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Let the specimen rotate uniformly with angular velocity $\mathbf{\Omega }=\Omega \hat{\mathbf{z}}$. In its rotating reference frame, the internal Hamiltonian containsHere $J_z$ is the internal angular momentum operator about the rotation axis, and ${\mathcal{H}}_0$ contains the internal interactions and any imposed magnetic field expressed in that frame. It does not include the separately treated rigid body's rotational kinetic energy. For a sector with magnetic moment $m_z=\gamma J_z$, the rotational term has exactly the form of a magnetic dipole potential energy:This Barnett equivalent magnetic field has tesla units, with the angular convention for the signed gyromagnetic ratio: $\Omega$ is in radians per second, not revolutions per second. Its sign matters. For an electronic contribution with $\gamma <0$, positive $\Omega$ favors positive $J_z$ and negative $m_z$. A sector with $\gamma >0$ has the opposite magnetic sign. Equal $\Omega$ need not correspond to equal Barnett fields for sectors having different gyromagnetic ratios.

A short derivation fixes the sign independently of any material model. Put $U(t)=\exp (-\mathrm{i}\Omega t{J}_z/\hslash )$ and transform a laboratory quantum state by $|{\psi }_{\mathrm{rot}}⟩=U^{†}|{\psi }_{\mathrm{lab}}⟩$. The Schrödinger equation givesThe transformed laboratory term is denoted by ${\mathcal{H}}_0$; a laboratory perturbation not stationary in the rotating frame can make it time dependent. Establishing a steady thermal response also requires exchange of internal angular momentum with the rotating body, for example through spin-lattice relaxation. A mere relabeling of coordinates cannot polarize isolated, unrelaxed spins.

The Barnett field is a useful description of spin-rotation coupling, not a claim that empty laboratory space acquires a uniform magnetic flux density $\Omega /\gamma$. In particular, inserting it into the Lorentz force on an unrelated test electric charge is unjustified. A nearby magnetometer detects the specimen's actual stray magnetic field and backgrounds, rather than directly detecting this internal Hamiltonian parameter.

Classical mechanics describes the motion of bodies by assigning positions and velocities and specifying how their interactions change those quantities. A point mass is the simplest model: its size and internal structure are ignored, leaving a mass and a moving position. A rigid body retains orientation and rotational kinetic energy. Choosing a model means deciding which features affect the question being asked; a sliding block may be treated as a point mass, whereas predicting its rolling without slipping requires its moment of inertia.

Three complementary questions organize the subject. Newton's second law asks how the net force determines the instantaneous acceleration. The work-energy theorem asks how that force, acting through a displacement, changes kinetic energy. Conservation of mechanical energy asks which combinations of position and velocity are accessible when the interactions admit a time-independent potential energy and no other force supplies net power. These descriptions are compatible, but answer different questions: an energy calculation may find a maximum speed without first solving for position at every time.

The harmonic oscillator below makes this unity visible. A restoring force produces a repeating position trace; the corresponding phase plane has a closed ellipse; kinetic energy and potential energy exchange while their sum stays fixed. Lagrangian mechanics and Hamiltonian mechanics provide further ways to organize the same dynamics, especially when coordinates or constraints make individual forces awkward to track.

The scope of classical mechanics also matters. Newtonian mechanics approximates motion at speeds small compared with the speed of light; special relativity changes that description when this approximation fails. Quantum mechanics is needed for phenomena for which a classical assignment of simultaneous position and momentum is inadequate. These limitations do not prevent classical mechanics from being a useful predictive model within its regime. ^[ref]

The local elastic strain energy and rotational kinetic energy per length are $u_s=G{J}_p{\theta }_x^2/2$ and $u_k=\rho J_p{\theta }_t^2/2$. Multiplying the torsional wave equation in a circular shaft by $\rho J_p{\theta }_t$ and integrating the product derivative gives ${\partial }_t(u_s+u_k)+{\partial }_x\mathcal{P}=0$. Thus the energy flux is torque times angular speed, with the sign set by the positive direction. The inequality $|\mathcal{P}|=2\sqrt{G/\rho }\sqrt{u_s{u}_k}\le \sqrt{G/\rho }(u_s+u_k)$ follows from a square. A one-way travelling wave has $u_s=u_k$; a steadily rotating twisted shaft need not. Stored energy divided by transit time is not a separate bound for each energy component in a general superposition.

The equipartition theorem says that each independent quadratic term in a Hamiltonian described by classical mechanics has mean energy $k_{B}T/2$ in a canonical ensemble. Here $k_B$ is the Boltzmann constant and $T>0$ the absolute temperature. Count terms in the energy, rather than vaguely counting everything that can move: one harmonic oscillator has one position degree of freedom but TWO quadratic terms, one in momentum and one in displacement.

