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For rotational motion about a fixed axis, let $r$ be cylindrical radius, $v$ the signed tangential velocity, and $dm$ the mass measure. Define the kinetic energy, axial angular momentum, and moment of inertia byThe Cauchy-Schwarz inequality gives $L^2\le (\int v^{2}dm)(\int r^{2}dm)=2KI$. More precisely, direct expansion gives the nonnegative energy gapThus the minimum at fixed $L$ and fixed mass distribution occurs exactly when $v=(L/I)r$ almost everywhere with respect to $dm$. This means constant angular velocity $\Omega =L/I$, or solid-body rotation. If the profile is continuous, the equality holds at every point in the mass support.

For a uniform unit-density cylinder of radius $a$, height $h$, and volume $V=\pi a^{2}h$, the moment of inertia is $I=V{a}^2/2$. Therefore $K\ge L^2/(V{a}^2)$, with equality for $\omega (r)=2L/(V{a}^2)$. Multiplying by a uniform nonunit density changes the mass measure and $I$ consistently.

A rigid rotor models rotational motion with fixed internal separations. Its use in the synthetic Stark effect figure omits vibrational motion and additional interactions in real molecules.

Rigidity fixes the molecular geometry, not the orientation, and does not require a spherical mass distribution. Remove motion of the centre of mass, choose principal axes of the inertia tensor, and write the classical rotational kinetic energy asHere $I_a,I_b,I_c$ are the principal moments of inertia, ${\omega }_a,{\omega }_b,{\omega }_c$ components of angular velocity, and $L_a,L_b,L_c$ components of angular momentum. Quantization replaces these last components by angular momentum operators. Different principal moments of inertia produce different spectra: the single-constant linear rigid rotor is a special case, not a formula for every molecule.^[ref]

If an inner race of radius $R$ rotates at angular velocity $\omega$ and an outer race is stationary, a ball of radius $r$ rolling at both contacts has centre speed $R\omega /2$ and spin $-R\omega /(2r)$. Its centre follows a circle of radius $R+r$. The angular momentum about the common race axis includes both orbital angular momentum and spin.

A straight rigidly rotating classical string has speed of light at a free endpoint. Its energy and angular momentum satisfy a Regge trajectory. A segment with one anchored and one free end has $J=E^2/(\pi T)$, twice the slope of the string with two free ends.

The angular momentum associated with end-over-end molecular rotation. In an open-shell diatomic molecule it is useful to distinguish $\mathbf{N}$ from the total $\mathbf{J}=\mathbf{N}+\mathbf{S}$, which also contains the electronic spin angular momentum.

A frequency change associated with rotation and exchange of angular momentum. If an optical element rotates at angular velocity $\Omega$ about a chosen axis and gains angular momentum $\Delta L$ from a photon, its mechanical energy gain is $\Omega \Delta L$ in the prescribed slow-rotation model. Energy conservation then gives photon frequency change $\Delta \omega =-\Omega \Delta L/\hslash$. The rotating half-wave plate frequency shift is a spin-polarization example.

For strictly positive principal moments of inertia and a fixed rotational kinetic energy $K_0>0$, the allowed angular velocity components form an ellipsoid:Its semiaxis lengths $s_i$ have units of angular speed. The longest semiaxis corresponds to the smallest principal moment of inertia: that direction permits the greatest angular speed for the same rotational kinetic energy. This is a surface in angular velocity space, not the physical outline of the body. If one principal moment of inertia vanishes in an ideal line-mass model, this constant-energy surface is unbounded in that direction and is no longer an ellipsoid.

The rotational energy ellipsoid gives the geometry behind Poinsot's ellipsoid, but a constant-energy surface alone does not determine a free-rotation trajectory. A torque-free trajectory must also satisfy the conserved angular momentum constraint and the Euler equations for a torque-free rigid body.

The net external torque about an inertial origin equals the time derivative of angular momentum. For rotation about a fixed axis with constant moment of inertia, its axial component reduces to $I\dot{\omega }=\tau$.

