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In electrostatics, Gauss's law isThe surface normal points outward. Spherical symmetry makes the field radial and constant in magnitude on a concentric Gaussian surface.

For a uniformly charged sphere, enclosed charge is $4\pi {\rho }_0{r}^3/3$ inside and $Q=4\pi {\rho }_0{R}^3/3$ outside. Thus the radial field isSet potential zero at infinity. Outside, $V=Q/(4\pi {ϵ}_{0}r)$. Inside, integrate $dV/dr=-E_r$ from the known boundary value, obtainingThe field rises linearly from zero to its maximum at $R$, then falls as $r^{-2}$. The potential falls quadratically from ${\rho }_0{R}^2/(2{ϵ}_0)$ to its surface value and then as $r^{-1}$. Both are continuous at the surface; there is no separate sheet of surface charge.

For the electric field of a hydrogen ground-state electron cloud, its enclosed charge isWriting $u=2s/a_0$ and using $\int u^2{e}^{-u}du=-e^{-u}(u^2+2u+2)$ givesThe total electron-cloud charge tends to $-e$, as required. Dividing by $4\pi {ϵ}_0{r}^2$ yieldsThe expression is negative and therefore points inward. Its apparently singular terms cancel at small radius, leaving $E_{e,r}\simeq -er/(3\pi {ϵ}_0{a}_0^3)$.

A distant external charge polarizes the neutral atom. Its electric field displaces the positive and negative charge distributions relative to one another, giving an induced dipole $\mathbf{p}=\alpha {\mathbf{E}}_{\mathrm{ext}}$ with static electric polarizability $\alpha >0$. The field is stronger on the nearer side, so the opposite-charge attraction exceeds the like-charge repulsion. This is charge-induced dipole attraction, not a net monopole charge on the atom.

Quantitatively, for external charge $Q=\pm e$ at separation $r$, the polarization energy is $U=-\alpha E_{\mathrm{ext}}^2/2$, henceHere negative $F_r$ means decreasing separation. Reversing the external charge reverses the induced dipole but leaves this leading long-range force attractive and of the same magnitude. The factor one half accounts for the energy cost of inducing the dipole.

The supplied charge density describes the unperturbed atom. Adding its electron field to the nucleus field leaves $E_{\mathrm{atom}}=e{e}^{-2r/a_0}(r^{-2}+2/(a_{0}r)+2/a_0^2)/(4\pi {ϵ}_0)$, an exponentially small residual because the cloud has an infinite tail. A completely frozen density would give only that tiny sign-dependent force. Allowing polarization is essential to obtain the dominant attractive power-law force for either sign of external charge when $r\gg a_0$.

A position-dependent force is a conservative force when its work depends only on endpoints. Equivalently, its work around any closed allowed path vanishes, and one can define a potential energy $U$ satisfyingThe potential is defined up to an additive constant. A vanishing curl, $\nabla \times \mathbf{F}=0$, is a local test; it guarantees a global single-valued potential on a simply connected region with suitable regularity. Holes or excluded singularities require checking the global path condition.

The work-energy theorem gives $dK/dt=\mathbf{F}\cdot \mathbf{v}$. For a static potential this implies $d(K+U)/dt=0$. If $U$ depends explicitly on time, then instead $d(K+U)/dt=\partial U/\partial t$ under that force alone. Friction and other nonconservative forces add their own power and can convert mechanical energy into heat.

Examples include the gravitational potential energy $-GMm/r$ outside a spherical mass and the electrostatic energy $U=qV$, where $\mathbf{E}=-\nabla V$. For a point source charge $Q$, $U=qQ/(4\pi {ϵ}_{0}r)$, giving attraction or repulsion through the sign of $qQ$. A spring has $U=k{x}^2/2$ and force $-kx$. An equilibrium has $\nabla U=0$; a strict local minimum is stable against small displacements when the quadratic curvature is positive. Expanding about that minimum explains the harmonic restoring force used in many small-oscillation models.

Magnetic force on a moving point charge does no work, but it is velocity-dependent and is not generally a position-only force $-\nabla U$. Likewise a time-varying magnetic flux induces a circulating electric field with $\oint \mathbf{E}\cdot d\mathbf{l}=-d{\Phi }_B/dt$, so that electric field cannot globally be represented solely by an electrostatic scalar potential. These distinctions prevent treating every field force as conservative simply because an energy description exists for the complete system.

