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The shell theorem gives enclosed halo mass $4\pi \rho r^3/3$. Define $\alpha =4\pi \rho /3$. Then $F_r=-GMm/r^2-\alpha Gmr$. Interior to the uniform halo its potential energy may be writtenOuter shells alter the additive constant, not this interior force. An actually infinite uniform mass distribution has no finite potential referenced to infinity; only local potential differences are needed here.

The central force has zero torque, so $\mathbf{J}=\mathbf{r}\times m\dot{\mathbf{r}}$ is conserved and the motion is planar. With $J=m{r}^2\dot{\varphi }$, the effective radial potential and equation areFor circular radius $r_0$, $J=m{r}_0^2\Omega$ givesSet $r=r_0+\xi$ at fixed $J$. The derivative of the right-hand side at equilibrium is $-3{\Omega }^2+2GM/r_0^3-\alpha G$, soBoth frequencies are positive, but radial and azimuthal motion no longer close with the Kepler ratio. At first order in eccentricity $e$, one may write $r=r_0(1-e\cos \omega t)$ and $\varphi =\Omega t+(2\Omega e/\omega )\sin \omega t$. For a weak halo this is a nearly Keplerian ellipse whose apsis rotates at angular rate $\Omega -\omega$.

Put ${\Omega }_K=\sqrt{GM/r_0^3}$ and $\epsilon =\alpha r_0^3/M\ll 1$. Expansion yieldsThe precession is retrograde. In cycles per unit time the precession frequency is $\dot{\varpi }/(2\pi )$; per radial period the apsidal shift is $2\pi (\Omega /\omega -1)\simeq -3\pi \epsilon$. This weak-halo interpretation avoids confusing a focus-centred Kepler ellipse with the centred ellipse of a purely harmonic potential.

A normal mode is a pattern in which all coordinates share one frequency and fixed relative amplitudes. For a conservative linear system, it solves a generalized eigenvalue equation with stiffness matrix $K$ and mass matrix $M$. For this free three-mass spring chain,Solving $K\mathbf{u}={\omega }^{2}M\mathbf{u}$, for $\alpha >0$, givesTheir overall normalization is arbitrary; distinct modes are orthogonal in the mass inner product. The zero mode is rigid translation because neither end is fixed. In the antisymmetric mode the middle mass stays still, so changing its mass cannot alter the frequency. The symmetric internal mode keeps the centre of mass fixed through its middle displacement $-2/\alpha$. Its frequency rises without bound as $\alpha \to 0$ and approaches $\sqrt{k/m}$ as $\alpha \to \infty$, becoming degenerate with the antisymmetric frequency in that limit.

For equal masses put ${\omega }_1=\sqrt{k/m}$ and ${\omega }_2=\sqrt{3k/m}$. Positions are equilibrium positions plus displacementsThe translational mode needs the linear-in-time term; a cosine at zero frequency alone would miss arbitrary centre-of-mass velocity.

Taking the springs initially at equilibrium, the impulse gives $\dot{\mathbf{q}}(0)=(v,0,0)=(v/3){\mathbf{u}}_0+(v/2){\mathbf{u}}_1+(v/6){\mathbf{u}}_2$. Their squared Euclidean norms are $3,2,6$, so their initial kinetic energies are $m{v}^2/6,m{v}^2/4,m{v}^2/12$. Relative to total $m{v}^2/2$,Subsequent motion conserves each mode's total energy, exchanging its kinetic and spring contributions. The zero-mode share remains centre-of-mass kinetic energy.

Use the rotating reference frame equilibrium model, in which the centrifugal potential energy of a particle of mass $m_i$ is $U_i(r)=-m_i{\omega }^2{r}^2/2$. The Boltzmann distribution then givesDefine the concentration ratio as $c=n_1/n_2$. At the axis,Thus the axis favours the lighter species. The expression assumes a dilute classical mixture at one temperature, equilibrium in the rotating reference frame, and sampling sufficiently close to $r=0$.