Rotational kinetic energy is the kinetic energy of a rigid body's rotation. For pure rotation about an axis fixed in an inertial frame,Here $r_{\perp }$ is the perpendicular distance of a mass element $dm$ from that axis, $I$ is the moment of inertia about that same axis, and $\omega$ is the common angular speed, measured in radians per second. The rigid body assumption fixes the relative positions of the mass elements; they share angular velocity but have different speeds $v=\omega r_{\perp }$. Summing their kinetic energies $dm{v}^2/2$ gives the formula. This is a nonrelativistic model; deformation or internal motion can contribute additional kinetic energy.^[ref]

For a computed example, take a uniform disc and an ideal thin ring, each of mass $M=2\mathrm{kg}$ and radius $R=0.30\mathrm{m}$, rotating about its central perpendicular axis at angular speed $10rad/s$. The moment of inertia of a uniform disc is $M{R}^2/2=0.090\mathrm{kg}{\mathrm{m}}^2$, while the ring has moment of inertia $M{R}^2=0.180\mathrm{kg}{\mathrm{m}}^2$. Their rotational kinetic energies are therefore $4.5\mathrm{J}$ and $9.0\mathrm{J}$. The ring stores twice the energy at the same angular speed, because more mass lies farther from the axis. At the same nonzero rotational kinetic energy, its angular speed would instead be $1/\sqrt{2}$ times the disc's. This makes radial placement useful for storing energy in a rotating wheel, although a real design must also withstand the stresses of rotation.

For fixed-axis rotation with constant moment of inertia, write the signed rotation rate as $\Omega =\dot{\theta }$, with angle $\theta$ in radians. The axial component of net external torque obeys ${\tau }_{\mathrm{axis}}=I\dot{\Omega }$, so the work-energy theorem becomesAn opposing torque removes rotational kinetic energy; bearing friction must be included in the net torque. The signed $\Omega$ matters when the direction reverses. Rotation about a fixed axis need not be about a principal axis: the scalar energy formula still holds, but transverse bearing torques can be needed to maintain that axis.

For general rigid body motion, use the inertia tensor ${\mathsf{I}}_G$ about the centre of mass $G$:This is the kinetic-energy decomposition about the centre of mass. In a principal body frame, the rotational term is $\frac{1}{2}(I_1{\omega }_1^2+I_2{\omega }_2^2+I_3{\omega }_3^2)$, using the principal moments of inertia. Away from a principal axis, angular momentum and angular velocity need not be parallel. A freely rotating asymmetric rigid body can change the direction of angular velocity relative to the body even while rotational kinetic energy remains constant; Euler equations for a torque-free rigid body govern that motion. One constant axial moment of inertia is then insufficient.^[ref]

A rough contact may transmit both normal and tangential impulses. Conservation of linear and angular momentum fixes totals but does not specify their division between translation and rotation. Predicting outgoing states requires constitutive data such as normal restitution and a friction or tangential restitution law. Rough contact is not synonymous with adhesive contact.

Use the convention $R=L-{\sum }_i{p}_i{\dot{q}}_i$ for the eliminated cyclic coordinates, expressing their velocities in terms of their fixed conjugate momenta. Terms involving derivatives of those eliminated velocities cancel because $\partial L/\partial {\dot{q}}_i=p_i$. Thus the remaining coordinates obey the usual Euler-Lagrange equations for R. For a particle of fixed angular momentum $\ell$ in a central potential, this gives $R=m{\dot{r}}^2/2-V(r)-{\ell }^2/(2m{r}^2)$, rather than the expression obtained by direct substitution into L alone. An opposite overall sign convention is also used, with correspondingly signed mixed Hamiltonian equations.

Ground-state electronic angular momentum $J=1/2$ couples to nuclear spin $I=3/2$, giving $F=1,2$. For contact hyperfine interaction $A\mathbf{I}\cdot \mathbf{J}/{\hslash }^2$, their zero-field energies are $-5A/4$ and $3A/4$. The splitting is $2A$. Electronic weak-field contributions give $g_{F=1}\simeq -1/2$ and $g_{F=2}\simeq +1/2$; the small nuclear contribution refines these numbers.