Directions. Choose $+x$ along the horizontal spindle from the pivot to the disk, and $+z$ vertically upward. With lever arm $\mathbf{r}=l\hat{\mathbf{x}}$ and weight ${\mathbf{F}}_g=-mg\hat{\mathbf{z}}$, the torque about the pivot isClockwise precession viewed from above has $\mathbf{\Omega }=-\Omega \hat{\mathbf{z}}$. The relation $\mathbf{\tau }=\mathbf{\Omega }\times \mathbf{L}$ therefore requires the spin angular momentum to point along $-x$, from the disk toward the pivot. The disk spins clockwise when viewed from its outer face looking along the spindle toward the pivot. Looking from the pivot reverses that apparent rotation. The right-hand rule fixes these signs unambiguously.

The drawing shows an instantaneous top view; the disk's angular momentum and the horizontal spindle turn clockwise together. The pivot force has zero torque about the pivot.

Rate of precession. For steady horizontal motion the spin contribution has magnitude $L=I\omega$ and turns through $d\varphi =\Omega dt$ in time $dt$. Its change is perpendicular to itself and has magnitude $dL=Ld\varphi$. HenceThis is the usual slow steady gyroscope precession relation. Any constant vertical angular momentum associated with steady precession has zero time derivative; the horizontal spin contribution supplies the changing component here. Transient nutation is excluded by the prescribed horizontal steady motion.

Clock design. Convert both rotation rates to angular frequencies:For the moment of inertia of a uniform disc, $I=m{R}^2/2$, so the mass cancels:

Spin friction. At fixed geometry, reducing $\omega$ increases $\Omega \propto 1/\omega$. The precession period, $T_p=2\pi /\Omega$, therefore decreases. This trend holds while slow steady precession remains a suitable description; it cannot be extrapolated to a stopped disk.

Counterweight. To oppose the disk's gravitational torque, place the counterweight on the other side of the pivot, at $-l_c\hat{\mathbf{x}}$. The net torque isA horizontal axle can remain fixed when this vanishes, givingFor the fixed spin direction already found, counterbalanced gyroscope precession has signed rateMoving the counterweight outward from the pivot first slows the clockwise motion, eliminates its driving torque at balance, and then reverses the steady motion to counterclockwise. A counterweight on the disk's side would add to the gravitational torque. These statements neglect axial spin moment of inertia of the counterweight and assume the stated horizontal motion, without a separately imposed precession at balance.

Axial moment of inertia. In the original diagram $R_1$ is the outer radius and $R_2$ the inner radius. For uniform mass density $\rho$, a coaxial annulus of thickness $dr$ has $dm=2\pi \rho Lrdr$. Therefore the axial moment of inertia of a hollow cylinder isSince $M=\pi \rho L(R_1^2-R_2^2)$, factorizing the fourth-power difference givesFor a solid cylinder $R_2\to 0$, giving $I=M{R}_1^2/2$. For a thin cylindrical shell with fixed mass and $R_1,R_2\to R$, $I=M{R}^2$. The latter limit agrees with every mass element being at radius $R$.

Undamped rolling oscillations. Let $x$ be axle displacement from the spring equilibrium. Rolling without slipping imposes $|{\omega }_{\mathrm{roll}}|=|\dot{x}|/R$. The sum of translational and rotational kinetic energy isThe potential energy is $U=k{x}^2/2$. Differentiating $T+U$ in the absence of damping, or applying the force and torque equations below, givesFor the thin shell $I=m{R}^2$, the effective inertia is $2m$, and the angular frequency of the simple harmonic motion isStatic contact friction supplies the required torque but does no work when the contact point is instantaneously stationary.

Damped force balance. Take rightward force as positive and counterclockwise rotation as positive. Write $f$ for the signed static friction at the bottom contact. ThenEliminate $f=-I\ddot{x}/R^2$ and use $I=m{R}^2$:Here $b$ has units $\mathrm{kg}{\mathrm{s}}^{-1}$, so $B$ has units ${\mathrm{s}}^{-1}$ and $C$ units ${\mathrm{s}}^{-2}$. The stated drag force is interpreted as acting through the axle, with no additional viscous torque. An oil bath could also exert such a torque physically, but its magnitude is not specified by this force-only model.

