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Let a horizontal gyroscope have signed spin angular momentum $L_x$ along the axle. The disk's centre of mass is at $+l\hat{\mathbf{x}}$ and a counterweight is at $-l_c\hat{\mathbf{x}}$. Gravity supplies torque $g(ml-m_c{l}_c)\hat{\mathbf{y}}$. Since $\mathbf{\tau }=\mathbf{\Omega }\times \mathbf{L}$, steady horizontal precession gives the displayed signed vertical angular velocity. Neglect the axle's mass and the counterweight's axial spin moment of inertia. The zero-torque choice $m_c{l}_c=ml$ allows the axle to remain fixed. Moving the counterweight through this position reverses the gravitational torque and the associated slow precession.

A gyrocompass uses a spinning gyroscope and Earth's rotation to identify geographical north. In an ideal model whose axle is constrained to the local horizontal plane but free to turn about the vertical, the northward direction of the spin angular momentum is stable away from the poles; the opposite spin direction is unstable. A real settling instrument also requires damping. This is different from merely preserving an initial inertial orientation with a freely mounted gyroscope.

A gyroscope is a rotating instrument whose rapidly spinning, nearly symmetric rotor provides a direction reference through its angular momentum. In an inertial frame, negligible external torque approximately preserves that direction by conservation of angular momentum. A transverse torque turns the angular momentum instead: the surprising sideways motion of a supported spinning wheel is precession. An ideal freely falling test gyroscope in general relativity has its spin undergo parallel transport along its timelike geodesic. Comparing that direction with distant reference axes measures Lense-Thirring precession; the direction need not remain fixed in arbitrary rotating coordinates.

A gyroscope direction reference is useful precisely because its direction can persist while its housing turns. A spinning-wheel precession experiment deliberately supplies gravitational torque to reveal how that persistence fails. The gyroscope vector geometry explains the direction, and slow steady gyroscope precession gives a numerical prediction with a check on its assumptions.

At linear order in $ϵ=GM/R$ and the leading rotation term, this coordinate change takes the interior shell metric to Minkowski spacetime. Consequently the retained interior Riemann curvature tensor vanishes although inertial axes rotate relative to distant reference axes. A freely moving particle has $\mathbf{X}={\mathbf{X}}_0+\mathbf{V}T$, while its original coordinates obey $d^2\mathbf{x}/d{t}^2=2\mathbf{\omega }\times d\mathbf{x}/dt$ to first rotation order. Terms of order ${\omega }^2$ would include centrifugal acceleration and are outside this approximation. A gyroscope at rest to the retained order has its parallel-transported spin precessing at $\mathbf{\omega }$ in the original coordinates.

For a rigid body with three distinct positive principal moments of inertia $0<I_1<I_2<I_3$ and zero external torque about its centre of mass, steady rotation about the first or third principal axis is Lyapunov stable in angular velocity, while rotation about the intermediate principal axis is unstable. Here steady rotation means $\mathbf{\omega }=\Omega {\hat{e}}_i$ in the principal body frame, with constant $\Omega \ne 0$. A small sideways perturbation remains small about the extreme principal moments of inertia, but generically grows about the intermediate one. Stable does not mean that the wobble decays: without damping, there is no asymptotic stability. Coincident principal moments of inertia and a stationary body are excluded.^[ref]

No external torque is needed for a body to tumble. Conservation of angular momentum fixes $\mathbf{L}$ in the fixed space frame, but the body can turn relative to $\mathbf{L}$, so its angular velocity components in the principal body frame change. The torque-free rigid-body invariants constrain this motion. A supported gyroscope with gravitational torque has different governing assumptions.

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.

Let $\mu =GM/c^2$, $f=1-2\mu /r$, $q=\sqrt{1-3\mu /r}=1/u^0$, and choose $\Omega >0$. A torque-free test gyroscope moving on a geodesic has a spin four-vector obeying parallel transport,The orthogonality condition says the temporal spin component vanishes in the body's rest frame, not necessarily in Schwarzschild coordinates. Here it givesIn addition to the radial Christoffel symbols in part (a), use ${\Gamma }_{tr}^t=\mu /(r^{2}f)$ and ${\Gamma }_{r\varphi }^{\varphi }=1/r$. On the equator the parallel transport equations becomeThe radial equation follows by substituting $s^t$ into $d{s}^r/dt+{\Gamma }_{tt}^r{s}^t+\Omega {\Gamma }_{\varphi \varphi }^r{s}^{\varphi }=0$. Eliminating $s^{\varphi }$ gives $d^2{s}^r/d{t}^2+{\Omega }^2{q}^2{s}^r=0$.