A thermodynamic potential packages the equilibrium state in variables suited to the imposed constraints. For a simple fluid, $U(S,V,N)$ has differential $dU=TdS-pdV+\mu dN$. Legendre transforms give $H=U+pV$, $F=U-TS$ and $G=U+pV-TS$, with natural variables of a thermodynamic potential $(S,p,N)$, $(T,V,N)$ and $(T,p,N)$ respectively. At fixed particle number, Helmholtz free energy is minimized at fixed $T,V$; Gibbs free energy at fixed $T,p$.

A flow vessel is an open control volume, even if its walls are enclosed. For each inlet, pressure forces carry flow work $pdV$ in addition to internal energy. The steady first-law balance isWith no heat exchange, shaft work, or significant kinetic and gravitational energy differences, the enthalpy fluxes are equal. Mass flow must also balance. This equality concerns mass-flow-weighted specific enthalpies, not an unweighted sum over pipes.

Use the liquefaction yield with separate warm-gas and liquid outlets. The wording specifies that the outgoing gas is at 14 K; it does not require the outgoing liquid to have that temperature. The two outlet streams need not be in mutual thermal equilibrium. Taking $H$ as the printed specific enthalpy, the steady-flow energy balance givesFor fixed outlet enthalpies the greatest yield occurs at the smallest inlet enthalpy. Among the tabulated pressures this is 30 atm. Approximating the atmospheric-pressure gas enthalpy by the tabulated zero-pressure limiting value givesThis uses a low-pressure approximation: $p_0=1$ atm is not literally the zero-pressure table entry. Linear interpolation to 1 atm would instead give $H(p_0)\simeq 8.651\times 10^4\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}$ and $\alpha \simeq 0.193$. The sparse table does not determine an exact continuous optimum between entries.

Atmospheric-pressure liquid helium must be much colder than the 14 K gas outlet: its normal boiling point is about 4.224 K.^[ref] A counterflow heat exchanger can return cold gas through the incoming stream before that gas leaves the overall control volume, while the collected liquid leaves separately at the cold end. This is a possible interpretation of the whole apparatus, rather than an extra diagram specified by the question. The enthalpy calculation needs only the stated inlet and outlet properties; it does not assume one equilibrium vessel containing both phases at 14 K. In particular, $H(p_0)-H_{\ell }$ is an outlet-stream enthalpy difference, not the latent heat of vaporization at one common temperature.

For isotropic planar motion, the planar effusion flux gives an escape rate through a gap of width $w$ is $\Gamma =w(N/\mathcal{A})⟨v⟩/\pi$, where $\mathcal{A}$ is tray area. At fixed equilibrium kinetic energy, or fixed shape of the scaled speed distribution, $v\propto m^{-1/2}$. Hence halving the mass increases the rate by $\sqrt{2}$. This planar effusion result assumes the same spatial density and an unrestricted small opening. Strictly, average energy alone fixes $⟨v^2⟩$, not $⟨v⟩$; an arbitrary change of distribution shape would need more information.

In irrotational flow, the velocity has a velocity potential locally, and globally in a simply connected flow domain. incompressibility then gives $\nabla \cdot \mathbf{v}={\nabla }^2\Phi =0$. For a sphere moving at speed $U$ along $z$ through fluid at rest far away, the potential of a translating sphere is a decaying axisymmetric dipole. Its solution and its impermeability condition giveThese are laboratory fluid velocities expressed relative to a moving origin. A velocity field measured in the sphere-rest frame instead requires adding $-Ur\cos \theta$ to the potential.

