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Support for the left curve comes from high laboratory friction and substantial in situ differential stress; support for the right curve comes from low inferred fault traction, weak fault materials or high pore pressure. No single observation directly measures the whole plotted envelope.

For the left, stronger shallow-crust model:

For the right, weak shallow-layer model:

To discriminate, measure the strength-controlling quantities through depth and across both faults and their surroundings. Obtain borehole stress, pore pressure and temperature profiles where feasible; use borehole breakouts and hydraulic tests to constrain effective stress. Recover fault and host rocks and test their friction and material creep at relevant fluid conditions. The right curve specifically predicts an end to the shallow weakening near its sharp rise: investigate changes in permeability of a porous medium, alteration and fault gouge with depth, rather than testing only a surface sample.

Combine those measurements with heat flux transects, magnetotelluric sounding for conductive fluids, accurate earthquake depths and focal mechanisms, and GNSS or InSAR observations of deformation. The latter can help constrain the depth distribution of locking and postseismic relaxation, but neither locking nor seismicity alone gives absolute strength. Earthquake stress drop is a change in stress, not the pre-event stress. Low heat flux can also reflect heat transport or transient weakening; high electrical conductivity can have causes other than free water. A joint explanation of all these data is needed to choose between the models locally.

Neglect upstream speed and hydraulic losses. If $H$ is the upstream free-surface height above the crest, the specific energy at the crest is $H=y+v^2/(2g)$ and flow per unit width is $Q=yv$. At the critical broad-crested control section, extremizing $Q^2=2g{y}^2(H-y)$ gives $y=2H/3$ and $v^2=gy$. ThereforeFinite approach speed, friction and contraction introduce measured discharge coefficients; the ideal result assumes a long enough crest for the near-hydrostatic critical-flow approximation.

Sadi Carnot's ideal engine was motivated by the practical problem of obtaining mechanical work economically from heat. Existing engines made temperature differences, boilers, condensers and the waste associated with cooling impossible to ignore. His use of a cyclic working substance and two thermal reservoirs therefore retained decisive features of real engine operation. An engine that has to be reset after each stroke cannot be assessed by considering expansion alone.

His crucial step was to strip away friction and uncontrolled heat transfer and ask for a limiting cycle that could be reversed. Combining a putatively better engine with a reversed ideal engine would produce an unacceptable result without a compensating thermal change; the comparison makes the maximum performance depend on reservoir temperatures rather than the identity of the working substance. This is a theoretical limit against which imperfect machines can be judged, not a blueprint that an actual engine can execute at finite power without losses.

There is a historically important separation between this reasoning and its modern reformulation. Carnot worked substantially within a caloric framework in which heat was treated as conserved in transfer. Modern thermodynamics instead distinguishes heat input, work output and heat rejected, with $W=Q_h-Q_c$ and reversible efficiency $1-T_c/T_h$. It would be anachronistic to attribute the complete modern energy-and-entropy analysis to his 1824 text.

Practical experience thus supplied the problem, relevant constraints and examples of irreversibility, while idealization supplied a result that experience alone could not establish by testing finitely many engines. The ideal engine's importance lies precisely in exceeding any particular machine without pretending that its limiting operation is practically attainable at useful power.^[ref]

William Thomson had to reconcile a successful constraint on engine performance with evidence that mechanical work could be converted into heat. Count Rumford's cannon-boring observations challenged a finite store of caloric by showing sustained production of heat through mechanical action. James Prescott Joule made the connection quantitative through controlled comparisons of mechanical or electrical input and thermal output. Neither result alone supplied the general limit on converting heat back into useful work.

Carnot's reversible-engine reasoning supplied that second ingredient, but its conservation-of-caloric interpretation could not simply coexist unchanged with the mechanical equivalent of heat. The synthesis retains energy conservation while replacing caloric conservation: in a cycle, the heat received equals work produced plus heat rejected. A second principle restricts what transformations of that conserved energy are possible. In the Thomson formulation, a cyclic device cannot have as its sole result the extraction of heat from one reservoir and its complete conversion into work. A reversible engine between two temperatures sets the upper bound; an irreversible one performs worse.