Let $s_0$ be the invariant spin magnitude. Initially radial spin means $s^r(0)=s_0\sqrt{f}$ and $s^t(0)=s^{\varphi }(0)=s^{\theta }(0)=0$. The complete solution isTo interpret the orientation, convert coordinate components into the orbiting orthonormal frame. The Lorentz factor relative to a Schwarzschild static observer is $\gamma =\sqrt{f}/q$, so the body's measured radial and tangential spin components areTheir squared sum is $s_0^2$, and their angle relative to the outward radial direction decreases uniformly at rateThe differing coordinate amplitudes in $s^r,s^{\varphi }$ must not be interpreted as a nonuniform physical spin rotation. After one orbital period the radial basis has turned by $2\pi$ while the spin has turned by $-2\pi q$ relative to it. The residual geodetic precession is $\Delta \psi =2\pi (1-q)$ in the orbital direction.

The geodetic precession per orbit isThe orbital position and four-velocity repeat after a full orbit, so the direction to distant Earth in the instantaneous rest frame repeats. The freely transported gyroscope does not: its beam axis has turned through $\Delta \psi$ relative to the original aiming direction. This applies to any initial in-plane spin direction by linearity of parallel transport, not only the radial initial condition used in part (b). The beam's angular diameter is $1^{\circ }$, giving a half-angle of $0.5^{\circ }$. Thus Earth lies outside the beam after one orbit; comparing the offset with the diameter would already show a miss, but the correct acceptance angle is the half-angle.

The initial radial ray must be outward from the black hole for a direct path to Earth. A precisely inward radial null geodesic falls into the black hole and cannot provide the assumed initial communication. For outgoing near-radial Schwarzschild light deflection, the angular correction is especially small. If a static observer at $R$ measures a small emission angle $\psi$ from the outward radial direction, its impact parameter is $b=R\sin \psi /\sqrt{f(R)}$. The outgoing null geodesic givesRelative to flat propagation, curvature changes the small pointing angle and beam radius by the fraction $f(R{)}^{-1/2}-1\simeq 0.00251$. At an offset of $1.35^{\circ }$ this is only about $0.0034^{\circ }$; the beam's $0.5^{\circ }$ radius changes by about $0.0013^{\circ }$. The exactly radial central ray has zero angular deflection. This is not the geometry of the usual $4GM/(b{c}^2)$ scattering formula, which assumes a ray coming from infinity, passing the black hole, and returning to infinity.

Relativistic aberration also distinguishes a coordinate-radial light ray from a beam aimed radially in the moving crew's frame. Here $v/c=1/\sqrt{398}\simeq 0.0501$. Successful initial aiming includes that fixed correction; the same correction applies at the repeated orbital position. More precisely, an outgoing static-radial ray initially has crew-frame angle ${\psi }_0^{\prime }=-\arcsin (v/c)$ from the radial axis. The local small-angle aberration factor about that aim is $d\psi /d{\psi }^{\prime }=\gamma =\sqrt{398/397}$; including the outgoing propagation factor gives $\gamma /\sqrt{f}=1/q\simeq 1.00377$. Thus the asymptotic miss is about $1.358^{\circ }$ and the beam half-angle about $0.502^{\circ }$, with the same conclusion. The small distortion of the outward beam cannot correct the much larger accumulated geodetic precession.