Rutherford scattering is elastic scattering by a Coulomb potential. In the repulsive classical case, a positively charged projectile bends around a positively charged atomic nucleus. A close approach can turn it backwards without a contact collision. The calculation connects an invisible impact parameter with a measurable outgoing angle, then with a differential scattering cross-section and a detector count.

Assume a projectile of mass $m$ and electric charge $ze$, a stationary point atomic nucleus of electric charge $Ze$, and positive $z,Z$. Here $e$ is the elementary charge. Neglect nuclear recoil, electron screening effect, energy loss, and the nuclear force. With vacuum permittivity ${\epsilon }_0$, defineHere $E$ and $v$ are the incident kinetic energy and speed far from the atomic nucleus, $b\ge 0$ is the impact parameter, and $L$ is the magnitude of angular momentum. The nonrelativistic limit requires $v\ll c$, with $c$ the speed of light. The fixed-target approximation requires the target mass to be much greater than $m$. In SI, $\kappa$ has units $\mathrm{J}\mathrm{m}$, $E$ has units $\mathrm{J}$, and $b$ has units $\mathrm{m}$.

For $V(r)=c/r^2$ with $c>0$, nonzero physical angular momentum $h$ and positive energy $E$, put $\beta =\sqrt{1+2mc/h^2}>1$ and $r_{\min }=\sqrt{(c+h^2/(2m))/E}$. Direct integration of the effective potential radial equation gives the displayed orbit. It has one radial minimum and two straight-line asymptotes separated by polar angle $\pi /\beta$. The repulsive scattering deflection is $\pi -\pi /\beta$; no bounded or circular orbit exists at finite positive energy.

A timelike geodesic with unit specific Killing energy can plunge from infinity into a Schwarzschild black hole only if its specific angular momentum satisfies $|h|<4M$. The limiting case $|h|=4M$ asymptotes to the unstable orbit at $r=4M$. A geodesic already inside the angular-momentum barrier can cross the event horizon with larger $|h|$, so the origin-at-infinity condition matters.

A secular tidal torque is the torque remaining after averaging the changing tidal interaction over fast orbital and rotational motion. The misaligned tidal deformation lets the companion's gravity change the spin angular momentum; the orbit receives the opposite torque. Tidal dissipation converts mechanical energy into heat, while total angular momentum is conserved in an isolated system. Thus the spin-orbit coupling exchanges angular momentum rather than destroying it. ^[ref].

For intuition near synchronous rotation on a circular orbit, hold mean motion $n$ and moment of inertia $I$ fixed and adopt the illustrative linear dissipative lawHere $\omega$ is spin angular velocity, $\theta$ is a marked surface direction in an inertial frame, and $K$ is a response coefficient, not a universal material constant. A faster spin is slowed and a slower prograde spin is accelerated. The solution and the associated mechanical power loss areThe orbital contribution to this power is $-n⟨\tau ⟩$, and the spin contribution is $\omega ⟨\tau ⟩$; their sum becomes heat. Holding $n$ fixed is a local approximation to the coupled evolution. Actual secular tidal torque depends on the deformation response and orbital eccentricity, so this simple law does not establish which spin-orbit resonance an eccentric body captures into.

Completing the square in $H^{\prime }=\frac{1}{2}(p_x^2+p_y^2+x^2+y^2)-e\mathcal{E}x$ shifts the oscillator centre to $(a,0)$ and lowers every energy by $a^2/2$. Translated eigenfunctions keep their normalization. The angular momentum about the new centre is $L^{\prime }=(x-a)p_y-y{p}_x$, which commutes with $H^{\prime }$, whereas the angular momentum about the old centre has $[L,H^{\prime }]=-i\hslash ay$. For $a\ne 0$, a common normalizable eigenfunction of $L,H^{\prime }$ would have $y\psi =0$ and hence vanish in $L^2({\mathbb{R}}^2)$. This stronger argument excludes individual common eigenstates; noncommutativity alone generally excludes only a complete common eigenbasis.

In the simplest nuclear shell model of an odd-mass number nucleus, paired nucleons contribute zero total angular momentum and magnetic dipole moment. The last unpaired nucleon supplies the nuclear spin $j$ and the stretched-state moment $\mu =g_{j}j{\mu }_N$, where $g_j$ is the Landé g-factor and ${\mu }_N$ the nuclear magneton.

