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With gauge pressure $p_g$ referenced to atmospheric pressure $p_{\mathrm{atm}}$, absolute pressure is $p=p_g+p_{\mathrm{atm}}$. Thus Boyle's law becomesEven a constant atmospheric pressure cannot be dropped from these sums. For the calculated $100\mathrm{kPa}$ to $200\mathrm{kPa}$ example, take $p_{\mathrm{atm}}=100\mathrm{kPa}$. The initial and final gauge pressures are $0$ and $100\mathrm{kPa}$; their products with the respective volume are $0$ and $5000\mathrm{kPa}\mathrm{c}{\mathrm{m}}^3$, which are unequal. Zero gauge pressure means balance with the surroundings, not absence of gas or absence of wall momentum transfer.

For a frictionless piston of surface area $A$ with its outer face exposed to atmospheric pressure, the additional inward force required for mechanical equilibrium is $F=(p-p_{\mathrm{atm}})A=p_{g}A$. This is why a mechanical load can reveal a pressure difference even though the equation of state requires absolute pressure. With piston friction, the load alone no longer determines $p_{g}A$; a direct pressure sensor avoids that particular inference.

Adiabatic compression decreases the volume of a sample without heat transfer across its boundary. For the fixed-composition ideal gas and endpoint assumptions in adiabatic process, compression work increases internal energy; positive heat capacity at constant volume then means a higher final temperature. The effect comes from supplied work, even though no heat enters. The amount of temperature increase depends on the work actually delivered to the gas, not on the adiabatic process condition alone.

The reversible ideal-gas adiabat supplies a particularly useful quantitative benchmark. Assume a fixed amount of substance $n$, a quasistatic process, negligible friction, only pressure-volume work, and constant heat capacity ratio $\gamma =C_p/C_V>1$. Here $C_p$ and $C_V$ are the whole sample's heat capacity at constant pressure and heat capacity at constant volume. With $R$ the molar gas constant, Mayer's relation gives $C_V=nR/(\gamma -1)$. The proof outline is $C_{V}dT=-pdV$ from the first law of thermodynamics, followed by $p=nRT/V$ from the ideal gas law and integration. The result isThese finite power laws belong to this reversible, constant-$\gamma$ ideal gas model. If heat capacity ratio varies, use the integral relation in heat capacity ratio; if the compression is irreversible, use the actual work balance, as in irreversible adiabatic piston compression. A short stroke by itself justifies neither power law. ^[ref]

Writing $r=V_1/V_2>1$ for the compression ratio, the reversible model predictsAll temperatures in these ratios are absolute. The larger pressure rise than in an isothermal compression, where $p_2/p_1=r$, explains the greater local stiffness described by isentropic compressibility.

For an original stipulated example, set $\gamma =7/5$, $p_1{V}_1=nR{T}_1=1\mathrm{kJ}$ and $r=2$. Then $T_2/T_1=2^{0.4}\simeq 1.31951$ and $p_2/p_1=2^{1.4}\simeq 2.63902$. The input work and internal energy increase are both $0.79877\mathrm{kJ}$, while $Q_{\mathrm{in}}=0$. No initial temperature or apparatus performance is implied by the chosen energy scale. Independent integration gives the signed outward workwhich is negative during compression and agrees with $-\Delta U$. This integral uses equilibrium gas pressure only because the path is reversible.

In the sensor experiment of adiabatic process, compare the pressure-volume integral with $C_V\Delta T$ as an energy check only when internal pressure is sufficiently uniform and piston dynamics permit that boundary-work estimate. Account for friction separately before comparing it with hand input work. PASCO's manual supplies a compression procedure and an illustrative logarithmic-slope result $1.39$, compared with the model value $1.40$ for a diatomic gas; that is its sample result, not new measurements here. ^[ref]

Adiabatic expansion increases the accessible volume without heat transfer. In a loaded piston expansion of the fixed-composition ideal gas considered in adiabatic process, positive outward work comes from internal energy, so the final temperature decreases. This provides a cooling mechanism without removing heat during the stroke. In the Carnot cycle, reversible adiabatic expansion connects two different temperatures between its isothermal processes.^[ref]