Use units $c=1$, signature $(+---)$, and the weak-field metric $g_{ab}={\eta }_{ab}+h_{ab}$. Define the trace-reversed metric perturbation by ${\bar{h}}_{ab}=h_{ab}-{\eta }_{ab}h/2$, with $h={\eta }^{ab}{h}_{ab}$. In the Lorenz gauge in linearized gravity, ${\partial }^a{\bar{h}}_{ab}=0$, the sign convention of Q2 givesFor completeness, the linearized Ricci tensor in this convention isSubstituting the Lorenz condition ${\partial }_c{h}^c{}_b={\partial }_{b}h/2$ and trace reversing gives the stated Einstein tensor. For a stationary source, $\square =-{\nabla }^2$, so ${\nabla }^2{\bar{h}}_{ab}=16\pi G{T}_{ab}$.

Keep the monopole terms of order $ϵ=GM/R$ and the leading term linear in the rotation rate. Since $\Omega R=O({ϵ}^{1/2})$, the latter metric term is of order $ϵ\Omega R=O({ϵ}^{3/2})$ on the shell. Thus it would disappear from a strictly order-$ϵ$ truncation; the requested approximation must retain the leading rotation correction as well. Terms $T_{ij}=\rho v_i{v}_j$ generate metric corrections of order $ϵ(\Omega R{)}^2=O({ϵ}^2)$ and are neglected. Corrections to four-velocity normalization and the supporting shell stresses affect the omitted higher orders. To retained order,The minus sign in $T_{0i}$ comes from lowering its spatial index; using $T^{0i}$ in its place would reverse the dragging direction.

The scalar Poisson solution regular at the centre and vanishing at spatial infinity is ${\bar{h}}_{00}=-4GM/r$ outside the shell and ${\bar{h}}_{00}=-4GM/R=-4ϵ$ inside. Continuity and the derivative jump ${\bar{h}}_{00}^{\prime }(R^{+})-{\bar{h}}_{00}^{\prime }(R^{-})=4GM/R^2$ verify the density $M\delta (r-R)/(4\pi R^2)$. Also ${\bar{h}}_{ij}=0$ at this order. Undoing trace reversal givesComparing with the scalar parts of the interior line element yields $A=-ϵ$ and $C=ϵ$.

For the mass-current contribution writeEach nonzero angular component is a degree-one spherical harmonic. The supplied Laplacian eigenvalue therefore givesAway from the shell, the two radial solutions are $r$ and $r^{-2}$. Regularity at $r=0$ and decay at infinity select $F_{\mathrm{in}}=ar$ and $F_{\mathrm{out}}=b/r^2$. Continuity at $R$ requires $b=a{R}^3$, and integrating the differential equation across the shell givesThus $a=4GM\Omega /(3R)=\omega$, with $F_{\mathrm{in}}=\omega r$. Trace reversal leaves mixed components unchanged, so $h_{0i}=\omega (\hat{\mathbf{z}}\times \mathbf{x}{)}_i=\omega (-y,x,0{)}_i$. The paper writes $g_{0i}=-B_i$, giving the frame dragging inside a slowly rotating spherical shellThis dipole matching is valid throughout $r<R$; it does not assume the observation point is very close to the centre. The stationary current is divergence-free, so this solution also satisfies the Lorenz gauge. The exterior result is $h_{0i}=2G(\mathbf{J}\times \mathbf{x}{)}_i/r^3$, where the thin shell has $\mathbf{J}=(2/3)M{R}^2\mathbf{\Omega }$, providing a check on the coefficient.

To identify the freely falling frames inside, let ${\mathcal{R}}_z(\theta )$ be the ordinary spatial rotation by angle $\theta$. Define the inertial coordinates inside a slowly rotating spherical shell byDifferentiating and dropping ${ϵ}^2$, products beyond the retained rotation order, and ${\omega }^2$ giveswhich is precisely the interior metric. Thus the retained interior curvature is zero, despite its nontrivial relation to the nonrotating coordinates anchored at infinity. Timelike geodesics in these local inertial coordinates are straight lines $\mathbf{X}={\mathbf{X}}_0+\mathbf{V}T$, with $|\mathbf{V}|<1$. In the original coordinates this isunderstood to the same perturbative order. Their coordinate equation is $d^2\mathbf{x}/d{t}^2=2\mathbf{\omega }\times d\mathbf{x}/dt$ to first rotation order. There is no interior Newtonian acceleration from the constant potential; the velocity-dependent term is the apparent Coriolis term relative to the locally dragged inertial axes.

