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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 $n$.

For the opposite internal torques on a black hole and its accretion disk, the total angular momentum ${\mathbf{J}}_t$ and the black-hole spin magnitude $J_h$ are constant. If $A=K_2|{\mathbf{J}}_h\times {\mathbf{J}}_d{|}^2$, then $d({\mathbf{J}}_h\cdot {\mathbf{J}}_t)/dt=A$ and $d{J}_d^2/dt=-2A$. The $K_1$ term produces precession; the positive $K_2$ term increases the alignment of the hole with the total vector while decreasing the disk angular-momentum magnitude. The exact direction equation is $\dot{c}=K_2{J}_h{J}_t(1-c^2)$ for $c=\cos \angle ({\mathbf{J}}_h,{\mathbf{J}}_t)$.

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.

An unwanted ellipse is an experimental warning: a finite-amplitude pendulum can develop additional orbit-axis precession even without Earth's rotation. A symmetrical, firm support and careful launch reduce this effect.^[ref] Distinguish that nonlinear effect from the tiny ellipse already present in the Exact linear Foucault pendulum trajectory. The linear calculation establishes a controlled approximation, not the complete motion of an arbitrary large swing.

Support anisotropy creates a second competing mechanism. As an illustrative modification, let the two horizontal restoring angular frequencies be ${\omega }_x$ and ${\omega }_y$. Even with $\alpha =0$, a diagonal release from rest givesWhen ${\omega }_x\ne {\omega }_y$, the relative phase changes on the scale $1/|{\omega }_x-{\omega }_y|$: the trace changes shape without any terrestrial Coriolis force. It need not be uniform precession of a Foucault pendulum. To cleanly resolve the terrestrial effect over times of order $1/|\alpha |$, a useful design target is $|{\omega }_x-{\omega }_y|\ll |\alpha |$. Near the equator that target becomes particularly demanding. Compare fast periods along perpendicular launch directions to look for this frequency splitting; repeat at several launch orientations to detect preferred suspension axes.

Other controls address distinct alternatives: shelter the apparatus from air currents, track the support against the building, measure both orbit axes, and compare early and late fitted directions as the amplitude decays. A sustaining drive, if used in a different apparatus, must supply mechanical energy without imposing an azimuthal bias; an undriven run provides a useful comparison. A change in swing direction alone cannot exclude these effects. Agreement of the signed rate with independently known geographical latitude and Earth's rotation, together with stable repeat launches and small competing effects, is the inference the experiment is designed to support.

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.

An equatorial coordinate system labels a viewing direction on the celestial sphere by right ascension $\alpha$, increasing eastward around the reference celestial equator, and declination $\delta$, increasing northward from that celestial equator. These two angles locate a direction, without specifying the object's Euclidean distance. In the traditional convention, zero right ascension points toward the reference equinox, the northward crossing of the Sun's annual path through the celestial equator. An hour circle through the direction supplies the north-south arc on which declination is measured. ^[ref].

For a concrete construction, choose Cartesian coordinates with $x$ pointing toward the reference equinox, $z$ toward the north reference celestial pole, and positive $y$ given by the cross product of the positive $z$ and $x$ directions. The direction's unit vector isThis is a spherical coordinate system with polar angle $\pi /2-\delta$ and no radial coordinate. The trigonometric functions require consistent angular units: convert right ascension in hours to radians before evaluating them. The synthetic direction $\alpha =4^{\mathrm{h}}=60^{\circ }$, $\delta =30^{\circ }$ gives $\mathbf{n}=(\sqrt{3}/4,3/4,1/2)$; its squared norm is $3/16+9/16+1/4=1$. A change of reference axes applies a rotation to this unit vector before recovering the two angles. Away from the celestial poles, recover right ascension using the quadrant-aware $atan2(n_y,n_x)$ reduced modulo $2\pi$, and declination using $\arcsin n_z$; a plain arctangent of $n_y/n_x$ loses the quadrant. Adding a common offset to every right ascension describes only the special case of a rotation about the same polar axis.

