tier 2 · Standard

3 marks

ORIGINAL

An energy-recovery flywheel of moment of inertia $40\,\text{kg m}^2$ stores $32\,\text{kJ}$ as rotational kinetic energy. Determine its angular speed.

Tier 3 · Hard

5 marks

ORIGINAL

A machine flywheel with moment of inertia $18\,\text{kg m}^2$ slows from $75\,\text{rad s}^{-1}$ to $45\,\text{rad s}^{-1}$ in $5.5\,\text{s}$ . During this interval, $72\%$ of the decrease in rotational kinetic energy is transferred usefully. Calculate the mean useful power.

3.11.1.3

Rotational motion

Angular displacement $\theta$ is measured in radians. Angular velocity $\omega=\Delta\theta/\Delta t$ is signed for a chosen positive rotation direction; angular speed is its magnitude, and $\alpha=\Delta\omega/\Delta t$ .
For uniform angular acceleration, use the linear-motion analogues: $\omega_2=\omega_1+\alpha t$ , $\theta=\frac12(\omega_1+\omega_2)t$ , $\theta=\omega_1t+\frac12\alpha t^2$ , and $\omega_2^2=\omega_1^2+2\alpha\theta$ .
On an $\omega$ - $t$ graph, gradient is angular acceleration and area is angular displacement; curved sections represent non-uniform angular acceleration.
Convert revolutions using $1\text{ revolution}=2\pi\,\text{rad}$ . A common error is to use rotational frequency in place of angular speed.

Tier 1 · Easy

2 marks

ORIGINAL

The angular speed of a mixer rotor rises uniformly from $5.0\,\text{rad s}^{-1}$ to $17\,\text{rad s}^{-1}$ in $4.0\,\text{s}$ . Calculate its angular acceleration.

Tier 2 · Standard

3 marks

ORIGINAL

A rotor starts from rest and has constant angular acceleration $4.2\,\text{rad s}^{-2}$ for $6.0\,\text{s}$ . Determine the number of revolutions completed.

Tier 3 · Hard

5 marks

ORIGINAL

A flywheel's angular speed rises uniformly from $12\,\text{rad s}^{-1}$ to $48\,\text{rad s}^{-1}$ in $6.0\,\text{s}$ , remains at $48\,\text{rad s}^{-1}$ for $5.0\,\text{s}$ , then falls uniformly to $18\,\text{rad s}^{-1}$ in $10\,\text{s}$ . Determine its total angular displacement and the corresponding number of revolutions.

3.11.1.4

Torque and angular acceleration

For a force perpendicular to the radius, torque is $T=Fr$ and is measured in $\text{N m}$ ; use the perpendicular distance from the axis to the force's line of action.
The rotational form of Newton's second law is $T=I\alpha$ , analogous to $F=ma$ ; $T$ is the resultant torque, not automatically the driving torque.
Opposing torques such as bearing friction must be assigned the opposite sign before calculating $\alpha$ .
A common error is to use the applied force itself in $T=I\alpha$ . First convert each force to a torque and then find the resultant.

Tier 1 · Easy

2 marks

ORIGINAL

A tangential force of $85\,\text{N}$ acts at the rim of a wheel of radius $0.16\,\text{m}$ . Calculate the torque about the axle.

Tier 2 · Standard

3 marks

ORIGINAL

A constant resultant torque of $24\,\text{N m}$ acts on a rotor with moment of inertia $3.2\,\text{kg m}^2$ . Its initial angular speed is $6.0\,\text{rad s}^{-1}$ . Determine its angular speed after $3.0\,\text{s}$ .

Tier 3 · Hard

5 marks

ORIGINAL

A drive supplies a constant torque of $54\,\text{N m}$ to a flywheel of moment of inertia $6.2\,\text{kg m}^2$ . Bearing friction provides a constant opposing torque of $7.5\,\text{N m}$ . The initial angular speed is $10\,\text{rad s}^{-1}$ . Calculate the angular speed after $4.5\,\text{s}$ .

3.11.1.5

Angular momentum

For a rigid body rotating about a fixed axis, angular momentum is $L=I\omega$ with unit $\text{kg m}^2\text{ s}^{-1}$ .
Angular momentum is conserved when the resultant external torque is zero; if $I$ changes, $\omega$ changes so that total $I\omega$ remains constant.
A constant torque acting for time $\Delta t$ gives angular impulse $T\Delta t=\Delta(I\omega)$ , analogous to $F\Delta t=\Delta(mv)$ .
Choose a positive rotation direction and retain signs. A common error is to conserve rotational kinetic energy when two rotating objects lock together; angular momentum, not kinetic energy, is conserved in that collision.