A freely transported gyroscope at rest in the interior to the retained order has constant spatial direction in the $\mathbf{X}$ frame, hence its direction in the original axes obeys $d\mathbf{S}/dt=\mathbf{\omega }\times \mathbf{S}$. Local inertial axes precess in the same sense as the shell, at $\omega =(4ϵ/3)\Omega$ relative to distant nonrotating axes. This is the interior Lense-Thirring precession. For a weak shell the dragging is only a small fraction of its angular velocity, not perfect corotation. It is compatible with the equivalence principle: rotation relative to distant axes is a comparison between frames, and local flatness at the retained order does not make that comparison vanish. Quadratic rotation terms, which would include centrifugal effects and additional curvature, are outside the requested leading approximation.

An inertial frame is a nonrotating frame in which all force-free test bodies follow straight trajectories with constant velocity. The qualification “all” matters: a specially chosen coordinate system can make one accelerated trajectory look straight without making the frame inertial. In practice, approximately isolated test bodies and gyroscopes help identify inertial motion; identifying the relevant forces and calibrating instruments are part of testing the theory. The definition does not select a privileged state of uniform rest.

In its usual coordinate formulation, Newtonian mechanics states $m\ddot{\mathbf{x}}=\mathbf{F}$ in an inertial frame. A Galilean transformation, $t^{\prime }=t+b$, ${\mathbf{x}}^{\prime }=R\mathbf{x}+\mathbf{V}t+\mathbf{a}$ with constant orthogonal $R$ and constant $\mathbf{V}$, maps the class of inertial frames to itself. It preserves zero acceleration; a time-dependent translation generally does not. In the translating frame ${\mathbf{x}}^{\prime }=\mathbf{x}-\mathbf{a}(t)$, the same physical law reads $m{\ddot{\mathbf{x}}}^{\prime }=\mathbf{F}-m\ddot{\mathbf{a}}$. The extra fictitious force records the frame's acceleration, rather than a new interaction.

It is possible to state Newtonian mechanics without mentioning an inertial frame explicitly. On Galilean spacetime, let $u$ be a trajectory tangent normalized by the absolute time covector, $t_a{u}^a=1$. Thenuses an affine connection to define acceleration. The free-particle equation is ${\nabla }_{u}u=0$. This formulation removes a chosen coordinate frame but retains the geometric distinction between inertial motion and accelerated motion. Omitting both the inertial frame language and all equivalent inertial structure would leave the law unable to specify which trajectories count as unforced.

The same distinction applies to special relativity. Einstein's original presentation uses a class of inertial frames and calibrated rods, clocks and light signals.^[ref] A geometric presentation instead uses Minkowski spacetime, its Lorentzian metric and its Levi-Civita connection. A free massive particle obeys ${\nabla }_{u}u=0$, now with proper time parametrization and $g(u,u)=-c^2$. There is no preferred inertial rest frame, but there is a distinguished family of inertial trajectories and a physical Minkowski interval.

Accelerating coordinates are allowed in special relativity. They produce nonzero connection coefficients even though the Riemann curvature tensor remains zero. General relativity is needed when spacetime curvature and dynamical gravitation matter, not merely because the chosen observer accelerates. Thus neither theory requires a particular inertial frame in its coordinate-free statement; each still needs the structure that determines inertial motion.

A freely rotating rigid body with no external torque conserves its inertial angular momentum and kinetic energy. In body-fixed principal axes, the Euler equations includeand cyclic permutations. The body can still change its orientation relative to its fixed angular-momentum vector; zero torque does not require every principal axis to remain fixed.

For a symmetric top, $I_1=I_2=I_{\perp }$, $I_3=I_{\parallel }$. The axial rate ${\omega }_3$ is constant, and ${\omega }_1+i{\omega }_2$ rotates in body coordinates with signed frequency $(I_{\parallel }-I_{\perp }){\omega }_3/I_{\perp }$. In space,Thus the symmetry axis precesses about fixed $\mathbf{L}$ at rate $L/I_{\perp }$ and constant inclination. This body-frame and space-frame frequencies of free symmetric-top precession distinction avoids confusing two different frame-dependent rates. A sphere has no dynamically distinguished symmetry axis even though a chosen body marker rotates.