This separation explains why the first law does not imply the second. An imagined perfect converter of heat from a single reservoir could conserve total energy and still violate the required directionality. Conversely, observing heat from friction does not prove that every reverse conversion can occur without additional change. Absolute thermodynamic temperature gives the reversible comparison an operational scale independent of any particular thermometer substance.

Thomson's contribution was a developing reconciliation, not the immediate acceptance of a finished package. Experimental work, revised theoretical commitments and collaboration and disagreement with contemporaries all mattered. The resulting laws harmonized work-heat equivalence with limits on cyclic conversion while preserving the useful core, rather than every premise, of Carnot's argument.^[ref][ref]

Equipartition gives mean energy $k_{B}T/2$ for each independent quadratic term in a classical canonical Hamiltonian. For example, $p^2/(2m)$ and $k{x}^2/2$ each contribute this amount in a harmonic oscillator. It is a thermal average, not an instantaneous equality; constraints, nonquadratic terms or quantum level spacings large compared with $k_{B}T$ require a different calculation.

The fluctuation-dissipation theorem connects equilibrium fluctuations with the dissipative part of linear response. If an applied force couples as $-f(t)x$, use response to $e^{-i\omega t}$ and define the two-sided angular-frequency spectrum $S_{xx}(\omega )=\int e^{i\omega t}⟨\delta x(t)\delta x(0)⟩dt$. The classical relation isFor a viscously damped particle, $m\dot{v}=-\zeta v+\xi (t)$, the corresponding white-noise force satisfies $⟨\xi (t)\xi (t^{\prime })⟩=2\zeta k_{B}T\delta (t-t^{\prime })$. Noise replenishes energy lost by friction; their strengths cannot be varied independently at fixed equilibrium temperature.

Electrically, the one-sided voltage-noise spectrum per hertz of an equilibrium resistor is $4k_{B}TR$. A capacitor in parallel filters it toThe same equilibrium voltage variance follows from equipartition. Increasing R makes the spectrum taller and narrower without changing its integrated area. This thermodynamic connection between dissipative conductance and thermal electrical fluctuations is the content of the classical noise calculation.^[ref]

The $31.7^{\circ }$ result predicts initial failure in a uniform material under the assumed principal-stress orientations. The actual Figure 3 is a present-day NW–SE cross section through a thick, heterogeneous Alpine stack, extending to about 50 km depth. It contains near-horizontal basal stretches, curved ramps, tightly bent contacts, and steep structures in the internal zones around Viso and the Dora Maira complex. It does not show a single undeformed conjugate pair newly formed in horizontal uniform rock.

Several processes explain the departures:

Finally, a section shows apparent dip unless it is perpendicular to fault strike. For an ideal plane, $\tan {\delta }_{\mathrm{app}}=\tan {\delta }_{\mathrm{true}}\sin \alpha$, where $\alpha$ is the angle between the section direction and fault strike. Obliquity can make a plane look shallower, but cannot on its own explain an apparent dip steeper than its true dip. Do not infer that each black curve is a currently active brittle earthquake fault.

A modern teaching experiment slows gravity-driven motion enough to resolve it with a photogate. PASCO scientific's Inclined Plane Dynamics Set, ME-6966, Instruction Manual 012-10874A describes “Acceleration Down an Incline” on printed pages 15–16. Its apparatus is a GOcar on a PAStrack with an adjustable incline support, a bracket-mounted photogate, a cart-mounted picket fence, and an interface with recording software. For the interface method, the fence's $1\mathrm{cm}$ pattern is aligned to interrupt the beam. Recording starts before release from rest; the cart is caught at the bottom bumper. The procedure specifies three runs per angle, starting at $14^{\circ }$ and reducing by $2^{\circ }$, with the release position fixed. The readout is velocity against time; its slope estimates acceleration. Mean accelerations are plotted against $\sin \theta$, testing the ideal slope $g$. An added-mass comparison is also requested. These are protocol settings, not observed results.^[ref]

The cart undergoes rolling without slipping at its wheels, whereas the worked block undergoes contact slip. Even ideal wheel moment of inertia changes the cart prediction. For total cart mass $M$, wheel radii $r_i$ and axial moments of inertia $I_i$, rigid wheels with fixed axles and no losses giveTo derive it, combine translational kinetic energy $M{v}^2/2$ with wheel rotational kinetic energy ${\sum }_i{I}_i(v/r_i{)}^2/2$, then equate their rate of change to the release of gravitational potential energy $Mg\sin \theta v$. This is the multiwheel counterpart of rolling acceleration with rotational inertia. Neglecting wheel moment of inertia yields $a\simeq g\sin \theta$; treating a cart as a solid rolling cylinder would give a different correction.