Tier 1 · Easy

2 marks

ORIGINAL

A turntable has moment of inertia $2.4\,\text{kg m}^2$ and rotates at $18\,\text{rad s}^{-1}$ . Calculate its angular momentum.

Tier 2 · Standard

3 marks

ORIGINAL

A flywheel with moment of inertia $0.80\,\text{kg m}^2$ rotates at $30\,\text{rad s}^{-1}$ . It is coupled to a stationary coaxial wheel of moment of inertia $1.20\,\text{kg m}^2$ , and the pair then rotate together. Determine their common angular speed.

Tier 3 · Hard

5 marks

ORIGINAL

A rotating platform and athlete initially have moment of inertia $4.8\,\text{kg m}^2$ and angular speed $9.0\,\text{rad s}^{-1}$ . An external constant opposing torque of $3.2\,\text{N m}$ acts for $1.5\,\text{s}$ . The athlete then changes position so that the combined moment of inertia is $3.0\,\text{kg m}^2$ . Determine the final angular speed, assuming the position change itself involves negligible external torque.

3.11.1.6

Work and power

Work done by a constant torque through angular displacement is $W=T\theta$ , analogous to $W=Fs$ ; $\theta$ must be in radians.
Rotational power is $P=T\omega$ , analogous to $P=Fv$ . Use the torque and angular speed at the same instant when either changes.
Frictional torque dissipates energy and must be included in power and energy balances for rotating machinery.
At steady angular speed the resultant torque is zero, but the driving torque can still do work against friction and a load. A common error is to conclude that the power is zero.

Tier 1 · Easy

2 marks

ORIGINAL

A constant torque of $18\,\text{N m}$ turns a shaft through $3.5\,\text{rad}$ . Calculate the work done by the torque.

Tier 2 · Standard

3 marks

ORIGINAL

A motor provides a torque of $42\,\text{N m}$ while its shaft rotates at $1200\,\text{rev min}^{-1}$ . Determine the power transferred by the shaft.

Tier 3 · Hard

5 marks

ORIGINAL

A motor maintains a constant angular speed of $32\,\text{rad s}^{-1}$ for $25\,\text{s}$ while supplying torque $84\,\text{N m}$ . Bearing friction opposes the motion with torque $9.0\,\text{N m}$ ; the remaining torque acts on the load. Calculate the useful power and the useful energy transferred to the load.

3.11.2.1

First law of thermodynamics

AQA uses $Q=\Delta U+W$ , where $Q$ is energy transferred to the gas by heating, $\Delta U$ is the increase in internal energy, and $W$ is work done by the gas.
With this convention, heating gives $Q>0$ , cooling gives $Q<0$ , expansion normally gives $W>0$ , and compression gives $W<0$ because work is done on the gas.
The equivalent rearrangement $\Delta U=Q-W$ is often safest when substituting signed values.
Internal energy is a state function, so $\Delta U=0$ over a complete cycle even though the net heat transfer and net work need not be zero.
A common error is to make work done on the gas positive in AQA's formula. Translate the wording into the stated sign convention before calculating.

Tier 1 · Easy

2 marks

ORIGINAL

A gas receives $520\,\text{J}$ by heating and does $180\,\text{J}$ of work. Calculate its change in internal energy using AQA's convention $Q=\Delta U+W$ .

Tier 2 · Standard

3 marks

ORIGINAL

A gas is compressed so that $240\,\text{J}$ of work is done on it. During the compression, $90\,\text{J}$ of energy leaves the gas by heating. Using $Q=\Delta U+W$ and taking work output from the gas as positive, determine $\Delta U$ .

Tier 3 · Hard

5 marks

ORIGINAL

A gas completes a two-process cycle. In process A it receives $3.2\,\text{kJ}$ by heating and its internal energy increases by $1.1\,\text{kJ}$ . In process B the gas returns to its initial state while $1.4\,\text{kJ}$ leaves it by heating. Determine the net work done by the gas during the cycle. State the signs used in $Q=\Delta U+W$ .