The vector identity printed in the PDF has its signs reversed. The product rule gives the consistent identity $v^2=\nabla \cdot (\Phi \mathbf{v})-\Phi \nabla \cdot \mathbf{v}$. With outward normal of the exterior fluid equal to $-\hat{r}$ at the sphere, the divergence theorem yieldsThe integral at infinity vanishes. HenceThis added mass of a sphere is half the displaced fluid mass. For a spring of unchanged stiffness, buoyancy only shifts the static equilibrium. Its oscillation frequency with fluid added mass isFor a small-angle simple pendulum, buoyancy also reduces restoring torque from $mg\ell \theta$ to $(m-{\rho }_{0}V)g\ell \theta$, while the inertia is $(m+m_a){\ell }^2$. ThusThe downward equilibrium is stable only for $\rho >{\rho }_0$. The ideal results assume unbounded fluid, negligible viscosity, small excursions, and no important free-surface or container corrections.

For two bodies interacting through $U({\mathbf{r}}_1-{\mathbf{r}}_2)$, defineTheir kinetic energy separates exactly:Without external forces, the centre of mass moves uniformly while $\mu \ddot{\mathbf{r}}=-\nabla U$. The relative orbit is a one-body problem with reduced mass $\mu$, not generally either individual mass. For $m_2\gg m_1$, $\mu \simeq m_1$; for equal masses $m$, $\mu =m/2$. If the interaction is central, relative angular momentum is conserved and the relative motion is planar.

A central potential conserves angular momentum and confines a nonradial orbit to a plane. With reduced mass $\mu$, the radial motion obeysTurning points satisfy $E=U_{\mathrm{eff}}$. A radially bounded allowed interval gives bound motion, whereas an interval reaching infinity permits escape. The energy sign alone classifies these only with suitable assumptions about $U(\infty )$ and barriers. Circular orbits satisfy $U_{\mathrm{eff}}^{\prime }=0$ and are radially stable when $U_{\mathrm{eff}}^{\prime \prime }>0$. Bound orbits in a general central potential need not close: successive periapses can precess.

For attractive inverse-square gravity, $U=-\kappa /r$ andEllipses have $E<0$, $e<1$, and the attracting centre at a focus; parabolic and hyperbolic escape have $e=1$ and $e>1$. Conservation of angular momentum gives equal swept areas in equal times. The inverse-square law additionally yields $T^2=4{\pi }^2{a}^3/[G(m_1+m_2)]$. These are Kepler's laws; a harmonic central potential is another example with closed bound orbits, while generic central laws produce precession.

In a gravitational assist, a spacecraft's asymptotic speed relative to a planet is approximately unchanged but its direction turns. Transforming to the frame in which the planet has velocity $\mathbf{V}$ givesThe spacecraft can gain or lose orbital energy; the planet supplies the compensating energy and momentum, with a very small velocity change because of its large mass.

Repulsive Coulomb scattering instead has $U={\kappa }_C/r$. For incident relative kinetic energy $E$ and impact parameter $b$, the hyperbolic orbit givesThis Rutherford scattering formula follows by equating the incident annulus $2\pi bdb$ to the scattered solid-angle ring. It assumes classical elastic scattering by a point Coulomb centre; screening, finite nuclear size or quantum effects can change the applicable regime.

The dynamic viscosity measures the transport of tangential momentum between adjacent layers. The kinetic estimates of gas transport coefficients givewhere $n$ is number density, $m$ the molecular mass, $\lambda$ the mean free path, and ${\sigma }_c$ a collision cross section. At fixed temperature, raising the pressure increases the number of momentum carriers but shortens their mean free path by the same factor. Consequently the dynamic viscosity of a dilute gas is approximately independent of pressure.

For a temperature-independent collision cross section, the same estimate gives $\eta \propto \sqrt{m{k}_{B}T}/{\sigma }_c$, so $\eta \propto T^{1/2}$ in the hard-sphere model. Faster molecules carry momentum across layers more rapidly. Real intermolecular interactions make the effective collision cross section temperature-dependent, changing this simple exponent. The pressure-independence also fails when wall collisions, rather than molecule–molecule collisions, set the transport distance; the continuum estimate assumes a small Knudsen number.