Useful additional controls are repeated releases without a push, checking the fence alignment and spacing, and checking that the recorded velocity trace is straight over the fitted interval. A gently moving cart on a level track can expose losses or unintended tilt. These checks help distinguish an angle or timing error from neglected wheel moment of inertia, bearing friction, or drag force. The three-run spread measures repeatability but cannot by itself reveal a shared calibration error. The inspected protocol makes the gravity component measurable; it does not measure the sliding block's coefficient of kinetic friction.

In a functional model, a driven piston reduces the chamber's volume. A surrounding cylinder carries the wall stress; a deformable packing or gasket closes a potential leak path around a moving or removable closure. A sample and a separate pressure sensor occupy the chamber. A transmitting liquid can load the sample on all sides; the sensor's signal passes to an external readout. This is a division of experimental functions, not a scale reconstruction of one historical machine.

For a slowly moving frictionless drive piston of effective surface area $A$, external pressure $p_{\mathrm{ext}}$ and inward applied force $F$, Newton's second law at mechanical balance givesThus $20\mathrm{kN}$ on a stipulated $20\mathrm{m}{\mathrm{m}}^2$ effective surface area corresponds ideally to $1\mathrm{GPa}$ of differential pressure. These are illustrative numbers, not apparatus specifications. The effective surface area is the one associated with the pressure difference, not necessarily an external ram's visible cross section. With resisting friction $F_f$ during inward motion, the balance becomes $p-p_{\mathrm{ext}}=(F-F_f)/A$. During outward motion the sign of the friction contribution reverses. An applied-load reading can therefore show hysteresis even when the chamber's actual pressure is well defined.

A resistance pressure gauge instead supplies an internal signal. In a generic electrical readout, measure an electric potential difference $V_e$ with a known electric current $I$ and use Ohm's law to infer electrical resistance $R=V_e/I$. This describes a possible readout principle; the lecture passage cited below identifies the sensor material and calibration landmarks without specifying this wiring.

A quasistatic process is an ideal limit in which a changing thermodynamic system passes through a sequence of states arbitrarily close to thermodynamic equilibrium. Its intermediate pressure, temperature and volume are therefore well defined and obey its equation of state. Driving must be slow compared with the system's own equilibration, rather than merely taking a long time on a clock. A reversible thermodynamic process additionally allows both system and surroundings to be restored without net change; friction can produce entropy even in a quasistatic process.^[ref]

For a computed example, take a fixed amount of substance $n$ of ideal gas in a sealed cylinder with a frictionless piston, in contact with a heat bath at absolute temperature $T$. Reduce the external load in infinitesimal steps so that the piston expands from $V_0$ to $2V_0$. In the reversible limit, the gas pressure exceeds the external pressure by an infinitesimal amount during expansion, and the heat transfer occurs across an infinitesimal temperature difference. Exact mechanical equilibrium would give no driving force; the limiting imbalance supplies motion while preserving equilibrium intermediate states.^[ref]

Along this reversible isothermal expansion, the ideal gas law givesHere $R$ is the molar gas constant; specifying $nRT$ fixes an energy scale, without choosing $n$ and $T$ separately. The initial and final equilibrium states are $(V_0,p_0,T)$ and $(2V_0,p_0/2,T)$. Count work delivered to the surroundings as positive. The reversible work is the shaded area beneath the pressure-volume curve:The integral of $1/V$ supplies the natural logarithm. Since internal energy of an ideal gas depends only on temperature, $\Delta U=0$; the first law of thermodynamics, $\Delta U=Q_{\mathrm{in}}-W_{\mathrm{out}}$, gives $Q_{\mathrm{in},\mathrm{rev}}=0.693147\mathrm{kJ}$ to the displayed precision. The heat bath replaces the energy delivered as work.^[ref]

