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Place two equal point masses of $1\mathrm{kg}$ each on opposite sides of a fixed origin, at equal radius $r$, and let them share a signed angular velocity $\omega =\dot{\theta }$ in a plane. Their total mass is $M=2\mathrm{kg}$ and their moment of inertia about the perpendicular axis is $I=M{r}^2$. Radial pulling forces have zero torque about that origin, soTake an initial radius $r_i=1\mathrm{m}$ and angular velocity ${\omega }_i=1rad/s$, and a final radius $r_f=0.5\mathrm{m}$. ThenandThese are endpoint states with ${\dot{r}}_i={\dot{r}}_f=0$. During the pull, the full kinetic energy also contains radial motion: $K=M{\dot{r}}^2/2+K_{\mathrm{rot}}$. Assume the pulling apparatus supplies all net work to the two point masses; the work-energy theorem then fixes its net work at $+3\mathrm{J}$, including any radial acceleration and braking. Each mass spirals inward during the pull, allowing its inward force to do positive work. Conservation of angular momentum therefore predicts the final angular velocity without requiring kinetic energy conservation. For an apparatus included in a larger system, its stored energy supplies the increase consistently with conservation of energy. OpenStax uses the same mechanism to explain a skater pulling her arms inward. ^[ref]

Consider an ideal circular vacuum path of laboratory radius $R$, a limit of closely spaced mirrors constraining light around the circle at laboratory speed $c$. The source and detector rotate uniformly together with signed angular velocity $\Omega$, where $|\Omega |R<c$, in flat spacetime. Let $L=2\pi R$ be the laboratory circumference and $A=\pi R^2$ the positive area. The $+$ pulse travels counterclockwise and the $-$ pulse clockwise, independently of the sign of $\Omega$. Both depart from the detector at $t=0$.

Unwrap the circle into a laboratory arc coordinate. The two first-return conditions areConsequently the exact laboratory arrival times and their difference areThe denominators $c\mp \Omega R$ are closing rates in these laboratory equations; they are not local speed of light measurements by a rotating observer. The two arrival spacetime events lie on the same detector world line. Its constant speed is $|\Omega |R$, so proper time along that world line givesThus laboratory coordinate-time and rotating-detector proper time delays agree only to leading order at small $|\Omega |R/c$. The radius and circumference here are laboratory quantities, not lengths obtained by laying comoving rods around the rim. Finite mirror polygons, transparent media, and optical dispersion require their own path and propagation assumptions before using an exact formula.

An exact dimensionless check takes $\Omega R/c=1/5$. Relative to $L/c$, the laboratory arrival times are $5/4$ and $5/6$, giving $\Delta t/(L/c)=5/12$. The detector instead records $\Delta \tau /(L/c)=\sqrt{6}/6\simeq 0.408248$. The leading Sagnac area law would give $\Delta t/(L/c)\simeq 2/5$. These are model values chosen to make the relativistic distinction visible, not attainable parameters of the historical apparatus.

The rotating loop does not constitute a universal measurement of the one-way speed of light. It compares completed circuits at a single readout; assigning a velocity to each remote leg additionally requires a clock synchronization convention. Local Einstein synchronisation around a rotating rim does not close into a single globally synchronized clock network: the accumulated mismatch is another expression of the rotating-path asymmetry. Local comoving vacuum measurements still give $c$ in both directions. Special relativity predicts a nonzero closed-loop delay because the receiver moves during propagation, while no single comoving inertial frame covers the whole circuit.

Release the upper rod at ${\theta }_1(0)=\alpha =0.06$ radians while keeping the lower rod vertical, ${\theta }_2(0)=0$, with both angular velocities zero. The initial condition is the sum $\mathbf{\theta }(0)=(\alpha /2){\mathbf{a}}_{-}+(\alpha /2){\mathbf{a}}_{+}$. The solution from linearization is thereforeAt release the lower angle vanishes by cancellation between the two normal modes. Their different angular frequencies then destroy that cancellation. Differentiating twice gives ${\ddot{\theta }}_1(0)=-2g\alpha /\ell$ and ${\ddot{\theta }}_2(0)=2g\alpha /\ell$, agreeing with the small-angle limit of the exact double pendulum equations of motion.