The reference specification matters. For an equatorial coordinate system tied to the Earth, precession changes the reference celestial equator and reference equinox, and nutation distinguishes their mean and true versions. The date specifying those axes is distinct from the astronomical epoch at which a catalogue's source directions apply: advancing a star's proper motion and changing reference axes are separate operations. A modern catalogue may instead use fixed reference frame axes, aligned approximately with the traditional J2000.0 axes, without a moving reference equinox. Thus a pair of numbers labelled merely "J2000" does not settle every convention. State the reference frame, the astronomical epoch, and whether a direction is catalogued or apparent. ^[ref].

During one night's elementary fixed-direction approximation, a star's right ascension and declination remain constant while its altitude, azimuth and hour angle change because the Earth turns. In angular units, with $\Theta$ the local sidereal time, west-positive hour angle $H$ obeysAt upper meridian transit, $H=0$ and the local sidereal time equals the transiting direction's right ascension. Pair apparent sidereal time with apparent right ascension referred to the true celestial equator and reference equinox of date; mean sidereal time refers to the mean reference equinox. Raw fixed-frame catalogue coordinates require conversion before precision pointing. ^[ref].

This separation makes the equatorial coordinate system useful: a catalogue can identify a target independently of the observer's current astronomical horizon, while local sidereal time tells the observer which hour circle is on the celestial meridian. Precision observations also account for proper motion, stellar parallax, aberration of light and atmospheric refraction; the fixed-direction construction above omits those effects. A reference-axis rotation preserves angles between directions, whereas changing the observer or correcting atmospheric refraction is a different operation. ^[ref].

For release from rest relative to the building, take $x(0)=A>0$, $y(0)=\dot{x}(0)=\dot{y}(0)=0$. The exact solution of the stated coupled linear differential equations isIndeed $u(0)=A$ and $\dot{u}(0)=i\alpha A$, so $\dot{z}(0)=0$. In the slowly rotating coordinates, the ellipse has semiaxes $A$ and $A|\alpha |/\beta$. A bob released from rest therefore does not describe a perfectly straight line even in this ideal linear model. Replacing the bracket by $\cos {\omega }_{0}t$ gives the familiar leading approximation but introduces a spurious initial transverse velocity $-i\alpha A$. The exact expression keeps the initial condition and separates that tiny orbit width from the slow rotation of its long axis.

For the illustrative $\ell =20\mathrm{m}$ and $g=9.81m/{\mathrm{s}}^2$, the fast period $2\pi /\beta$ is approximately $8.9714\mathrm{s}$. At $45^{\circ }$ north, $|\alpha |/\beta \simeq 7.3624\times 10^{-5}$: the transverse semiaxis is too small to distinguish at the figure's scale. Nevertheless the long-axis orientation changes by about $-63.8138^{\circ }$ in six hours. Changing the sign of geographical latitude conjugates the entire solution, reflecting the trajectory across the eastward axis; setting $\alpha =0$ gives $z=A\cos {\omega }_{0}t$ exactly. These are stronger checks than merely drawing a rotating line.

The coupled equations give an additional independent constraint. Multiplying them by $\dot{x}$ and $\dot{y}$ and adding cancels the Coriolis force's contributions:This is mechanical energy per unit mass in the reduced model; precession does not require a loss of mechanical energy. Also, direct differentiation givesFor this release the constant is $\alpha A^2$. The exact trajectory satisfies both identities. They test the coupled dynamics separately from the visible slow rotation and express the fact that the Coriolis force does no work.

A Foucault pendulum is a long simple pendulum whose suspension allows the bob to swing in both independent horizontal directions. Its slowly changing swing direction relative to the building reveals Earth's rotation. A hinge restricting the bob to one vertical plane would prevent this measurement: the support would impose the very direction we want the pendulum to select freely. Léon Foucault demonstrated the effect in 1851.^[ref]

The bob's back-and-forth oscillation and the swing direction's precession are different motions. The bob traverses its arc in seconds, while successive arcs change direction over hours. In a conical pendulum the bob itself circulates around the vertical; a nearly planar Foucault pendulum instead reverses along an almost straight horizontal trace, whose orientation changes slowly. The measured quantity is that orientation against a fixed floor or wall reference, rather than the bob's instantaneous azimuth.

