Engineering physics (A-level only)

Explanation

• Moment of inertia $I$ measures an object's resistance to angular acceleration about a specified axis, so it is the rotational analogue of mass. For a point mass, $I=mr^2$; for several point masses, $I=\sum mr^2$, where $r$ is the perpendicular distance from the axis.
• Its unit is $\text{kg m}^2$.
• Both the total mass and its distribution matter: moving mass farther from the axis increases $I$ strongly because distance is squared.
• For an extended object, the supplied expression represents the sum over all its small mass elements.
• Examiners expect the stated axis to be used and every separate contribution to be included.

Explanation

• A rigid rotating object stores kinetic energy $E_k=\frac12I\omega^2$, directly analogous to translational energy $\frac12mv^2$. Increasing moment of inertia increases the stored energy in direct proportion, whereas increasing angular speed has a squared effect.
• A flywheel therefore stores substantial energy when mass is placed far from its axis and it rotates rapidly.
• In machinery, a flywheel absorbs energy when driving torque exceeds demand and releases energy when demand exceeds the drive, smoothing changes in torque and speed.
• Flywheels can also store energy in vehicles and production machines.
• Calculations require $I$ in $\text{kg m}^2$ and $\omega$ in $\text{rad s}^{-1}$.

Explanation

• Angular displacement $\theta$ is measured in radians. Angular velocity is $\omega=\Delta\theta/\Delta t$ for a chosen positive direction; angular speed is its magnitude.
• Angular acceleration is $\alpha=\Delta\omega/\Delta t$. For uniform angular acceleration, the translational equations have direct rotational 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 angular-velocity–time graph, gradient gives angular acceleration and signed area gives angular displacement. A changing gradient represents non-uniform angular acceleration.
• A straight graph section therefore represents constant angular acceleration over that interval.
• Examiners may require both graphical interpretation and equation selection.

Explanation

• Torque is the turning effect of a force. For a force perpendicular to the radius, $T=Fr$, where $r$ is the perpendicular distance from the axis to the force's line of action; torque has unit $\text{N m}$.
• The rotational form of Newton's second law is $T=I\alpha$, so the resultant torque produces angular acceleration in proportion to $1/I$.
• Driving and resisting torques act in opposite rotational senses and must be combined with signs before applying this equation.
• If the force is not perpendicular, the perpendicular force component or perpendicular moment arm is required.
• Examiners expect a resultant torque, not merely the applied driving torque.

Explanation

• For a rigid body rotating about a fixed axis, angular momentum is $L=I\omega$, measured in $\text{kg m}^2\text{ s}^{-1}$. Total angular momentum is conserved when the resultant external torque is zero.
• Consequently, reducing $I$ increases $\omega$ so that the signed total $I\omega$ remains constant, as in a spinning athlete drawing in the arms.
• A constant torque acting for time $\Delta t$ provides angular impulse $T\Delta t=\Delta(I\omega)$.
• Direction matters: opposite rotations have opposite angular momenta.
• When rotating objects couple and move together, angular momentum can be conserved even though rotational kinetic energy decreases, because the coupling is inelastic.

Explanation

• A constant torque acting through angular displacement transfers work $W=T\theta$, with $\theta$ in radians. The instantaneous rate of energy transfer is rotational power $P=T\omega$.
• These equations parallel $W=Fs$ and $P=Fv$.
• In real rotating machinery, frictional torque removes mechanical energy and must appear in the torque, work or power balance.
• At constant angular speed, resultant torque is zero, but the driving torque can still balance load and friction torques and therefore transfer power.
• When torque or angular speed varies, $P=T\omega$ uses simultaneous values; total work may instead be found from the energy change or an appropriate torque-angle area.