3.11.2.2

Non-flow processes

The named non-flow processes are isothermal (constant temperature), adiabatic (no heat transfer), constant pressure, and constant volume.
For an ideal gas, $pV=nRT$ . Isothermal change obeys $pV=\text{constant}$ , while adiabatic change obeys $pV^\gamma=\text{constant}$ .
At constant pressure, $W=p\Delta V$ . At constant volume, $\Delta V=0$ so $W=0$ ; apply $Q=\Delta U+W$ with AQA's work-by-the-gas sign convention.
For an ideal gas in an isothermal process, internal energy does not change, so $\Delta U=0$ and $Q=W$ . In an adiabatic process, $Q=0$ so $\Delta U=-W$ .
A common error is to call a rapid change isothermal. Rapid compression or expansion is more nearly adiabatic because there is little time for heat transfer.

Tier 1 · Easy

2 marks

ORIGINAL

A rigid sealed vessel receives $450\,\text{J}$ by heating. State the work done by the gas and calculate its increase in internal energy.

Tier 2 · Standard

3 marks

ORIGINAL

An ideal gas expands isothermally from pressure $240\,\text{kPa}$ and volume $2.0\times10^{-3}\,\text{m}^3$ to volume $5.0\times10^{-3}\,\text{m}^3$ . Determine its final pressure and state its change in internal energy.

Tier 3 · Hard

5 marks

ORIGINAL

A gas is compressed adiabatically from $p_1=100\,\text{kPa}$ and $V_1=4.8\times10^{-3}\,\text{m}^3$ to $V_2=1.8\times10^{-3}\,\text{m}^3$ . Take $\gamma=1.40$ . During the compression, $720\,\text{J}$ of work is done on the gas. Calculate the final pressure and the change in internal energy.

3.11.2.3

The p-V diagram

A process is represented as a path on a $p$ - $V$ diagram; movement to larger $V$ is expansion and movement to smaller $V$ is compression.
The work done by the gas is the area under the process curve. For constant pressure, $W=p\Delta V$ ; expansion gives positive work and compression negative work.
For a cyclic process, the enclosed loop area is the net work per cycle. A clockwise loop gives positive net work by the gas; an anticlockwise loop requires net work input.
Convert pressure to pascals and volume to $\text{m}^3$ so that the area has unit joules. A common error is to use the whole area under one branch instead of the area enclosed by a cycle.

Tier 1 · Easy

2 marks

ORIGINAL

A gas expands at constant pressure $180\,\text{kPa}$ from volume $1.2\times10^{-3}\,\text{m}^3$ to $3.7\times10^{-3}\,\text{m}^3$ . Calculate the work done by the gas.

Tier 2 · Standard

3 marks

ORIGINAL

A gas follows a clockwise rectangular cycle on a $p$ - $V$ diagram between pressures $150\,\text{kPa}$ and $400\,\text{kPa}$ and between volumes $1.0\times10^{-3}\,\text{m}^3$ and $3.5\times10^{-3}\,\text{m}^3$ . Determine the net work done per cycle.

Tier 3 · Hard

5 marks

ORIGINAL

On a $p$ - $V$ diagram, a gas expands along a straight line from $(0.8\times10^{-3}\,\text{m}^3,600\,\text{kPa})$ to $(3.2\times10^{-3}\,\text{m}^3,200\,\text{kPa})$ . It then compresses at constant pressure $200\,\text{kPa}$ to its initial volume before returning at constant volume to its initial state. Calculate the net work done by the gas in the cycle.

3.11.2.4

Engine cycles

A four-stroke engine completes induction, compression, power and exhaust strokes. A petrol engine compresses a fuel-air mixture before spark ignition, whereas a diesel engine compresses air before injected fuel ignites; their indicator diagrams show the corresponding $p$ - $V$ paths without requiring constructional detail.
Indicated power is $(\text{loop area})(\text{cycles per second per cylinder})(\text{number of cylinders})$ . A four-stroke cylinder completes one cycle per two crankshaft revolutions.
Input power is calorific value multiplied by fuel mass-flow rate. Brake power is $P=T\omega$ , and friction power is indicated power minus brake power.
Overall efficiency is brake power divided by input power, thermal efficiency is indicated power divided by input power, and mechanical efficiency is brake power divided by indicated power.
A common error is to use revolutions per second as cycles per second for a four-stroke engine, doubling the indicated power.

Tier 1 · Easy

2 marks

ORIGINAL

An engine produces a steady crankshaft torque of $140\,\text{N m}$ at $3000\,\text{rev min}^{-1}$ . Calculate its brake power.