A precise sufficient setting is $H(y,\xi )=a{y}^2/2+H_0(\xi )$ with $a>0$, where $y$ is one of the canonical variables, $\xi$ denotes all remaining variables, and the phase space integration measure factors as $dyd\xi$. The range of $y$ is the whole real line, independently of $\xi$; there is no constraint coupling its allowed values to other variables. The partition function must be finite, and the relevant expected values must exist. The Boltzmann factor then givesThe probability density function ${\rho }_y$ is a normal distribution with variance $k_{B}T/a$. A larger coefficient narrows the fluctuations enough to cancel its greater energy cost. This is why the mean contribution is independent of mass or spring constant.^[ref]

For a proof outline, the one-variable Gaussian integral is $Z_y={\int }_{-\infty }^{\infty }{e}^{-\beta a{y}^2/2}dy=\sqrt{2\pi /(\beta a)}$, so $-{\partial }_{\beta }\ln Z_y=1/(2\beta )$. More generally, integration by parts in a canonical variable gives $⟨y{\partial }_{y}H⟩=k_{B}T$ when its boundary term $[y{e}^{-\beta H}]$ vanishes and the integrals converge. For $a{y}^2/2$ this is twice its mean energy. At a finite hard boundary the boundary term need not vanish; hard-wall correction to harmonic equipartition illustrates the resulting departure. A constraint, a nonquadratic potential energy, or a noncanonical integration measure therefore requires its own calculation. The theorem is an ensemble average, not equality of instantaneous energies, and applying it to a recorded trajectory requires equilibration and sufficient sampling.

For a classical one-dimensional harmonic oscillator, with positive mass $m$ and spring constant $k$,The origin is the equilibrium position. Its heat capacity is $k_B$ at fixed $m,k$, not $k_B/2$. Although the two mean energies agree, their measured amplitudes do not: $⟨p^2⟩=m{k}_{B}T$ and $⟨x^2⟩=k_{B}T/k$. In a harmonic optical trap, the position measurement observes the potential energy contribution; it does not measure both contributions by observing $x$ alone.^[ref]

For $N$ particles in a three-dimensional classical monatomic gas, neglect interactions and thermally accessible internal excitations. Its three translational momentum terms give internal energy $U=3N{k}_{B}T/2$, heat capacity at constant volume $C_V=3N{k}_B/2$, and heat capacity at constant pressure $C_p=C_V+N{k}_B=5N{k}_B/2$. Per mole, replace $N{k}_B$ by the molar gas constant $R$. For example, using stipulated $R=8.314\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$, the molar heat capacity at fixed volume is $12.471\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$. These results assume a dilute, nonrelativistic ideal gas with classical translational statistics, rather than a quantum degeneracy regime.

For an ideal gas of diatomic molecules, modeled as two chemical atoms with a bond, there are three translational quadratic terms, two accessible rotational kinetic energy terms for changes of bond orientation, and one bond-stretching harmonic oscillator with two quadratic terms. Rotation about the bond axis is not an additional active mode in this linear-molecule model. The three familiar limits of heat capacity at constant volume areCorresponding molar heat capacities are $3R/2$, $5R/2$ and $7R/2$. Counting the stretching harmonic oscillator only once as $k_{B}T/2$ would give the wrong last value.^[ref]

The frozen regimes require quantum mechanics. In the simple molecular rotation model, $E_j=k_B{\theta }_{r}j(j+1)$ with state degeneracy $2j+1$, ${\theta }_r={\hslash }^2/(2I{k}_B)$, and moment of inertia $I$. For the bond-stretching quantum harmonic oscillator, ${\theta }_v=\hslash \omega /k_B$ with angular frequency $\omega$ and $\hslash =h/(2\pi )$ from the Planck constant. When ${\theta }_r\ll {\theta }_v$, the three limits above correspond to $T\ll {\theta }_r$, ${\theta }_r\ll T\ll {\theta }_v$, and $T\gg {\theta }_v$, provided translational motion remains classical. They are smooth crossovers, not abrupt switches. Real molecules can have different allowed rotational populations; at high temperature, thermal dissociation, internal excitations and nonharmonic bond motion can invalidate the fixed-molecule model before its last limit is reached.^[ref]

For a quantum harmonic oscillator with fixed $\omega$, put $z={\theta }_v/T$. Its exact mean energy and heat capacity areAt $T\gg {\theta }_v$, the mean energy approaches $k_{B}T$ and $C_{\mathrm{osc}}$ approaches $k_B$. At $T\ll {\theta }_v$, thermal excitation is exponentially suppressed and $C_{\mathrm{osc}}$ tends to zero, although the zero-point energy $\hslash \omega /2$ survives. Frozen thus means little thermal response, not zero motion or zero energy.^[ref]

A motivating measurement is thermal calibration of an optical trap. A focused laser through a microscope objective confines a bead in water; imaging or a quadrant photodetector records a position time series. Gieseler and colleagues describe a polystyrene bead of radius $1.03\mu \mathrm{m}$ at $293.15\mathrm{K}$, a $1064\mathrm{nm}$ trapping laser, and photodetector acquisition at $10^5\mathrm{Hz}$. Their recorded bead continues undergoing Brownian motion; higher trapping power produces a narrower equilibrium position distribution. Infer the centered variance, not a squared distance from an arbitrary camera origin:The normal distribution position histogram and agreement with dynamical calibration methods support the local harmonic optical trap interpretation. For a stipulated example, $T=300\mathrm{K}$ and ${\sigma }_x=20\mathrm{nm}$ imply $k\simeq 1.04\times 10^{-5}N/m=0.0104pN/nm$. This calculated scale is not an experimental reading.^[ref]

