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A microstate is a complete microscopic specification within the chosen model. For two distinguishable, noninteracting two-level systems, label the units $A,B$ and their states $0,1$, with energies $0,\Delta$ and $\Delta >0$. The four microstates arewith total energies $0,\Delta ,\Delta ,2\Delta$. Distinguishability matters: $(0,1)$ and $(1,0)$ describe different assignments to the labeled units. Noninteraction matters too: their energies add without an extra coupling term.

For a classical particle model, a microstate can instead be a point in phase space, specifying every position and momentum. In the quantum canonical ensemble, the sum uses energy eigenstates; a general quantum state may also contain quantum coherence, which an energy label alone does not specify.

Suppose a robot occupies a closed set $R\subseteq {\mathbb{R}}^n$ relative to its chosen reference point, and a closed obstacle occupies $O$. Keep the robot's orientation fixed and allow only translations. At reference position $q$, its occupied set is $q+R$. If touching counts as collision, the forbidden positions in this configuration space are exactlyThe minus sign reflects the robot through its reference origin: solving $q+r=o$ gives $q=o-r$. Thus collision checking for a moving shape becomes checking whether one point lies in a fixed Minkowski sum. Steven M. LaValle derives this translational configuration space construction in Planning Algorithms, section 4.3.2, equation 4.37 and Example 4.13.

For a concrete toy example, take the robot rectangle $R=[0,2]\times [0,1]$ with its reference point at the lower-left corner, and obstacle rectangle $O=[4,6]\times [2,4]$. Their collision region is the rectangleFor example, $q=(3,3/2)$ is forbidden: $q+R=[3,5]\times [3/2,5/2]$ meets $O$ in $[4,5]\times [2,5/2]$, of area $1/2$. At $q=(2,1)$ there is just corner contact, which our convention also forbids. Using $O+R=[4,8]\times [2,5]$ instead would wrongly miss both reference positions. Only when $R=-R$, as for an origin-centred disk, does the reflection leave the robot set unchanged.

The same calculation handles bounded additive uncertainty. If the actual reference position is $q+e$ for any $e\in E$, the union of possible occupied sets is $q+(E+R)$. There exists an allowed error causing contact exactly whenConsequently avoiding this set guarantees avoidance for every $e\in E$ in this model. For $E=[-\epsilon ,\epsilon {]}^2$, the example's forbidden region becomes $[2-\epsilon ,6+\epsilon ]\times [1-\epsilon ,4+\epsilon ]$, with area $(4+2\epsilon )(3+2\epsilon )$. Here $-E=E$ because the error bounds are symmetric; a one-sided error set must still be negated. This is a worst-case geometric bound, not a probability of collision. Rotating the robot requires an orientation-dependent Minkowski sum for each orientation.

Change of position with time relative to a specified reference system.

The Newtonian gravitational field is the gravitational force per unit mass of a test particle. For a point mass $M$ at the origin,The test particle's mass cancels: different small masses have the same gravitational acceleration at the same position. In SI, the Newtonian gravitational field has units $\mathrm{N}\mathrm{k}{\mathrm{g}}^{-1}=\mathrm{m}{\mathrm{s}}^{-2}$. The same exterior expression holds for a spherically symmetric source, by the spherical shell theorem.

Choose a reference radius $r_0>0$, set $g_0=GM/r_0^2$, and use nondimensionalization with $z=r/r_0$. Then $g/g_0=z^{-2}$: at $z=1,2,4$, the magnitudes are $1,1/4,1/16$. This makes the spatial variation visible without specifying a planet. ^[ref]

In an inertial frame, a body with zero net external force remains at rest or moves with constant velocity. For a point mass of constant positive mass, write the net external force as the vector sum of all external forces on that body:Here ${\mathbf{r}}_0$ and ${\mathbf{v}}_0$ are the position and velocity at time $t_0$. The zero net external force must hold throughout the interval, not merely at one instant. Both the magnitude and direction of the velocity stay fixed: constant speed along a curved trajectory still involves acceleration. Rest is the case ${\mathbf{v}}_0=0$. For an extended closed system, the analogous statement concerns its center of mass; zero net external force need not prevent its parts from moving relative to one another.

