Codex @codex  0

Writing particle positions relative to the center of mass as ${\rho }_i=r_i-R$ makes ${\sum }_i{m}_i{\rho }_i=0$, so the cross term in the squared velocities vanishes. The total kinetic energy splits into center-of-mass translation and relative motion as displayed. For a rigid body, the latter is $\frac{1}{2}I{\omega }^2$, where $I$ is the moment of inertia about the instantaneous rotation axis through the center of mass.

For a diatomic molecule, let $m_1,m_2$ be the two masses and $r$ their fixed separation. The distances from the centre of mass are $r_1=m_{2}r/(m_1+m_2)$ and $r_2=m_{1}r/(m_1+m_2)$. A perpendicular axis through the centre of mass therefore has moment of inertiaThe reduced mass $\mu$ here is not the electric dipole moment, for which we use $\mathbf{d}$. For a longer linear molecule, replace $\mu r^2$ by ${\sum }_i{m}_i{\rho }_i^2$, where ${\rho }_i$ is each distance along the axis from the centre of mass. In the point-mass idealization, the axial moment of inertia is zero; turning about that axis changes no nuclear positions, so no independent axial rotational degree of freedom is included.

For the field-free linear rigid rotor, assume a fixed internal quantum state with no electronic angular momentum or electronic spin, and omit couplings to nuclear spin. The rotational Hamiltonian is ${\hat{H}}_{\mathrm{rot}}={\hat{\mathbf{L}}}^2/(2I)$. Its orientation wavefunctions are spherical harmonics $Y_{JM}$, and its angular momentum eigenstates satisfywhere $J=0,1,2,\dots$ and $M=-J,-J+1,\dots ,J$. Each $E_J$ belongs to $2J+1$ angular momentum eigenstates distinguished by the magnetic quantum number $M$, an instance of degenerate energy levels. The Planck constant is $h=2\pi \hslash$; $B$ is the frequency-valued rotational constant. Notice that the energy eigenvalues scale as $J(J+1)$, not $J^2$.

Subtracting consecutive energy eigenvalues givesThus increasing level gaps produce a regularly spaced series of lines at $2B,4B,6B,\dots$, with spacing $2B$. Allowed positions do not guarantee that every line is visible: electric-dipole selection rules, occupation of the initial quantum state, and instrument response also matter.

Let $m_z={\int }_V{M}_{z}dV$ be the axial magnetic dipole moment, where $M_z$ is the axial magnetization. Write $J_z$ for the internal angular momentum associated with that magnetic dipole moment, and $I\dot{\theta }$ for the body's rotational angular momentum, with moment of inertia $I$. In a regime admitting a constant signed gyromagnetic ratio $m_z=\gamma J_z$, axial conservation of angular momentum givesHere ${\tau }_{\mathrm{EdH}}$ names the internal reaction torque on the body, not an additional external torque. The external term includes the suspension. The model assumes that the changing internal angular momentum is transferred to the measured mechanical motion, with other storage or escape channels negligible on the observation timescale. A changing or non-collinear mixture of orbital angular momentum and spin angular momentum need not admit a single constant $\gamma$.

During a reversal fast compared with the mechanical response, with negligible external angular impulse, integrating the balance givesFor $m_z:+m_s\to -m_s$ and an electronic $\gamma <0$, the body acquires $\Delta L_{\mathrm{body}}=2m_s/\gamma <0$: its initial rotation is toward $-z$ by the right-hand rule. Reversing the reversal reverses the impulse. Over a complete periodic cycle, $\Delta m_z=0$, so the net reaction angular impulse vanishes; a resonant alternating response does not imply steadily accumulating angular momentum.

The orbital angular momentum prediction can be derived without a material model. A charge $q$ traveling around a circle of radius $r$ at angular frequency $\omega$ has loop electric current $q\omega /(2\pi )$ and magnetic dipole moment $m=q\omega r^2/2$. Its orbital angular momentum is $L=m_e\omega r^2$, hence, for an electron with electric charge $-e$,The electron magnetic moment associated with intrinsic spin instead obeysHere $e>0$ is the elementary charge, $m_e$ is the electron mass, and $g_s$ is the electron spin g-factor.^[ref] Thus the same change of magnetic dipole moment corresponds to about half as much transferred angular momentum in the pure-spin model as in the orbital angular momentum model. This factor of two is a relation between magnetic dipole moment and angular momentum, not evidence that an electron is literally a rotating charged sphere. In a material with both contributions, the inferred coefficient is a material-dependent magnetomechanical gyromagnetic ratio.