Position calibration is essential: if detector output $v$ has conversion $x=s(v-v_0)$, its physical variance is $s^2\mathrm{Var}(v)$. An unknown scale $s$ cannot be determined together with $k$ from that one variance. A complementary calibration demonstrated by Tolić-Nørrelykke and colleagues combines the power spectrum of thermal motion with the bead response to known sinusoidal motion of a translation stage. The driven response supplies an additional length scale, and their method does not require bead size, liquid dynamic viscosity or wall distance as inputs. It still uses a thermal dynamical model; agreement between methods is a consistency check, not a theory-free proof of equipartition theorem.^[ref]

The readout itself can hide fluctuations. Wong and Halvorsen used a $1064\mathrm{nm}$ optical trap and adjustable-exposure CCD imaging of a bead in water. They varied exposure and trapping power, held the bead away from the nearest wall, adjusted illumination to maintain comparable images, and measured tracking noise using a stuck bead. The measured variance fell as exposure increased; corrected results agreed with their time-averaging model. For a stationary overdamped harmonic optical trap, with optical trap relaxation time $\tau$, an exposure of length $W$ measures ${\bar{x}}_W=W^{-1}{\int }_0^{W}x(t)dt$. Its ideal noise-free variance isFor $\alpha >0$, $S(\alpha )<1$: using the blurred variance in the uncorrected formula overestimates $k$ by $1/S(\alpha )$. At $W=\tau$, the calculated factor is $e/2\simeq 1.36$. Additive tracking noise instead raises the measured variance and biases the inferred $k$ downward if left uncorrected.^[ref]

Useful controls are to record over many optical trap relaxation times, compare blocks for drift, calibrate the position scale, measure fixed-bead noise, vary exposure, and check the position histogram for departures from a normal distribution. A sparse sequence of accurate instantaneous positions can estimate an equilibrium variance; a long exposure averages away motion even if many frames are collected. Flow, appreciable laser heating, escape from the harmonic region or nonconservative forces can break the assumed thermal equilibrium. Inferring $k$ by imposing equipartition theorem is a calibration, whereas testing the relation requires an independently established stiffness and position scale. The beautiful practical connection is that microscopic randomness can set a force scale: once calibrated, a small constant applied force displaces the mean by $f/k$. The same quadratic-term counting predicts macroscopic heat capacity and makes microscopic fluctuations useful measurements.

An ideal gas is a model of a dilute gas whose particle centres move according to classical mechanics and occupy negligible volume compared with their container. Treating the particles as point masses removes the molecular-size correction; neglecting interactions between separated particles removes the intermolecular contribution to potential energy. The usual kinetic theory model retains brief elastic collisions with the walls and, optionally, with other particles: these transfer momentum while conserving the total kinetic energy of the colliding bodies. There is no appreciable interaction energy stored between collisions. These assumptions lead to the ideal gas law, $PV=N{k}_{B}T$. ^{[ref] [ref]}

The model's attraction is its universality: at the same temperature and number density, different classical ideal gases have the same pressure, even when their particle masses differ. In ideal gas pressure from wall collisions, heavier particles move more slowly at a given temperature, and the two effects cancel. For example, one mole of any ideal gas at $300\mathrm{K}$ in $0.025{\mathrm{m}}^3$ has pressure approximately $99.774\mathrm{kPa}$, as computed under the ideal gas law.

The ideal gas law describes the translational contribution to pressure; it does not specify all of the internal energy or the heat capacity. A monatomic gas has three classical translational degrees of freedom, whereas a gas of molecules can also store rotational kinetic energy and energy in vibrational normal modes. An equation of state alone therefore does not determine how much heat a chosen temperature increase requires. It also does not select a process: isothermal expansion and adiabatic expansion need different additional constraints. ^[ref]

A real gas departs from this approximation when particle size or interactions become important, especially at high number density or low temperature near a phase transition. The Van der Waals equation adds such corrections. Independently, at sufficiently low temperature or high number density, quantum statistics can become essential: even particles with negligible interactions need not obey the classical ideal gas law. The classical statistical assumption is the nondegenerate gas regime, which is distinct from neglecting interactions. ^{[ref] [ref]}

The inertia tensor packages the moments of inertia about every axis through a specified point $O$ into one symmetric matrix. An axis still needs a direction: if $\mathbf{n}$ is a unit vector, its scalar moment of inertia is $I_{O,\mathbf{n}}={\mathbf{n}}^T{\mathbf{I}}_O\mathbf{n}$. The same mass layout can therefore be easy to spin about one direction and harder about another. The inertia tensor also determines rotational kinetic energy and relates angular velocity to angular momentum in the appropriate rigid body model.