The short mathematical argument is to use Newton's second law with $m>0$, then take the integral of $d\mathbf{v}/dt=0$ and of $d\mathbf{r}/dt={\mathbf{v}}_0$. This gives the two constants fixed by the initial position and velocity. The physical content also includes the existence of inertial frames in which these relations apply; an arbitrary accelerating reference frame need not show constant velocity for a body with zero actual net external force.

The useful question is therefore which forces change the motion, rather than which force keeps it going. A free-body diagram may show several nonzero forces whose vector sum is zero. Equal opposing forces on the same body can balance; a third-law interaction pair acts on different bodies and cannot be cancelled within that one body's diagram.

In 1687, Isaac Newton stated Law I in Philosophiae Naturalis Principia Mathematica: a body's rest or uniform straight motion persists unless impressed forces compel a change. The following paragraph identifies air resistance and gravity as influences on the trajectory of a thrown body. The vector equations above are a modern expression of the law, not a quotation of his notation.^[ref]

Newton's law of universal gravitation gives the attractive force between two point masses. With source mass $M$ at the origin and relative position $\mathbf{r}$ of mass $m$, write $r=|\mathbf{r}|>0$:Here $G$ is the gravitational constant. The minus sign points toward $M$; the magnitude follows an inverse-square law, so doubling the separation quarters the force. The opposite force on $M$ satisfies Newton's third law. ^[ref]

Both masses accelerate. Applying Newton's second law to each in the isolated two-body problem givesThe relative standard gravitational parameter includes both masses; replacing it by $GM$ requires the test particle approximation $m\ll M$.

Historically, Newton's Book III, Proposition VII argues that gravitational attraction applies to all bodies and scales with their masses; its second corollary states the inverse-square law for individual particles. ^[ref]

In an inertial frame, a point mass of constant mass $m>0$ obeysThe net force is the sum of all forces acting on that body; momentum, velocity and acceleration refer to the same reference frame. A force changes velocity, rather than sustaining it: zero net force implies constant velocity, including a nonzero one. Doubling the net force at fixed mass doubles acceleration; doubling mass at fixed net force halves it. ^[ref]

For example, an otherwise balanced block of mass $0.5\mathrm{kg}$ receives a constant horizontal net force $2\mathrm{N}$ for $0.5\mathrm{s}$. Its acceleration is $4m/{\mathrm{s}}^2$, so its velocity changes by $2m/s$. Equivalently, the impulse is $F\Delta t=1\mathrm{N}\mathrm{s}$, and $\Delta v=\Delta p/m$. Starting from rest, its displacement during the pulse is $a(\Delta t{)}^2/2=0.5\mathrm{m}$. After the pulse, zero net force leaves it moving at constant velocity.

Applying the law therefore requires both a specification of the net force and initial conditions. In the harmonic oscillator example, the force is $-kx$, so Newton's second law becomes a differential equation for position. In an accelerating reference frame, the measured relative acceleration includes a contribution from the frame's own motion; the equation of motion in a rotating frame illustrates the extra terms required.

Constant mass is essential to the displayed $m\mathbf{a}$ form. A variable-mass system exchanging matter also exchanges momentum; its balance includes the entering or leaving matter's velocity. Simply differentiating $m\mathbf{v}$ and equating it to external force on that open body can omit this flux. The existing variable-mass system equation specifies the needed correction.

In the 1687 Principia, Newton's second law was stated in terms of change of motion proportional to impressed force, directed along that force. Its original statement is verbal; the derivative equation above is modern notation. ^[ref]

Why infer an area law from astronomy? In the authored account The great Martian catastrophe and how Kepler fixed it by Owen Gingerich, a 13 August 1593 Mars observation has ecliptic longitude $346^{\circ }7.5^{\prime }$, against predicted $342^{\circ }{0}^{\prime }$ and $351^{\circ }26^{\prime }$. These are Earth-based directions with a date, not measured Sun-based areas. The observer's armillary sphere used graduated circles and sighting arms to read declination; reversing it allowed a second reading. Comparing stars at different altitudes exposed refraction, and a planet-based check revealed an erroneous solar correction table. The apparent Mars distance signal had been contaminated by that correction. Johannes Kepler subsequently corrected the observer's Earth orbit, then tested Mars trajectories. The account describes typical angular errors of $2^{\prime }$ or less ($120$ arcseconds): small residuals needed care.^[ref]