Explanation

• AQA writes the First Law as $Q=\Delta U+W$. Here $Q$ is energy transferred to the system by heating, $\Delta U$ is the increase in internal energy, and $W$ is work done by the system.
• Thus heating gives positive $Q$, cooling negative $Q$, expansion normally positive $W$, and compression negative $W$.
• Rearranging to $\Delta U=Q-W$ helps preserve this sign convention.
• Internal energy is a state function, so its change depends only on the initial and final states.
• Over a complete cycle $\Delta U=0$, although non-zero net heat transfer can equal the net work done by the gas.

Explanation

• The required non-flow changes are isothermal, adiabatic, constant pressure and constant volume. An ideal gas obeys $pV=nRT$.
• During an isothermal change, temperature is constant and $pV=\text{constant}$; for an ideal gas, $\Delta U=0$, so $Q=W$. During an adiabatic change, $Q=0$ and $pV^\gamma=\text{constant}$, so $\Delta U=-W$.
• At constant pressure, work done by the gas is $W=p\Delta V$. At constant volume, $\Delta V=0$ and $W=0$.
• Each process must also satisfy $Q=\Delta U+W$.
• Rapid well-insulated changes are commonly close to adiabatic because little energy transfers by heating.

Explanation

• A thermodynamic process is shown as a path on a pressure–volume diagram. Motion to larger volume is expansion; motion to smaller volume is compression.
• The work done by the gas is the signed area under the path. For constant pressure, $W=p\Delta V$; an expansion gives positive work and a compression negative work.
• For a cycle, the net work per cycle is the area enclosed by the loop, not the total area beneath one branch.
• A clockwise loop represents positive net work by the gas, while an anticlockwise loop requires net work input.
• Pressure must be in pascals and volume in $\text{m}^3$ for the area to be in joules.

Explanation

• A four-stroke petrol engine completes induction, compression, power and exhaust strokes; it compresses a fuel–air mixture before spark ignition. A diesel engine compresses air before fuel injection and ignition.
• Their indicator diagrams show measured pressure–volume cycles, which can be compared with theoretical cycles without requiring constructional detail. Indicated power equals loop area multiplied by cycles per second per cylinder and cylinder count.
• Input power is calorific value times fuel mass-flow rate. Brake power is $T\omega$, and friction power is indicated power minus brake power.
• Overall, thermal and mechanical efficiencies are respectively brake/input, indicated/input and brake/indicated power.
• Other supplied cycles require interpretation using the same ideas.

Explanation

• The First Law alone does not forbid complete conversion of heat into work, but the Second Law requires a heat engine to operate between a hot source and a colder sink.
• Each cycle receives $Q_H$, produces work $W$ and rejects $Q_C$, so $Q_H=W+Q_C$ and $\eta=W/Q_H=(Q_H-Q_C)/Q_H$.
• The maximum theoretical efficiency is $\eta_{\max}=(T_H-T_C)/T_H$, with temperatures in kelvin.
• Practical efficiency is lower because of friction, unwanted heat transfer, incomplete combustion and irreversible finite-rate processes.
• Combined heat and power schemes improve total energy use by using both $W$ and useful rejected heat, but this utilisation fraction is not the engine's heat-to-work efficiency.

Explanation

• A reversed heat engine uses work input $W$ to transfer energy $Q_C$ from a cold region to a hot region, delivering $Q_H=Q_C+W$. A refrigerator's useful transfer is $Q_C$, so $\mathrm{COP}_{\text{ref}}=Q_C/W=Q_C/(Q_H-Q_C)$.
• A heat pump's useful transfer is $Q_H$, so $\mathrm{COP}_{\text{hp}}=Q_H/W=Q_H/(Q_H-Q_C)$. Hence $\mathrm{COP}_{\text{hp}}=\mathrm{COP}_{\text{ref}}+1$.
• Ideal limits are $T_C/(T_H-T_C)$ and $T_H/(T_H-T_C)$ respectively, using kelvin. COP can exceed one because it compares heat transferred with work input; it is not a conversion efficiency.
• Only the basic principles and uses of refrigerators and heat pumps are required.
• Detailed device cycles are not required.