Now start from the same initial state, but abruptly lower the external pressure to the constant value $p_{\mathrm{ext}}=p_0/2=0.5\mathrm{MPa}$. There is a finite initial pressure imbalance, so this is an irreversible expansion rather than a quasistatic process. Assume the piston begins and ends at rest, has no net change in stored energy, and has no external friction; relaxation occurs within the gas and its heat bath. After motion has settled and the heat bath has restored temperature $T$, mechanical equilibrium and the ideal gas law give $V_f=nRT/p_{\mathrm{ext}}=2V_0$. The work delivered to the constant load isThis work follows directly from load force times piston displacement, using $dV=Adx$ for piston area $A$. Intermediate gas states need not have a single equilibrium pressure or temperature. In particular, neither $p(V)=nRT/V$ at the bath temperature nor $p(V)=p_{\mathrm{ext}}$ describes the actual gas during this irreversible motion. The rectangle below plots the known external load; its upper edge is not an equilibrium gas path.

The entropy accounting makes the difference precise. Because entropy is a state function, both expansions have the same gas entropy change, computed along the reversible isothermal expansion:The thermal reservoir stays at $T$ and loses heat $Q_{\mathrm{in}}=W_{\mathrm{out}}$, so $\Delta S_{\mathrm{bath}}=-W_{\mathrm{out}}/T$. Assume the ideal load stores work without an entropy change. The combined entropy production is thereforeNo numerical entropy unit is implied by $nRT=1\mathrm{kJ}$ alone: $nR$ or $T$ must also be specified. The lost work opportunity is $T\Delta S_{\mathrm{total}}=0.193147\mathrm{kJ}$ for this comparison. This calculation uses equilibrium endpoints and the reservoir's temperature, without assigning equilibrium intermediate states to the irreversible gas. ^[ref]

For another boundary choice, let the same ideal gas expand into an initially evacuated compartment of volume $V_0$ inside a rigid, insulated vessel of total volume $2V_0$. Removing the partition causes free expansion: no load is moved and no heat crosses the outer boundary, so $W_{\mathrm{out}}=Q_{\mathrm{in}}=0$ and $\Delta U=0$. After equilibration the temperature is again $T$, the pressure is $p_0/2$, and the gas entropy has increased by $nR\ln 2$. The outer vessel has fixed volume; the gas's accessible volume doubles. Restoring the initial gas state requires compression and removal of heat, changing the surroundings. Equal endpoints do not make free expansion reversible. ^[ref]

A slowly moving piston with dry friction supplies a different caution: its gas can remain arbitrarily close to thermodynamic equilibrium, yet rubbing still converts work into heat and produces entropy. If the frictional force has a nonzero slow-motion limit, its dissipated energy over a fixed travel does not vanish with speed. Reversing the motion reverses friction, so infinitesimally changing the external load cannot retrace the same operation and restore the surroundings. The heat capacity calculations additionally exclude friction and other dissipative work.^[ref]

The usefulness of a quasistatic process is that it replaces detailed motion by a readable sequence of equilibrium states. An equation of state and an integral then turn a pressure-volume drawing into an energy budget. The beautiful distinction is that internal energy and entropy depend on the endpoints, while heat and work record how the change was performed. Reversible benchmarks expose how much work an irreversible device sacrifices, while entropy production explains why simply making it slower need not recover that opportunity.

A resistance pressure gauge is a pressure sensor whose electrical resistance is calibrated against pressure. Even a simple local response needs both pressure and temperature. For an original illustrative model definewhere $R_0$ is the reference electrical resistance at $p=0$, $T=T_0$, and zero history offset. The derivatives $\alpha$ and $\beta$ are local sensitivities with units $\mathrm{P}{\mathrm{a}}^{-1}$ and ${\mathrm{K}}^{-1}$, respectively; $h$ is a dimensionless history offset. Constant sensitivities are an approximation over a chosen range, not material constants valid at every pressure.