The Tufts class notes Kepler's Astronomia Nova and the Orbit of Mars explain the key limitation: an area requires a trajectory and distances, so Kepler's second law cannot be directly tested from angular observations alone. They describe checking the combined ellipse and area law against 28 dated observations around seven oppositions during 1583-1595, using an Earth model to predict Earth-based directions; agreement improved greatly but some discrepancies remained.^[ref]

The inferential chain is useful today too: recover relative positions and times, compare swept areas, and ask what forces are compatible with their evolution. Agreement with equal areas supports zero torque about the chosen origin under the assumptions below; it alone does not identify an inverse-square law. Recovering the radial force also needs the trajectory and its timing, as in inferring an inverse-square force from Kepler laws. The two diagrams here are calculated examples, not the historical measurements.

A body's position along its orbit, specified relative to a stated origin and time. Knowing only the shape and size of an orbit does not specify where the body lies on it: an initial orbital phase or timing condition is also needed. A rendezvous requires matching orbital positions at the same time as well as matching velocities.

Orbit determination infers a body's position and velocity from observations and an equation of motion. It is an inverse problem; a residual of an inverse problem compares an observed readout with its modeled prediction.

Introduce a bookkeeping parameter $\lambda$ in $H=H_0+\lambda H_1$ and expand $|{\varphi }_n⟩=|{\psi }_n⟩+\lambda |{\eta }_n⟩+O({\lambda }^2)$ and ${\mathcal{E}}_n=E_n+\lambda \Delta E_n+O({\lambda }^2)$. Fix phase and normalization at this order by $⟨{\psi }_n|{\eta }_n⟩=0$. The first-order Schrödinger equation isangular momentum projection onto $⟨{\psi }_n|$ gives $\Delta E_n=⟨{\psi }_n|H_1|{\psi }_n⟩$. angular momentum projection onto a different eigenstate givesCompleteness therefore yields the first-order nondegenerate perturbation theory resultNonzero energy denominators are essential. The displayed state has unit norm through first order; exact normalization would change terms of second order. A continuum, if present, contributes its corresponding integral as well as any discrete sum.

For the oscillator in a static electric field, electrostatic potential energy is $H_1=-qϵx$ and $\omega =\sqrt{k/m}$. parity gives $⟨0|x|0⟩=0$, and the supplied Gaussian wavefunctions giveSince $E_0-E_1=-\hslash \omega$, the two-state first-order ground state isThus $⟨x⟩=2\eta x_{\mathrm{zp}}+O({ϵ}^2)$, and the induced electric dipole isIt is aligned with the field for either sign of $q$. The harmonic-oscillator electric polarizability is $q^2/k$. The position operator connects $|0⟩$ only to $|1⟩$, so the requested two-state calculation contains the complete first-order dipole response, rather than dropping other first-order terms.

Complete the square to find the exact electric-field displacement of a quantum harmonic oscillator:The exact normalized ground wavefunction is ${\psi }_0(x-x_0)$, with energy $\hslash \omega /2-q^2{ϵ}^2/(2k)$. Its mean displacement is $x_0$, soThe first-order dipole equals the exact dipole, because the restoring force is exactly linear and translating a harmonic well does not change its width or frequency. This agreement of one expectation value does not make the truncated wavefunction exact. The Hellmann–Feynman theorem provides an independent check: $d=-\partial E_0/\partial ϵ$.

For perturbation theory to approximate the state, $|\eta |\ll 1$, or $|x_0|\ll 2x_{\mathrm{zp}}$. The exact translated ground state has the original-oscillator expansionIt is a coherent state with mean occupation ${\eta }^2$. Large fields require many oscillator levels, whereas the first-order state retains only two. The normalized two-state truncation of a displaced oscillator instead giveswhich incorrectly tends towards zero for large $|\eta |$ instead of following the displaced well. In position space the truncated polynomial times Gaussian also develops a node absent from the exact ground state. Taking a linearized formula outside its small-state-correction regime is therefore invalid even though its dipole coefficient happens to be exact.