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.

Choose a fixed support point $O$ and nonrotating axes in an inertial frame. All angular momenta, torques, and moments of inertia below refer to $O$. Let $\hat{\mathbf{e}}$ point along the axle from $O$ toward the centre of mass, and let $\hat{\mathbf{z}}$ point vertically upward. Write $\theta$ for their angle. For a rapidly spinning symmetric top, the dominant angular momentum is ${\mathbf{L}}_O\simeq L_s\hat{\mathbf{e}}$, where $L_s=I_3{\omega }_3$ is signed and $I_3$ is the axial principal moment of inertia. The sign of the axial angular velocity ${\omega }_3$ follows the right-hand rule.

With mass $M$, centre of mass distance $d$, and constant gravitational acceleration $g$, the gravitational force and its torque areAn ideal support applies a force at $O$, so its lever arm and torque about $O$ vanish. At the instant the axle lies in the $xz$ plane with a positive $x$ component, the gravitational torque points along $+y$. It adds $d{\mathbf{L}}_O={\mathbf{\tau }}_{O}dt$ toward $+y$, rather than along the downward force. The dot product identity $d|{\mathbf{L}}_O{|}^2/dt=2{\mathbf{L}}_O\cdot {\mathbf{\tau }}_O$ shows why perpendicular torque changes direction without instantly changing magnitude. This vector balance and the slow-turning relation ${\mathbf{\tau }}_O=\mathbf{\Omega }\times {\mathbf{L}}_O$ are developed in Chapter 20: Rotation in space (sections 20-2 and 20-3, equations 20.13–20.15).

For steady precession about the vertical, $\mathbf{\Omega }=\Omega \hat{\mathbf{z}}$ and $d\hat{\mathbf{e}}/dt=\Omega \hat{\mathbf{z}}\times \hat{\mathbf{e}}$. Equating the two cross products givesThe angular momentum tip traces a circle of radius $|L_s|\sin \theta$: its sideways speed is $|\Omega L_s|\sin \theta$. Both that radius and the gravitational lever arm contain $\sin \theta$, explaining its cancellation for $0<\theta <\pi$. Exactly vertical alignment has zero gravitational torque and no distinguishable azimuthal precession to infer by this cancellation. Positive $L_s$ gives positive precession, counterclockwise viewed from above; reversing the spin reverses the slow precession. ^[ref]

A flywheel's spin angular momentum ${\mathbf{L}}_f$ must turn with the vehicle. The vehicle supplies torque $\mathbf{\Omega }\times {\mathbf{L}}_f$ to maintain this precession, and experiences the opposite reaction torque. With right-handed axes $x$ forward, $y$ left and $z$ up, a left turn has $\mathbf{\Omega }=+\Omega \hat{\mathbf{z}}$. A transverse flywheel with ${\mathbf{L}}_f=-L_f\hat{\mathbf{y}}$ then produces body torque $-\Omega L_f\hat{\mathbf{x}}$, opposing outward roll. Its magnitude is $L_{f}v/R_b$ for turn radius $R_b$. Reversing the flywheel reverses the effect.