For the same constant-$\gamma$ reversible ideal-gas adiabat, put $b=V_2/V_1>1$. The scaling and first law of thermodynamics giveWith the figure's stipulated $\gamma =1.4$, $p_1{V}_1=1\mathrm{kJ}$ and $b=2$, the final ratios are $T_2/T_1\simeq 0.75786$ and $p_2/p_1\simeq 0.37893$. Outward work is $0.60535\mathrm{kJ}$ and $\Delta U=-0.60535\mathrm{kJ}$. Reversing this particular path from its cooled endpoint would require the same work input; the figure's separate compression starts at the original state, so its input is different. Endpoint entropy supplies another independent check:The two logarithm terms cancel using $C_V(\gamma -1)=nR$, confirming the reversible path is an isentropic process. ^[ref]

PASCO's expansion demonstration first compresses trapped air and waits for thermal equilibrium before the rapid expansion. Its manual warns that friction-warmed cylinder regions contact the gas during expansion, compromising quantitative comparison. ^[ref] The hold separates cooling caused by expansion from simply undoing the earlier hot compression. As proposed controls, observe the temperature dip and subsequent warming, compare rapid and slow strokes from matched initial states, and assess wall heat transfer and probe lag before fitting a reversible model.

Cooling is not a universal consequence of adiabatic expansion: free expansion of an ideal gas has no outward work and returns to the same equilibrium temperature. Its entropy increases, so the reversible power laws do not describe it. Reuse free expansion, Joule's free-expansion experiment and temperature sensitivity of a free-expansion experiment for that apparatus, its energy budget and the real-gas qualifications. The useful distinction is the load: expansion cools this ideal gas when it must spend internal energy doing work.

An adiabatic process has no heat transfer across the chosen boundary throughout the process: $\delta Q=0$, rather than merely zero net heat after incoming and outgoing transfers cancel. It is a condition on energy transfer, whereas an isothermal process holds temperature constant. A sealed cylinder can approach an adiabatic process through insulation or through motion short compared with the time needed for appreciable heat transfer. Its temperature can nevertheless change because work crosses the boundary.^[ref]

Take the trapped gas as a closed system, with fixed amount of substance and composition and no net change in bulk kinetic energy or potential energy between the endpoints. Define $Q_{\mathrm{in}}>0$ as heat entering and $W_{\mathrm{out}}>0$ as work delivered to the surroundings. The first law of thermodynamics givesHere $U$ is internal energy. An adiabatic compression normally supplies work, so $W_{\mathrm{out}}<0$; an adiabatic expansion against a resisting load delivers work, so $W_{\mathrm{out}}>0$. An adiabatic process need not preserve internal energy and need not make the closed system an isolated system. ^[ref]

For a reversible thermodynamic process, $dS=\delta Q_{\mathrm{rev}}/T$ at positive absolute temperature $T$. A reversible adiabatic process therefore has $dS=0$: it is an isentropic process. Irreversible adiabatic processes can instead generate entropy; free expansion is the existing counterexample, with its calculation in entropy change in free expansion. Conversely, constant entropy alone does not certify an adiabatic process or reversibility: outward heat can export the entropy generated internally. For a closed system the macroscopic entropy balance iswhere $T_b$ is the absolute boundary temperature where each heat transfer occurs and $S_{\mathrm{gen}}$ is internal entropy production. With no heat transfer, this reduces to $\Delta S=S_{\mathrm{gen}}$; reversibility makes $S_{\mathrm{gen}}=0$. ^[ref]

A motivating experiment measures the gas temperature directly during compression and expansion, instead of inferring it from how warm a cylinder feels. PASCO's authored TD-8565 manual describes a sealed piston cylinder, a fine nickel-wire thermometer, an absolute pressure sensor and a position readout for volume. A computer records all three channels together. The wire's electrical resistance changes with temperature; its small size permits a rapid response. ^[ref]

Use ordinary trapped air and the apparatus's designed travel stops; mount the base securely, close the valve during a stroke, and record the actual volume limits. John Hart's university demonstration uses a large transparent cylinder with a low-heat capacity thermocouple, a pressure sensor, a travel gauge and a safety stop. It explicitly keeps the release valve closed during the demonstration. The basic observable is a rise in temperature on compression and a fall on loaded expansion, alongside the corresponding pressure and volume changes. Hart discusses wall thermal conduction, sensor response and leakage as reasons real records differ from the reversible model. ^[ref]