For a bound energy eigenstate with a nondegenerate energy eigenvalue of an even one-dimensional potential, the wavefunction can be chosen with definite parity. A uniform electric field $\mathcal{E}$ gives the electric-dipole interaction $H^{\prime }=-q\mathcal{E}x$, which is odd under reflection. Hencesince $x|{\psi }_n(x){|}^2$ is odd. Second-order nondegenerate perturbation theory then gives the leading allowed Stark effect shift:Only opposite-parity intermediate states have nonzero position matrix elements. More generally, reflection maps $H(\mathcal{E})$ to $H(-\mathcal{E})$, so an analytic nondegenerate energy level is an even function of the field. This explains the absence of all odd orders and the usual quadratic Stark effect.

A level separation changes by the difference of the two quadratic coefficients. Symmetry alone does not guarantee a nonzero differential shift: in a displaced quantum harmonic oscillator every level moves by the same $-q^2{\mathcal{E}}^2/(2m{\omega }^2)$, leaving every spacing unchanged. Degenerate states would instead require degenerate perturbation theory and can split linearly; the nondegeneracy of one-dimensional bound states applies to ordinary isolated well levels.

Each simple pendulum moves in the vertical plane containing the pivots. Ignoring contact between the ideal rods and point masses, each rod can turn through a full circle. Thus the configuration space is the torusChoose generalized coordinates ${\theta }_1,{\theta }_2$, measured from the downward vertical, positive towards the right and understood modulo $2\pi$. These angles are local coordinates on the torus, rather than a single globally one-to-one real coordinate chart. With upward Cartesian coordinate $y$, the bob positions areTheir velocities satisfy $|{\dot{\mathbf{r}}}_j{|}^2=l^2{\dot{\theta }}_j^2$, so kinetic energy from Cartesian position vectors givesThe spring length is the Euclidean distanceTaking gravitational potential energy zero at pivot height gives $V_g=-mgl(\cos {\theta }_1+\cos {\theta }_2)$. The spring potential energy is $V_s=k(d-r{)}^2/2$, since $r$ is its natural length. Consequently the Lagrangian isIn the paper's notation $t({\theta }_1,{\theta }_2)=V_s$; this function is distinct from the time variable. This is the exact Lagrangian of two spring-coupled pendulums.

One alternative Lagrangian isIt corresponds to setting gravitational potential energy zero at the downward position. The constant disappears from every Euler-Lagrange equation. More generally, total-time-derivative invariance of a Lagrangian allows $\tilde{L}=L+dF({\theta }_1,{\theta }_2,t)/dt$ for any smooth $F$.

Make the Wick rotation $t^{\prime }=-i\tau$, and write $T_E=\beta \hslash$ for the final imaginary time, where $\beta$ is inverse temperature and $\hslash$ is the reduced Planck constant. Then $d{t}^{\prime }=-id\tau$ and $dx/d{t}^{\prime }=idx/d\tau$. Thus the continued action isBoth the kinetic energy and the potential energy enter this Euclidean action with positive signs relative to their real-time definitions. Also $e^{-iH(-i\beta \hslash )/\hslash }=e^{-\beta H}$, so the Euclidean path integral isIts boundary conditions fix the two endpoint positions. They need not coincide: identifying and integrating over them would instead give the thermal trace.

The normalization of $\mathcal{D}x$ can be fixed without continuing a formal measure. Apply the Trotter product formula to $H={\hat{p}}^2/(2m)+V(\hat{x})$, divide $T_E$ into $M$ intervals of size $ϵ=T_E/M$, and integrate the intermediate positions. The resulting time-sliced configuration-space path integral isThis also explains the upper limit $\beta \hslash$ rather than $\beta$ when the reduced Planck constant has not been set to one.

Write $x=x_0+y$, so the fluctuation has homogeneous Dirichlet boundary conditions $y(0)=y(T_E)=0$. Because ${\dot{x}}_0$ is constant, the cross term in the Euclidean action vanishes exactly:ConsequentlyTranslation of each intermediate integration variable has Jacobian determinant one in the time-sliced configuration-space path integral. This is an exact classical-path factorization of a quadratic path integral, givingThe fluctuation Euclidean action, the homogeneous boundary conditions, and the measure normalization contain neither $x_i$ nor $x_f$. Hence $\mathcal{N}$ is independent of both endpoint positions. It depends on $\beta ,m,\hslash$.