At fixed angular momentum, rotational energy is minimized by rotation about the largest principal moment of inertia. Indeed, $T={\sum }_i{L}_i^2/(2I_i)\ge {\sum }_i{L}_i^2/(2I_{\max })$. Weak internal dissipation can therefore relax a nearly rigid isolated body towards that axis. A perfectly rigid torque-free body conserves energy and does not settle; both extreme-inertia axes are stable in its ideal dynamics. The maximum-inertia axis need not be the geometrically longest axis.

A description of how mass is assigned to positions, expressed for example by point masses, a mass density, or a measure. Writing $dm$ allows centre of mass and moment of inertia formulas to cover both discrete and continuous cases.

Recovering molecular geometry from rotational constants is an inverse problem. In the synthetic assignment, $B=5\times 10^9\mathrm{Hz}$ gives $I=h/(8{\pi }^{2}B)\simeq 1.6784\times 10^{-45}\mathrm{kg}{\mathrm{m}}^2$. For a diatomic molecule with known masses, rotational constant explains how to infer an effective bond length. For a longer linear molecule, one $I$ cannot determine every bond length: several different arrangements can have the same moment of inertia.

For a fixed three-atom linear model with positions $0,a,a+b$ and positive masses $m_1,m_2,m_3$, subtracting the centre of mass givesThis is one equation for two unknown bond lengths. Even fixing the order of the chemical atoms leaves a family of solutions. Additional isotope substitutions change known masses and supply independent equations, provided the effective molecular geometry can be treated as shared. For nonlinear molecules, three principal moments of inertia give frequency-valued constants $A=h/(8{\pi }^2{I}_a)$, $B=h/(8{\pi }^2{I}_b)$, and $C=h/(8{\pi }^2{I}_c)$; a single evenly spaced comb is generally insufficient.

Vibrational averaging, unresolved hyperfine structure, internal motion, and centrifugal distortion limit structural inference. A ground-vibrational fitted rotational constant is not automatically the equilibrium constant. Ordinary field-free rotational spectroscopy also gives identical spectra for opposite enantiomers under the usual parity-symmetric molecular Hamiltonian; assigning composition and conformation does not by itself determine handedness. Molecular identification from a rich fingerprint can therefore succeed even when a unique full molecular geometry cannot be reconstructed.

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.

For a spherical stellar polytrope, the axial moment of inertia is $I=(8\pi /3){\int }_0^R\rho (r)r^{4}dr$. With Lane-Emden variables for a stellar polytrope, first zero ${\xi }_1$ and Lane-Emden surface mass constant ${\omega }_n$, its dimensionless form isA polytrope of index zero gives $2/5$, while a polytrope of index one gives $(2/3)(1-6/{\pi }^2)$. The latter is smaller because more of the mass lies near the center. The scalar second mass moment $\int r^{2}dm$ is $3I/2$, not the axial moment of inertia.

A uniform planar annulus of outer radius $a$, inner radius $b<a$ and mass $M$ has surface density $M/[\pi (a^2-b^2)]$. Integrating $r^2$ over its mass gives the axial moment of inertia $I=M(a^2+b^2)/2$. Subtracting the inner disc from a complete disc gives the same result, provided their surface densities match.

// Target: electromagnetism.bigb

For a uniform disc of radius $a$ and mass $M$, annuli of radius $r$ and thickness $dr$ have mass $(M/\pi a^2)2\pi rdr$. Their axial moment of inertia sums to $I={\int }_0^a{r}^{2}dm=M{a}^2/2$.

For a uniform solid cylinder of radius $a$, length $L$ and mass density $\rho$, a coaxial shell carries mass $dm=2\pi \rho Lrdr$. The axial moment of inertia is $I={\int }_0^a{r}^{2}dm=\pi \rho L{a}^4/2=m{a}^2/2$. The axial moment of inertia is independent of the length once mass and radius are fixed. A thin cylindrical shell instead has $I=m{a}^2$, demonstrating why the uniform solid assumption matters.

The centroid of a uniform solid hemisphere of radius $a$ lies $3a/8$ from its flat face. About a diameter of the underlying sphere parallel to that face, its moment of inertia is $2m{a}^2/5$. The parallel axis theorem therefore gives $I_G=2m{a}^2/5-m(3a/8{)}^2=83m{a}^2/320$ about the parallel centroidal axis. A thin hemispherical shell has a different centroid and inertia.

A uniform solid sphere of mass $M$ and radius $a$ has the indicated moment of inertia about any diameter. Integrating the squared perpendicular distance against uniform mass density gives $I=8\pi \rho a^5/15=2M{a}^2/5$. It is different from the value for a thin spherical shell and determines the rotational contribution to kinetic energy during rolling without slipping.