Choose synthetic coefficients $\alpha =0.020\mathrm{GP}{\mathrm{a}}^{-1}$ and $\beta =2.0\times 10^{-4}{\mathrm{K}}^{-1}$, with $T-T_0=10\mathrm{K}$ and $h=0$. A measured ratio $R/R_0=1.022$ then gives $p=(0.022-0.002)/0.020=1.00\mathrm{GPa}$. Ignoring the temperature change gives $1.10\mathrm{GPa}$, a $0.10\mathrm{GPa}$ error. This model is deliberately stipulated; these numbers are not measured manganin coefficients or Bridgman's data.

If two histories differ by $\Delta h=0.0004$, their outputs at the same pressure differ by $0.0004R_0$; interpreting both with a single history-free curve gives an apparent pressure separation $\Delta h/\alpha =0.020\mathrm{GPa}$. The half separation is $0.010\mathrm{GPa}$. Reversing the drive can additionally cause mechanical friction hysteresis, independently of this sensor model. Record both the electrical calibration history and the loading direction.

For independent small standard measurement uncertainties $u_y,u_T,u_h$ and exact stipulated $\alpha ,\beta$, first-order propagation givesFor $u_y=10^{-4}$, $u_T=0.5\mathrm{K}$, and $u_h=10^{-4}$, this is $u_p=0.00866\mathrm{GPa}$. Real calibration also includes coefficient measurement uncertainties, correlations, reference-point measurement uncertainty, and any unresolved nonhydrostatic loading. A sensor calibrates its local environment; it cannot by itself certify the absence of deviatoric stress throughout the sample.

The motivating observation is a bob going around at almost constant height while its string traces a cone. The University of Iowa's demonstration places a camera above a short pendulum to reveal this motion. ^[ref] The following proposed measurement procedure develops that observation into a test of the conical pendulum model; it is not a report of collected data.

Suspend a compact bob from a rigid, fixed support using a light string, and measure $\ell$ from the pivot to the bob's centre. Put a stationary marker beside the orbit as a timing reference. Displace the bob sideways and give it a horizontal push tangent to the intended circle, then let go. For a chosen starting angle $\theta$, the predicted initial speed is $v=\sqrt{g\ell {\sin }^2\theta /\cos \theta }$. Adjust the launch between trials until a short observation interval has nearly constant orbit radius and height. Keep the pivot fixed while recording: circling the support by hand changes the problem, and a maintained drive would introduce another force.

Record from above with the camera looking approximately along the vertical axis. Check that the bob's centre follows a circle and that successive returns to the marker take nearly equal times. A separate side recording checks whether the bob's height stays constant. A markedly elliptical orbit or visibly rising and falling bob is not the uniform circular motion assumed by the conical pendulum equations. Measure the orbit diameter with a scale at the orbit's height; in a level side view, use the two extreme horizontal positions, not one arbitrary frame. Halve the diameter to obtain $r$, or measure the vertical drop $h$ directly in a calibrated side view. An oblique camera view and a scale at a different depth distort apparent lengths, so arrange the geometry before interpreting the images.

Time $N$ complete returns to the same marker in the same direction and estimate $P=\Delta t/N$. Counting ten revolutions, when the orbit stays sufficiently steady, spreads an endpoint timing error over ten periods. Repeat the timing and compare the first and last few revolutions to detect drift; using a longer interval helps only if the motion remains close to uniform. From independently measured geometry, compute $\sin \theta =r/\ell$ or $\cos \theta =h/\ell$, and check $r^2+h^2\simeq {\ell }^2$ if both were measured. Then compare the measured oscillation period with $P^2=4{\pi }^{2}h/g$. MIT's 8.01X experimental assignment explicitly asks students to compare the conical pendulum's period, length, and orbit radius with theory. ^[ref]

Repeat at several launch radii with the same $\ell$. Larger $r$ means smaller $h=\sqrt{{\ell }^2-r^2}$, so the ideal oscillation period decreases. A plot of measured $P^2$ against independently measured $h$ should follow a straight line through the origin with slope $4{\pi }^2/g$; conversely, $g=4{\pi }^{2}h/P^2$ estimates gravitational acceleration. Do not infer $h$ from the timing equation and then use that same equation as a test. Compare discrepancies with the spread of repeated timings and the resolution of the length measurements. String stretch, a moving pivot, and appreciable bob size challenge the stated geometry; drag force and pivot friction challenge the assumption of undriven uniform motion. Observe a short interval and record any shrinking or drifting orbit rather than silently treating it as an exact conical pendulum.