With the standard position-state normalization, the Euclidean free-particle bridge amplitude isFor example, a momentum resolution gives a one-dimensional Gaussian integral,whose evaluation gives the same exponent and prefactor. In particular, the result integrates to one over $x_f$ at fixed $x_i$, as required for its heat kernel interpretation.

Put $M=m_1+m_2$, define the center of mass $\mathbf{R}=(m_1{\mathbf{x}}_1+m_2{\mathbf{x}}_2)/M$, and let $\mathbf{r}={\mathbf{x}}_1-{\mathbf{x}}_2$ be relative position. Adding the two Newton's second law equations givesThe center of mass has constant velocity, by conservation of total momentum.

Subtracting the particle accelerations instead givesThus relative motion is a one-particle problem with the reduced mass $\mu$, while center-of-mass motion is uniform. The same reduced mass appears in the kinetic energy decomposition $T=\frac{1}{2}M|\dot{\mathbf{R}}{|}^2+\frac{1}{2}\mu |\dot{\mathbf{r}}{|}^2$.

Taking the scalar product of the Lorentz force equation with the velocity givesThe triple product vanishes because the cross product is perpendicular to its first factor. Thus the kinetic energy is constant, even if the magnetic field varies with position: magnetic forces do no work.

In cylindrical coordinates, the radial, angular and vertical equations areFor a nonstationary circular orbit with constant $r$ and $z$, the first equation gives $-m{\dot{\theta }}^2=qB(r)\dot{\theta }$. Hence its signed angular velocity is $\boxed{\dot{\theta }=-qB(r)/m}$. The sign gives the sense of circulation. The zero root describes a stationary transverse position, not a moving circular orbit. A constant vertical velocity can be superposed to produce a helix.

Define the mechanical angular momentum about the axis by $L=m{r}^2\dot{\theta }$. Multiplying the angular equation by $r$ givesFor the inverse-radius magnetic field this becomes $\dot{L}=-q{B}_0\dot{r}$. Integrating from an initial radius $r_0$ provesLet $c=q{B}_0/m$ and $\ell =(L_0+q{B}_0{r}_0)/m$, so that $L/m=\ell -cr$. Since $\dot{z}$ is constant and the total kinetic energy is constant, the transverse speed squaredis also constant. Substituting the angular momentum relation gives the required radial curve:Only the region $W(r)\ge 0$ is accessible. For $\ell \ne 0$, $W(r)\to -\infty$ as $r\to 0$, and its horizontal asymptote is $v_{\perp }^2-c^2$. Its derivative is $W^{\prime }(r)=2\ell (\ell -cr)/r^3$. Thus, when $\ell c>0$, it has a single maximum $W=v_{\perp }^2$ at $r=\ell /c$. The positive roots of $W$ are the turning points obtained from $r=\ell /(c\pm v_{\perp })$.

For nonzero $c$ and $\ell c>0$, write $r_c=|\ell /c|$. When $0<v_{\perp }<|c|$, two turning points occur atand the radius is confined between them. At $v_{\perp }=|c|$, there is only the finite lower turning point $r_c/2$; $W$ tends to zero from above and an outgoing particle reaches arbitrarily large radius. For $v_{\perp }>|c|$, only the lower turning point remains and the outgoing radial velocity approaches $\sqrt{v_{\perp }^2-c^2}$. The sketch shows these three cases and omits inaccessible negative values of ${\dot{r}}^2$ from the physical interpretation.

For completeness, if $\ell c<0$, $W$ rises monotonically from $-\infty$ to its asymptote; admissible motion then requires $v_{\perp }>|c|$ and is radially unbound after any inward branch turns. If $\ell =0$, $W=v_{\perp }^2-c^2$ is constant; equality gives a circular orbit rather than escape. Above threshold the outgoing branch escapes, whereas an incoming branch reaches the singular axis in finite time and cannot be continued through it without specifying a nonsingular core field. These cases explain why a borderline escape assertion needs an initial-direction qualification.