For a Newtonian fluid with velocity $u(z)\hat{\mathbf{x}}$, $\tau =\eta {\partial }_{z}u$ is the $x$ component of traction on a plane whose normal points along $+z$. The opposite face has the opposite traction. The coefficient $\eta$ is dynamic viscosity, with units $\mathrm{Pa}\mathrm{s}$. Faster layers transfer positive $x$ momentum toward slower neighbours; the internal shear stress resists relative sliding.

Take a thin fluid element with horizontal area $A$ and thickness $dz$. The difference between its upper and lower viscous tractions is $A\eta u_{zz}dz$. The pressure gradient supplies $-A{p}_{x}dz$. Incompressibility is satisfied because $u$ has no $x$ dependence, and the convective acceleration vanishes: $(\mathbf{u}\cdot \nabla )\mathbf{u}=u{\partial }_x\mathbf{u}=0$. Equating net force to mass times acceleration therefore gives the relevant component of the Navier-Stokes equation:For steady flow, integrate $\eta u^{\prime \prime }=p^{\prime }$ twice and impose the no-slip boundary condition at each plate, $u(0)=0$ and $u(a)=u_0$. The Couette-Poiseuille flow profile isThe linear term is Couette flow and the parabolic term is plane Poiseuille flow. A negative $p^{\prime }$ drives flow in $+x$. Integrating the profile across the gap gives the volumetric flow per unit width,It has units ${\mathrm{m}}^2{\mathrm{s}}^{-1}$; multiplying by the plate width gives the actual volume per time.

With zero pressure gradient, substituting $u=\mathrm{Re}[A{e}^{i(kz-\omega t)}]$ gives $-i\rho \omega =-\eta k^2$. Select the root that decays into fluid occupying $z>0$:The other sign applies to decay in the opposite half-space. With a real wall velocity amplitude $U_0$, the established periodic solution of Stokes' second problem isThe oscillation's amplitude decays exponentially and its phase lags with distance from the plate. This is diffusion of tangential momentum, not a freely propagating undamped elastic wave. Initial transients must have decayed before this purely periodic expression applies.

The fluid-displacement amplitude and the wall-displacement amplitude have the same ratio as their velocity amplitudes, since both are divided by the common $\omega$. Thus for the specified amplitudes,Using the standard approximate water density $\rho \simeq 10^3\mathrm{kg}{\mathrm{m}}^{-3}$, explicitly an approximation rather than another value printed on the paper, gives $\delta =0.564\mathrm{mm}$ and $z\simeq 2.21\mathrm{mm}$ from either face of a sufficiently large plate. The viscous penetration depth varies as ${\rho }^{-1/2}$ if a more precise water density is used. The ideal parallel-flow solution ignores finite plate edges and tank walls.

For the attractive inverse-square force, choose zero potential energy at infinity, so $U=-A/r$. At either radial turning point $r_p,r_a$, the kinetic energy is purely tangential, with conserved angular momentum $J$. ThusIts two roots are the periapsis and apoapsis distances. Their sum is $r_p+r_a=-A/E$, whereas the ellipse has $r_p+r_a=2a$. Therefore $a=-A/(2E)$, with $E<0$ for a nondegenerate bound ellipse. The circular case follows by the double-root limit.

For the two-body problem, define $\mathbf{r}={\mathbf{r}}_B-{\mathbf{r}}_A$, total mass $M=m_A+m_B$, reduced mass $\mu =m_A{m}_B/M$, and center of mass $\mathbf{R}$. The relative equation is $\mu \ddot{\mathbf{r}}=\mathbf{F}$, while the isolated center of mass has $\ddot{\mathbf{R}}=0$. Once the relative orbit is known,The relative energy uses $\mu$ in its kinetic energy and $A=G{m}_A{m}_B$ in the gravitational potential energy.

In the equal-mass case the initial relative speed is $u$, $\mu =m/2$, and the center-of-mass velocity is half B's initial velocity. ThusNegative relative energy requires $u^2<4Gm/a_1$. The escape threshold is $u_E=2\sqrt{Gm/a_1}$; equality is marginal parabolic escape, so a strictly bound pair has $u<u_E$.