For an illustrative calculation, assume $\ell =1.00\mathrm{m}$, $m=0.050\mathrm{kg}$, $g=9.81m/{\mathrm{s}}^2$, and $\theta =40^{\circ }$. The model gives $r\approx 0.643\mathrm{m}$, $h\approx 0.766\mathrm{m}$, $P\approx 1.756\mathrm{s}$, $v\approx 2.300m/s$, and $T\approx 0.640\mathrm{N}$. Its vertical tension component is $0.4905\mathrm{N}$, equal to the gravitational weight; its inward component is approximately $0.412\mathrm{N}$. These are predictions from chosen parameters, not measured data or measurements of local $g$. The next figure uses the same stipulated $\ell$ and $g$, including illustrative angles $15^{\circ }$, $40^{\circ }$, and $60^{\circ }$.

The Technische Universität Darmstadt demonstration uses a steel ball launched outward from the centre of a rotating disc. A fixed lecture-room camera records a straight path, while a camera carried above the disc records a curved path. The instructions specify launching toward the fixed camera so that the laboratory motion is clearly visible. The rotating iPod transmits its view wirelessly through Apple TV; two vertical support rods of length $1\mathrm{m}$ are specified for the slow-motion arrangement. The paired video views are the readout: one physical motion is represented in two reference frames.^[ref]

A useful preparation control appears in the Universität Freiburg instructions: with the disc stationary, check that the ball travels radially toward its edge. A camera is mounted on supports attached to the disc; turn the disc by hand as the ball leaves its centre and observe the camera view.^[ref]

The inference is a dependence of the observed trajectory on the rotating axes; a curved camera trace alone does not isolate the Coriolis force. To compare with the constant-angular velocity free particle in rotating coordinates, measure the disc angular velocity, calibrate lengths in the camera image, and compare launches at rest and with opposite rotation directions. These are checks of the ideal model, not reported measurements from the manuals. A rolling ball can exchange forces with the surface through friction, and hand rotation need not have constant angular velocity; an Euler acceleration then enters too. The ideal free-ball drawing deliberately omits those effects. Neither manual supplies a numerical uncertainty or a precision measurement of the Coriolis acceleration.

PASCO's PASPORT Rotary Motion Sensor Experiment Guide, Experiment 1, specifies a PS-2120A sensor, a rod carrying two equal-radius weights, a three-step drive wheel, a thread and hanging mass, a clamp-on guide wheel, a support, a balance and calipers. The thread leaves the drive wheel tangentially. The sensor records angular velocity against time; a linear regression over the accelerating interval supplies angular acceleration $\alpha$. The guide measures the drive radius with calipers and repeats the run with the weights removed to subtract the apparatus's own moment of inertia. Its tables are blank: it prescribes a measurement and does not report an observed numerical agreement. ^[ref]

With hanging mass $m_h$ including its hanger, effective drive radius $b$, no thread slip and negligible guide-wheel inertia, Newton's second law gives thread tension $F=m_h(g-b\alpha )$. The axial torque is $bF$. Under negligible friction, the inference is thereforeRepeat the inference for the unloaded assembly and subtract: $I_{\mathrm{weights}}=I_{\mathrm{loaded}}-I_{\mathrm{empty}}$. Omitting the $m_h{b}^2$ term incorrectly treats thread tension as the hanging weight. Comparing with $(M_1+M_2)R^2$ tests the point mass approximation; finite-sized weights also have their own moments of inertia about their centres of mass. ^[ref]

Useful proposed checks go beyond that ideal inference. Verify the rotary encoder's angle scale by turning the shaft through a marked full revolution; check the recorded time against an independent clock and the balance against a known mass. Measure $b$ at the thread's running surface, including thread thickness; avoid overlapping wraps that change $b$. Release without a push, inspect the angular velocity record for a straight interval, and stop before the hanger lands or the thread runs out. Repeat with several hanging masses, weight radii and rotation directions. A consistent inferred moment of inertia across drive torques supports the model; curvature, direction dependence or drift indicates a limitation rather than a new principal moment of inertia.