For a proposed rapid-versus-slow comparison, let the air and walls return to the same initial thermal equilibrium before each run, then compare the same starting and finishing volume values with different stroke durations. Read temperature, absolute pressure and position continuously, including the hold after motion stops. A rapid compression gives a temporary temperature rise; holding the smaller volume lets heat leave and the temperature approach the surroundings again. Slow compression in good thermal contact allows that heat transfer during the stroke and approaches an isothermal compression. Loaded expansion reverses the signs: the rapid stroke cools, while heat entering during slow motion reduces the cooling. PASCO explicitly supports rapid nearly adiabatic processes and slow isothermal processes; the matched-run controls here are a proposed extension, not a reported set of measurements. ^[ref]

Insulation suppresses thermal conduction; it does not prove that heat transfer vanishes. As proposed controls, repeat the matched strokes with and without a removable insulating jacket, record room and wall temperatures, measure fixed-position drift, and check valve closure and seal leakage. Calibrate the thermometer and position readout independently, including any connected dead volume; account for the heat capacity of the probe rather than treating it as a passive observer. Hart explicitly identifies the small air sample relative to the probe and walls, sensor inertia and seal leakage as limitations. A fast display cannot eliminate those physical effects. ^[ref]

There are two distinct timing requirements. Let ${\tau }_{\mathrm{eq}}$ describe internal equilibration, ${\tau }_{\mathrm{probe}}$ the thermometer response, $t_{\mathrm{move}}$ the stroke duration and ${\tau }_{\mathrm{heat}}$ the effective heat transfer time constant. A useful sufficient window for a resolved, nearly reversible adiabatic process iswith negligible friction. The left inequality permits a quasistatic process and a usable readout; the right limits heat leakage. A violent stroke may satisfy only the right inequality. PASCO reports a roughly $30$–$50\mathrm{ms}$ lag of its temperature channel behind the other channels, so apparent simultaneous samples require care. If no such window exists, analyze equilibrium endpoints with the first law of thermodynamics, and regard the path as approximate. ^[ref]

The diagram's model is deliberately a hold after the stroke, when boundary work is zero. For constant whole-sample heat capacity at constant volume $C_V$ and effective thermal conductance $K>0$, it assumesThis is Newton's law of cooling with one uniform gas temperature and a fixed surrounding temperature; actual wall and probe dynamics can require more than one time constant. At $t/{\tau }_{\mathrm{heat}}=0.1$, the fraction of the initial temperature anomaly erased is $1-e^{-0.1}\simeq 0.0952$; at one time constant it is about $0.6321$. These are analytical fractions, not experimental tolerances or evidence that the preceding stroke was reversible. Measuring the hold helps estimate heat leakage independently of the motion.

The usefulness of an adiabatic process is that it separates mechanical work from heat transfer. Its reversible limit connects temperature changes, mechanical stiffness and entropy in one model: Rüchardt experiment and Clément-Desormes experiment already use that connection to measure heat capacity ratio. The contrast with free expansion explains the deeper point: suppressing heat does not determine how much work is exchanged or how much entropy is produced.

An air track supports a cart on a cushion of air to reduce contact friction. It provides a nearly one-dimensional platform for a damped harmonic oscillator; residual drag remains part of the baseline damping.

An actual teaching experiment in Arbor Scientific's The Level Air Track uses one glider, an air track, two photogates, and a ruler. Turn on the blower and place the glider near the middle. Adjust the feet until it does not drift when released. Put the photogates at least half a track length apart and align them with the glider's flag. Measure the length $L$ that interrupts the beams. Give a gentle push, let go, and record each beam-blocking time $\Delta t_i$. The readout is the pair of speeds $s_i=L/\Delta t_i$. The worksheet asks how those speeds compare, which forces remain after the push, and which interaction the air track minimizes.^[ref]

The motivating contrast is between a brief push and the subsequent glide. The air cushion reduces contact friction; it does not remove the downward gravitational weight or the upward support force. Similar successive speeds after release are consistent with a small horizontal net external force, rather than with a continuing push. The experiment approximates the zero net external force model over a finite track and observation interval.