Take position vectors $\mathbf{r}=(x,y,z)$ relative to $O$, expressed in an orthonormal basis. For point masses $m_a$, or a continuous distribution with mass density $\rho$ and $dm=\rho dV$,Here ${\delta }_{ij}$ is the Kronecker delta. Equivalently, using the identity matrix $1$ and an outer product,Every entry has SI units $\mathrm{kg}{\mathrm{m}}^2$. The diagonal entries are axial moments of inertia; the off-diagonal entries carry a minus sign in this convention. Some texts instead define the integrals without the prefixed minus sign $\int xydm$, $\int xzdm$, $\int yzdm$ as their named off-diagonal quantities and put the minus signs into the displayed matrix. Check that convention before combining formulas. ^{[ref] [ref]}

The symmetric matrix is positive semidefinite, because its quadratic form isThe dot product and cross product identify this as a sum of squared distances when $\mathbf{u}$ is a unit vector. A zero moment of inertia occurs precisely when all the mass lies on that axis, up to sets of zero mass. A genuine three-dimensional body with positive mass density in a volume has strictly positive axial moments of inertia. Do not confuse this inertia tensor with the mass second-moment tensor $\mathbf{S}=\int \mathbf{r}{\mathbf{r}}^{T}dm$: they obey ${\mathbf{I}}_O=\mathrm{tr}(\mathbf{S})1-\mathbf{S}$, where $\mathrm{tr}$ denotes the matrix trace.

If $O$ is fixed in an inertial frame and each velocity is $\mathbf{v}=\mathbf{\omega }\times \mathbf{r}$, summing the individual kinetic energies and angular momenta givesThese formulas also describe angular momentum and rotational kinetic energy relative to the centre of mass $G$ of a translating rigid body. In that case the total kinetic energy in the inertial frame is $K_{\mathrm{total}}=\frac{1}{2}M|{\mathbf{V}}_G{|}^2+\frac{1}{2}{\mathbf{\omega }}^T{\mathbf{I}}_G\mathbf{\omega }$. For a general moving point $O$, the body's actual velocities include ${\mathbf{V}}_O$, so ${\mathbf{I}}_O\mathbf{\omega }$ alone need not be its angular momentum about $O$. ^{[ref] [ref]}

Changing the orthonormal basis changes components, rather than the physical distribution: if the columns of an orthogonal matrix $Q$ are the new basis vectors expressed in the old basis, then ${\mathbf{I}}_O^{\prime }=Q^T{\mathbf{I}}_{O}Q$. Changing the reference point is a different operation, governed by the tensor parallel-axis theorem. Both principal axes and principal moments of inertia must therefore be quoted with their reference point.

An exact planar example makes the off-diagonal entries tangible. Join four equal point masses $m$ with negligible-mass rigid connectors at $a(2,1,0)$, $a(-2,-1,0)$, $a(1,-1,0)$ and $a(-1,1,0)$, where $a>0$ is a length. Their centre of mass is $O=(0,0,0)$. Summing $x^2$, $y^2$ and $xy$ gives $10a^2$, $4a^2$ and $2a^2$, respectively, soThe $z$ axis is a principal axis, and $I_{zz}=I_{xx}+I_{yy}$ illustrates the perpendicular axis theorem. The $x$ axis is not a principal axis: angular velocity $\mathbf{\omega }=(\Omega ,0,0)$ gives angular momentum $m{a}^2\Omega (4,-2,0)$, which points in a different direction. A bearing can enforce fixed-axis rotation despite this difference, supplying the necessary transverse torque.

For an in-plane unit vector $\mathbf{n}=(\cos \theta ,\sin \theta ,0)$, the planar model hasIts extrema are the principal moments of inertia $(7-\sqrt{13})m{a}^2$ and $(7+\sqrt{13})m{a}^2$; the remaining principal moment of inertia is $14m{a}^2$. These values follow from the two-by-two characteristic polynomial $(4-\lambda )(10-\lambda )-4$, so they can be checked without trusting a numerical eigenvalue routine. At a stipulated angular speed $\Omega =1rad/s$, with $m=1\mathrm{kg}$ and $a=1\mathrm{m}$, rotation about $x$, $y$ or $z$ has rotational kinetic energy $2$, $5$ or $7\mathrm{J}$, respectively.

A truly three-dimensional example adds two more equal point masses at $a(0,0,1)$ and $a(0,0,-1)$. Each contributes $m{a}^2\mathrm{diag}(1,1,0)$ to the inertia tensor, givingThe in-plane principal axes stay the same while their principal moments of inertia each increase by $2m{a}^2$. The $z$ moment of inertia stays unchanged because the added point masses lie on that axis. This comparison changes the total mass from $4m$ to $6m$; it holds the original four point masses fixed rather than redistributing a fixed total mass.

For fixed particle masses and positive total mass, write ${\mathbf{x}}_i=\mathbf{R}+{\mathbf{y}}_i$ with $\mathbf{R}$ the centre of mass. The weighted relative velocities sum to zero, so expanding the kinetic energy eliminates the mixed term and yields the displayed decomposition. For a rigid body, the relative contribution is the rotational kinetic energy $I{\omega }^2/2$ about the chosen centre-of-mass axis.