Use local east, north, and up unit vectors, with $x$ eastward, $y$ northward, and positive orientation $\psi$ anticlockwise as viewed from above. Let $\phi$ be signed geographical latitude, positive north, and $\Omega >0$ the magnitude of Earth's angular velocity. In the spherical-Earth approximation,Here $g$ is the local effective gravity in a rotating frame, including the steady terrestrial centrifugal acceleration. For small horizontal displacements compared with $\ell$, the horizontal restoring acceleration from tension is $-{\omega }_0^2(x,y)$. Vertical displacement is second order in the swing amplitude, so vertical velocity can be omitted at this order. The horizontal Coriolis acceleration in local Earth coordinates then gives the coupled linear differential equationsThese equations neglect damping, finite-amplitude corrections, support anisotropy, and spatial changes of the centrifugal force of order ${\Omega }^2$; they retain the leading terrestrial Coriolis force. A related derivation uses south/east axes instead.^[ref]

Introduce the complex number $z=x+iy$. Multiplying the second equation by $i$ and adding givesSet $z=e^{-i\alpha t}u$. Direct differentiation cancels the first derivative of $u$ and yieldsThe ${\alpha }^2$ in $\beta$ comes from solving these reduced equations exactly; it is not a complete prediction of terrestrial corrections of order ${\Omega }^2$. Thus the horizontal trajectory is an ellipse in coordinates rotating at angular velocity $-\alpha$ relative to the building. When $|\alpha |\ll {\omega }_0$, each rapid swing samples an almost fixed long axis; its slow precession isThe sidereal day $T_{\mathrm{sid}}=2\pi /\Omega$ is a rotation relative to distant directions. $T_{\mathrm{p}}$ denotes a full $2\pi$ turn of a continuously followed orientation; an unlabelled swing line repeats after $T_{\mathrm{p}}/2$. At the equator there is no finite precession period in this model. The Coriolis parameter is $f=2\alpha$, so the plane's rate is $-f/2$, not $-f$.

The limiting cases check the sign and scale. At the north pole $\alpha =\Omega$: the swing direction turns clockwise once per sidereal day. At the equator $\alpha =0$: both horizontal equations reduce to independent simple harmonic motions and a line launch stays on its original line. At the south pole $\alpha =-\Omega$: the direction turns anticlockwise once per sidereal day. Vanishing horizontal precession at the equator does not mean that every component of Coriolis acceleration vanishes there.

For a stipulated $\Omega =7.292115\times 10^{-5}{\mathrm{s}}^{-1}$, the sidereal day is approximately $23.9345\mathrm{h}$. At $45^{\circ }$ north the predicted rate is $-10.6356^{\circ }/\mathrm{h}$ and $T_{\mathrm{p}}\simeq 33.8485\mathrm{h}$; at $45^{\circ }$ south the rate changes sign. At the illustrative geographical latitude $34^{\circ }$ south, the prediction is $+8.4109^{\circ }/\mathrm{h}$. These calculations are model predictions, not new observations of the UNSW apparatus.

A swing plane fixed among the stars is generally only a polar picture. A north–south swing at the equator turns with Earth in an inertial frame, while having no precession against the building.^[ref]

Variation in the inclination of a rotating body's axis, distinguished from its azimuthal precession. For a heavy symmetric top, the cyclic angular momenta reduce the motion to ${\dot{\mu }}^2=F(\mu )$ with $\mu =\cos \theta$ and $F$ cubic. Its admissible interval and turning points describe the nutation.

Reflection symmetry gives ${\Phi }_z(R,0)=0$. In a vertically stable thin disk, expand the vertical force as ${\Phi }_z(R,z)={\Omega }_z^{2}z+O(z^3)$, where ${\Omega }_z^2={\Phi }_{zz}(R,0)$ is the vertical epicyclic frequency squared. For local hydrostatic equilibrium and height-independent isothermal sound speed, $p=c_s^2\rho$ andThus $\rho ={\rho }_0\exp [-z^2/(2H^2)]$ with $H=c_s/{\Omega }_z$. Integrating the pressure and mass density through the disk gives the vertically integrated pressure of an isothermal disk:The force expansion is approximate in $H/R$, while $P=c_s^2\Sigma$ is exact for a height-independent sound speed.