The quadratic mechanical energy at release is $E_2=mg\ell {\alpha }^2$. Each normal mode contributes half, because $(4+2\sqrt{2})(2-\sqrt{2})=(4-2\sqrt{2})(2+\sqrt{2})=4$. Equal coordinate amplitudes need not mean equal energies in general; here the modal inertias and squared angular frequencies happen to compensate. Releasing instead along $\mathbf{\theta }(0)=\alpha {\mathbf{a}}_{-}$ or $\alpha {\mathbf{a}}_{+}$, with zero angular velocities, excites just one normal mode.

Expanding the squared velocity $v_G+\omega \times r$ and integrating over mass removes the mixed term because the centre-of-mass first moment is zero. The remaining rotational integral is half the scalar product of angular velocity and angular momentum about the centre of mass. Principal axes diagonalize that rotational quadratic form.

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 round body of radius $R_0$ in a fixed cylinder of radius $R_1$ has center of mass path radius $a=R_1-R_0$. With orbital angle $\theta$ and spin angular velocity $\Omega$ both positive about the same axis, rolling without slipping gives $a\dot{\theta }+R_0\Omega =0$. For moment of inertia $I=\gamma m{R}_0^2$, its kinetic energy is $\frac{1}{2}m{a}^2(1+\gamma ){\dot{\theta }}^2$. Release from rest at angle $\alpha$ gives, by conservation of energy, a descent time $\sqrt{1+\gamma }$ times that of frictionless sliding along the same center of mass path. The spin sign reverses if its positive direction is chosen opposite to the orbital angle.

If a frame rotates with angular velocity $\mathbf{\Omega }$ relative to an inertial frame, the derivatives of any vector $\mathbf{a}(t)$ satisfyApplying this formula twice to a position gives the Coriolis acceleration, Euler acceleration, and centrifugal acceleration terms.

For a rigid frame with constant angular velocity $\mathbf{\Omega }$, add the Coriolis force and centrifugal force to the inviscid magnetohydrodynamic momentum equation. A constant magnetic permeability and solenoidal magnetic-field constraint split the Lorentz force density into magnetic pressure and magnetic tension. This nonrelativistic conducting-fluid approximation neglects the displacement current, net electric-charge force and viscosity. Differential rotation belongs to the fluid's velocity field, not to a spatially varying rigid-frame rotation vector. A time-dependent frame requires the additional Euler force.

For identical point vortices on the plane, use $\omega ={\sum }_{a}d{x}_a\wedge d{y}_a$, ${\iota }_{X_H}\omega =dH$, and $L=\frac{1}{2}{\sum }_a(x_a^2+y_a^2)$. The generator $Y$ of simultaneous counterclockwise rotations is $-X_L$, and $L$ is its moment map with the convention ${\iota }_Y\omega =-dL$. In coordinates rotating with angular velocity $\Omega$, the Hamiltonian vector field is $X_H-\Omega Y=X_{H+\Omega L}$. Hence a collision-free configuration rigidly rotating at that velocity is exactly a critical point of the augmented Hamiltonian function $H+\Omega L$. For $H=-{\sum }_{a<b}\log |{\mathbf{r}}_a-{\mathbf{r}}_b{|}^2$, scaling all positions shows ${\sum }_a{\mathbf{r}}_a\cdot {\nabla }_{a}H=-N(N-1)$; a rotating relative equilibrium therefore obeys $2\Omega L=N(N-1)$. For $N\ge 2$ this implies positive $\Omega$ in the stated circulation convention. Reversing the contraction convention requires consistent changes to the Hamiltonian and rotation signs.

The Technische Universität Darmstadt demonstration uses a steel ball launched outward from the centre of a rotating disc. A fixed lecture-room camera records a straight path, while a camera carried above the disc records a curved path. The instructions specify launching toward the fixed camera so that the laboratory motion is clearly visible. The rotating iPod transmits its view wirelessly through Apple TV; two vertical support rods of length $1\mathrm{m}$ are specified for the slow-motion arrangement. The paired video views are the readout: one physical motion is represented in two reference frames.^[ref]

A useful preparation control appears in the Universität Freiburg instructions: with the disc stationary, check that the ball travels radially toward its edge. A camera is mounted on supports attached to the disc; turn the disc by hand as the ball leaves its centre and observe the camera view.^[ref]

The inference is a dependence of the observed trajectory on the rotating axes; a curved camera trace alone does not isolate the Coriolis force. To compare with the constant-angular velocity free particle in rotating coordinates, measure the disc angular velocity, calibrate lengths in the camera image, and compare launches at rest and with opposite rotation directions. These are checks of the ideal model, not reported measurements from the manuals. A rolling ball can exchange forces with the surface through friction, and hand rotation need not have constant angular velocity; an Euler acceleration then enters too. The ideal free-ball drawing deliberately omits those effects. Neither manual supplies a numerical uncertainty or a precision measurement of the Coriolis acceleration.