For $0<I_1<I_2<I_3$, the transverse linear stability rates about steady rotation with signed component $\Omega \ne 0$ areAxes one and three have transverse eigenvalues $\pm i{\nu }_1$ and $\pm i{\nu }_3$, producing bounded oscillations. Axis two has transverse eigenvalues $\pm {\lambda }_2$, giving growing and decaying exponential functions. A neutral longitudinal perturbation corresponds to a small change of $\Omega$. Generic intermediate-axis disturbances contain a growing component; a precisely chosen decaying direction is exceptional. For the normalized intermediate-axis model with $|\Omega |=1$, the values are ${\nu }_1=1/\sqrt{3}$, ${\nu }_3=1$ and ${\lambda }_2=1/\sqrt{3}$.

Proof outline: the linearization of the Euler equations for a torque-free rigid body about the third principal axis gives transverse perturbations obeyingand henceCyclic permutation gives the other rates. For nonlinear Lyapunov stability of the extreme rotations, use the torque-free rigid-body invariants: the conserved quadratic forms $L^2-2I_{1}E$ and $2I_{3}E-L^2$ have strictly positive transverse coefficients about axes one and three respectively. They bound transverse angular velocity, while $E$ bounds its longitudinal change near nonzero rotation. Imaginary transverse eigenvalues alone would not establish this nonlinear conclusion.^[ref]

The practical lesson is to choose a stable principal axis when using rotation to maintain an instrument's pointing direction. Both extreme principal moments of inertia are stable in the ideal torque-free model; if internal damping lowers rotational kinetic energy while approximately preserving angular momentum, the minimum energy $E=L^2/(2I_3)$ instead favors the largest principal moment of inertia. A control system must account for torque, changing mass distributions and flexibility before applying this ideal result.

When material sticks to a rotor, the incoming angular momentum flux ${\dot{L}}_{\mathrm{in}}$ and external torque ${\tau }_{\mathrm{ext}}$ obey $d(I\omega )/dt={\dot{L}}_{\mathrm{in}}+{\tau }_{\mathrm{ext}}$. With zero incoming axial angular momentum and zero external axial torque, $I\omega$ is constant. For deposition at fixed radius $a$ at mass flow rate $\alpha$, $\dot{I}=\alpha a^2$ and the rotational kinetic energy satisfies $\dot{K}=-\alpha a^2{\omega }^2/2$. This loss is the irreversible cost of bringing the incoming material up to tangential speed; it does not violate conservation of angular momentum.

Molecular rotation changes a molecule's orientation. Its rotational kinetic energy depends on the arrangement and masses of its chemical atoms. A rigid rotor freezes internal distances to isolate this motion; rotational spectroscopy tests the resulting energy eigenvalues through transitions between quantum states. The molecular beam electric resonance experiment provides a complementary probe of the coupling to an electric field.

The moment of inertia about a specified axis measures how a mass distribution resists changes of angular velocity about that axis. For point masses $m_i$ at perpendicular distances $r_{\perp i}$, or a continuous distribution with mass density $\rho$,Its SI unit is $\mathrm{kg}{\mathrm{m}}^2$. The distance is to the entire axis, not to one point on it: moving a point mass along the axis leaves its contribution unchanged. The moment of inertia depends on both the distribution and the chosen axis; it is not determined by total mass alone. About an axis fixed in an inertial frame, a rigid body with larger moment of inertia has smaller angular acceleration under the same net axial torque, and greater rotational kinetic energy at the same angular speed.^[ref]

For a simple model, put four equal point masses of total mass $M$ on light rigid spokes at radius $a$. Their moment of inertia is $M{a}^2$. Doubling $a$ quadruples the moment of inertia without changing $M$. The illustrated placements $a=R/3$ and $a=R$ give $M{R}^2/9$ and $M{R}^2$. Actual spokes and finite-sized weights contribute their own moments of inertia; the point mass model neglects them. Repositioning weights during rotation also breaks the fixed-distribution assumption, so it requires a separate work and angular momentum calculation.

The moment of inertia of a uniform disc follows by dividing it into concentric annuli: $dm=(2M/R^2)rdr$ and $I={\int }_0^R{r}^2(2M/R^2)rdr=M{R}^2/2$. For an ideal thin ring every mass element is at $R$, so $I=M{R}^2$. A ring of finite radial thickness instead uses the moment of inertia of a uniform annulus, $I=M(R^2+b^2)/2$, with inner radius $b$. A moment of inertia is thus a compact summary of radial placement rather than a complete description of shape. For several joined pieces, their moments of inertia add only when each is calculated about the same axis.^[ref]

A physical pendulum makes the moment of inertia measurable through timing. Martín Monteiro, Cecilia Stari, Cecilia Cabeza and Arturo C. Martí used a metal bar with suspension holes $1.0\mathrm{cm}$ apart and an attached smartphone. They report bar length $1.199\mathrm{m}$, bar mass $0.2518\mathrm{kg}$ and phone mass $0.1311\mathrm{kg}$. Changing the suspension hole changes the pivot axis. Their readouts were phone gyroscope angular velocity records, analyzed by autocorrelation or a fit to a sinusoid, and an independent photogate; the logged records for the sinusoid fits lasted $10\mathrm{s}$. They observed agreement between the three timing methods and the geometry-based model, including a minimum of oscillation period as the suspension point varied. These are reported apparatus details and observations, not measurements made for this article.^[ref]