For $u=u_E/2=\sqrt{Gm/a_1}$, $E_{\mathrm{rel}}=-3G{m}^2/(4a_1)$ and the relative semimajor axis is $a=2a_1/3$. The initial transverse speed is below the relative circular value $\sqrt{2Gm/a_1}$, so the initial turning point is apoapsis. HenceChoose the initial separation along $+x$ from A to B, with B's initial velocity $+u\hat{\mathbf{y}}$. Conservation of angular momentum gives relative periapsis speed $3u$. At the opposite end of the ellipse the direction is $-\hat{\mathbf{y}}$, soThus B has the same speed as initially but the opposite velocity. A has $2u\hat{\mathbf{y}}$ at that instant. A value $3u/2$ for B would be its center-of-mass-frame speed, not the requested initial laboratory frame result.

The relative orbital period follows from Kepler's third law with total gravitating mass $2m$:In the center-of-mass frame each star follows an ellipse of semimajor axis $a_1/3$ and eccentricity $1/2$, with the center of mass at a focus; they remain opposite on their simultaneous position vectors. In the laboratory frame add the uniform drift $ut/2$ along $y$. The laboratory paths of an equal-mass binary are therefore not closed ellipses, even though their relative motion repeats.

A stationary person relative to the rim has centripetal acceleration $a{\omega }^2=g$. If $u$ is a signed walking speed relative to the rim, the inertial tangential speed is $a\omega +u$, so the inward normal reaction satisfiesConsequently the support force per mass, and hence apparent weight, is $4g$ for $u=+a\omega$ and zero for $u=-a\omega$. In the latter case the walker has zero inertial speed and needs no centripetal support; ordinary weight-bearing contact cannot be assumed to persist without a restraint. For artificial gravity in a rotating ring, the centrifugal force and the Coriolis force alone are insufficient: the walker also has relative-path centripetal acceleration $u^2/a$. Omitting it would give the wrong apparent weights at these large walking speeds.

The method of images replaces induced charges or currents on a boundary by fictitious sources outside the physical region, chosen to satisfy that boundary's conditions. For example, a point charge $q$ a height $d$ above a grounded conducting plane has an image $-q$ at the reflected point. The sum has zero potential on the plane, and the electrostatic uniqueness theorem identifies it with the physical solution in the upper half-space. Images do not describe a second physical charge below the metal.

For the superconducting plane, use the different boundary condition $B_z=0$ at its surface: magnetic flux cannot enter the ideal bulk. Choose the real ring moment $\mathbf{m}=m\hat{z}$, where $m=I\pi R^2$, at $z=h$. Its image has moment $-m\hat{z}$ at $z=-h$. This superconducting plane image dipole makes the normal fields cancel at $z=0$, while their tangential fields add. The image is a computational continuation; the physical field inside the plane is zero. The original sketch shows the dipole-scale field, appropriate to $h\gg R$; the fine field close to the ring wire is outside that approximation.

In the vacuum region the magnetic scalar potential isAt the surface $z=0$, it reduces to $-mh/[2\pi ({\rho }^2+h^2{)}^{3/2}]$. Its tangential derivative therefore givesBoth $B_z$ and $B_{\varphi }$ vanish there; the sign is inward radially for the chosen positive moment. The expression is the leading dipole result, not an exact finite-ring formula at arbitrary $R/h$.

For the force, calculate the external image field on the axis near the real dipole:Differentiate this field with the image location held fixed, then evaluate at $z=h$. The dipole levitation force above a superconducting plane isdirected upward. Differentiating a substituted separation $2h$ while moving both dipoles in the force derivative would incorrectly double the result.

