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Two uniform discs of mass m and radius a stick at first edge contact. One initially translates at speed v and the other rests, with neither initially spinning. If the incoming velocity makes angle $\theta$ to the line of centres, the composite axial moment of inertia about its centre of mass is $3m{a}^2$. Conserved angular-momentum magnitude is $mav|\sin \theta |$, so the angular velocity magnitude is $v|\sin \theta |/(3a)$. The centre-of-mass speed is v/2. Splitting kinetic energy into translation and rotation gives $K_f=m{v}^2(1/4+{\sin }^2\theta /6)$. The difference from $m{v}^2/2$ is dissipated internally; it cannot be removed from angular-momentum conservation.

Angular acceleration is the time derivative of angular velocity, $\mathbf{\alpha }=d\mathbf{\omega }/dt$. For rotation about a fixed axis, its signed scalar component obeys $\tau =I\alpha$ when the axial moment of inertia is constant and $\tau$ is the total applied torque about that axis.

The signed accumulated rotation about a specified fixed axis. Unlike an angle specified only modulo $2\pi$, angular displacement counts complete revolutions as well as the final orientation. Its derivative with respect to time is angular velocity. The displayed formula therefore follows from the fundamental theorem of calculus.

Angular momentum measures how a point mass's momentum passes around a chosen origin. Motion straight toward or away from that origin contributes zero; transverse momentum contributes more when it passes farther away. The picture shows the sign convention before the calculation.

Fix an origin $O$ and nonrotating axes in an inertial frame. For a nonrelativistic point mass of constant mass $m>0$, let $\mathbf{r}$ be its position from $O$ and $\mathbf{p}=m\mathbf{v}$ its mechanical momentum in that same reference frame. Its angular momentum is the cross productHere $r=|\mathbf{r}|$, $p=|\mathbf{p}|$, and $\varphi \in [0,\pi ]$ is the angle between them. The right-hand rule gives the direction: curl from $\mathbf{r}$ toward $\mathbf{p}$, and the thumb points along ${\mathbf{L}}_O$. Thus angular momentum is an axial vector, perpendicular to both position and momentum. Its International System of Units unit is $\mathrm{kg}{\mathrm{m}}^2/s$. A point mass need not follow a circle to have angular momentum.^[ref]

With $x$ rightward, $y$ upward, and $+z$ out of the page, the cross product gives $L_z=x{p}_y-y{p}_x$. TakeThen ${\mathbf{L}}_O=(0,0,6)\mathrm{kg}{\mathrm{m}}^2/s$. Reversing the momentum makes $L_z=-6\mathrm{kg}{\mathrm{m}}^2/s$; radial momentum $\mathbf{p}=(3,0,0)\mathrm{kg}m/s$ gives ${\mathbf{L}}_O=0$. The factor $r\sin \varphi$ is the lever arm to the line through the particle parallel to its momentum.

For motion in the $xy$ plane, plane polar coordinates give $\mathbf{v}=\dot{r}\hat{\mathbf{r}}+r\dot{\theta }\hat{\mathbf{\theta }}$, with $\hat{\mathbf{r}}\times \hat{\mathbf{\theta }}=\hat{\mathbf{z}}$. ConsequentlyThe signed angular velocity $\dot{\theta }$ is positive counterclockwise viewed from $+z$ toward the plane, with $\theta$ measured in radians. Radial velocity does not enter the cross product. Under conservation of angular momentum, this relation predicts faster angular motion at smaller $r$ and constant areal velocity $\dot{A}=L_z/(2m)$ for consistently oriented swept area. That is the geometric content of Kepler's second law.

For several point masses, add their angular momenta about the same origin: ${\mathbf{L}}_O={\sum }_i{\mathbf{r}}_i\times {\mathbf{p}}_i$. The origin matters: angular momentum under an origin shift explains why the same moving particle can have different values about two fixed points. The central axis of a momentum system and angular momentum decomposition about the centre of mass make this dependence useful.

A transformed radial equation containing $L^2/(2q^2)$ becomes a full planar central-force orbit by choosing $d\bar{\theta }/d\tau =L/q^2$. For $d\tau /dt=(q/r)dq/dr$ and old angular velocity $L/r^2$, this gives $d\bar{\theta }/d\theta =(r/q)dq/dr$. Keeping the old angle generally fails to preserve the stated transformed angular momentum.

Angular speed is the nonnegative magnitude of angular velocity: $\omega =|\mathbf{\omega }|$. It measures the rate of rotation about an instantaneous axis without specifying its direction. For uniform circular motion of radius $R$, the ordinary speed is $v=R\omega$.