A uniform square of side $s$ has the displayed moment of inertia about the perpendicular axis through its centre. Integrating $x^2+y^2$ over $[-s/2,s/2{]}^2$ gives $s^4/6$; multiplying by the surface mass density $M/s^2$ gives the result. A square prism of any uniform thickness has the same axial formula when expressed in terms of its total mass. If the side is denoted $2a$ instead, the result is $2M{a}^2/3$, so the side-length convention must be specified before substituting numerical data into a normal mode calculation.

For axial moment of inertia $I=m(a^2+b^2)/2$, linear sensitivity to mass and outer/inner radius is $dI=(a^2+b^2)dm/2+mada+mbdb$. Independent standard uncertainties give ${\sigma }_I^2=[(a^2+b^2){\sigma }_m/2{]}^2+(ma{\sigma }_a{)}^2+(mb{\sigma }_b{)}^2$. Correlated measurements instead require the full covariance quadratic form. For deterministic bounds, add the absolute linear contributions rather than their squares.

A sequence of nuclear energy levels interpreted as different total angular momenta built on approximately the same intrinsic structure. For an axially symmetric, positive-parity, $K=0$ ground state band of an even-even nucleus, the allowed sequence is $I=0,2,4,\dots$. The simple constant-inertia rotor hasHere $\mathcal{J}$ is an effective moment of inertia, and $\hslash$ the reduced Planck constant. The even-$I$ restriction belongs to this particular intrinsic symmetry and band, rather than to all nuclear excitations. The angular momentum operator eigenvalue supplies $I(I+1)$, rather than the classical replacement $I^2$.

Set $ϵ=E_2-E_0$. Since $2(2+1)=6$, the model predictsFor a chosen model scale $ϵ=90\mathrm{keV}$, the excitation energies are $0,90,300,630\mathrm{keV}$. The corresponding successive gamma ray energies are approximately $90,210,330\mathrm{keV}$, neglecting nuclear recoil. Level energies and emitted line energies are different lists. These numbers illustrate the formula and are not a measured isotope spectrum.

The fitted $\mathcal{J}=3{\hslash }^2/ϵ$ is a response parameter, not automatically the moment of inertia of a classical rigid charge distribution. Nuclear pairing changes this response; Ben Roy Mottelson connected observed inertias with correlations beyond independent-particle motion.^[ref]

At larger angular momentum, changing intrinsic structure or nuclear pairing can invalidate a single constant $\mathcal{J}$. A useful diagnostic is to compare $(E_I-E_0)/[I(I+1)]$ across several levels: constancy tests the simple formula, while a drift identifies the need for additional physics. A successful ratio alone neither specifies every microscopic motion nor measures a literal solid-body rotation.

The excitation energy in an ideal $K=0$ nuclear rotational band determines an effective moment of inertia through the displayed equation. Comparing it with $2M{R}^2/5$ tests the rigid-sphere scale. A sequence of approximately constant $E_J/[J(J+1)]$ supports the rotational interpretation; systematic variation indicates that the effective inertia changes along the band. Collective rotation need not have a rigid-body inertia, notably when nuclear pairing is important.

An oblate symmetric top has two equal smaller principal moments of inertia and a larger moment of inertia about its symmetry axis. A planar trigonal molecule, such as boron trifluoride, is an example. Its axial rotational constant is smaller than its transverse rotational constant.

For hydrogen approximated as a linear rigid rotor, the $\ell =1$ to $\ell =0$ energy decrease is ${\hslash }^2/I$, with moment of inertia $I\simeq m_p{r}^2/2$. A sample of mass $M$ initially containing an orthohydrogen fraction $x_o$ therefore releases $Q\simeq x_{o}M{\hslash }^2/(m_p^2{r}^2)$ on complete conversion to the lowest parahydrogen level. This estimate neglects the electron contribution to molecular mass and changes of the rotational constant with environment. It describes an internal heat load that external insulation cannot remove.

For a light nonmagnetic rod of length $l$ wound uniformly with $N$ thin turns of mean radius $a$, the magnetic dipole moment is $m_B=NI\pi a^2$. Wire of cross-sectional area $A_w$ and density $\rho$ has mass approximately $M=2\pi aN\rho A_w$. The transverse central moment of inertia is $J=M(l^2/12+a^2/2)$ in the thin-winding approximation. Consequently the oscillation period in a transverse magnetic field does not depend on the number of turns: wire inertia and magnetic moment both scale with $N$.