To levitate ring mass $M$, balance this force against $Mg$:A light, thin wire and the smallest practical gap reduce current. At fixed gap and wire cross-section $s$, $M=2\pi Rs{\rho }_w$ gives $I_{\mathrm{lev}}\propto h^2\sqrt{s{\rho }_w}/R^{3/2}$, favouring a larger ring until the dipole-limit constraint on $R/h$ matters. There is no unique least current without geometric and material constraints. As a geometrically consistent order-of-magnitude example, take $R=1\mathrm{mm}$, $h=10\mathrm{mm}$, wire radius $50\mu \mathrm{m}$ and an assumed metallic-scale density ${\rho }_w=6.0\times 10^3\mathrm{kg}{\mathrm{m}}^{-3}$. Then $M=2.96\times 10^{-7}\mathrm{kg}$ and $I_{\mathrm{lev}}\simeq 2.8\times 10^2\mathrm{A}$. The large value reflects the weak $h^{-4}$ force in the stipulated far-field geometry. Uniformly shrinking this geometrical example reduces current as size to the power $3/2$, but increases required electric current density; arbitrarily thin wire is therefore not a free optimization.

This balance estimate is not a verified superconducting design. The actual wire must carry that current without quenching, its self-field counts toward the limit, the plane must remain in its assumed flux-excluding state, and leads or supports add load. Transport critical-current measurements specify temperature, applied field and the voltage criterion rather than a material-independent limiting current.^[ref]

For a body-fixed mass element at position $\mathbf{r}$ relative to the fixed tip, $\mathbf{v}=\mathbf{\omega }\times \mathbf{r}$. ThereforeThe inertia tensor is real symmetric. Its orthogonal eigenvectors define principal axes and its eigenvalues are principal moments of inertia. angular momentum is parallel to angular velocity when rotation is along a principal axis, but generally not for arbitrary rotation.

Write $\mathbf{n}={\hat{\mathbf{e}}}_3$ and $s=I_3{\omega }_3$. Axial symmetry gives $\mathbf{L}=I_1\mathbf{\omega }+(I_3-I_1){\omega }_3\mathbf{n}$. Since $\dot{\mathbf{n}}=\mathbf{\omega }\times \mathbf{n}$, the vector triple-product identity gives $\mathbf{n}\times \dot{\mathbf{n}}=\mathbf{\omega }-{\omega }_3\mathbf{n}$. HenceDifferentiate in the inertial frame, using constant $s$ and $\dot{\mathbf{n}}\times \dot{\mathbf{n}}=0$. The gravitational torque about the fixed tip is $(\ell \mathbf{n})\times (-mg\hat{\mathbf{z}})$, yielding the symmetry-axis equation of a heavy symmetric top:The tip reaction exerts no moment about that tip.

For steady precession, $\mathbf{n}=(\sin \theta \cos {\Omega }_{P}t,\sin \theta \sin {\Omega }_{P}t,\cos \theta )$. At azimuth zero, put $s_{\theta }=\sin \theta$ and $c_{\theta }=\cos \theta$. Then $\dot{\mathbf{n}}=(0,{\Omega }_P{s}_{\theta },0)$, $\ddot{\mathbf{n}}=(-{\Omega }_P^2{s}_{\theta },0,0)$, and the $y$ component gives $s{\Omega }_P{s}_{\theta }-I_1{\Omega }_P^2{s}_{\theta }{c}_{\theta }=mg\ell s_{\theta }$. For a nonvertical axis this becomesReal steady branches require a nonnegative discriminant. In particular, for a top pointing above the horizontal there is a spin threshold for these fixed-angle solutions. At $\cos \theta =0$ use the quadratic's linear limit ${\Omega }_P=mg\ell /s$, rather than dividing by zero. For a vertical axis, azimuth does not describe a physical conical motion and the division by $\sin \theta$ is inappropriate.

For rapid positive spin $s$ with $s^2\gg 4mg\ell I_1\cos \theta$, the slow and fast steady precession roots areThe approximation assumes a nonsingular fixed inclination. For $g=0$, the quadratic instead gives ${\Omega }_P=0$ or ${\Omega }_P=s/(I_1\cos \theta )$. The zero branch has a stationary symmetry axis and pure axial spin. On the other branch angular momentum is constant and parallel or antiparallel to the chosen $z$ axis; $L_z=I_1{\Omega }_P$ and $s=L_z\cos \theta$. This is torque-free precession in space, not the different rate at which transverse components rotate in body coordinates.