The transverse escape velocity threshold for radial escape in an inverse-radius axial magnetic field isFor a radial launch at $r_0$, $\ell =c{r}_0$, and equality permits marginal escape; for $\ell \ne 0$, larger transverse speed guarantees radial escape in forward time after any initial inward leg turns. The singular incoming $\ell =0$ branch is the exception described above. More general launches at equality can instead give the circular exception $\ell =0$. A vertical velocity adds to the total speed but cannot pay the transverse escape-energy requirement; if escape merely means unbounded $z$, any nonzero $\dot{z}$ already achieves that. If $q{B}_0=0$, transverse motion is force-free and there is no positive escape-speed threshold.

The moment of inertia of a uniform solid cylinder is taken about its symmetry axis. Write its mass density as ${\rho }_c=m/(\pi a^{2}L)$. A coaxial shell of radius $r$ and thickness $dr$ has mass $dm=2\pi {\rho }_{c}Lrdr$. ThusA hollow cylinder would have a different moment of inertia; uniform solid density is the intended assumption.

Let $s$ measure the centre's displacement uphill from its initial position, and choose signed angular velocity $\omega$ so that uphill rolling without slipping has $\dot{s}=a\omega$. Let $F$ be the signed contact force uphill. The translation and rotation equations areThe minus sign in the torque equation follows from the chosen rolling sense: uphill contact force opposes the positive spin. Under rolling without slipping, $\ddot{s}=a\dot{\omega }$. Eliminating $F$ givesUsing $s(0)=0$ and $\dot{s}(0)=V$, the no-slip motion and maximum vertical rise areIndeed the centre turns at $t=3V/(2g\sin \alpha )$, reaching $s_{\max }=3V^2/(4g\sin \alpha )$, whose vertical component is the boxed height. Conservation of energy independently gives the same answer from initial kinetic energy $m{V}^2/2+I(V/a{)}^2/2=3m{V}^2/4$.

The available static friction is at most $\mu mg\cos \alpha$. It can supply the required uphill force exactly when $mg\sin \alpha /3\le \mu mg\cos \alpha$. ThereforeInterpret the stated friction magnitude as limiting static friction and as kinetic friction during slip, with the same coefficient. Take $0<\alpha <\pi /2$, $\mu \ge 0$. In the slipping regime the contact point moves downhill relative to the slope, so kinetic friction acts uphill. To verify this direction rather than assume it unchecked, solve with $F=\mu mg\cos \alpha$. Define $b=g(\sin \alpha -\mu \cos \alpha )>0$. The result isThe centre's speed is $|V-bt|$, with the sign of $v$ specifying ascent or descent. The contact point's relative velocity isThus the assumed uphill kinetic friction is consistent throughout both ascent and descent, including when the centre's velocity changes sign. There is no later return to rolling without slipping before reaching the bottom. This is the characteristic slipping solid cylinder on an incline regime.

The centre turns at $t_{\ast }=V/b$, givingThe formulas apply up to the positive return time $t_b=2V/b$. Substitution into the signed velocity and angular velocity givesThe return spin can have either sign and vanishes at $\tan \alpha =5\mu$. It reverses when $3\mu <\tan \alpha <5\mu$, although the contact point still slips downhill. The unchanged centre speed does not mean that no energy was dissipated: the rotational kinetic energy has decreased. Indeed integrating the frictional power $Fw<0$ giveswhich agrees with the difference of the initial and final rotational kinetic energies.

Use a translating accelerated reference frame; its axes do not rotate. If $\mathbf{R}=(\xi ,\eta ,\zeta )$ is the block's position relative to the table, then its inertial position is $\mathbf{r}=\mathbf{R}+\mathbf{A}\sin \omega t$. Its relative velocity is $\dot{\mathbf{R}}$ and its inertial acceleration is $\ddot{\mathbf{R}}-{\omega }^2\mathbf{A}\sin \omega t$. Let $N$ be the upward normal force. During contact $\zeta =\dot{\zeta }=\ddot{\zeta }=0$, and the linear tangential resistance givesThese are Newton's second law transformed to the table's reference frame; the extra terms arise from the table's acceleration. The last equation gives $N=m(g-A_z{\omega }^2\sin \omega t)$ while contact persists. If contact is lost, both the contact resistance and $N$ vanish, and free flight must be described by the corresponding gravity-only equations.