Light damping. Substitution of $z=A{e}^{-pt}$ into the differential equation gives $p^2-Bp+C=0$. In the underdamped regime $B^2<4C$,Taking real linear combinations gives the general damped harmonic oscillator displacementThe constants are fixed by the initial displacement and velocity. For example, release from rest at $x_0$ gives $A_1=x_0$ and $A_2=b{x}_0/(4m{\omega }_d)$. Light damping means $b^2\ll 8mk$, in which case ${\omega }_d\simeq {\omega }_0$ and the envelope decays slowly compared with an oscillation.

Field of discrete masses. Newton's law of universal gravitation and superposition give the gravitational field away from the source points:It is the force per unit test mass, with units $\mathrm{N}\mathrm{k}{\mathrm{g}}^{-1}=\mathrm{m}{\mathrm{s}}^{-2}$, and points toward each positive source mass.

Infinite circular cylindrical rod. Let $b$ be the radius of the circular cross-section, so $A=\pi b^2$, and write $\lambda =\rho A$ for the linear mass density. Choose a coaxial cylindrical Gaussian surface with radius $r>b$ and length $L$. Translation and rotation symmetry make $\mathbf{g}=g_r(r)\hat{\mathbf{r}}$. The end caps have zero field flux because their normals are axial; on the curved surface the outward-normal component is the constant negative value $g_r$. Gauss's law for gravity therefore givesThe gravitational field of an infinite uniform cylinder inside the material instead has enclosed mass $\rho \pi r^{2}L$, giving $g_r=-2\pi G\rho r$. The exterior result assumes a circular cylindrical rod; area by itself would not specify the exact near field of a noncircular cross-section.

In the left panel the dashed circle is a transverse section of the coaxial Gaussian surface; the field points inward and the surface normals outward. Its side area is $2\pi rL$.

For an exterior circular orbit, Newton's second law equates the force required for centripetal acceleration to gravity:The orbital speed is independent of radius and of the orbiting object's mass, for orbits outside the infinite rod.

Finite rod. Taking the supplied field as the gravitational field of a finite uniform line segment, the corresponding balance at radius $R$ isSince $M=2a\rho A=2a\lambda$, its two limits areThe first reproduces the infinite-rod result: a nearby observer sees a very long source. The second is the familiar circular orbit speed about a point mass, since a distant observer cannot resolve the rod's length. The dimensionless curve isIt starts at a constant plateau and decreases smoothly, approaching an $R^{-1/2}$ dependence far away. For $R>0$, $dv/dR=-\sqrt{GM}R/[2(R^2+a^2{)}^{5/4}]<0$, consistent with the sketch.

Derivation of the finite-rod field. Put the source along $-a<z<a$ on the $z$ axis. In its central perpendicular plane, an element $dm=\lambda dz$ is at distance $s=\sqrt{r^2+z^2}$. Its field magnitude is $G\lambda dz/s^2$, and its inward radial component has the extra factor $r/s$. The axial components cancel between $z$ and $-z$. ThusReplacing $2a\lambda$ by $M$ givesThe integral treats the source as a line mass. For a finite cylinder of nonzero cross-section, the displayed finite-rod formula is a thin-rod approximation, not an exact field for all $r$. Its near-rod limit should therefore be read as $b\ll R\ll a$, keeping the orbit outside the material; the apparent line-source singularity at $r=0$ does not describe the interior of a real cylinder.

An electric dipole is a localized pair of opposite electric charges, or more generally a neutral electric charge distribution with nonzero first electric moment. For charges $-q$ and $+q$ separated by displacement $\mathbf{d}$ directed from negative to positive, the electric dipole moment isA neutral molecule can have a permanent electric dipole moment when its positive and negative charge centres do not coincide. An applied electric field can also produce an induced electric dipole moment by distorting an initially nonpolar electric charge distribution.

At distance $r\gg |\mathbf{d}|$ in vacuum, expand the difference of the two point-charge potentials to first order in the separation. The leading electric potential isThe monopole term cancels because the net electric charge is zero. Taking $\mathbf{E}=-\nabla \Phi$ gives the electric dipole fieldOn the dipole axis its magnitude is $2p/(4\pi {\epsilon }_0{r}^3)$ and its direction is along $\mathbf{p}$; in the equatorial plane the magnitude is $p/(4\pi {\epsilon }_0{r}^3)$ and the field points opposite to $\mathbf{p}$. The $r^{-3}$ decay is faster than the $r^{-2}$ field of an isolated point charge.

