For this gyrocompass, let $\mathbf{n},\mathbf{w},\mathbf{z}$ be local north, west and upward unit vectors, forming a right-handed basis. Earth's rotation has local componentsLet the spin axis point along $\mathbf{u}=\cos \theta \mathbf{n}+\sin \theta \mathbf{w}$, and write its dominant angular momentum as $J_s\mathbf{u}$, where $J_s=I_1\omega >0$. In mechanical equilibrium relative to Earth, this leading angular momentum is fixed in the terrestrial rotating reference frame. The bearings can exert horizontal torques to keep the axle horizontal, but free rotation about the vertical implies $G_z=0$. The vertical component of the rotating-frame derivative formula therefore requiresAway from either geographical pole, $\cos \lambda \ne 0$, henceThis is a necessary equilibrium condition; the next part distinguishes its stable and unstable spin orientations. The rapid-spin approximation neglects angular momentum terms of order $I_2{\Omega }_E$ compared with $I_1\omega$. At either pole the horizontal component of Earth's rotation vanishes, and this north-seeking mechanism gives no preferred horizontal direction. The paper explicitly excludes $90^{\circ }$; if latitude is signed, $-90^{\circ }$ must likewise be excluded.

One application is the slow steady precession of a rapidly spinning gyroscope supported at one point. Let its centre of mass be a distance $\ell$ from the pivot along its spin axis, at angle $\alpha$ to the upward vertical. Write $J\simeq I_1\omega$ for its dominant spin angular momentum. In a reference frame precessing steadily about the vertical with angular velocity ${\mathbf{\Omega }}_p$, the spin angular momentum is approximately stationary, soThe gravitational torque has magnitude $mg\ell \sin \alpha$. The cross product on the right has magnitude ${\Omega }_{p}J\sin \alpha$, giving, for a nonvertical axis,Its direction follows from the torque equation. This is the slow precession approximation: the additional angular momentum associated with precession is small compared with the spin contribution.

Use orthonormal basis vectors ${\mathbf{e}}_i(t)$ fixed in the rotating reference frame, and expand the angular momentum as $\mathbf{J}={\sum }_i{J}_i{\mathbf{e}}_i$. Over time $dt$, the basis undergoes infinitesimal rotation $\mathbf{\Omega }dt$, soThe product rule gives the rotating-frame derivative formula:Since the applied couple is the rate of change of angular momentum in the inertial frame,The second term accounts for the changing basis, even when the components of angular momentum are constant in the rotating reference frame.

Using $Z_{\mathrm{rot}}\simeq CT=C/(k_B\beta )$ givesThe internal-energy identity from (b), followed by differentiation with respect to temperature at fixed moment of inertia, yieldsFor one mole of independent rotors, the rotational energy is $RT$ and the rotational molar heat capacity is $R\simeq 8.31\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$.

This agrees with the equipartition theorem. The given spectrum describes a linear rotor with two independent rotational degrees of freedom, each contributing $k_{B}T/2$ to the high-temperature mean energy. Rotation about the molecular axis is not an additional degree of freedom of this ideal linear-rotor model, so the result is $k_B$ rather than $3k_B/2$ per rotor. These are rotational contributions alone; translational and vibrational heat capacities have not been included. At low temperature the discrete excitation gap freezes rotational excitation out, so the rotational heat capacity tends to zero instead of remaining at its classical value.

At low temperature, $T\ll {\theta }_r$, retain the ground level and the first excited level. Their energies are $0$ and ${\hslash }^2/I$, with degeneracies $1$ and $3$. Thereforeso in the notation of the question,$A$ is dimensionless and $B$ has units of kelvins. The first omitted level has three times the first excitation energy, explaining the exponent in the error term.

At high temperature, $T\gg {\theta }_r$, many rotational levels are occupied and the sum can be replaced to leading order by an integral:Set $u=J(J+1)$, so that $du=(2J+1)dJ$. ThenThusThe high-temperature approximation here concerns the rotational spectrum. For a real molecule it should still be used below temperatures where neglected vibrational excitation or dissociation invalidates the rigid-rotor model.

For the supplied linear rigid rotor spectrum, sum over every nonnegative integer $J$ and include all $2J+1$ magnetic substates:It is convenient to introduce the rotational temperatureso that the first terms are $Z_{\mathrm{rot}}=1+3e^{-2{\theta }_r/T}+5e^{-6{\theta }_r/T}+\cdots$. Here $I$ is the moment of inertia and $\hslash$ is the reduced Planck constant, as confirmed in the original PDF. The stipulated degeneracies are used as given, without adding nuclear-spin weights or omitting selected $J$ values.