To infer a moment of inertia from such a measurement, determine the total mass $M_{\mathrm{tot}}$ including every attached part, locate the assembly's centre of mass $G$, and measure its distance $h>0$ below the horizontal pivot $O$. Record successive crossings in the same direction, or the elapsed time $t_N$ for $N$ complete oscillations, giving oscillation period $P=t_N/N$. The physical pendulum model with a fixed pivot, a uniform gravitational field, negligible friction and the small-angle approximation givesThe crossing times, masses and geometric distances are the measured inputs; $I_{O,\mathrm{inferred}}$ is an inference conditional on this model, while an integral over the assumed shape supplies an independent model prediction. The parallel axis theorem converts the inferred pivot moment of inertia to one about a parallel axis through $G$. The controls and interpretation of that comparison are described there. An axial measurement determines one scalar moment of inertia, not the complete inertia tensor needed for general rotation.

The parallel axis theorem transfers a moment of inertia from an axis through the centre of mass $G$ to a parallel axis through $O$:Here $M$ is the total mass and $d$ is the shortest perpendicular distance between the two axes. Both moments of inertia concern the same mass distribution and the same axis direction. The axis for $I_G$ must pass through the centre of mass; $I_O$ may refer to an axis outside the body. No uniform-density or planar-shape assumption is needed. Within any family of parallel axes, the centre of mass axis minimizes the moment of inertia.^[ref]

A short proof isolates why the centre of mass matters. Choose parallel axes perpendicular to an $xy$ plane, put $G$ at the origin, and place $O$ at $(d,0)$. Thenbecause $\int xdm=0$ in centre of mass coordinates. Individual mass elements can move closer to the new axis, but their weighted cross terms cancel. For arbitrary parallel axes $A$ and $B$, use $I_B-I_A=M(d_B^2-d_A^2)$, where each distance is measured from the parallel centre of mass axis; adding $M$ times the squared separation of $A$ and $B$ is generally incorrect. A displacement along the axis has $d=0$. Changing the axis direction instead requires the inertia tensor or, in its applicable planar setting, the perpendicular axis theorem.

For a uniform disc, mass $M$ and radius $R$, the central perpendicular moment of inertia is $I_G=M{R}^2/2$. Shifting that axis to the rim gives $d=R$ and $I_O=3M{R}^2/2$, three times $I_G$. In the computed $M=2\mathrm{kg}$, $R=0.30\mathrm{m}$ example, $I_G=0.090\mathrm{kg}{\mathrm{m}}^2$, $M{d}^2=0.180\mathrm{kg}{\mathrm{m}}^2$ and $I_O=0.270\mathrm{kg}{\mathrm{m}}^2$. At angular speed $10rad/s$ the rim-pivoted disc has rotational kinetic energy $13.5\mathrm{J}$. Its centre of mass now moves, so $\frac{1}{2}{I}_O{\omega }^2$ already includes that translation; adding a separate centre of mass translation term to it would count that contribution twice.

The physical pendulum measurement described under moment of inertia provides a practical check: infer $I_O=Mgh(P/2\pi {)}^2$ at several pivot positions, then subtract $M{h}^2$ to obtain $I_G$. Keep all pivot axes parallel and measure $h$ to the centre of mass of the entire moving assembly. The same assembly should yield the same $I_G$ within measurement precision. If a smartphone is attached, its mass and its own moment of inertia belong in both the geometric prediction and the centre of mass calculation. The published bar-and-phone study explicitly adds both contributions using the parallel axis theorem.^[ref]

For a careful replication, release small swings without a push, repeat timings at several small amplitudes, and check that the extrapolated oscillation period is stable as amplitude decreases. Time many complete oscillations and repeat the run; two crossings of a photogate per cycle must not be mistaken for two cycles. Inspect decay of amplitude and compare early and late timings to detect significant pivot friction or air drag. Mark the bar and phone mount to detect relative slip, tighten the attachment, and check that the pivot support does not rock or migrate between holes. Repeat with an independent photogate or video readout to check the phone's timing. These are proposed controls for the replication, not additional results reported by the authors. The ideal model excludes appreciable damping, large swings and flexing; subtraction of $M{h}^2$ also magnifies relative errors when $I_G$ is much smaller than $I_O$.

The oscillation period need not increase monotonically with pivot distance: the shift raises the moment of inertia but also increases the gravitational restoring torque. For a computed slender uniform bar model of length $L$, $I_G=M{L}^2/12$ andThis expression has its minimum at $h=L/\sqrt{12}$. With $L=1\mathrm{m}$, $h=0.50\mathrm{m}$ and stipulated $g=9.81m/{\mathrm{s}}^2$, the prediction is $P\simeq 1.638\mathrm{s}$; it is not a recorded timing or a prediction for the bar with a phone attached. The competing terms explain why a timing experiment can reveal a mass distribution rather than merely its weight.