In a uniform external electric field, the opposite forces on the two charges cancel in their resultant but form a couple. Its torque isFor a fixed permanent electric dipole moment, the interaction potential energy isDifferentiating with respect to orientation recovers the torque. Parallel alignment minimizes $U$ and is stable against a small rotation; antiparallel alignment maximizes $U$ and is unstable. Rotating a fixed dipole slowly from parallel to antiparallel requires external work $2pE$.

A nonuniform electrostatic electric field can also exert a net force. Expanding the two charge forces gives $\mathbf{F}\simeq (\mathbf{p}\cdot \nabla )\mathbf{E}$, which equals $\nabla (\mathbf{p}\cdot \mathbf{E})$ for a fixed dipole in a curl-free electrostatic field. Thus a suitably aligned dipole is attracted toward stronger field. An induced electric dipole moment requires care in computing energy: when $\mathbf{p}=\alpha \mathbf{E}$ for constant isotropic electric polarizability, building the dipole costs energy and the effective interaction is $U=-\alpha E^2/2$, rather than substituting directly into the fixed-dipole formula. These field, torque and energy effects explain molecular orientation and electric polarization.

A conservative vector field has path-independent work: the integral ${\int }_A^B\mathbf{F}\cdot d\mathbf{r}$ depends only on its endpoints. Equivalently, the line integral around every closed path in its domain vanishes. One can then define a single-valued potential energy $U$ byThe arbitrary additive constant in $U$ has no effect on the force. A useful zero is often chosen at infinity for a localized attractive interaction.

A differentiable conservative force has $\nabla \times \mathbf{F}=0$ because the curl of a gradient vanishes. Conversely, a smooth curl-free field on a simply connected domain admits a global scalar potential. The domain condition matters: on the punctured plane, $\mathbf{F}=(C/r)\hat{\mathbf{\theta }}$ has zero curl away from the missing origin but closed-loop work $2\pi C$ around it. It is not globally conservative there. The existence of a local potential does not establish a single-valued potential throughout a domain with holes.

The work-energy theorem gives $\Delta T=W$ for kinetic energy $T$. If all forces are conservative and $U$ has no explicit time dependence, thenThis is conservation of mechanical energy. In one dimension, an equilibrium occurs at $dU/dx=0$; a local minimum with $U^{\prime \prime }>0$ is stable, while a maximum with $U^{\prime \prime }<0$ is unstable. Expanding about a minimum gives $U\simeq U_0+U^{\prime \prime }(x_0)(x-x_0{)}^2/2$, producing simple harmonic motion with ${\omega }^2=U^{\prime \prime }(x_0)/m$.

Examples include the spring force $F_x=-kx$ with $U=k{x}^2/2$, the near-surface gravitational force $-mg\hat{\mathbf{z}}$ with $U=mgz$, and the central Newtonian gravitational field with $U=-GMm/r$. An electrostatic force on a fixed electric charge $q$ has $U=q\Phi$, where $\Phi$ is the electric potential.

Sliding friction provides a contrasting physical test: traversing a closed path still dissipates mechanical energy, since the frictional force opposes the local motion on both outward and return legs. Its work depends on path length and cannot be represented by a position-only $U$. Total energy can remain conserved when the resulting internal energy is included. Also, even if $\mathbf{F}=-\nabla U(\mathbf{r},t)$ at each instant, explicit time dependence gives $d(T+U)/dt=\partial U/\partial t$; a static conservative-field argument alone does not guarantee constant mechanical energy in that situation.

For a light test particle of mass $m$ outside a spherical source of mass $M$, Newton's law of universal gravitation gives $\mathbf{F}=-GMm\hat{\mathbf{r}}/r^2$. It is a central force, so its torque about the source vanishes and angular momentum $\mathbf{L}=m\mathbf{r}\times \mathbf{v}$ is conserved. The motion therefore lies in a fixed plane when $L\ne 0$. Its gravitational potential energy is $-GMm/r$, and conservation of mechanical energy givesHere $ϵ$ is specific orbital energy and $h$ is specific angular momentum. A Kepler orbit has polar equationThe angle is measured from periapsis. The conic section is an ellipse for $ϵ<0$ and $0\le e<1$, a parabola for $ϵ=0$ and $e=1$, and a hyperbola for $ϵ>0$ and $e>1$. The circle is the $e=0$ elliptical case. The special case $h=0$ is radial motion and must be handled as a degenerate limit. Bound ellipses have the source at one focus, which is Kepler's first law.