For a general body, rotations about the largest and smallest principal inertias are stable to small perturbations, while rotation about the intermediate inertia is unstable. With $I_1<I_2<I_3$ and rotation ${\omega }_2=\Omega$, linearization gives ${\ddot{\omega }}_1={\Omega }^2(I_3-I_2)(I_2-I_1){\omega }_1/(I_1{I}_3)$, showing exponential growth. This is torque-free motion, distinct from the gravity-driven precession of a supported gyroscope.

A rapidly spinning gyroscope approximately has angular momentum $\mathbf{L}=I_3{\omega }_3\hat{\mathbf{e}}$ along its axle. With no torque in an inertial frame, this vector retains its direction. A transverse torque changes its direction according to $d\mathbf{L}/dt=\mathbf{\tau }$; it does not initially push the axle in the direction of the applied force.

For a fixed pivot, centre of mass $l\hat{\mathbf{e}}$ from it and gravity $-Mg\hat{\mathbf{z}}$, the torque is $Mgl\hat{\mathbf{z}}\times \hat{\mathbf{e}}$. In slow steady precession about the vertical, $d\mathbf{L}/dt=\Omega \hat{\mathbf{z}}\times \mathbf{L}$, givingThe factor $\sin \theta$ appears both in torque and in the radius traced by the angular-momentum tip, so it cancels for a nonvertical axle. Reversing spin reverses this precession. The full symmetric-top steady condition is $I_1\cos \theta {\Omega }^2-I_3{\omega }_3\Omega +Mgl=0$; the approximation neglects its quadratic term. Sudden release can also produce nutation, so large spin alone does not guarantee a steady sweep.

A wheel-on-pivot experiment can compare precession periods at different measured spin speeds and reverse the spin. Keep the pivot and centre of mass fixed; bearing friction, spin decay, an initial push and support motion can mimic departures from the predicted inverse-spin relation. A freely gimballed direction reference and a gravity-loaded precessing wheel test different torque conditions.^[ref]

For torque-free rotation, angular momentum is fixed in inertial space, but angular velocity need not point along it. In principal body axes,and cyclic permutations. A symmetric top with transverse moment $A$ and axial moment $C$ has constant ${\omega }_3$, while ${\omega }_1+i{\omega }_2$ rotates in body coordinates at signed frequency $(C-A){\omega }_3/A$. The symmetry axis precesses around the fixed angular momentum in space at rate $L/A$. These body-frame and space-frame rates describe different motions and need not coincide.

Poinsot construction expresses the same torque-free motion geometrically. The energy ellipsoid $\sum I_i{\omega }_i^2=2T$ is fixed in the body. Its normal at the angular-velocity point is parallel to $\mathbf{L}$, and $\mathbf{L}\cdot \mathbf{\omega }=2T$ places that point in a fixed plane perpendicular to $\mathbf{L}$. The ellipsoid rolls on this invariable plane without slipping: the instantaneous velocity of its contact point is proportional to $\mathbf{\omega }\times \mathbf{\omega }=0$. The contact curve is the polhode on the body ellipsoid and the herpolhode on the plane.^[ref]

For $I_1<I_2<I_3$, linearization about spin $\Omega$ on the third axis givesThe extreme-moment axes are stable; the intermediate one is unstable. The major-axis relaxation theorem adds weak internal dissipation while retaining negligible external torque. At fixed $L$,Energy loss favours rotation about the axis of largest principal moment. Here “major axis” means maximum inertia, not necessarily the body's longest geometric direction. A perfectly rigid torque-free body does not dissipatively settle; its energy is conserved. Distinguishing stability from relaxation also explains why minimum-inertia rotation can be stable in the ideal rigid model.

Forced precession occurs when an external torque changes $\mathbf{L}$, with $\dot{\mathbf{L}}=\mathbf{\tau }$. For steady precession, $\mathbf{\tau }={\mathbf{\Omega }}_p\times \mathbf{L}$. A rapidly spinning, pivoted heavy gyroscope has $L\simeq C{\omega }_3$ along its symmetry axis, givingThe torque is perpendicular to the nearly axial angular momentum, so it mainly changes its direction. For a steady inclined symmetric top, the more complete condition isIt exhibits slow and fast precession branches when real solutions exist. A generic release excites nutation as well: inclination oscillates instead of staying fixed. Axis constraints and bearing forces must be included when specifying which angular momentum and pivot are being used.