To strengthen the documented procedure, use independent measurement calibration: check the flag length and each timer against a known length and time reference; measuring $L$ sets the length scale but does not calibrate the timer. Repeat launches in both directions without contact during the timed passage, and check for drift at several track locations. These are additional controls, not reported steps or measured results in the worksheet. A residual inclined plane produces a downhill component of gravitational weight with the same laboratory direction on both runs; friction and drag force oppose the direction of travel. Reversing the launch can therefore help distinguish these effects.

A photogate directly measures a beam-blocking time; $L/\Delta t_i$ is an average speed during that passage, not an exact instantaneous velocity. Assign the direction separately. Equal readings at two gates do not prove zero acceleration everywhere between them: opposing changes could cancel. Measurement uncertainty from timer resolution, an incorrectly measured flag length, residual slope, airflow disturbances, and remaining friction or drag force limit the inference. No measured glider speeds are assumed here.

Consider an ideal pressure-driven closure plug with a large chamber-facing projected area $A_p$. Its opposite side seats against a smaller annular packing surface area $A_{\mathrm{seal}}$. The remaining projected part $A_u=A_p-A_{\mathrm{seal}}>0$ is the unsupported area: it has no opposing packing contact in that plane. Here $p$ is the chamber pressure relative to the exterior. Let $q$ denote the mean axial compressive contact stress on the packing, not the chamber's hydrostatic stress.

Neglect friction, the plug's weight, and other axial reactions. The plug's Newton's second law in static balance isThe load is the same on both sides; its smaller supporting surface area raises the packing's mean contact stress. For a stipulated ratio $A_p/A_{\mathrm{seal}}=1.25$ and $p=1\mathrm{GPa}$, the ideal seating stress is $q=1.25\mathrm{GPa}$. Its excess over chamber pressure is $0.25\mathrm{GPa}$. Reducing $A_u$ to zero removes this amplification. This is the elementary force balance behind self-energizing seating, consistent with the lecture's Figure 1 caption.^[ref]

The relation is a model, not a universal sealing law. The mean axial contact stress need not equal the local contact stress along every possible leakage path. Confinement and plasticity in continuum mechanics determine whether the packing closes a gap or instead extrudes into it. Friction diverts load into other surfaces; material plasticity in continuum mechanics changes contact geometry; the vessel must carry its own stress. A finite initial seating load may also be needed before chamber pressure can develop. An arbitrarily small $A_{\mathrm{seal}}$ does not yield arbitrarily useful sealing: the assumptions fail when local deformation, extrusion or structural failure takes over.

The Chézy law relates mean open-channel speed to hydraulic radius and energy slope through a friction coefficient. Equivalently one can use bed shear ${\tau }_b=\rho fu|u|$; for positive uniform flow, balance $\rho gAI={\tau }_b{P}_b$ gives $u=\sqrt{gI/f}\sqrt{A/P_b}$. The absolute value makes drag oppose motion if reverse flows are admitted.

The dimensionless ratio between an appropriate tangential friction force and normal load in a specified contact model. Static limiting and sliding coefficients differ, and nanoscale adhesion can invalidate a simple load-independent ratio.

A conical pendulum is a pendulum whose bob executes uniform horizontal circular motion beneath a fixed pivot, with its string at a constant angle $\theta$ to the downward vertical. The string sweeps a right circular conical surface; the bob follows its circular rim, without any physical cone supporting it. The ideal model has a point mass $m>0$, a taut, massless, inextensible string of length $\ell >0$, uniform gravitational acceleration of magnitude $g>0$, and negligible drag force and pivot friction. A nonzero circular motion requires $0<\theta <\pi /2$. This geometry and the corresponding forces are described in Richard Fitzpatrick, The conical pendulum, equations 262–267.

The right triangle formed by the pivot, the orbit centre, and the bob gives $r=\ell \sin \theta$ and $h=\ell \cos \theta$. Here $r$ is the orbit radius and $h$ is the vertical drop from the pivot to the bob's horizontal plane, not the bob's height above a table. Write $T$ for tension, $P$ for the oscillation period of one revolution, and $\omega =2\pi /P$ for the angular speed; then the bob's speed is $v=\omega r$. A side view compresses the horizontal circle to a line segment; the top view reveals the circle and its tangent velocity.