For a symmetric top with axial unit vector $\mathbf{e}$ and angular velocity $\mathbf{\omega }$, the classical axial spin is the scalar $n=\mathbf{\omega }\cdot \mathbf{e}$. The identity $\dot{\mathbf{e}}=\mathbf{\omega }\times \mathbf{e}$ implies $\mathbf{\omega }=\mathbf{e}\times \dot{\mathbf{e}}+n\mathbf{e}$. If the transverse and axial principal moments of inertia are $A$ and $C$, then the angular momentum about the fixed vertex is $\mathbf{L}=A\mathbf{e}\times \dot{\mathbf{e}}+Cn\mathbf{e}$. A torque perpendicular to $\mathbf{e}$ conserves $n$. During steady precession at rate $\Omega$ and inclination $\alpha$, the general angular velocity is $\Omega \mathbf{k}+(n-\Omega \cos \alpha )\mathbf{e}$; the precession rate by itself does not fix $n$.

A ball bearing transfers loads through rolling spherical elements between two races. The no-slip conditions at both contacts determine the translation and spin of each ball; these are generally not the same angular velocity.

Exchange of angular momentum between a stellar bar and a dark matter halo through orbital resonances. If the unperturbed halo distribution function depends only on energy and decreases with it, the resonant torque from a rotating perturbation is nonnegative for positive pattern angular velocity. Strict transfer requires nonzero resonant overlap. The reaction torque slows a rigid bar of constant positive moment of inertia; evolving bar structure and halo redistribution require updating the response.

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.

The 2015 experiment used an air-driven nonmagnetic rotor containing polycrystalline gadolinium, with magnetic shielding and an external fluxgate magnetometer. Air drive reduced motor backgrounds. Measurements at 297–305 K were above the Curie temperature, at rotation frequencies up to 1.5 kHz. Reversing airflow reversed rotation and the magnetic signal; an empty capsule showed no speed-dependent signal. The readout measured stray magnetic field; geometry converted it to magnetization. Sample temperature was checked immediately after stopping. Heating and the conversion model limit inference.^[ref]

An additional analysis check follows from rotation symmetry. For signed sensor output $S$, separateAn ideal linear Barnett effect appears in the odd component; a fixed offset appears in the even component. This is a proposed diagnostic, not a claim about the paper's processing. A background correlated with direction can also be odd: reversal alone is insufficient. Empty-rotor checks, zero-speed readings, calibration of sensor polarity, and controlled temperature changes test different errors. Equatorial stray-field sign must also be converted to magnetization sign: for an axial magnetic dipole moment, the magnetic dipole field at its equator points opposite to that magnetic moment. The apparatus diagram makes this otherwise easy-to-miss sign change visible.

For a smooth horizontal circular hoop of radius $a$, rotated at constant angular velocity $\omega$ about a vertical axis through a rim point, measure $\theta$ at that pivot from the diameter toward the bead. In rotating Cartesian coordinate system coordinates, $\mathbf{r}=a(1+\cos 2\theta ,\sin 2\theta ,0)$. The tangential component of the centrifugal acceleration gives the displayed equation; the Coriolis acceleration is normal to the hoop. The farthest point $\theta =0$ is stable for $\omega \ne 0$, with small oscillation angular frequency $|\omega |$. The pivot itself is unstable. This geometry differs from a vertical rotating hoop and has no tangential gravitational term.

A smooth wire has instantaneous shape $z=f(r)$ in cylindrical coordinates and rotates at constant angular velocity $\dot{\theta }=\omega$. The tangent vector at fixed time is ${\mathbf{e}}_r+f^{\prime }(r){\mathbf{e}}_z$. Its dot product with the constraint force vanishes. Projecting Newton's second law along this tangent vector gives the equation in the title.

Multiplying that equation by $\dot{r}$ proves conservation ofThis is a conserved quantity in the rotating description; the motor can exchange ordinary mechanical energy with the bead. A relative equilibrium at $a>0$ satisfies $U^{\prime }(a)=g{f}^{\prime }(a)-{\omega }^{2}a=0$. Its linearized equation is $(1+f^{\prime }(a{)}^2)\ddot{ϵ}=-U^{\prime \prime }(a)ϵ$. A strict local minimum with $U^{\prime \prime }(a)>0$ gives Lyapunov stability, while $U^{\prime \prime }(a)<0$ gives an unstable equilibrium. If $U^{\prime \prime }(a)=0$, higher orders are needed.

If a vector $k$ is fixed in space while a body frame rotates with angular velocity $\omega$, its body-frame components satisfy

For angular velocity depending only on cylindrical radius, the centrifugal acceleration is a gradient. Steady barotropic balance is $\nabla p/\rho =-\nabla \Psi$, hence specific enthalpy satisfies $h(\rho )+\Psi =\text{constant}$. On an invertible equation-of-state branch, mass density and pressure are functions of $\Psi$. Cylindrical rotation is essential to this scalar centrifugal-potential representation.