Take the origin at the centre of mass, so ${\int }_B\rho \mathbf{r}{d}^{3}r=0$, and write $r_i=\mathbf{r}\cdot {\mathbf{e}}_i$ in the specified orthonormal basis. The inertia tensor isHere ${\delta }_{ij}$ is the Kronecker delta. For a rigid body with angular velocity $\mathbf{\omega }$, this tensor gives $\mathbf{L}=\mathbf{I}\mathbf{\omega }$ and rotational kinetic energy $T=\mathbf{\omega }\cdot \mathbf{I}\mathbf{\omega }/2$.

The inertia tensor is a real symmetric matrix, so it admits an orthonormal basis of eigenvectors. The lines in these eigenvector directions through the centre of mass are its principal axes; the corresponding eigenvalues $I_1,I_2,I_3$ are its principal moments of inertia:If an eigenvalue repeats, the corresponding principal axes need not be unique.

The magnetohydrodynamic total pressure isThe Lorentz force has been split into the gradient of magnetic pressure and the remaining magnetic tension term. An appropriate boundary model is impermeable, perfectly conducting stationary cylindrical walls with no initial normal magnetic flux: $u_r=B_r=0$ at $r=a,b$. In an inviscid fluid this is a slip condition, not a no-slip condition on $u_{\varphi }$ or $u_z$. It also makes the tangential ideal electric field vanish at the walls. Take perturbations periodic in $\varphi$ and periodic or Fourier-resolved in $z$; the wall reaction supplies the normal pressure force without an extra prescribed pressure value.

Write the steady velocity and magnetic field as $U(r){\mathbf{e}}_{\varphi }$ and $B(r){\mathbf{e}}_{\varphi }$. They satisfy both divergence constraints. Their material accelerations have only radial curvature terms, and the steady ideal magnetohydrodynamic induction equation is identically satisfied because the parallel azimuthal flow and field have cancelling curvature terms. The radial momentum equation isHence the most general such equilibrium isIts gas pressure is $p=\Pi -B^2/(2{\mu }_0)$; on a bounded annulus the additive constant can be chosen to keep it positive. Neither solid-body rotation nor a current-free field is imposed by the inviscid equilibrium equations.

For the linear stability calculation, put $\Omega =U/r$, $V=B/\sqrt{{\mu }_0\rho }$, and define $h=\delta \Pi /\rho$. Use axisymmetric normal modes proportional to $e^{ikz-i\omega t}$. Denote the velocity amplitudes by $(u,v,w)$ and the magnetic amplitudes in velocity units by $(b_r,b_{\varphi },b_z)=\delta \mathbf{B}/\sqrt{{\mu }_0\rho }$. Linearization of the cylindrical momentum equations givesThe linearized ideal magnetohydrodynamic induction equation and divergence equations areThe curvature terms are essential: although the perturbations have no azimuthal dependence, the cylindrical unit vectors do.

For $\omega \ne 0$, $b_r=b_z=0$. Introduce the radial Lagrangian fluid displacement by $u=-i\omega \xi$. The azimuthal amplitudes reduce toConsequently the radial and vertical equations, combined with the incompressible flow constraint, givewhere the Michael criterion for axisymmetric toroidal-field interchange coefficient isFor $k\ne 0$, the required single displacement equation is thereforeThe same differential equation holds for $u$, which is a constant multiple of $\xi$.

Set $y=r\xi$. Multiplication by $y^{\ast }$ and integration by parts with the wall conditions gives the global variational form of the Michael criterion:The denominator is strictly positive for a nonzero displacement, so every mode's ${\omega }^2$ is real. If $\mathcal{D}\ge 0$ everywhere, no exponentially growing mode exists. Conversely, if the continuous coefficient is negative somewhere, choose a smooth test function supported in a negative interval; its quotient is negative. For sufficiency, let $A=-d(r^{-1}d/dr)/dr+k^2/r$ be the positive Sturm-Liouville operator with Dirichlet boundary conditions and let $C$ multiply by $k^2\mathcal{D}/r$. The generalized mode equation ${\omega }^{2}Ay=Cy$ is equivalent to the compact operator and self-adjoint operator problem $A^{-1/2}C{A}^{-1/2}f={\omega }^{2}f$. The negative test quotient guarantees a negative eigenvalue and therefore a growing solution $\omega =i\gamma$. This argument remains valid when the coefficient changes sign, without incorrectly assuming a positive weight in Sturm-Liouville theory.

Thus, for any fixed nonzero vertical wavenumber,Equality is marginal. When $k=0$, the incompressible flow constraint and both rigid-wall conditions force the radial velocity to vanish, so that special case has no radial interchange mode. When the field vanishes, the coefficient reduces to ${\kappa }^2=r^{-3}(r^4{\Omega }^2{)}^{\prime }$, recovering the Rayleigh centrifugal stability criterion.

For a Keplerian disk, ${\kappa }^2={\Omega }^2$. If $B_{\varphi }\propto r^p$, the toroidal interchange field threshold in a thin Keplerian disk isFor ordinary radial gradients on the scale $r$, the Alfvén speed must therefore be comparable to the orbital speed. A field on the gas pressure scale has $V\lesssim c_s\sim H\Omega \ll r\Omega$ in a thin disk, and cannot normally meet this condition. Exceptionally sharp field gradients must be evaluated with the exact derivative criterion; the orbital-speed estimate is not independent of the field length scale.