The two actual forces on the bob are tension toward the pivot and gravitational weight vertically downward. The vertical acceleration is zero, while the horizontal centripetal acceleration is $v^2/r={\omega }^{2}r$. Thus Newton's second law givesThe inward net force is the horizontal component of tension, rather than an extra interaction to add to these two forces. Constant speed still requires acceleration because the velocity changes direction. ^[ref]

Dividing the two component equations and using $r=\ell \sin \theta$ gives the useful collection of predictionsIf $\varphi$ is the azimuthal angle, then $\dot{\varphi }=\pm \omega$; either direction of rotation satisfies ${\dot{\varphi }}^2\cos \theta =g/\ell$. The mass cancels from the angular speed and oscillation period, but not from the tension. The motion is a relative equilibrium: in a rotating frame following the bob at angular speed $\omega$, its position is fixed. This is a concrete way to see how rotational symmetry can turn an apparently moving solution into a stationary one.

Use local east, north, and up unit vectors $({\mathbf{e}}_E,{\mathbf{e}}_N,{\mathbf{e}}_U)$, with ${\mathbf{e}}_E\times {\mathbf{e}}_N={\mathbf{e}}_U$. In a spherical-Earth approximation at geographical latitude $\phi$, the angular velocity has components $\mathbf{\Omega }=(0,\Omega \cos \phi ,\Omega \sin \phi )$. Expanding the cross product for relative velocity $(v_E,v_N,v_U)$ givesThese are Coriolis acceleration components, not the total acceleration. For horizontal motion, the horizontal part is $(f{v}_N,-f{v}_E)$, where the Coriolis parameter is $f=2\Omega \sin \phi$. Thus northward motion in the northern hemisphere has an eastward Coriolis acceleration, to the right of its velocity; in the southern hemisphere the horizontal turning is to the left. At the equator $f=0$, but the full Coriolis acceleration need not vanish: vertical or eastward motion still couples through $\Omega \cos \phi$.

For a numerical example, take the Earth rotation rate $\Omega =7.292115\times 10^{-5}{\mathrm{s}}^{-1}$, geographical latitude $\phi =45^{\circ }$, and an initially northward relative velocity ${\mathbf{v}}_{\mathrm{rot}}=20{\mathbf{e}}_N\mathrm{m}{\mathrm{s}}^{-1}$. Then

For a local, short-time estimate with no other horizontal force and zero initial eastward velocity, hold the initial northward velocity in the leading term: $\Delta x_E\simeq \frac{1}{2}fu{t}^2$. At $t=10\mathrm{s}$ this gives about $0.1031\mathrm{m}$ eastward. It assumes nearly fixed local axes and geographical latitude, negligible changes of height, and $|ft|\ll 1$. It is not a complete wind trajectory: pressure gradients, friction, and evolving velocity must enter an atmospheric equation of motion. The effective gravity in a rotating frame already combines ordinary gravity with the steady terrestrial centrifugal acceleration.

A damped pendulum is a simple pendulum whose mechanical energy is removed by a resisting torque. Here the model is undriven, with a fixed pivot, a point mass $m>0$ at distance $\ell >0$, uniform gravitational acceleration $g>0$, and resisting torque $-bd\theta /dt$ with constant $b>0$. The angle $\theta$ is measured from the downward vertical in radians. An ideal viscous rotary dashpot realizes this torque law; tangential linear drag $-\alpha v$ on the bob gives the same equation with $b=\alpha {\ell }^2$. The moment of inertia is $m{\ell }^2$, soHere $t$ is physical time, ${\omega }_0$ is the undamped small-amplitude angular frequency, and $\gamma$ has units of inverse time. In the normalized equation $\ddot{\theta }+c\dot{\theta }+\sin \theta =0$, dots mean differentiation with respect to $\tau$, not $t$. The parameter $c$ is dimensionless. This separates the chosen damping law from the small-angle approximation: retaining $\sin \theta$ allows finite angles even when the resisting torque is linear in angular velocity.