In a spherically symmetric gravitational potential $\Phi (r)$, a circular orbit has $r{\Omega }^2={\Phi }^{\prime }(r)$, specific orbital energy $\epsilon =\Phi +r^2{\Omega }^2/2$ and specific angular momentum $\mathbf{h}=r^2\Omega \mathbf{n}$, with $|\mathbf{n}|=1$. Differentiating gives ${\epsilon }^{\prime }=2r{\Omega }^2+r^2\Omega {\Omega }^{\prime }=\Omega (r^2\Omega {)}^{\prime }$. Changes in orbital orientation have $\mathbf{n}\cdot d\mathbf{n}=0$, so $\mathbf{\Omega }\cdot d\mathbf{h}=\Omega d(r^2\Omega )=d\epsilon$. For a particle of variable mass $M$ and total angular momentum $\mathbf{H}=M\mathbf{h}$, this implies $d(M\epsilon )=(\epsilon -\Omega h)dM+\mathbf{\Omega }\cdot d\mathbf{H}$. Moreover $d(\epsilon -\Omega h)/dr=-h{\Omega }^{\prime }$, which is positive when the orbital angular velocity decreases outward.

Chapter 3 of Classical Dynamics by David Tong. www.damtp.cam.ac.uk/user/tong/dynamics/dynhtml/S3.html. Section 3.5.2 resolves angular velocity in material-fixed axes; sections 3.6–3.6.3 treat the heavy symmetric top, uniform precession and upright spin stability. This is an authored lecture-note chapter, not an answer set for the assigned exam.

Conservation of angular momentum says that total angular momentum stays fixed when the net external torque about the chosen fixed inertial origin vanishes. Individual angular momenta and angular velocities can change. The radius change at constant angular momentum below makes this distinction visible: pulling two rotating point masses inward lowers the moment of inertia, raises the angular velocity, and supplies work that increases the kinetic energy.

Follow the same constant-mass point masses throughout, in an inertial frame, with all positions measured from a fixed origin $O$. If the internal pair forces are opposite and central, meaning directed along each pair's separation, thenThus zero net external torque throughout an interval implies constant total angular momentum throughout that interval. A vanishing external torque component, such as ${\tau }_z^{\mathrm{ext}}=0$, conserves the corresponding $L_z$ even when the other components change. This is the balance law for a fixed collection; angular-momentum flux must also be accounted for when matter crosses its boundary. ^[ref]

The central qualification is essential. Write ${\mathbf{F}}_{ij}$ for the internal force on member $i$ from member $j$. Under Newton's third law, ${\mathbf{F}}_{ji}=-{\mathbf{F}}_{ij}$, but the sum of their torques isThis vanishes when ${\mathbf{F}}_{ij}$ is parallel to ${\mathbf{r}}_i-{\mathbf{r}}_j$. Equal-and-opposite forces alone do not suffice: two opposite transverse forces with separated lines of action form a couple with nonzero torque. Cancellation of the internal pair torques, together with Newton's second law, proves the total balance above.

For one point mass under a central force about $O$, $\mathbf{F}=f(r)\hat{\mathbf{r}}$ gives ${\mathbf{\tau }}_O=\mathbf{r}\times \mathbf{F}=0$. The force may still change the speed and kinetic energy; its zero torque preserves angular momentum. With nonzero ${\mathbf{L}}_O$, $\mathbf{r}\cdot {\mathbf{L}}_O=0$ keeps the motion in a fixed plane. Choosing the $z$ axis perpendicular to that plane, $r^2\dot{\theta }=L_z/m$ governs its angular progress. This removes an unknown from the central-force radial turning point calculation and helps reduce the Kepler orbit to a radial problem.

There is a historical geometric version of this result. In Book I, Section II, Proposition I of the Principia, Newton proves that motion under a central force toward a fixed centre stays in a plane and sweeps equal areas in equal times, using successive central impulses and equal triangle areas. ^[ref] In the modern notation above, constant areal velocity is $\dot{A}=L_z/(2m)$: the old geometric proof and the angular momentum calculation express the same invariant.

For positive angular velocity, a motor delivering constant mechanical power $P$ produces torque $P/\omega$. With constant opposing kinetic friction torque ${\tau }_f$, angular momentum balance gives the displayed equation and a steady speed ${\omega }_{\ast }=P/{\tau }_f$. A small increase in speed reduces the driving torque below the friction torque, making this equilibrium stable within the model. The constant-power idealization is singular at exact rest and requires a separate starting regime in a real motor.

The instantaneous velocity of the material point of a rigid body touching another body. In the displayed formula, ${\mathbf{r}}_P$ is its position relative to the centre of mass, and $\mathbf{\Omega }$ is the body's angular velocity. The rotational term is a cross product. This material-point velocity differs from the motion of the geometrical contact location: in rolling without slipping on a stationary plane it vanishes even while the contact location advances. For a sphere touching a horizontal plane, with rightward centre velocity $v$ and anticlockwise spin $\Omega$, its tangential component is $v+a\Omega$.