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Parahydrogen uses the antisymmetric spin-one-half singlet state,Here $I_{\mathrm{n}}$ labels total nuclear spin; it is not the moment of inertia $I$ in the linear rigid rotor formula. There is one nuclear-spin state for every rotational quantum state. The lowest allowed rotational level is $J=0$, and the next is $J=2$. The description “opposite spins” alone is insufficient: the relative minus sign distinguishes this singlet state from the zero-projection member of the triplet state.

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.

For an axis through the center of mass perpendicular to the bond, the point-mass moment of inertia isThe prefactor is the reduced mass $\mu$; this is the standard reduction of a two-body linear rigid rotor to $I=\mu r^2$.

Low rotational energy spacings can be excited in cold interstellar gas when electronic and most vibrational excitations are frozen out. A polar molecule such as carbon monosulfide has allowed electric dipole transition lines whose frequencies depend on its moment of inertia, providing a distinctive pattern for rotational spectral line assignment. Several matching spectral lines and their isotope shifts give a much stronger identification than one coincidence. The method is especially effective for polar species; a molecule without a permanent electric dipole moment lacks an ordinary pure electric-dipole rotational spectrum.

Use rovibrational spectroscopy combination differences so that the band origin cancels. With the assignments above, $R(2)$ and $P(4)$ end at the same upper level $J^{\prime }=3$, givingLikewise $R(3)$ and $P(3)$ start at the same lower level $J=3$, soOnly the printed wavenumbers are needed:Thus $B^{\prime }-B^{\prime \prime }\simeq 0.00429{\mathrm{cm}}^{-1}$. The rotational constants differ only slightly, so the average moment of inertia changes little on vibrational excitation. The upper value is slightly larger, corresponding to a slightly smaller effective moment of inertia. This sign is possible for a bending excitation; the intuition that excitation must always lengthen a diatomic bond is not a general rule for a polyatomic bend.

The rounded printed values are internally consistent: fitting all six positions to the same elementary model gives approximately ${\tilde{\nu }}_{\ast }=711.9945{\mathrm{cm}}^{-1}$, $B^{\prime \prime }=1.47797{\mathrm{cm}}^{-1}$ and $B^{\prime }=1.48203{\mathrm{cm}}^{-1}$, with residuals below $0.002{\mathrm{cm}}^{-1}$. The displayed combination-difference values have more digits than their physical accuracy; the meaningful scale is about $1.48{\mathrm{cm}}^{-1}$.

Set $x=r\varphi$ and $a=2\pi r$. The wrapped spatial form isHence the differential expression for the quantum rotor acts asThe box expression becomes $E_n=n^2{\hslash }^2/(8m{r}^2)$, so consistency givesthe moment of inertia of a point mass a distance $r$ from the axis. Equivalently, ${\partial }_x=r^{-1}{\partial }_{\varphi }$ transforms the kinetic operator directly. Being an eigenfunction of this differential expression is not yet sufficient: the ring boundary conditions must also hold.

For a wavefunction normalized with the angular measure $\mathrm{d}\varphi$, use ${\Phi }_n(\varphi )=\sqrt{r}{\Psi }_n(r\varphi )={\pi }^{-1/2}\sin (n\varphi /2)$. This change of normalization convention does not change the energy eigenvalue.

The substituted molecule has the larger entropy. Increased mass and moment of inertia reduce translational and rotational spacings and make more states thermally accessible. The largest contribution here is the loss of the twofold rotational indistinguishability of the oxygen ends, contributing $R\log 2$. It is a rotational symmetry number correction for one isotope-defined species, not an entropy of mixing between different molecules.

Write $M=m_A+m_B$ and $\mu =m_A{m}_B/M$. An atom has no rotational or vibrational internal factor under the stated electronic assumption. The transition-state moment of inertia is $I=\mu R^2$, so the high-temperature rigid rotor result isThe translational factors therefore giveSubstituting into the transition state theory expression and simplifying,For distinct atoms, $\sigma =1$. This is a number-density rate coefficient, with units of volume per time. A rate law in molar concentrations uses $N_A{k}_{2nd}$. The derivation assumes a classical high-temperature rotor at the dividing surface and a transmission coefficient of one.

With integer fluorine mass $m_B=19m_{\mathrm{u}}$ and $R=1.5\mathring{\mathrm{A}}$,This angle is below the planar limit $90^{\circ }$, consistent with the proposed pyramid. The sulfur mass does not enter this axial moment of inertia.