For a bound orbit of semi-major axis $a$, $ϵ=-GM/(2a)$. Combining this with the energy equation gives the vis-viva equation:The particle therefore moves faster near periapsis and slower near apoapsis. Conservation of angular momentum also gives constant areal velocity, $dA/dt=h/2$, expressing Kepler's second law. Using the ellipse's area and $h^2=GMa(1-e^2)$ gives the orbital periodwhich is Kepler's third law.

A circular orbit follows from centripetal acceleration $v_c^2/r=GM/r^2$, giving $v_c=\sqrt{GM/r}$ and $T=2\pi \sqrt{r^3/(GM)}$. The minimum escape velocity from radius $r$ follows from the marginally unbound condition $ϵ=0$:These ideal orbits neglect drag force, other gravitating bodies and changes in the central source. For two comparable masses, the relative two-body problem uses $G(M+m)$ instead of $GM$. An orbiting object is continuously in free fall; the inward gravitational force changes its direction even when its speed is constant. These inverse-square results assume the physical variation of gravity with radius, rather than extending a near-surface constant-$g$ approximation to an entire orbit.

Definitions. The torque of a force $\mathbf{F}$ about an origin is $\mathbf{\tau }=\mathbf{r}\times \mathbf{F}$, where $\mathbf{r}$ runs from that origin to the force's point of application. Its component about a fixed axis changes angular momentum about that axis. The moment of inertia about that axis iswhere $s$ is perpendicular distance to the axis. It measures resistance to angular acceleration; for fixed-axis rigid rotation, ${\tau }_{\mathrm{net}}=I\alpha$. The units are $\mathrm{N}\mathrm{m}$ for torque and $\mathrm{kg}{\mathrm{m}}^2$ for moment of inertia.

Mass and axial inertia. Slice the shaft perpendicular to its common axis. A slice of radius $r(z)$ has mass $dm=\rho \pi r(z{)}^{2}dz$ and moment of inertia $dI=\frac{1}{2}r(z{)}^{2}dm$, using the moment of inertia of a uniform disc. Thus the axial moment of inertia of a solid of revolution uses the second and fourth powers of its radius:For the tapered section, take $z$ from its narrow end toward its wide end, so $r(z)=a(1+z/l)$ for $0\le z\le l$. Its contributions areThe two cylinders contribute $4\rho \pi a^{2}l$ and $\rho \pi a^{2}l$ to mass, and $8\rho \pi a^{4}l$ and $\frac{1}{2}\rho \pi a^{4}l$ to moment of inertia. Adding the three sections givesAll slice rotation axes coincide with the shaft axis, so the moments add directly.

Frictional torque. The normal load on the narrow circular end is $Mg$. With the stipulated uniform pressure, $p_N=Mg/(\pi a^2)$. A contact annulus of radius $s$ and width $ds$ has area $dA=2\pi sds$, normal force $p_{N}dA$ and tangential kinetic friction $dF=\mu p_{N}dA$. Multiplying by its lever arm $s$ and integrating gives the uniform-pressure friction torque on a circular contact:Here $G_f$ is the positive magnitude; its torque opposes the angular velocity.

Steady speed and stopping rotations. Mechanical power in rotation is $P=\tau \omega$. At steady angular velocity, the driving and frictional torques balance, so constant-power rotation against dry friction givesUsing $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$ and $a=0.100\mathrm{m}$,Once the motor is removed, the speed-independent kinetic friction gives constant angular acceleration $-G_f/I$. Equivalently, the initial rotational kinetic energy is dissipated as frictional work. The stopping angle under constant resisting torque is thereforeThe total angular displacement corresponds toHence the shaft completesand then another $0.16668$ of a rotation. Its stopping time is $I{\omega }_{\ast }/G_f=2.05461\mathrm{s}$. Using the rounded value $g=9.8\mathrm{m}{\mathrm{s}}^{-2}$ changes the fractional rotation slightly but leaves four complete rotations.

Equilibrium and undamped motion. Resolve forces along the incline, taking positive displacement down the slope. At equilibrium, Hooke's law balances the downhill component of weight:If $x$ is measured from that equilibrium position, Newton's second law becomesThus the motion is a harmonic oscillator withGravity shifts the equilibrium extension but drops out of this oscillation frequency because the downhill force component is constant.