To motivate the model, release a pendulum without a push and record how its swing shrinks. A published example by Gintautas and Hubler uses a rotary sensor built from a computer mouse to measure the frequency and damping coefficient of a physical pendulum. ^[ref] The following is a proposed experiment with a chosen viscous rotary dashpot; the plotted observations are predictions, not a reproduction of their measurements.

Mount a compact bob on a light rigid rod with a fixed pivot, couple the pivot to an adjustable viscous rotary dashpot, and measure $\ell$ from the pivot to the bob's centre. First record decay with the dashpot disconnected, then repeat with it connected at several settings; this establishes how much decay is already due to other losses. Keep the bob, rod, camera, and release angle unchanged between these trials. The simple model requires rod and damper-rotor moments of inertia to be negligible compared with $m{\ell }^2$. Otherwise measure the total moment of inertia $I$ and the distance $d$ to the assembly's centre of mass: the physical pendulum equation becomes $I{d}^2\theta /d{t}^2+bd\theta /dt+Mgd\sin \theta =0$, where $M$ is the total moving mass. A damper that also exerts an elastic restoring torque would require another term.

Place a camera perpendicular to the swing plane, include the pivot and bob centre in every frame, and determine their coordinates relative to the resting vertical. For horizontal and downward separations $x,y$, use $\theta =atan2(x,y)$; record the frame timestamps rather than assuming a frame rate. Release from rest at about $10^{\circ }$, record through several complete swings, and repeat the release. For every same-side turning point, tabulate its time $t_n$ and angular amplitude $A_n=|\theta (t_n)|$. Measure the oscillation period from differences $t_{n+1}-t_n$; comparing several periods exposes drift. Interpolating around each observed extremum reduces frame discretization error. Check that late peaks still exceed the angular resolution; taking the natural logarithm of tiny, uncertain peaks amplifies their relative errors.

In the small-angle approximation, the damped pendulum becomes a damped harmonic oscillator. For $0<\gamma <2{\omega }_0$ and release from rest at ${\theta }_0$, its exact linear-model solution isThe decaying sinusoidal envelope has coefficient $|{\theta }_0|{\omega }_0/{\omega }_d$; because release is from rest, its successive same-side extrema occur at $t_n=n{T}_d$ and have $A_n=|{\theta }_0|e^{-\gamma n{T}_d/2}$. Thus the envelope and the sequence of actual peak heights are different descriptions of the same trace. The linear damping regimes and this solution are derived in Richard Fitzpatrick, Damped Oscillatory Motion, equations 82–93.

The experiment tests two predictions independently: nearly equal same-side peak spacing and a straight line when plotting $\ln (A_n/A_0)$ against $t_n-t_0$. Its predicted slope is $-\gamma /2$, givingUse several peaks and repeated trials rather than a single ratio, and compare the frequency identity with independently measured $\ell$ and $g$. For the illustrative choices $\ell =1.00\mathrm{m}$, $g=9.81m/{\mathrm{s}}^2$, $c=0.20$, and ${\theta }_0=10^{\circ }$, the linear model gives $\gamma \approx 0.6264{\mathrm{s}}^{-1}$, $T_d\approx 2.0162\mathrm{s}$, and $A_{n+1}/A_n\approx 0.5318$. The first two positive peaks after release are therefore approximately $5.318^{\circ }$ and $2.828^{\circ }$. The undamped small-amplitude oscillation period is about $2.0061\mathrm{s}$: substantial loss per swing need not imply a large change in period.

Repeat at a larger release angle to distinguish two limitations. The relative error in replacing $\sin \theta$ by $\theta$ is about ${\theta }^2/6$ for small angles: about $0.51\%$ at $10^{\circ }$, but about $17.3\%$ at $60^{\circ }$ when compared with $\theta$. Finite-amplitude swing spacing can therefore change as the swing shrinks even if $b$ is constant. Separately, friction at the pivot or quadratic drag can invalidate the assumed linear resisting torque; exponential peak decay should be checked, not imposed on all recorded motion. Peters discusses why real pendulum damping can require nonlinear terms. ^[ref] Repeatability, residual pivot motion, changes in damper temperature, and finite camera resolution are useful checks on this proposed measurement.