For a linear rigid rotor, dividing the angular momentum squared eigenvalue by twice the moment of inertia givesThe quantum degeneracy counts the values $M=-J,\dots ,J$ with the same energy. It excludes nuclear-spin multiplicity and any fine or hyperfine splitting absent from this model.

Apply the chain rule to $f_0=f_0(E)$ with $E=H_0(\mathbf{J})$:On the resonance selected by the Dirac delta function, $\mathbf{k}\cdot \mathbf{\Omega }=k_{\varphi }{\Omega }_{\mathrm{p}}$. ThusEvery weight is nonnegative, and $-f_0^{\prime }(E)>0$. Strict positivity additionally requires a populated nonaxisymmetric resonance with nonzero coupling on a set of positive surface measure. The unconditional statement is nonnegativity, rather than positive definiteness. As a concrete counterexample to strict positivity of this formula, retain one nonzero conjugate harmonic pair with $k_{\varphi }\ne 0$ and choose its smooth coefficient ${\Psi }_{\mathbf{k}}$ supported in an open region $|{\Delta }_{\mathbf{k}}|>d>0$. The perturbation is nonzero but has zero resonant coupling, so $\mathcal{T}=0$ even when $f_0^{\prime }(E)<0$. The even simpler zero-perturbation case also gives zero. Whether a particular physical stellar bar has nonzero resonant overlap is additional information.

For a coupled stellar bar and halo, the halo gains positive angular momentum and exerts the opposite torque on the stellar bar. If the halo mass is $M_h$ and the rigid stellar bar has constant positive moment of inertia $I_b$, thenThis bar-halo angular-momentum transfer is a form of dynamical friction: a resonantly coupled stellar bar tends to slow down while the dark matter halo spins up. Actual long-term evolution changes both the stellar bar pattern and the halo distribution function, so the fixed-pattern response must then be updated. If the stellar bar changes its moment of inertia, the general relation is $d(I_b{\Omega }_{\mathrm{p}})/dt=-M_h\mathcal{T}$; positive moment of inertia alone would not fix the sign of $d{\Omega }_{\mathrm{p}}/dt$ without that rigidity assumption.

In the quadratic form, the two equal off-diagonal contributions together supply the mixed term. Comparing coefficients givesFor a nontrivial physical moment of inertia scale take $I_0>0$; then this inertia tensor is positive definite.

The moment of inertia of a uniform length-$2l$ rod about its transverse central axis isThe kinetic energy decomposition about the center of mass gives $\boxed{T=M{V}^2/2+M{l}^2{\dot{\theta }}^2/6}$ for center speed $V$ and rod angle $\theta$.

For a falling arch with $0<\alpha <\pi /2$, the symmetric two hinged rods with both ends sliding on a floor have the following motion. The hinge remains vertically over the arch center. Their centers are ${\mathbf{R}}_{\pm }=(\pm l\cos \theta ,l\sin \theta )$, so each has squared speed $l^2{\dot{\theta }}^2$. Total kinetic energy is $T=4M{l}^2{\dot{\theta }}^2/3$, and total gravitational potential energy is $U=2Mgl\sin \theta$. Conservation of energy from rest at angle $\alpha$ gives ${\dot{\theta }}^2=3g(\sin \alpha -\sin \theta )/(2l)$. At the hinge impact $\theta =0$, its height derivative is $2l\cos \theta \dot{\theta }$, givingThe floor-contact assumption is consistent: differentiating the energy relation gives $\ddot{\theta }=-3g\cos \theta /(4l)$, and the normal reaction on each outer endpoint is $N=Mg[1+9{\sin }^2\theta -6\sin \alpha \sin \theta ]/4$. Completing the square shows $N\ge Mg{\cos }^2\alpha /4\ge 0$ throughout the fall.

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.

The rim-mass wheel has moment of inertia $I=M{R}^2$. A uniform solid cylinder has axial moment $\frac{1}{2}m{S}^2$, so the roller of mass $M/2$ has $J=M{S}^2/4$. No slip gives equal contact speeds, $R\omega =S\sigma$, when $\omega ,\sigma$ denote angular-speed magnitudes. ThereforeWith parallel signed axes the two angular velocities have opposite signs. Their kinetic energies add regardless of that convention:The reflected inertia of a no-slip wheel and roller must be used because accelerating the wheel also accelerates the roller. The transmitted contact forces redistribute energy without dissipating it. For an applied wheel torque $T$, the input power is $T\omega$, so the work-energy theorem gives $K\omega \dot{\omega }=T\omega$, or $K\dot{\omega }=T$; this also holds at rest by continuity or the constrained equation of motion.