Take $\zeta$ as signed distance from the neutral axis within the beam cross-section. Bending into a large radius $R$ changes the length of a fibre proportionally to $\zeta /R$, so linear elasticity gives $\sigma =E\zeta /R$, up to the curvature sign convention. The moment magnitude isHere $I$ is the second moment of area about the neutral axis, with units of length to the fourth power, not the mass moment of inertia. A homogeneous cross-section has its neutral axis through the centroid under pure bending.

For the straight pin-ended beam, the compressive dead load produces bending moment $-Fy$ at transverse offset $y$. In the small-slope approximation, curvature is $y^{\prime \prime }$ with the corresponding sign convention, givingThe stated endpoints fix positions rather than slopes: this interpretation is consistent with the illustrated end forces and the later sinusoidal unstressed shape. Truly clamped endpoints would have a different threshold. A nonzero solution $y=A_n\sin (n\pi x/L)$ requires $F_n=n^2{\pi }^{2}EI/L^2$. Thus the first Euler load is $F_c={\pi }^{2}EI/L^2$. Below it the perfect straight state is stable, at it the first mode is marginal in linear theory, and above it the straight state is unstable. The linear equation selects thresholds but cannot set the amplitude or describe a finite postbuckled equilibrium.

For the initially curved compressed beam, elastic stress responds to the change from its stress-free curvature. With $y=y_0+\eta$, the moment is $EI(y^{\prime \prime }-y_0^{\prime \prime })=EI{\eta }^{\prime \prime }$, so force balance givesPut $q=\pi /L$ and seek the branch continuously connected to the unloaded shape. Since $y_0=Y\sin qx$, substitute $\eta =C\sin qx$ to obtain $-EI{q}^{2}C=-F(Y+C)$, orAway from eigenvalues, the homogeneous solution $A\sin (\sqrt{F/(EI)}x)+B\cos (\sqrt{F/(EI)}x)$ vanishes under the two end conditions, so this particular solution is the unique linear equilibrium. For $0\le F<F_c$ and $Y>0$, the maximum additional displacement is ${\eta }_{\max }=YF/(F_c-F)$ and maximum total displacement is $y_{\max }=Y/(1-F/F_c)$, both at midspan.

This sinusoidal imperfection amplification of a column produces deflection at every positive load and strong amplification near $F_c$. The apparent divergence signals loss of the small-slope model before infinite displacement; nonlinear geometry is needed there. The formal continuation above the first threshold is not a stable physical loading branch. Both total and additional displacement are distinguished in the graph.

Coupled oscillators have more than one collective pattern of motion. For two equal masses $m$, each attached to a fixed support by a spring of stiffness $k$ and linked to each other by stiffness $\kappa$, small displacements obeyHere equal linear damping $b$ is included to make finite steady-state responses at resonance. The normal mode coordinates $q_{+}=(x_1+x_2)/\sqrt{2}$ and $q_{-}=(x_1-x_2)/\sqrt{2}$ diagonalize these equations. Their natural angular frequencies areThe symmetric normal mode leaves the coupling spring unextended; the antisymmetric normal mode stretches it and therefore has the higher frequency. With $F_{\pm }=(F_1\pm F_2)/\sqrt{2}$, their complex steady-state responses areThus the drive's spatial pattern matters as well as its frequency. Equal in-phase forces drive only the symmetric normal mode; opposite forces drive only the antisymmetric one; driving a single mass generally excites both. Finite damping broadens the resonance peaks and produces a phase lag. In the ideal undamped system, a resonant drive with nonzero projection onto a normal mode produces secular growth rather than a bounded resonant steady state.

The displacement of a particular mass adds contributions from both normal modes. They can cancel at an antiresonance. For $F_2=0$ and no damping, $x_1$ vanishes at ${\Omega }^2=(k+\kappa )/m$, while the partner still moves. This differs from the two nearby resonances, where a mode responds strongly; a small measured response need not mean that no motion or stored energy exists elsewhere.