The normalized equilibrium points are $(\theta ,\dot{\theta })=(k\pi ,0)$. Even multiples of $\pi$ are attracting: their linearization has eigenvalues $(-c\pm \sqrt{c^2-4})/2$, giving a stable focus for $0<c<2$, repeated negative eigenvalues at $c=2$, and a stable node for $c>2$. These are the underdamped, critically damped, and overdamped small-displacement regimes. Odd multiples of $\pi$ have eigenvalues $(-c\pm \sqrt{c^2+4})/2$ of opposite signs and are saddle equilibria. The energy dissipation of a damped pendulum explains why trajectories in the phase portrait spiral inward near a stable focus instead of closing into the recurring orbit of an undamped simple pendulum.

A double pendulum consists of two pendula coupled because the second pivot moves with the first mass. It has two degrees of freedom: even when the upper pivot is fixed, moving the upper arm carries the lower pivot along. This makes the double pendulum a compact example of how mechanical constraints couple motion, and how Lagrangian mechanics converts geometry into dynamics.

Use two equal point masses $m>0$ and two massless rigid rods of equal length $\ell >0$, with joints that allow free rotation in one vertical plane. The upper pivot is fixed; gravity is uniform, with gravitational acceleration $g>0$. Neglect joint friction and drag force. Both angles ${\theta }_1,{\theta }_2$ are measured from the downward vertical, positive toward the right, in radians. In particular, ${\theta }_2$ is an absolute angle, not the bend between the rods: that bend is ${\theta }_2-{\theta }_1$.

The double pendulum Lagrangian below gives the exact dynamics of this ideal model. Near the downward stable equilibrium, normal modes of a double pendulum replace the coupled motion with two independent harmonic oscillators. Their fixed amplitude ratios explain which releases make the rods swing together and which make them swing in opposite directions.

An ideal elastic collision between a sphere and a stationary plane reverses normal approach velocity. On the fully rough, tangentially elastic branch, the tangential contact-point velocity reverses too. If $K_t$ is the signed tangential impulse, translation and spin obey $M\Delta v=K_t$ and $I\Delta \Omega =a{K}_t$, with positive spin anticlockwise. Restoring tangential translational plus rotational kinetic energy givesThe nonzero root is the displayed impulse; the other root is the smooth-contact branch. For $I=2M{a}^2/5$, the map is $v^{\prime }=(3v-4a\Omega )/7$ and $a{\Omega }^{\prime }=-(10v+3a\Omega )/7$. It preserves kinetic energy and the angular momentum about the contact point, $I\Omega -Mav$. An unspecified rough surface need not realize this idealization: finite friction or a dissipative tangential contact law requires additional constitutive data. Setting the outgoing contact slip to zero instead loses $M(v+a\Omega {)}^2/7$ of kinetic energy for this uniform sphere.

Enthalpy is the state function obtained by adding pressure times volume to internal energy:Here $U$ denotes internal energy, also written $E$ in some references. Like internal energy, enthalpy is measured in joules and is an extensive variable. Its definition packages the internal energy change and expansion work into one useful quantity for fixed-pressure calculations. It does not define a body's stored heat. ^[ref]

For a fixed amount and composition of a simple compressible system, the relation between neighboring equilibrium states isHere $T$ is absolute temperature and $S$ is entropy. Thus entropy and pressure are the natural variables of this thermodynamic potential, with partial derivatives $T=(\partial H/\partial S{)}_p$ and $V=(\partial H/\partial p{)}_S$. Fixed composition matters: changing composition adds further terms. The formula follows in one line from the equilibrium first law of thermodynamics identity:The cancellation of $pdV$ is the reason enthalpy replaces volume with pressure as a natural variable. This identity concerns equilibrium states; it does not identify $TdS$ with actual irreversible heat transfer. ^[ref]

For a closed system undergoing a quasistatic process at constant pressure, require fixed amount and composition, only pressure-volume work, no friction, and unchanged bulk kinetic energy and potential energy. Then the actual first law of thermodynamics balance givesThe symbol $Q_p$ means transferred heat for this fixed-pressure process. If the piston example in heat additionally meets these conditions, its $20\mathrm{kJ}$ of incoming heat equals $\Delta H$, while $\Delta U=15\mathrm{kJ}$ and $p\Delta V=W=5\mathrm{kJ}$. The example shows exactly what enthalpy adds to internal energy: the expansion work required at the imposed pressure.