In a nearly Keplerian disk, the vertical oscillation and radial epicyclic motion frequencies are close to the orbital frequency. The exact leading orbital differences are $\Omega -{\Omega }_z$ and $\Omega -\kappa$, soThus ${\omega }_1$ is the nodal precession rate of a slightly tilted orbit, and ${\omega }_2$ is the apsidal precession rate of a slightly eccentric orbit. Their signs distinguish prograde from retrograde precession. This also agrees with the sign convention in the complex tilt: a freely precessing ring obeys $W_t=i{\omega }_{1}W$.

For exactly Keplerian motion, ${\Omega }_z=\kappa =\Omega$, so ${\omega }_1={\omega }_2=0$ and $P/\Sigma =c_s^2$. In the local approximation, freeze the slowly varying coefficients and take $W,G\propto e^{i(kR-\omega t)}$. The inviscid warp equations of a nearly Keplerian disk becomeEliminating $G$ gives ${\omega }^2=k^{2}P/(4\Sigma )=k^2{c}_s^2/4$. HenceThese are bending waves of a Keplerian disk, carrying changes in orbital tilt. The local approximation needs wavelength small compared with $R$; the secular thin-disk warp equations also describe the long-vertical-wavelength regime $H\ll \lambda \ll R$.

For a nonspinning hole, $a=0$ gives ${\omega }_1=0$. A steady disk warp then satisfies $G_R=0$. The zero inner torque forces $G=0$ throughout, and the second warp equation, with positive $P$, forces $W_R=0$. The outer orientation fixes $W(R)=W_{\infty }$ everywhere. An apsidal precession term by itself therefore does not require a tilt warp.

For nonzero spin, set the time derivatives to zero. The two equations giveEliminating the internal torque yieldsUse ${\omega }_2=3\Omega /r$, ${\omega }_1=2a\Omega r^{-3/2}$, ${\Omega }^2{R}^3=GM$, and $P\simeq {\Omega }^2\Sigma H^2$. The first coefficient becomes $GM\Sigma H^{2}r/12$ and the second becomes $2aGM\Sigma r^{-3/2}$. ThereforeThis is the stationary disk warp equation under the stated weak-precession approximations. The inner condition is $W_R(R_{\mathrm{in}})=0$ when its coefficient is finite and nonzero, because $G=iGM\Sigma H^{2}r{W}_R/12$.

Let $R_g=GM/c^2$, so $R=R_{g}r$, and impose $H=ϵR$, $\Sigma \propto R^{-3/4}$. Converting both radial derivatives carefully to $r$ and dividing out the common positive factor givesor $W_{rr}+9W_r/(4r)+(24a/{ϵ}^2)r^{-9/2}W=0$. For $a>0$, introduceSince $x_r$ is proportional to $r^{-9/4}$, the first-derivative term cancels exactly, leaving $W_{xx}+W=0$. Thus $W=A\cos x+B\sin x$.

The outer boundary $r\to \infty$ is $x\to 0$, giving $A=W_{\infty }$. The specified inner-radius relation is exactly $x_{\mathrm{in}}=\pi$. The inner zero-torque condition $W_x(\pi )=0$ then gives $B=0$. Hence the stationary bending warp of a Kerr disk isThe tilt magnitude is $\beta (r)=|W_{\infty }||\cos x|$. Far out it tends to the imposed tilt. Moving inward, it decreases to zero atthen grows again with opposite signed tilt, reaching $W(r_{\mathrm{in}})=-W_{\infty }$. All tilt vectors lie on the same line in the complex $W$ plane. Their azimuth is $\arg W_{\infty }$ outside the node and differs by $\pi$ inside; azimuth is undefined at the aligned node itself. The tilt vector remains smooth there. Thus this is an oscillatory standing bending warp, with equal outer and inner tilt magnitudes and reversed inner orientation, rather than monotone inner alignment.

The prescribed real inner radius assumes $a>0$: for $a<0$, its square root is not real, so that particular boundary prescription does not apply. With a separately chosen positive inner radius and negative spin, put $x=4\sqrt{24|a|}{r}^{-5/4}/(5ϵ)$. The equation becomes $W_{xx}-W=0$, and the same outer tilt and zero inner torque giveThis negative-spin solution decreases smoothly toward the inner disk without a sign reversal. The specified positive-spin solution is the cosine profile above; the $a=0$ case is the flat solution already derived.