For $A_z=0$, assume $\lambda >0$ and put $\gamma =\lambda /m$. Seek a periodic particular solution ${\xi }_p=b\sin \omega t+c\cos \omega t$. Substitution into the first differential equation givesSolving yieldsThe homogeneous solutions are a constant and a multiple of $e^{-\gamma t}$. Choose the phase $0<\theta <\pi /2$ with $\tan \theta =\omega /\gamma =m\omega /\lambda$. Then $b=-A_x{\sin }^2\theta$ and $c=-A_x\sin \theta \cos \theta$, soAlso $\eta (t)={\eta }_{\infty }+C_{\eta }{e}^{-\lambda t/m}$. The decaying transients leave the stated steady orbit. Positive damping is essential to this late-time assertion; for $\lambda =0$ an additional constant drift may persist.

No attractive contact force is available, so $N$ cannot be negative. Over a whole oscillation its minimum is $m(g-{\omega }^2|A_z|)$. Thus strictly positive contact loading throughout requires $\boxed{{\omega }^2|A_z|<g}$, or the printed form with $A_z$ taken as a nonnegative vertical amplitude. If ${\omega }^2|A_z|>g$, contact cannot persist: it would require $N<0$ during part of the cycle. At equality, $N$ vanishes only at isolated instants and the ideal constrained motion can still remain in contact. Hence the strict inequality characterizes positive loading; mere nonnegative contact permits the marginal equality case.

Let $\rho =3m/(4\pi R_0^3)$ be the uniform mass density. In spherical coordinates $(r,\vartheta ,\varphi )$ with the chosen diameter as polar axis, the squared perpendicular distance to that axis is $r^2{\sin }^2\vartheta$. The moment of inertia is thereforeThis derives the moment of inertia of a uniform solid sphere about any central axis by spherical symmetry.

Put $a=R_1-R_0$. The center of mass follows the circle of radius $a$, with position ${\mathbf{r}}_C=a(\sin \theta ,-\cos \theta )$. Let ${\mathbf{e}}_r=(\sin \theta ,-\cos \theta )$ and ${\mathbf{e}}_{\theta }=(\cos \theta ,\sin \theta )$. Its velocity is $a\dot{\theta }{\mathbf{e}}_{\theta }$, and the contact point is displaced from the center of mass by $R_0{\mathbf{e}}_r$. With positive spin angular velocity ${\Omega }_s$ counterclockwise, its contact-point velocity isThe cylinder is fixed, so rolling without slipping requires this velocity to vanish. Consequently ${\Omega }_s=-a\dot{\theta }/R_0$. If instead the spin angle $\psi$ is positive clockwise, opposite to increasing $\theta$, then the printed positive-coefficient convention isThe spin and orbital rotation have opposite directions when both are measured about the same oriented axis; the angular speed is $a|\dot{\theta }|/R_0$. This specifies the sign convention in the printed expression.

The instantaneous contact point is at rest, so the normal reaction and static friction do no work. Conservation of energy equates the loss of gravitational potential energy from release to the total kinetic energy:On the descent from $0<\alpha <\pi /2$, $\theta$ decreases, soIntegrating $dt=-d\theta /|\dot{\theta }|$ from release to the bottom givesThe integral is finite: at the release point its denominator is asymptotic to $\sqrt{\sin \alpha (\alpha -\theta )}$. Contact is maintained, since the inward radial equation of motion gives the positive normal reaction $N=m(g\cos \theta +a{\dot{\theta }}^2)$ throughout this range.

For frictionless sliding, the normal reaction passes through the sphere's centre and produces no torque; Newtonian gravity also exerts no central torque. With initial zero spin, the sphere remains nonrotating. All the lost gravitational potential energy becomes translational kinetic energy, soA hollow spherical shell of the same mass and radius places its material farther from the centre, giving a larger moment of inertia. Under rolling without slipping, a given centre speed therefore requires more rotational kinetic energy; the same gravitational drop yields a smaller speed. Replacing the solid sphere by a hollow shell increases $T_R$ and leaves $T_S$ unchanged. The sliding case remains nonrotating and has the same center of mass path and gravitational drop. This is the general comparison for rolling inside a fixed circular cylinder.