A rigid pendulum tilted from the upward vertical and driven at fixed azimuthal angular velocity $\Omega$ has effective potential $\mu {\theta }^2/2+mga\cos \theta -m{a}^2{\Omega }^2{\sin }^2\theta /2$. With $\mu =2mga$, its upright equilibrium loses linear stability when ${\Omega }^2$ passes $g/a$. At equality the leading restoring torque is cubic and the potential has a positive quartic minimum. This gives a stable quartic oscillator with zero infinitesimal harmonic frequency, rather than a freely drifting neutral state.

For a sphere of radius $a$ in an otherwise stationary, unbounded Newtonian fluid with dynamic viscosity $\eta$, impose the no-slip boundary condition and the slow-flow assumptions of Stokes flow. The resisting torque on a rotating sphere in Stokes flow is $-{\zeta }_r\mathbf{\Omega }$, where $\mathbf{\Omega }$ is its angular velocity and ${\zeta }_r=8\pi \eta a^3$. Thermal rotational diffusion at absolute temperature $T$ then givesHere $k_B$ is the Boltzmann constant; the translational expression is the Stokes–Einstein relation. The ratio has units of inverse length squared, so comparing the numerical values of $D_r$ and $D_t$ without their units is meaningless. At fixed temperature and dynamic viscosity, decreasing radius tenfold increases $D_r$ a thousandfold but $D_t$ only tenfold. This strong size dependence explains why small particles can lose directional memory extremely quickly.

Choose $T=298\mathrm{K}$ and $\eta =0.001\mathrm{Pa}\mathrm{s}$ as illustrative model inputs. Using $k_B=1.380649\times 10^{-23}\mathrm{J}{\mathrm{K}}^{-1}$ gives $k_{B}T=4.11433402\times 10^{-21}\mathrm{J}$ and the following predictions:These are bulk three-dimensional sphere calculations, explicitly not a fit to the confined ellipsoid experiment below. Walls, nonspherical shape and departure from the no-slip boundary condition can change rotational resistance. A visible surface mark is still necessary to read the sphere's turning in an image, even when its rotational diffusion coefficient is well defined physically.

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.

PASCO's PASPORT Rotary Motion Sensor Experiment Guide, Experiment 1, specifies a PS-2120A sensor, a rod carrying two equal-radius weights, a three-step drive wheel, a thread and hanging mass, a clamp-on guide wheel, a support, a balance and calipers. The thread leaves the drive wheel tangentially. The sensor records angular velocity against time; a linear regression over the accelerating interval supplies angular acceleration $\alpha$. The guide measures the drive radius with calipers and repeats the run with the weights removed to subtract the apparatus's own moment of inertia. Its tables are blank: it prescribes a measurement and does not report an observed numerical agreement. ^[ref]

With hanging mass $m_h$ including its hanger, effective drive radius $b$, no thread slip and negligible guide-wheel inertia, Newton's second law gives thread tension $F=m_h(g-b\alpha )$. The axial torque is $bF$. Under negligible friction, the inference is thereforeRepeat the inference for the unloaded assembly and subtract: $I_{\mathrm{weights}}=I_{\mathrm{loaded}}-I_{\mathrm{empty}}$. Omitting the $m_h{b}^2$ term incorrectly treats thread tension as the hanging weight. Comparing with $(M_1+M_2)R^2$ tests the point mass approximation; finite-sized weights also have their own moments of inertia about their centres of mass. ^[ref]

Useful proposed checks go beyond that ideal inference. Verify the rotary encoder's angle scale by turning the shaft through a marked full revolution; check the recorded time against an independent clock and the balance against a known mass. Measure $b$ at the thread's running surface, including thread thickness; avoid overlapping wraps that change $b$. Release without a push, inspect the angular velocity record for a straight interval, and stop before the hanger lands or the thread runs out. Repeat with several hanging masses, weight radii and rotation directions. A consistent inferred moment of inertia across drive torques supports the model; curvature, direction dependence or drift indicates a limitation rather than a new principal moment of inertia.