The internal energy is the canonical mean of the energy:Holding the external mechanical parameters fixed, so that the spectrum and degeneracies do not change when differentiating with respect to $\beta$, givesComparison yieldsAn explicitly temperature-dependent effective Hamiltonian would require care: differentiating its partition sum would also differentiate the assigned energy levels. The stated result uses the fixed physical spectrum of the canonical system.

The expression is the canonical partition function for a system in thermal equilibrium with a large heat bath at fixed temperature $T$. The system exchanges energy with the bath, while its particle number and external mechanical parameters, such as volume, are fixed. The bath is large enough that its temperature is not appreciably changed by this exchange.

The system has allowed energy levels ${ϵ}_j$. The index $j$ labels distinct energy values, and the integer degeneracy factor $g_j$ counts distinct quantum states with that same energy. $k_B$ is the Boltzmann constant, and $\exp [-{ϵ}_j/(k_{B}T)]$ is a state's Boltzmann factor. The dimensionless sum $Z$ normalizes the probabilities:Thus the degeneracy multiplies the probability of occupying the level, not the probability assigned to each individual state within it.

The canonical model is not restricted to a classical ideal gas. It applies to a quantum rotor, for example, provided the sum counts all of that system's allowed states with their proper degeneracies. For a many-particle system, ${ϵ}_j$ would be the energy of a many-particle state, with any quantum-statistical restrictions already built into the state count.

The phase diagram must contain three coexistence curves meeting at the triple point: a solid-gas sublimation curve, a solid-liquid melting curve and a liquid-gas vaporization curve. The vaporization curve ends at a liquid-gas critical point; beyond it liquid and gas are continuously connected as a supercritical fluid.

The imposed contraction on freezing means $v_{m,s}<v_{m,l}$. Melting therefore has $\Delta v_m=v_{m,l}-v_{m,s}>0$ and $L_{\mathrm{fusion}}>0$. The Clapeyron equation then requiresThus raising the pressure raises the melting temperature. The liquid-gas and solid-gas slopes are also positive because the transformations into gas increase both entropy and volume. The liquid region lies between the positively sloping melting line and the vaporization curve, while gas occupies the low-pressure side. The figure is a qualitative solution sketch, not a reproduction or fit of the paper's vapor-pressure measurements; its axes are schematic.

Taking a derivative of the logarithmic form gives a direct way to use the plot's base-ten vertical axis:Alternatively, use two nearby readings and eliminate the unknown $p_0$:Reading the original PDF graph approximately gives ${\log }_{10}(p/\mathrm{Pa})\simeq -0.6$ at $300\mathrm{K}$ and $\simeq 3.8$ at $500\mathrm{K}$. These bracket the requested temperature and give an average latent heat near $400\mathrm{K}$:Thus a suitable graph-based estimate isThe graph does not warrant several significant figures; a local tangent or different reasonable nearby readings give a comparable estimate. The factor $\ln 10$ is essential when using a base-ten logarithmic axis.

For liquid-vapor equilibrium, use molar quantities. At low vapor pressure the gas is dilute, approximately ideal, and has a much larger molar volume than the liquid. HenceSubstitution into the Clapeyron equation givesIf the molar latent heat varies sufficiently slowly to be treated as constant over the temperature interval, integration givesThe positive constant $p_0$ has pressure units and is an integration constant. It is not necessarily a real infinite-temperature pressure, since the approximations fail before that limit.

For mercury between $300$ and $500\mathrm{K}$, the original plot places this range above the triple point and below the normal boiling and critical temperatures. Its vapor pressure is low relative to the critical pressure, making a dilute-gas description reasonable; the vapor's molar volume greatly exceeds that of the dense liquid. The interval is also far from the critical region where $L$ tends to zero, so a roughly constant latent heat is a useful first approximation. The approximations are not exact: Kirchhoff's law of thermochemistry gives, at fixed pressure, the temperature derivative of a transformation enthalpy as the difference between the two molar heat capacities, so a precise fit over a wide interval would allow $L$ to vary.

At coexistence, the two phases of a pure substance have equal chemical potential, or equivalently equal molar Gibbs free energy:For either phase, $d\mu =-s_{m}dT+v_{m}dp$. Move an infinitesimal distance along the coexistence curve, where the equality must remain true. Subtracting the two differentials givesA reversible transformation at this common temperature absorbs latent heat $L=T\Delta s_m$. ThusThe derivation assumes thermodynamic equilibrium between two bulk phases of a single pure substance, with the same temperature and pressure, and only ordinary pressure-volume work. Surface-curvature shifts, additional external-field work and composition changes are omitted. It applies to a first-order phase boundary where the latent heat and volume discontinuity are defined. No ideal-gas assumption and no constant-$L$ assumption are needed for this differential relation. At the critical endpoint both discontinuities vanish, so their ratio requires a limiting treatment rather than direct substitution of two zeros.