Assume the driving constants $Q,\Omega$ are positive. With the speed-dependent torque,Separating this linear equation givesFor the fan, its resisting roller torque dissipates power $\gamma {\sigma }^3$. Expressing this in the wheel coordinate gives a resisting wheel torque $\gamma (R/S{)}^3{\omega }^2$. Thus, with $\Gamma =\gamma (R/S{)}^3$,The right-hand side is positive at zero, strictly decreasing for nonnegative $\omega$, and has a unique positive zero; the wheel starting at rest approaches this terminal angular speed. HenceFor a nonzero fan drag this is less than $\Omega$, and as $\gamma \to 0$ it tends to $\Omega$. The cubed radius ratio is essential: one factor arises from converting roller power to wheel torque, in addition to the two factors in ${\sigma }^2$.

Label the initially left-hand rod $A$ and the right-hand rod $B$. At impact their centres are $(0,\ell )$ and $(0,-\ell )$, and their contact points are both the origin. Each rod has total mass $2m$ and moment of inertiaLet the impulse on $B$ be $+J{\mathbf{e}}_x$, so the impulse on $A$ is $-J{\mathbf{e}}_x$. The impulsive form of Newton's second law givesThe contact lever arms from the centres are $(0,-\ell )$ for $A$ and $(0,+\ell )$ for $B$. Both angular impulses are $-\ell J{\mathbf{e}}_z$. Thus, taking counterclockwise rotation as positive,Equal and opposite linear impulses therefore produce equal clockwise rotations, not opposite rotations.

The restitution law uses the velocities of the contacting endpoints, including the rotational terms. For $A$ that velocity isand for $B$ it is $u_B=-V+J/m$. The approach speed before impact is $2V$. The coefficient of restitution $e$ therefore givesSubstituting into the centre and angular velocities yields the endpoint collision of two freely rotating equal rods:For the first, perfectly elastic case, $e=1$, soAll initial translational energy has become rotational energy. Indeed the total initial kinetic energy is $2m{V}^2$, while the final energy for general $e$ isThe loss is $m{V}^2(1-e^2)$, zero for an elastic collision. Conservation of total angular momentum about the origin provides another check: initially it is $-4m\ell V$; after collision the centre-motion contribution is $-2m\ell V(1-e)$ and the spins contribute $-2m\ell V(1+e)$, with the same total. The initial distance $d$ determines when the collision occurs but does not enter the impulsive velocities.

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.

For the complete uniform disc, the surface density is $\sigma =M/(\pi a^2)$. An annular strip of radius $r$ and thickness $dr$ has mass $dm=2\pi \sigma rdr$, so its contribution to the axial moment of inertia is $r^{2}dm$. HenceThis derives the moment of inertia of a uniform disc rather than assuming it.

Removing a concentric disc of radius $a/2$ at the same density removes mass $M/4$ and inertia $\frac{1}{2}(M/4)(a/2{)}^2=M{a}^2/32$. The remaining annulus therefore has mass $m_A=3M/4$ andThe moment of inertia of a uniform annulus gives the same value directly.

Let $A_c$ be the downhill centre-of-mass acceleration and let $F$ denote the magnitude of the uphill static friction. Translation, rotation about the centre, and rolling without slipping giveThus $F=I_A{A}_c/a^2=(5/8)m_A{A}_c$. Eliminating $F$ gives the rolling acceleration with rotational inertiaSubstitution gives the contact forcesThe normal force balances the perpendicular component of gravity because the centre remains at a constant distance from the plane. Although the annulus accelerates downhill, friction must act uphill to supply the torque that increases its rolling angular speed. For a plane with $0\le \alpha <\pi /2$, the assumed no-slip motion requires a coefficient of static friction satisfying ${\mu }_s\ge F/N=(5/13)\tan \alpha$. With that condition, contact friction does no work because the instantaneous contact point is stationary; it redistributes the gravitational energy between translational and rotational kinetic energy.

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.

The moment of inertia of the uniform disc about its central perpendicular axis isThe parallel axis theorem gives $I_A=I_O+m{a}^2=3m{a}^2/2$. For angle $\beta$ from the downward resting position, the gravitational torque is $-mga\sin \beta$, so $I_A\ddot{\beta }=-mga\sin \beta$. Linearizing at $\beta =0$ gives $\ddot{\beta }+(2g/(3a))\beta =0$. Thus the uniform disc pivoted at its rim is a physical pendulum with period