For approximately constant resisting torque ${\tau }_f\ge 0$ in one rotation direction, an original calibration model isFit the left-hand quantity against $\alpha$ across several drives: the slope estimates the axial moment of inertia and the intercept estimates ${\tau }_f$. This calibration is conditional on friction remaining constant over the fitted angular speeds; a freely slowing run can expose speed dependence. Sensor differentiation noise, guide-wheel moment of inertia, thread slip, flexing connectors and uncertain weight radii limit the inference. Removing weights can also change bearing loads, so empty-apparatus subtraction does not automatically cancel friction. These are proposed controls and a model extension, not results claimed by PASCO.

A fixed-axis readout measures ${\mathbf{n}}^T{\mathbf{I}}_O\mathbf{n}$, not the full inertia tensor. In a proposed generalization, repeat about $x$, $y$, $z$ and the three diagonal directions $({\mathbf{e}}_x+{\mathbf{e}}_y)/\sqrt{2}$, $({\mathbf{e}}_x+{\mathbf{e}}_z)/\sqrt{2}$, $({\mathbf{e}}_y+{\mathbf{e}}_z)/\sqrt{2}$, all through the same $O$. For example,Six independent scalar readings then determine the six entries of a symmetric matrix. Each mounting needs its own apparatus correction, and different reference points require the tensor parallel-axis theorem before the readings can be combined.

Rotational kinetic energy is the kinetic energy of a rigid body's rotation. For pure rotation about an axis fixed in an inertial frame,Here $r_{\perp }$ is the perpendicular distance of a mass element $dm$ from that axis, $I$ is the moment of inertia about that same axis, and $\omega$ is the common angular speed, measured in radians per second. The rigid body assumption fixes the relative positions of the mass elements; they share angular velocity but have different speeds $v=\omega r_{\perp }$. Summing their kinetic energies $dm{v}^2/2$ gives the formula. This is a nonrelativistic model; deformation or internal motion can contribute additional kinetic energy.^[ref]

For a computed example, take a uniform disc and an ideal thin ring, each of mass $M=2\mathrm{kg}$ and radius $R=0.30\mathrm{m}$, rotating about its central perpendicular axis at angular speed $10rad/s$. The moment of inertia of a uniform disc is $M{R}^2/2=0.090\mathrm{kg}{\mathrm{m}}^2$, while the ring has moment of inertia $M{R}^2=0.180\mathrm{kg}{\mathrm{m}}^2$. Their rotational kinetic energies are therefore $4.5\mathrm{J}$ and $9.0\mathrm{J}$. The ring stores twice the energy at the same angular speed, because more mass lies farther from the axis. At the same nonzero rotational kinetic energy, its angular speed would instead be $1/\sqrt{2}$ times the disc's. This makes radial placement useful for storing energy in a rotating wheel, although a real design must also withstand the stresses of rotation.

For fixed-axis rotation with constant moment of inertia, write the signed rotation rate as $\Omega =\dot{\theta }$, with angle $\theta$ in radians. The axial component of net external torque obeys ${\tau }_{\mathrm{axis}}=I\dot{\Omega }$, so the work-energy theorem becomesAn opposing torque removes rotational kinetic energy; bearing friction must be included in the net torque. The signed $\Omega$ matters when the direction reverses. Rotation about a fixed axis need not be about a principal axis: the scalar energy formula still holds, but transverse bearing torques can be needed to maintain that axis.

For general rigid body motion, use the inertia tensor ${\mathsf{I}}_G$ about the centre of mass $G$:This is the kinetic-energy decomposition about the centre of mass. In a principal body frame, the rotational term is $\frac{1}{2}(I_1{\omega }_1^2+I_2{\omega }_2^2+I_3{\omega }_3^2)$, using the principal moments of inertia. Away from a principal axis, angular momentum and angular velocity need not be parallel. A freely rotating asymmetric rigid body can change the direction of angular velocity relative to the body even while rotational kinetic energy remains constant; Euler equations for a torque-free rigid body govern that motion. One constant axial moment of inertia is then insufficient.^[ref]

A rough contact may transmit both normal and tangential impulses. Conservation of linear and angular momentum fixes totals but does not specify their division between translation and rotation. Predicting outgoing states requires constitutive data such as normal restitution and a friction or tangential restitution law. Rough contact is not synonymous with adhesive contact.