This is a toroidal magnetic interchange driven by the field/current distribution, opposed by restoration set by the radial epicyclic frequency. It is not the usual weak vertical-field magnetorotational instability. In the vertical-field problem, a nonzero $k{B}_z$ bends magnetic field lines and couples displaced fluid elements, allowing decreasing $\Omega$ to supply rotational kinetic energy even when ${\kappa }^2>0$. Here the axisymmetric perturbation has no scalar variation along the purely toroidal background field, and no such weak-field coupling: a sufficiently adverse toroidal-field gradient, rather than merely $d\Omega /dr<0$, is required. Both are magnetic disk instabilities, but their available energy and instability criteria differ.

For the moment of inertia of a uniform solid sphere, write the mass density as $\rho =3m/(4\pi a^3)$. In spherical coordinates about the chosen axis, the squared perpendicular distance is $R^2{\sin }^2\vartheta$. The volume element is $R^2\sin \vartheta dRd\vartheta d\phi$, so

Take $0<\alpha <\pi /2$ and $\ell >0$, with a fixed plane. The normal force has magnitude $N=mg\cos \alpha$ and is perpendicular to the plane. On the perfectly smooth plane there is no friction. Both gravity and the normal force have zero torque about the marble's centre, so a marble released from rest does not rotate. The component of Newton's second law down the plane is $m{A}_1=mg\sin \alpha$, henceThe normal force does no work because the velocity is tangent to the plane. Gravity is conservative, so mechanical energy, the sum of translational kinetic energy and gravitational potential energy, is conserved.

For rolling without slipping, let the static friction force have magnitude $f$, pointing up the plane. It supplies the torque that spins the marble while reducing its translational acceleration. If $\Omega$ is the angular velocity and $A_2$ the acceleration down the plane, the rolling constraint is $a\Omega =\dot{s}$, hence $a\dot{\Omega }=A_2$. The translation and rotation equations areThus $f=I{A}_2/a^2$ andThese are the equations for rolling acceleration with rotational inertia; the plane must be rough enough to supply this static friction.

The point of contact is instantaneously at rest relative to the fixed plane, so static friction does no total work on the rolling marble. Equivalently, its translational power is $-f\dot{s}$ and its rotational power is $fa\Omega =f\dot{s}$, which cancel. The normal force also does no work. Consequently mechanical energy is conserved here too, now with rotational kinetic energy included:where $H$ is the centre's height. No energy loss is implied by the presence of ideal static friction. Constant acceleration from rest gives

For the hollow marble, put $I=\lambda m{a}^2$ in the same equations. The smooth-plane time is unchanged, while the rolling acceleration becomes $g\sin \alpha /(1+\lambda )$. ThereforeThis again assumes rolling without slipping and enough static friction; the ratio depends on the mass distribution through its moment of inertia.

In a frame centered at the fixed point or centre of mass, a rotating material point has velocity $\dot{r}=\omega \times r$. The identity $|\omega \times r{|}^2=|\omega {|}^2|r{|}^2-(\omega \cdot r{)}^2$ converts the rotational kinetic energy into $T=\frac{1}{2}{\omega }_i{I}_{ij}{\omega }_j$, where the inertia tensor isLikewise integrating $r\times (\omega \times r)$ gives the angular momentum $\boxed{L=I\omega }$. These expressions refer to rotational motion about the chosen origin.

Assume the particle masses $m_i$ are fixed and their total mass $M={\sum }_i{m}_i$ is positive. The centre of mass and the relative position vectors areTheir weighted sum is zero, so differentiation gives ${\sum }_i{m}_i{\dot{\mathbf{y}}}_i=0$. Expanding the kinetic energy with ${\dot{\mathbf{x}}}_i=\dot{\mathbf{R}}+{\dot{\mathbf{y}}}_i$ makes the mixed term vanish:This kinetic-energy decomposition about the centre of mass separates translation from internal motion without assuming any particular force law.

For the rigid body, its relative velocity is ${\dot{\mathbf{y}}}_i=\omega \mathbf{n}\times {\mathbf{y}}_i$. The cross product identity for a unit vector gives $|\mathbf{n}\times {\mathbf{y}}_i{|}^2=|{\mathbf{y}}_i{|}^2-(\mathbf{n}\cdot {\mathbf{y}}_i{)}^2$, the squared perpendicular distance to the rotation axis. Hence the moment of inertia of the particle is $I_i=m_i[|{\mathbf{y}}_i{|}^2-(\mathbf{n}\cdot {\mathbf{y}}_i{)}^2]$, and summing the rotational kinetic energy yields

Let $q=\sigma /L$, with $L>0$, and let $Z=X^1+i{X}^2$. Direct differentiation gives $\ddot{Z}=Z^{\prime \prime }=-Z/L^2$, while the time coordinate also satisfies the wave equation. Using the Minkowski metric,Thus both Virasoro constraints hold, and the induced worldsheet metric is $g_{\mu \nu }={\cos }^{2}q\mathrm{diag}(-1,1)$. The configuration solves the Nambu–Goto action equations in its interior.