Using addition of angular momentum with $s=1/2$ givesFor $j=\ell +1/2$, this reduces to $\mu /{\mu }_N=g_{\ell }\ell +g_s/2$. For $j=\ell -1/2$, it givesThe free-nucleon orbital factors are $g_{\ell }=1$ for a proton and $0$ for a neutron. Residual nuclear interactions and configuration mixing can shift measured moments away from this ideal prediction. A single hole in an otherwise closed $j$ shell has the same stretched-state moment as a particle in that shell: removing the $m_j=-j$ particle leaves total $J_z=j$ and moment $g_{j}j{\mu }_N$.

Finite optical pumping time lets an atom climb a light-shift hill before transfer into a lower magnetic sublevel. Spontaneous emission dissipates the gained potential energy; climbing alone is conservative.^[ref]

To make the geometry and signs explicit, take an original choice of phase origin for equal opposing waves:The real electric field is $\mathrm{Re}[\mathcal{E}(z)e^{-i{\omega }_{L}t}]$; $\lambda$ is wavelength. The fractions $w_{\pm }$ are squared projections onto the defined circular basis vectors, divided by total intensity. Their sum is one: interference changes electromagnetic polarization despite constant total intensity. At $z=0,\lambda /4,\lambda /2$ the field is respectively ${\mathbf{ϵ}}_{+},{\mathbf{ϵ}}_{-},{\mathbf{ϵ}}_{+}$; halfway between these positions it is linearly polarized. Specifying the vectors avoids an observer-dependent verbal handedness convention.

For an illustrative $J_g=1/2\to J_e=3/2$ angular momentum transition, use the squared Clebsch-Gordan coefficients $1$ and $1/3$ for the strong and weak circular couplings from each lower magnetic sublevel. Interchanging circular components interchanges these strengths. The weighted sums are $w_{+}+w_{-}/3=(2+\cos (2kz))/3$ and $w_{-}+w_{+}/3=(2-\cos (2kz))/3$. Rescaling their common negative shift givesBoth potential energies are negative relative to the unilluminated level. Each hill has height $U_0$, and opposite hills are displaced by $\lambda /4$. An atom starting at the $U_{+}$ minimum at $z=0$ loses $U_0$ of kinetic energy if it reaches $z=\lambda /4$ on that branch. A pump transfer there lowers its potential energy by $U_0$, leaving it at a $U_{-}$ minimum. The next uphill segment ends at $z=\lambda /2$, where the favoured transfer is reversed. Actual transfers are stochastic and need not occur exactly at these extrema.

For one completed climb and pump, conservation of energy gives $\Delta K=-U_0$, while the ground-state potential energy returns to its starting value. Neglecting recoil and treating the scattered photon as the sole outgoing radiation change, $\hslash ({\omega }_{\mathrm{out}}-{\omega }_L)=U_0$. A potential drop is an internal-state reset, not an automatic equal upward kick in kinetic energy. Random emission nevertheless adds momentum fluctuations and recoil energy, so useful cooling requires the dissipated energy to compete with that heating.

The recoil energy $E_R=(\hslash k{)}^2/(2m)$ sets a different scale from $k_B{T}_D=\hslash \Gamma /2$. Conventional Sisyphus cooling reaches the recoil regime when reducing the hill height also weakens cooling; it is not a claim of arbitrarily small temperature at vanishing intensity. Subrecoil techniques need further physics.^[ref]

For a slowly evolving circular orbit with tangential deceleration $D>0$, the specific angular momentum is $j=r{V}_c(r)$ and its torque is $\dot{j}=-rD$. Differentiating $j$ gives the displayed drift when $1+q\ne 0$. The positive local decay time $r/(2|\dot{r}|)$ is $(1+q)V_c/(2D)$. In a uniform-density core $q=1$, so it is $V_c/D$; for a flat galaxy rotation curve $q=0$, it is $V_c/(2D)$. Simply setting ${\dot{V}}_c=-D$ neglects the radial contribution to angular momentum change.