The relation describes the slope of the phase coexistence curve of a pure substance in the pressure-temperature plane:Here $p(T)$ is the common equilibrium pressure of the two phases, $T$ is their common absolute temperature in kelvins, and the derivative is taken along their coexistence curve. Choose a direction of transformation from phase 1 to phase 2. Then $L$ is the reversible latent heat for that transformation and $\Delta V=V_2-V_1$ is the corresponding volume change, using the same amount of substance for both quantities.

In the molar convention used below, $L=h_{m,2}-h_{m,1}$ has units $\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}$ and $\Delta V=v_{m,2}-v_{m,1}$ has units ${\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$, so the right-hand side has units $\mathrm{Pa}{\mathrm{K}}^{-1}$. For vaporization, both $L$ and $\Delta V$ are positive. The displayed general relation is also called the Clapeyron equation; the ideal-vapor approximation used in (c) gives its usual Clausius form.

The Magnus effect is the transverse force produced when a spinning body moves relative to a fluid. Rotation changes the flow and circulation around the body; the resulting asymmetric pressure and wake deflect the fluid, with the opposite transverse force on the body. Examples include the curved flight of spinning balls and the lift of rotating-cylinder sails.

A useful quantitative model is steady two-dimensional incompressible flow past a cylinder of radius $a$, with fluid velocity $\mathbf{U}=U\hat{\mathbf{x}}$ far away. Let circulation $\Gamma$ be positive anticlockwise as viewed from $+z$. The potential-flow velocity at the cylinder surface isThe Bernoulli equation outside the boundary layer givesIntegrating the pressure force around the cylinder, per unit span, yieldsThus the Kutta–Joukowski theorem givesThe sign here refers to the fluid velocity relative to the cylinder. For a body moving through otherwise stationary fluid, that relative fluid velocity points opposite to the body's translational velocity.

Real viscosity and the no-slip boundary condition transmit the surface rotation to the fluid and help set the circulation and separation pattern. Pure inviscid flow with only an impermeability condition does not determine $\Gamma$ from the cylinder's angular speed. Therefore the circulation must not be assigned a universal spin-rate formula without additional assumptions. The cylinder result illustrates the lift mechanism; quantitative forces on a finite spinning ball also depend on its geometry, Reynolds number, rotation rate, surface roughness and wake.

For a slender elastic beam, the Euler-Bernoulli beam equation assumes small strain, small deflection slopes and cross-sections that remain plane and normal to the deflected centreline. A fibre at signed distance $y$ from the neutral axis has longitudinal strain ${ϵ}_{xx}=-y\kappa$, where $\kappa$ is the centreline curvature. Hooke's law gives ${\sigma }_{xx}=-Ey\kappa$, with Young's modulus $E$.

Choose the internal bending-moment sign so that $M=-\int y{\sigma }_{xx}dA$. ThenHere $I_A$ is the second moment of area, with units ${\mathrm{m}}^4$, not the mass moment of inertia of a rotating body. For a uniform material under pure bending, zero axial resultant places the neutral axis through the section's centroid.

For small slopes, $\kappa \simeq w^{\prime \prime }(x)$. Combining the moment-curvature relation with transverse force and moment balance gives, with a consistent load sign convention,For constant stiffness this becomes $E{I}_A{w}^{(4)}=q$. The bending stiffness $E{I}_A$ has units $\mathrm{N}{\mathrm{m}}^2$. Boundary conditions specify displacement and slope at a clamped end, or bending moment and shear force at a free end.

For a cantilever of length $\ell$ carrying a transverse end load $F$, take deflection positive towards the load. The moment magnitude is $M(x)=F(\ell -x)$, and integration with $w(0)=w^{\prime }(0)=0$ givesThe free-end moment vanishes and the tip shear supplies the applied load. For a rectangular section of width $b$ and bending thickness $h$, $I_A=b{h}^3/12$, so increasing thickness strongly increases bending resistance. The model ceases to be reliable for large deflections, plastic strains or short thick beams where shear deformation matters.

Lagrangian mechanics expresses dynamics through generalized coordinates $q_i$ that respect the constraints. For a system with conservative, velocity-independent forces and standard kinetic energy, the Lagrangian is $L=T-V$. Its equations follow from the principle of stationary action:for variations that vanish at the endpoints. Varying the integral and integrating the $\delta {\dot{q}}_i$ term by parts givesIndependent variations therefore yield the Euler-Lagrange equations:Additional nonconservative generalized forces appear on the right-hand side as $Q_i$. The action is stationary, not necessarily a minimum. For ideal holonomic constraints, using suitable generalized coordinates removes the need to solve explicitly for constraint forces.