For approximately constant resisting torque ${\tau }_f\ge 0$ in one rotation direction, an original calibration model isFit the left-hand quantity against $\alpha$ across several drives: the slope estimates the axial moment of inertia and the intercept estimates ${\tau }_f$. This calibration is conditional on friction remaining constant over the fitted angular speeds; a freely slowing run can expose speed dependence. Sensor differentiation noise, guide-wheel moment of inertia, thread slip, flexing connectors and uncertain weight radii limit the inference. Removing weights can also change bearing loads, so empty-apparatus subtraction does not automatically cancel friction. These are proposed controls and a model extension, not results claimed by PASCO.

A fixed-axis readout measures ${\mathbf{n}}^T{\mathbf{I}}_O\mathbf{n}$, not the full inertia tensor. In a proposed generalization, repeat about $x$, $y$, $z$ and the three diagonal directions $({\mathbf{e}}_x+{\mathbf{e}}_y)/\sqrt{2}$, $({\mathbf{e}}_x+{\mathbf{e}}_z)/\sqrt{2}$, $({\mathbf{e}}_y+{\mathbf{e}}_z)/\sqrt{2}$, all through the same $O$. For example,Six independent scalar readings then determine the six entries of a symmetric matrix. Each mounting needs its own apparatus correction, and different reference points require the tensor parallel-axis theorem before the readings can be combined.

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]

If the otherwise force-free second sphere is held against rotation, its angular velocity relative to the incident fluid is $-3a({\mathbf{U}}_0\times \mathbf{n})/(4R^2)$. The rotlet reflected back to the first sphere addsThis follows by evaluating the rotating sphere in Stokes flow at displacement $-R\mathbf{n}$ and using the vector triple-product identity. The holding couple does no work because the held sphere's angular velocity is zero.

For a rigidly rotating prescribed vacuum path, use spatial coordinates attached to the apparatus and laboratory time $t$. At a path point $\mathbf{r}$, the local apparatus velocity is $\mathbf{v}=\mathbf{\Omega }\times \mathbf{r}$. A light displacement has instantaneous laboratory displacement $d\mathbf{r}+\mathbf{v}dt$, up to the common rotation of the spatial axes. The null condition is thereforeWrite $d\ell =|d\mathbf{r}|$. To first order in $|\mathbf{v}|/c$, the positive root isReversing the path reverses the dot product term but leaves the leading path-length term unchanged. Subtract the two return times and use the scalar triple product identity:This is a line integral version of the Sagnac area law; the vector area identity also follows from Stokes theorem. For a mirror-defined polygon the actual ray paths move slightly as the mirrors rotate; the stationary optical path at zero rotation gives the same first-order result. The prescribed-path calculation is an idealization, not an exact finite-speed formula for every moving-mirror layout.

The dot product explains the geometry: only the projection of the oriented vector area along the angular velocity contributes. A planar loop whose normal vector is perpendicular to $\mathbf{\Omega }$ has zero leading delay. For repeated traversals with the same orientation, the signed vector areas add; oppositely oriented lobes can cancel even when the total unsigned area is large. Equal perimeter alone does not imply equal response. A constant translational velocity contributes nothing to this leading closed-path line integral, since ${\oint }_{C}d\mathbf{r}=0$.

The Sagnac effect is a difference in return times, or an associated interference fringe displacement, for light sent in opposite directions around a closed optical path in a rotating apparatus. The light leaves one beam splitter, travels around the same loop in opposite senses, and returns to the rotating beam splitter and readout. For a slowly rotating apparatus in flat spacetime, the leading laboratory arrival-time difference isHere $c$ is the vacuum speed of light, $\mathbf{\Omega }$ is the apparatus's angular velocity relative to an inertial frame, and $\mathbf{A}$ is the vector area oriented by the chosen positive direction around the closed curve $C$. With positive rotation about the upward normal vector, the counterclockwise beam is labelled $+$ and arrives later. Reversing rotation, or reversing the orientation convention alone, changes the sign of $\Delta t$.

The surprise is that a small loop can reveal its rotation by comparing two beams locally, even though each local vacuum speed of light measurement gives $c$. A rotating reference frame is not a single inertial frame: a beam must catch the moving receiver after a full circuit. The Sagnac area law makes this difference a practical way to measure angular velocity, while the Relativistic Sagnac delay on a circular ring distinguishes laboratory coordinate time from the receiver's proper time.