The Rüchardt experiment infers heat capacity ratio from the period of an oscillation of a piston supported by trapped gas. Use a vertical precision cylinder with a close-fitting movable piston, known moving mass $M$, face area $A$, trapped equilibrium volume $V_0$, and local atmospheric pressure $p_{\mathrm{atm}}$. With negligible static friction, force balance gives the absolute gas pressure $p_0=p_{\mathrm{atm}}+Mg/A$, where $g$ is gravitational acceleration. Displace the piston slightly and release it: compression raises pressure, and expansion lowers it, producing a restoring force. The oscillation is a mechanical probe of thermal response, without separately measuring two heat inputs. ^[ref]

Let $x$ be upward displacement from equilibrium, so $V=V_0+Ax$. For constant heat capacity ratio and a reversible adiabatic process, the net upward force, including the ambient load and weight, isThe linear approximation requires $|Ax|/V_0\ll 1$. Applying Newton's second law gives simple harmonic motion withHere ${\omega }_0$ is the undamped angular frequency and $\mathcal{T}$ the undamped oscillation period, rather than temperature. Record repeated cycles, measure $M,A,V_0,p_{\mathrm{atm}}$ independently, and include all connected trapped volume. An illustrative model with $M=0.020\mathrm{kg}$, $A=2.0\times 10^{-4}{\mathrm{m}}^2$, $V_0=0.010{\mathrm{m}}^3$, $p_{\mathrm{atm}}=100000\mathrm{Pa}$, $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$ and $\gamma =1.4$ gives $p_0=100981\mathrm{Pa}$ and $\mathcal{T}\simeq 1.182\mathrm{s}$. These stipulated values are not measurements.

In Shearwood and Sloan's implementation, piston-mounted magnets induce an electromotive force in a stationary coil through Faraday's law of induction; an oscilloscope records the signal. They observed faster oscillations at smaller trapped volume values and stronger damping at small cylinder volume values. The signal tracks velocity locally, so its amplitude is not directly piston displacement. Calibration also identified inlet dead space beyond the graduated cylinder volume. ^[ref]

Repeat at smaller initial amplitudes to test the linear approximation, and compare several $V_0$ values. Heat transfer weakens the restoring force toward the isothermal process limit, biasing the inference toward $1$. Check leakage through resting-piston drift and before-and-after trapped volume readings. For viscous damping with envelope $e^{-t/\tau }$, the measured angular frequency obeys ${\omega }_0^2={\omega }_d^2+{\tau }^{-2}$; this correction does not remove thermal leakage. Internal equilibration must be faster than an oscillation period, with heat transfer and leakage slow over a run. Large amplitude or sticking friction require a richer model. ^[ref]

For an isolated system, the second law requires $\Delta S\ge 0$. More generally, include all affected surroundings to obtainEquality describes the reversible limit; irreversible macroscopic processes produce entropy. A closed system can lose entropy by exporting heat, so nondecrease does not apply to that system alone. The Clausius statement of the second law forbids a cyclic device whose sole effect is transferring heat from cold to hot. The Kelvin-Planck statement of the second law forbids a cyclic device whose sole effect is converting heat from one thermal reservoir completely into work. "Sole effect" matters: a refrigerator moves heat from cold to hot by consuming work. ^[ref]

For direct transfer of $Q=1000\mathrm{J}$ from a thermal reservoir at $600\mathrm{K}$ to one at $300\mathrm{K}$, with no work extracted,The reverse transfer alone would have negative total entropy change. Both directions conserve energy, but only the hot-to-cold direction is spontaneous. The finite-body example under entropy replaces these constant temperatures by changing ones.

Friction, free expansion and heat transfer across a finite temperature difference all introduce irreversibility. Moving arbitrarily slowly is insufficient if friction remains. A reversible thermodynamic process must permit restoration of both the system and surroundings. This gives engineers a useful ideal benchmark: irreversibility spends an opportunity to obtain work, even though energy itself is conserved. ^[ref]