Each annulus has a torque proportional to ${\mathbf{J}}_h\times \delta {\mathbf{J}}_d$, so its scalar product with ${\mathbf{J}}_h$ is zero. Summing over annuli preserves this property even when the proportionality coefficient varies with radius. Therefore $\boxed{\mathbf{T}\cdot {\mathbf{J}}_h=0}$ for the total torque due to Lense-Thirring precession.

The torque specified is the torque on the disk. Assuming a closed disk-plus-hole angular-momentum budget with no external torque, its reaction on the black hole has the opposite sign:Thus ${\dot{\mathbf{J}}}_t=0$. The $K_1$ term describes precession, because it is perpendicular to both spins. Since ${\mathbf{J}}_h\times {\mathbf{J}}_d={\mathbf{J}}_h\times {\mathbf{J}}_t$, the hole precesses about the fixed total vector. The $K_2$ term changes the spin directions dissipatively. Writing ${\mathbf{J}}_{d,\perp }$ for the disk component perpendicular to the hole, the vector triple product identity givesA positive $K_2$ removes this component from the disk and transfers it into the hole's changing direction. This is black-hole and disk alignment under internal torque.

Since $\mathbf{T}$ is perpendicular to ${\mathbf{J}}_h$,Hence $\boxed{J_h\text{ is constant}}$, even if the torque coefficients vary in time.

Now differentiate the scalar product with the conserved total vector. The precession term contributes zero, and the vector triple product identity givesAlso ${\mathbf{J}}_h\cdot {\mathbf{J}}_t=J_h^2+{\mathbf{J}}_h\cdot {\mathbf{J}}_d$, so constancy of $J_h$ givesFinally, $J_t^2=J_h^2+J_d^2+2{\mathbf{J}}_h\cdot {\mathbf{J}}_d$ is constant. Differentiating provesThus the disk's spin magnitude decreases while the hole's magnitude is fixed; their vector sum remains fixed as well.

For the alignment conclusion, assume $J_h,J_t>0$ and let $c=\cos \theta ={\mathbf{J}}_h\cdot {\mathbf{J}}_t/(J_h{J}_t)$. Since ${\mathbf{J}}_h\times {\mathbf{J}}_d={\mathbf{J}}_h\times {\mathbf{J}}_t$,For a noncollinear initial state, $|c(0)|<1$, this integrates exactly toConsequently a fixed positive $K_2$, or more generally a positive coefficient of divergent integrated strength, gives $c(t)\to 1$. In this persistent dissipative evolution, $\boxed{{\mathbf{J}}_h\to J_h{\hat{\mathbf{J}}}_t}$: the hole aligns asymptotically with the total angular momentum. The exactly antiparallel configuration $c=-1$ is a stationary, nongeneric exception; when $J_t=0$ there is no total-vector direction to align with.

If the phrase $K_2>0$ allows arbitrary time dependence, positivity alone does not logically force complete eventual alignment. The persistent-torque criterion for asymptotic spin alignment is the needed qualification. For example, take $J_h=J_t=1$, $K_1=0$, $K_2(t)=e^{-t}$ and $c(0)=0$. The same torque equations give $c(t)=\tanh (1-e^{-t})$, with limit $\tanh 1<1$. The final configurations below use the intended persistent-torque regime, including constant positive $K_2$.

Once the hole has aligned, conservation of angular momentum fixes the disk without any additional torque calculation:The disk and hole therefore finish parallel only if $J_t>J_h$. If $J_t<J_h$, they finish antiparallel: this is counteralignment of a black hole and accretion disk. In terms of the initial disk-hole angle ${\theta }_0$, squaring $J_t$ shows that counteralignment occurs whenIf $J_t=J_h$, the final disk angular momentum vanishes and has no spin direction. Thus alignment of the hole with the total vector does not imply alignment of both component spins.