A Newtonian fluid exerts viscous shear stress proportional to the rate of strain. For an incompressible fluid of constant dynamic viscosity $\eta$,The local momentum balance is the Navier-Stokes equation:The left side is density times material acceleration. A steady velocity field can still have convective acceleration when a moving parcel changes speed or direction; setting $\partial \mathbf{u}/\partial t=0$ does not eliminate it.

The Reynolds number $\mathrm{Re}=\rho UL/\eta$ compares inertia with viscous effects. At low Reynolds number, the fluid's own inertia can often be neglected, giving Stokes flow. This does not say that every immersed body has zero acceleration: its mass and the net pressure, viscous and body forces still enter its own equation of motion.

For a small rigid sphere moving through a slowly varying viscous background, Stokes drag is $-6\pi \eta a(\mathbf{v}-\mathbf{u})$. In still fluid, a simple settling model includes weight, buoyancy and drag:Ignoring fluid added mass and unsteady history forces, this gives relaxation time $m/(6\pi \eta a)$ and terminal velocityAt the terminal velocity the net force and acceleration vanish, while viscous drag and gravity remain individually nonzero. The dissipated power $6\pi \eta a{v}_t^2$ is supplied by the loss of gravitational potential energy relative to buoyant support. The creeping-flow and quasisteady-drag assumptions must be checked before applying this expression at large speed or immediately after a rapid start.

For two bodies interacting through a potential energy that depends only on their separation, introduce the centre of mass coordinate and the two-particle relative coordinate:Their kinetic energy separates exactly:The centre of mass moves uniformly if there is no external force, while the internal two-body problem reduces to one effective particle of reduced mass $\mu$:The original trajectories follow from ${\mathbf{r}}_1=\mathbf{R}+(m_2/M)\mathbf{r}$ and ${\mathbf{r}}_2=\mathbf{R}-(m_1/M)\mathbf{r}$.

For Newtonian gravity, $U=-G{m}_1{m}_2/r$, so the relative acceleration is $\ddot{\mathbf{r}}=-GM\mathbf{r}/r^3$. Conserved relative angular momentum and energy then give the usual conic sections. For a bound orbit of relative semimajor axis $a$,The two bodies share this orbital period and orbit their common centre of mass with semimajor axes in inverse proportion to their masses. If $m_2\gg m_1$, then $\mu \simeq m_1$, recovering the fixed-heavy-centre approximation. Using $M$ as the internal inertial mass would confuse centre-of-mass translation with relative motion.

Let $s(t)$ be the length of the free straight segment from the current contact point to the mass, and let $\theta$ measure the contact point's rotation from first contact. For an inextensible taut rope, winding through angle $\theta$ uses length $R\theta$:The point mass follows an involute of the pole's circular cross-section. Its instantaneous velocity is perpendicular to the free rope. To verify the geometry explicitly, use planar polar unit vectors at the contact point:The rope's tension therefore does no work, and the speed stays at its initial value $v$. The geometry gives $v=s\dot{\theta }$. ConsequentlyThe distance from the pole's axis is $\sqrt{R^2+s^2}$, so a point mass first reaches the pole when $s=0$. The winding time around a circular pole is thusThis has units of time. The model assumes a light inextensible rope, no energy loss at contact and a point mass. The required tension $m{v}^2/s$ diverges at the ideal final contact, so finite strength and size matter very close to impact.

For a cold, unmagnetized, collisionless plasma, take an equilibrium electron density $n_e$, electrons of mass $m_e$ and charge $-e$, and approximately stationary ions. Use fields proportional to $e^{-i\omega t}$. The electron displacement satisfiesIts electric polarization is $\mathbf{P}=-n_{e}e\mathbf{x}$, hence the cold collisionless plasma dielectric responseThe negative susceptibility arises from the inertial electron response, rather than a restoring force binding an electron to a particular atom.