The conjugate momentum is $p_i=\partial L/\partial {\dot{q}}_i$. If a coordinate is cyclic, $\partial L/\partial q_i=0$ and its conjugate momentum is conserved. If $L$ has no explicit time dependence, the energy functionis conserved. It equals $T+V$ for the usual quadratic kinetic energy with time-independent geometry, but that identification is not universal for every Lagrangian or constraint.

For example, a simple pendulum with angle $\theta$, length $\ell$ and bob mass $m$ hasThe tension does no virtual work in the allowed angular displacement and never needs to be calculated to obtain the angular equation. The method scales naturally to coupled systems and coordinates with complicated geometry.

For torque-free precession of a symmetric top, take principal moments of inertia $I_1=I_2=I_{\perp }$ and $I_3=I_{\parallel }$, with no external torque. In an inertial frame moving with its centre of mass, its angular momentum $\mathbf{L}$ is constant in an inertial frame, and its rotational kinetic energy is conserved.

In the principal body axes, Euler equations for a torque-free rigid body reduce toTherefore ${\omega }_3$ is constant andThus the transverse angular velocity rotates uniformly in the body frame; its signed rate reverses between oblate and prolate inertia tensors. This body-frame wobble rate should be distinguished from the space-frame precession of the symmetry axis.

Let ${\mathbf{e}}_3$ be the body's symmetry axis. Sinceits angle $\theta$ to $\mathbf{L}$ is constant, with $\cos \theta =I_{\parallel }{\omega }_3/L$, andHence the symmetry axis describes a cone about fixed $\mathbf{L}$ at the space-frame precession rateThere is no nutation of this angle for torque-free motion of an exactly symmetric top. In the usual Euler angles, ${\omega }_3=\dot{\psi }+\dot{\varphi }\cos \theta$, so $\dot{\psi }=-{\Omega }_b$. Both rates describe the same motion in different axes. A freely tossed axisymmetric disc illustrates the effect; a top supported at a point under gravity is a different, externally torqued problem.

Changing shape changes the demagnetizing field, and hence the magnetization induced by the same applied magnetic field intensity. For a small linear sample in the ellipsoid approximation, the force magnitude scales withFor $\mu >1$, a more elongated body aligned with the field has smaller $N_d$ and a stronger response. A sphere has $N_d=1/3$, giving ${\chi }_{\mathrm{eff}}=3(\mu -1)/(\mu +2)$; a very thin plate with field normal to it has $N_d\to 1$, giving $(\mu -1)/\mu$. Thus an axially slender rod generally experiences a stronger attraction than a sphere or a transverse thin plate of the same material and volume, within the linear uniform-applied-field approximation.

The shape response is precisely why the $1/\mu$ factor in part (c) cannot apply universally to every small magnetic body. Orientation also matters through the appropriate principal demagnetizing factor; for general nonellipsoidal bodies the internal field is not uniform and a single scalar correction need not be exact.

For a linear induced response, reversing $I$ reverses $H_0$ and the sample magnetization, while also reversing the field gradient. Their product in $F_z={\mu }_{0}V{\chi }_{\mathrm{eff}}{H}_0{H}_0^{\prime }$ therefore keeps the same sign and magnitude:This is also explicit in the $I^2$ dependence in part (d), for both the shape-consistent and printed alternatives. The conclusion assumes no independent permanent moment, remanence or hysteresis; those would require a different response model.

In the small-body approximation, write ${\chi }_{\mathrm{eff}}=\chi /(1+N_d\chi )$, with $\chi =\mu -1$. The induced magnetic dipole moment is $\mathbf{m}=VM\hat{\mathbf{z}}=V{\chi }_{\mathrm{eff}}{H}_0\hat{\mathbf{z}}$. Its effective interaction potential energy at fixed source current isThe factor $1/2$ accounts for inducing the moment: differentiating $-m(z)B_0(z)$ as if the moment were permanent would double the force. Equivalently, the instantaneous dipole force is $F_z=md{B}_0/dz$. Thus the force on a small linearly magnetizable body isFrom part (b), $d{H}_0/dz=-3a^{2}Iz/[2(a^2+z^2{)}^{5/2}]$. HenceFor the ideal slender rod, ${\chi }_{\mathrm{eff}}\simeq \mu -1$. If the source's printed magnetisation is imposed instead, its corresponding conditional answer isFor $\mu >1$ and $z>0$, both expressions point toward the loop, where the applied magnetic field is stronger; for $z<0$ the direction reverses toward the loop plane. A material with diamagnetism, ${\chi }_{\mathrm{eff}}<0$, reverses this attraction. The force is in newtons, since a magnetic dipole moment times a field gradient has units $\mathrm{A}{\mathrm{m}}^2\mathrm{T}{\mathrm{m}}^{-1}$. Both formulas assume linear response, negligible saturation and approximately uniform applied field across the sample.