For an arbitrary smooth variation $q_i\mapsto q_i+\epsilon {\eta }_i$ vanishing at both endpoints,The endpoint term in integration by parts is zero. stationarity for every variation and the fundamental lemma of the calculus of variations imply the Euler-Lagrange equations $d(L_{{\dot{q}}_i})/dt=L_{q_i}$. Least action here means stationary action, not necessarily a global minimum. The conjugate momentum is $p_i=L_{{\dot{q}}_i}$; a cyclic coordinate, $L_{q_i}=0$, makes $p_i$ constant.

For the symmetric top, $\psi$ and $\varphi$ are cyclic. Thusare constant, and consequently so is ${\omega }_3$. Conservation of energy from time-translation invariance also supplies the conserved energyThese give the requested additional conserved quantities $p_{\varphi }$ and $E$. The $\theta$ equation isFor constant $\theta$ and constant precession rate $\dot{\varphi }$, it reduces toConversely, choose constants satisfying this balance and set $\dot{\psi }={\omega }_3-\dot{\varphi }\cos \theta$. The cyclic equations and $\theta$ equation are then all satisfied, so this constructs uniform precession wherever the Euler-angle chart is regular.

For nonvertical uniform precession of a heavy symmetric top, write $\dot{\varphi }=\Omega$ and fix $\theta$. The $\theta$ Euler-Lagrange equation reduces toFor $\cos \theta \ne 0$ its solutions areFor $0<\theta <\pi /2$, uniform precession therefore requires $|{\omega }_3|\ge 2\sqrt{I_{1}Mgl\cos \theta }/I_3$. For $\pi /2<\theta <\pi$, the discriminant is positive for every spin, including zero. At $\theta =\pi /2$, the condition is $\Omega =Mgl/(I_3{\omega }_3)$, requiring nonzero spin. Once this condition holds, choose $L_z=I_1\Omega {\sin }^2\theta +L_3\cos \theta$ and $\dot{\psi }={\omega }_3-\Omega \cos \theta$; both cyclic equations are then satisfied. The exactly vertical configurations are also solutions for arbitrary spin, but the azimuthal member of the Euler angles is degenerate there, so their existence is separate from this nonvertical discriminant condition.

Put $L_z=L_3(1-{ϵ}^2/2)$ and seek $\theta =O(ϵ)$. Using $\cos \theta =1-{\theta }^2/2+O({\theta }^4)$ and ${\sin }^2\theta ={\theta }^2+O({\theta }^4)$, the $\theta$-dependent part of the effective potential isIt has a local minimum precisely whenwhich is the gyroscopic stabilization of an inverted symmetric top condition. Differentiating the displayed approximation givesand henceTaking $L_3>0$ without loss of orientation, the steady precession rate isso

For steady precession at rate $\Omega$ about the fixed vertical unit vector $\mathbf{k}$,Consequently $\mathbf{e}\wedge \ddot{\mathbf{e}}={\Omega }^2\cos \alpha \mathbf{e}\wedge \mathbf{k}$ and $\dot{\mathbf{e}}=-\Omega \mathbf{e}\wedge \mathbf{k}$. Since the axial angular velocity of a symmetric top $n$ is constant, substitution in the equation of motion givesAs $0<\alpha <\pi /2$, the cross product can be cancelled. For a specified axial spin, the required value isThe corresponding total angular velocity isIf the stated angular velocity $\Omega \mathbf{k}$ is the total angular velocity of the top, then $n=\Omega \cos \alpha$ and the requested value specializes toThis specialization permits a direct check: ${\mathbf{L}}_Q=A\Omega \mathbf{k}+(C-A)\Omega \cos \alpha \mathbf{e}$, whose time derivative is $(A-C){\Omega }^2\cos \alpha \mathbf{e}\wedge \mathbf{k}$, exactly the required torque. The first formula keeps the separate axial spin explicit when $\Omega$ is understood as the rate of precession.

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.