One application is the slow steady precession of a rapidly spinning gyroscope supported at one point. Let its centre of mass be a distance $\ell$ from the pivot along its spin axis, at angle $\alpha$ to the upward vertical. Write $J\simeq I_1\omega$ for its dominant spin angular momentum. In a reference frame precessing steadily about the vertical with angular velocity ${\mathbf{\Omega }}_p$, the spin angular momentum is approximately stationary, soThe gravitational torque has magnitude $mg\ell \sin \alpha$. The cross product on the right has magnitude ${\Omega }_{p}J\sin \alpha$, giving, for a nonvertical axis,Its direction follows from the torque equation. This is the slow precession approximation: the additional angular momentum associated with precession is small compared with the spin contribution.

Translation and rotation. For particles constituting a rigid body, the total linear momentum isThus the total external force governs the motion of the centre of mass. About that centre of mass, the angular momentum and applied torque arewhere primes denote relative positions and velocities. The derivative of the vector is taken relative to nonrotating axes. The inertia tensor relates angular momentum to angular velocity, ${\mathbf{L}}_{\mathrm{CM}}=\mathsf{I}\mathbf{\omega }$. For rotation about a fixed unit direction $\mathbf{n}$, the moment of inertia is $I_{\mathbf{n}}=\int r_{\perp }^{2}dm=\mathbf{n}\cdot \mathsf{I}\mathbf{n}$; it measures the resistance to changing that rotation and gives rotational kinetic energy $I_{\mathbf{n}}{\omega }^2/2$. The simpler vector relation $\mathbf{L}=I\mathbf{\omega }$ applies when the rotation axis is a principal axis, not for every orientation of an arbitrary rigid body. These translational and rotational balances must be applied together; a nonzero torque does not replace the net-force equation.

Gyroscope precession. A rapidly spinning gyroscope has a large angular momentum approximately along its spin axis. The torque component parallel to this vector changes its magnitude, while the perpendicular component turns its direction. If the vector undergoes precession with angular velocity $\mathbf{\Omega }$, then over $dt$ its change is the infinitesimal rotation $d\mathbf{L}=(\mathbf{\Omega }dt)\times \mathbf{L}$. HenceThis describes direction-changing precession at constant $|\mathbf{L}|$; any spin-up contribution must be added separately. For perpendicular torque and angular momentum, the precessional component perpendicular to $\mathbf{L}$ is ${\mathbf{\Omega }}_{\perp }=\mathbf{L}\times \mathbf{G}/L^2$. The familiar tendency of a supported gyroscope to turn sideways is this vector balance, not the absence of gravitational force.

Wheel-lift threshold without the flywheel. Let $h=0.50\mathrm{m}$ be the centre of mass height, $a=1.0\mathrm{m}$ half the wheel separation, and $R_b=10\mathrm{m}$ the bend radius. The required inward force is $F=M{v}^2/R_b$, provided by static friction. Let $N_o,N_i$ be the summed outer and inner wheel normal forces. Vertical balance gives $N_o+N_i=Mg$. Moment balance about the centre of mass givesThe inner wheels lift when $N_i=0$. Thus the vehicle rollover threshold isThe stated mass cancels. This assumes the rigid geometry and sufficient static friction specified in the paper; the threshold is set by loss of contact, not by sliding.

Engine flywheel. The moment of inertia of a uniform disc givesIts spin angular speed, spin angular momentum, and turn angular speed at $40\mathrm{km}{\mathrm{h}}^{-1}$ areThe transverse flywheel axis is perpendicular to the vertical turn axis, soTo specify its direction, use right-handed axes $+x$ forward, $+y$ to the car's left, and $+z$ upward, and consider a left turn. Then $\mathbf{\Omega }=+\Omega \hat{\mathbf{z}}$. Choose the flywheel angular momentum to the car's right, ${\mathbf{L}}_f=-L_f\hat{\mathbf{y}}$. The car must applyto obtain the required precession. This gyroscopic roll torque in a turning vehicle opposes the outward roll tendency in a left turn. By the right-hand rule, the flywheel must turn counterclockwise viewed from the car's right side toward its centre, opposite to the rotation of wheels rolling forward. Reversing flywheel spin would aggravate the outward roll. Reversing the turn reverses both roll and gyroscopic torques, so the same flywheel spin remains helpful.