Within a single thermodynamic phase, heat capacity at constant pressure giveswith the amount and composition fixed. Constant $C_p$ gives $\Delta H=C_p\Delta T$; dividing by mass gives specific heat capacity at constant pressure. At equilibrium melting, enthalpy can instead change through latent heat while temperature stays fixed. The sensible heat example therefore adds two single-thermodynamic phase enthalpy changes and one latent heat contribution.

The usefulness of enthalpy lies in matching the imposed constraint: a fixed-pressure experiment naturally reports $\Delta H$, while fixed-volume heating with no other work reports $\Delta U$. Neither equality is universal. Additional work or changing pressure alters the relation between supplied heat and $\Delta H$; irreversible entropy production also prevents identifying actual $\delta Q$ with $TdS$. The separate Joule-Thomson coefficient branch uses enthalpy to describe throttling under its own assumptions.

Fine-grained fault rock produced by comminution and alteration; its mineralogy and fluid state can strongly affect friction.

At the limiting release height for a solid cylinder, the loop normal force is $N=(7mg/3)(1+\cos \theta )$ and the required static-friction magnitude is $mg|\sin \theta |/3$. Thus $|f|/N=|\sin \theta |/[7(1+\cos \theta )]$ diverges near the top, even though both forces vanish there. A finite friction coefficient therefore requires a height above the ideal contact threshold to maintain no-slip rolling throughout. A question imposing rolling but giving no coefficient determines the ideal limiting bound, not a realistic friction-dependent minimum.

Why must a conducting piston move slowly? A simple model couples the first law of thermodynamics to Newton's law of cooling. Assume a fixed amount of ideal gas with uniform temperature $T(t)$ and constant whole-sample heat capacity at constant volume $C_V$, a heat bath maintained at $T_b$, negligible friction, and mechanically slow motion. Let $G>0$ be the effective thermal conductance, the reciprocal of the thermal resistance, in $\mathrm{W}{\mathrm{K}}^{-1}$. ThenPositive heat transfer must offset expansion work to keep $T$ near $T_b$. A finite $G$ and nonzero piston speed therefore require a temperature difference. This uniform-temperature model neglects spatial temperature gradients, sensor response, cylinder heat capacity and any change of $G$ with piston position; it is an illustrative calculation, not a fit to the isothermal gas syringe experiment.

For a prescribed exponential volume change $V=V_i{e}^{\lambda t}$, define $\tau =C_V/G$, $s=t/\tau$ and $ϵ=nR\lambda /G$. The dimensionless temperature $\theta =T/T_b$ obeys a linear ordinary differential equation with initial value $\theta (0)=1$:The plotted expansion uses stipulated $C_V=(5/2)nR$ and stops at $V_f=2V_i$, at $s_f=\ln 2/[(5/2)ϵ]$. During expansion, the larger $ϵ$ produces the larger temperature deficit; once the piston is held fixed, the deficit undergoes exponential decay as $e^{-(s-s_f)}$. Reducing $|ϵ|$ makes the path more nearly an isothermal process, while increasing its duration. Reversing the piston with $\lambda <0$ produces a temperature excess and outward heat transfer; the small-$|ϵ|$ regime is the intended near-isothermal approximation. Mechanical slowness alone does not ensure sufficient heat transfer.

Friction is a contact force tangent to the surfaces. Kinetic friction opposes contact slip; static friction maintains contact without contact slip, including rolling without slipping. The perpendicular contact component is the normal force. These two components answer different questions: how strongly the surfaces press together, and how strongly they resist sliding past one another.

The ideal Coulomb dry-contact model combines a bound on static friction with a sliding law for kinetic friction. Its parameters are the coefficient of static friction and the coefficient of kinetic friction; constant coefficients are an empirical approximation, rather than a consequence of Newton's laws of motion. Surface condition matters, and the static coefficient need not universally exceed the kinetic coefficient. ^[ref]

The model's historical setting was practical machinery: Coulomb's Théorie des machines simples investigated friction and the stiffness of ropes. The Académie's bibliography records its 1781 prize and its appearance in volume 10, pages 161–332, of the Recueil des savants étrangers. ^[ref]

Relative tangential motion along a contact or fault resisted by friction.