Tier 2 · Standard

3 marks

ORIGINAL

The area enclosed by the indicator diagram for one cylinder of a four-stroke engine is $620\,\text{J}$ per cycle. The four-cylinder engine runs at $2400\,\text{rev min}^{-1}$ . Determine the indicated power.

Tier 3 · Hard

6 marks

ORIGINAL

A four-cylinder, four-stroke engine runs at $3000\,\text{rev min}^{-1}$ . Each cylinder's indicator-loop area is $380\,\text{J}$ . The brake torque is $105\,\text{N m}$ . Fuel of calorific value $44\,\text{MJ kg}^{-1}$ is supplied at $1.60\times10^{-3}\,\text{kg s}^{-1}$ . Determine the indicated power, brake power, friction power, and thermal efficiency.

3.11.2.5

Second Law and engines

The First Law alone would allow complete conversion of heat to work, but the Second Law requires a heat engine to operate between a hot source and a colder sink.
For each cycle, $Q_H=W+Q_C$ and efficiency is $\eta=W/Q_H=(Q_H-Q_C)/Q_H$ .
The maximum theoretical efficiency is $\eta_{\max}=(T_H-T_C)/T_H$ , with both temperatures in kelvin; it is below $100\%$ whenever $T_C>0$ .
Practical engines are less efficient than the theoretical maximum because of friction, heat transfer to surroundings, finite-rate processes, and incomplete combustion.
Combined heat and power schemes use both $W$ and otherwise-wasted heat. A common error is to call the sum an engine efficiency; it is an energy-utilisation fraction.

Tier 1 · Easy

2 marks

ORIGINAL

A heat engine receives $5.0\,\text{kJ}$ from its hot source and rejects $3.2\,\text{kJ}$ to its sink each cycle. Calculate its efficiency.

Tier 2 · Standard

3 marks

ORIGINAL

An engine operates between a source at $720\,\text{K}$ and a sink at $310\,\text{K}$ . Determine its maximum theoretical efficiency and explain why a practical engine operating at these temperatures has a lower efficiency.

Tier 3 · Hard

5 marks

ORIGINAL

A combined heat and power plant receives thermal energy at $120\,\text{MW}$ and produces $45\,\text{MW}$ of electrical work. It supplies $60\,\text{MW}$ of otherwise rejected heat to nearby buildings. The engine source and sink temperatures are $850\,\text{K}$ and $300\,\text{K}$ . Calculate the engine efficiency, the total useful-energy fraction, and the maximum theoretical efficiency. Explain the difference between the last two quantities.

3.11.2.6

Reversed heat engines

A reversed heat engine uses work input $W$ to transfer energy $Q_C$ from a cold region to a hot region, with $Q_H=Q_C+W$ .
For a refrigerator, $\mathrm{COP}_{\text{ref}}=Q_C/W=Q_C/(Q_H-Q_C)$ because the desired transfer is out of the cold space.
For a heat pump, $\mathrm{COP}_{\text{hp}}=Q_H/W=Q_H/(Q_H-Q_C)$ because the desired transfer is into the warm space; therefore $\mathrm{COP}_{\text{hp}}=\mathrm{COP}_{\text{ref}}+1$ .
For ideal devices, $\mathrm{COP}_{\text{ref}}=T_C/(T_H-T_C)$ and $\mathrm{COP}_{\text{hp}}=T_H/(T_H-T_C)$ , using kelvin.
COP can exceed $1$ because it measures heat transferred per unit work input, not conversion efficiency. A common error is to use $Q_C$ for a heat pump or $Q_H$ for a refrigerator.

Tier 1 · Easy

2 marks

ORIGINAL

A refrigerator removes $360\,\text{J}$ from its cold compartment for every $120\,\text{J}$ of electrical work supplied. Calculate its coefficient of performance.

Tier 2 · Standard

3 marks

ORIGINAL

A heat pump delivers $14\,\text{kW}$ to a building while consuming $4.0\,\text{kW}$ of electrical power. Determine its heat-pump coefficient of performance and the rate at which it extracts energy from outside.

Tier 3 · Hard

5 marks

ORIGINAL

A heat pump keeps a building at $293\,\text{K}$ when the outside temperature is $268\,\text{K}$ . Its actual coefficient of performance is $42\%$ of the maximum theoretical value. Calculate the electrical power required when the heat delivered to the building is $18\,\text{kW}$ .