For comparison, the outward inertial roll moment at the stated speed is $Mh{v}^2/R_b=6173\mathrm{N}\mathrm{m}$, and the limiting gravitational restoring moment is $Mga=9810\mathrm{N}\mathrm{m}$. The gyroscopic reaction is about $30\%$ of the former and $19\%$ of the latter, so it is appreciable for this large flywheel. If this ideal model is extended to include the favorable gyroscopic reaction in the wheel-lift condition, it gives $Mh{v}^2/R_b-L_{f}v/R_b=Mga$, or about $56.8\mathrm{km}{\mathrm{h}}^{-1}$ rather than $50.4\mathrm{km}{\mathrm{h}}^{-1}$. This quantifies a useful but finite increase in the rollover threshold.

The spin of a small torque-free test gyroscope is represented by a spacelike vector orthogonal to its four-velocity, $s^{\mu }{u}_{\mu }=0$. Its temporal component vanishes in its own rest frame. Along a geodesic, $D{s}^{\mu }/d\tau =0$: parallel transport preserves the orthogonality and the invariant spin magnitude. For an accelerated observer the corresponding nonrotating transport is Fermi-Walker transport.

Torque measures a force's turning effect about a specified origin. The geometry in the angular momentum diagram applies with force replacing momentum: transverse forces give positive or negative torque, while a radial force gives zero. A longer perpendicular lever arm makes the same force more effective, explaining why pushing near the end of a handle is useful.

For a force $\mathbf{F}$ applied at position $\mathbf{r}$ from a fixed origin $O$ in an inertial frame, its torque, or moment of force, isHere $F=|\mathbf{F}|$, $\varphi$ is the angle between position and force, and $\ell =r\sin \varphi$ is the perpendicular lever arm. Direction follows the right-hand rule; in the $xy$ plane ${\tau }_z=x{F}_y-y{F}_x$, positive counterclockwise as viewed from $+z$. The International System of Units unit is $\mathrm{N}\mathrm{m}=\mathrm{kg}{\mathrm{m}}^2/{\mathrm{s}}^2$. Although torque and energy have the same physical dimensions, torque is an axial vector and energy is a scalar; their roles differ.

For $\mathbf{r}=(2,0,0)\mathrm{m}$ and $\mathbf{F}=(0,3,0)\mathrm{N}$, ${\tau }_z=+6\mathrm{N}\mathrm{m}$. Applying the same force at $(1,0,0)\mathrm{m}$ gives $+3\mathrm{N}\mathrm{m}$; reversing it gives a negative torque; directing it along $\mathbf{r}$ gives zero. More generally, sum ${\mathbf{r}}_i\times {\mathbf{F}}_i$ to find the total torque about one common origin.

For a constant-mass point mass, the product rule and Newton's second law show why net torque changes angular momentum:The first cross product vanishes because its factors are parallel. For a collection with canceling internal torques, the remaining sum is external torque, giving conservation of angular momentum when it vanishes.^[ref]

A nonzero torque can turn the direction of angular momentum as well as change its magnitude. Indeed, the dot product gives $d|{\mathbf{L}}_O{|}^2/dt=2{\mathbf{L}}_O\cdot {\mathbf{\tau }}_O$: a perpendicular torque changes direction without instantaneously changing the magnitude. This is the geometry behind gyroscope precession. For a rapidly spinning symmetric top with a fixed tip, centre of mass distance $d$ from the tip, mass $M$, and spin-dominated angular momentum magnitude $L\simeq I\omega$, the slow steady precession approximation gives $\Omega \simeq Mgd/L$, where $g$ is gravitational acceleration. ^[ref] This is an approximation with $\Omega \ll \omega$; the uniform precession of a heavy symmetric top retains contributions omitted by the spin-dominated estimate.