Viscous damping. For a damped harmonic oscillator, put $\beta =b/(2m)$. Substitution of $z=A{e}^{-pt}$ in the supplied linear ordinary differential equation givesIn the underdamped regime $\beta <{\omega }_0$, define ${\omega }_d=\sqrt{{\omega }_0^2-{\beta }^2}$. The two characteristic roots are $p=\beta \pm i{\omega }_d$, and the general real displacement isHere $C,D$, or equivalently $A_{\mathrm{env}}$ and $\varphi$, are fixed by the initial conditions. For release from rest at $x_0$, the condition $\dot{x}(0)=0$ requires $D=\beta x_0/{\omega }_d$, not $D=0$:The curve oscillates about zero between exponentially shrinking envelopes $\pm A_{\mathrm{env}}{e}^{-\beta t}$, where $A_{\mathrm{env}}=|x_0|\sqrt{1+(\beta /{\omega }_d{)}^2}$ for this release. The upper panel illustrates that shape; its damping strength is chosen to make the envelope visible.

Frequency change. Viscous damping reduces the angular frequency and lengthens the oscillation period:For light damping, ${\omega }_d\simeq {\omega }_0[1-b^2/(8mk)]$, so the frequency shift is second order in $b$, even though the envelope decay rate is first order in $b$.

Will the block reach the stop? The given parameters giveFor $x_0=-0.260\mathrm{m}$, differentiating the release-from-rest solution givesIt is positive throughout the first half-cycle. The first positive turning point therefore occurs at $t_{+}=\pi /{\omega }_d=0.993460\mathrm{s}$. The turning points of an underdamped release from rest give its unobstructed height:Since $x_{+}>0.250\mathrm{m}$ and the displacement increases continuously up to that point,Solving $x(t)=0.250\mathrm{m}$ gives first contact at approximately $0.91117\mathrm{s}$. The plotted dashed continuation is the hypothetical free motion after that time; the post-impact motion is not determined by the supplied damping model.

The original diagram shows two equal balls, each of mass $m$, attached to the fixed upper pivot and light sliding collar by four equal rods. Each ball describes a horizontal circle of radius $r=L\sin \theta$. The pivot-to-collar distance is $2L\cos \theta$, so the upward collar displacement compresses the spring byThus Hooke's law gives a downward spring force $F_s=k\delta$ on the collar. Let $T_u,T_l$ be the tensions in the upper and lower rods attached to one ball. Vertical force balance on the light collar and then on either ball givesThe radial force on one ball supplies its centripetal acceleration:For $\theta \ne 0$, adding the vertical-balance results and eliminating the rod tensions givesFor the given centrifugal governor, $\cos 30^{\circ }=\sqrt{3}/2$ andHenceThe factor of two in the collar balance matters: both lower rods support the same light collar. The spring force is not an extra downward force of this full magnitude on each ball; its effect on each ball is transmitted through its own lower rod. The formula is the steady spring-loaded centrifugal governor balance.

For injected velocity $\mathbf{v}=v\hat{\mathbf{x}}$, the Lorentz force isFor nonzero electric charge and nonzero operating current, the velocity selector condition isThe requested nonrelativistic kinetic energy is consequentlyBoth electric current and electric charge cancel. Reversing the sign of $q$ reverses both component forces and preserves their cancellation. The dimensions are consistent because $R/{\mu }_0$ has units of speed, while $w/d$ is dimensionless. The assumed nonrelativistic particle regime requires $Rw/({\mu }_{0}d)\ll c$; the problem supplies no values with which to evaluate a numerical energy.

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 force law separates into two identical harmonic oscillator equations:For positive $m,k$, the general real solution isSince $\mathrm{Re}(e^{i\alpha })=\cos \alpha$ by Euler's formula, these are exactly the stated real parts of complex exponentials. The amplitudes $x_0,y_0$ and phases ${\varphi }_1,{\varphi }_2$ are fixed by the four initial position and velocity components. This is two-dimensional isotropic harmonic oscillator flow: the same angular frequency acts in every direction.

The force is conservative, with $\mathbf{F}=-\nabla U$ and potential energy zero at the origin:Released from rest at $(c,c)$, the trajectory has $x=y=c\cos \omega t$, givingThe right panel of the figure shows this nonnegative oscillation: it starts at a maximum $k{c}^2$, reaches zero at $t=\pi /(2\omega )$, and repeats after $\pi /\omega$. Its frequency is twice the coordinate frequency. Conservation of mechanical energy gives $K(t)=k{c}^2{\sin }^2\omega t$, so the missing potential energy is kinetic energy, not an energy loss.