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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.

Measure $\theta$ from the upward vertical, let ${\mathbf{e}}_3$ point from the support along the symmetry axis, and use the support as the reference point for all moments of inertia. Choose ${\mathbf{e}}_2=(\hat{\mathbf{z}}-\cos \theta {\mathbf{e}}_3)/\sin \theta$ and ${\mathbf{e}}_1={\mathbf{e}}_2\times {\mathbf{e}}_3$. These form an instantaneous principal-axis frame; its transverse axes need not spin with marks painted on the body. The three infinitesimal Euler angles rotations add as vectors:Resolving the vertical direction givesA convention matters here: in genuinely material-fixed transverse axes, the first two components are rotated by the spin angle $\chi$. For one conventional choice they are $\dot{\theta }\cos \chi +\dot{\varphi }\sin \theta \sin \chi$ and $-\dot{\theta }\sin \chi +\dot{\varphi }\sin \theta \cos \chi$. Since $I_1=I_2$, that rotation changes neither kinetic energy nor the axial and vertical conserved momenta. The simplified displayed components use the instantaneous transverse frame just defined.^[ref]

For a frictionless fixed support, the support does no work and gravity is conservative, so the total energy is constant. The heavy symmetric top Lagrangian isThe two cyclic coordinates $\chi ,\varphi$ giveEquivalently, gravitational torque has no component along the vertical or symmetry axis. Eliminate $\dot{\varphi }=(J_z-J_3\cos \theta )/(I_1{\sin }^2\theta )$ and the axial spin from $E=T+V$:This is the heavy symmetric top reduction at fixed $J_3,J_z$. A constant inclination must satisfy the reduced equation of motion $I_1\ddot{\theta }=-U_{\mathrm{eff}}^{\prime }$, not merely have zero instantaneous velocity. Differentiation at fixed conserved momenta gives, for $0<\theta <\pi$ and $\Omega =\dot{\varphi }$,Therefore uniform precession of a heavy symmetric top requiresFor a prescribed upward inclination $0<\theta <\pi /2$, real precession rates exist only if the quadratic discriminant is nonnegative:For positive spin the absolute-value sign can be dropped. The two roots are the slow and fast branches and coalesce at the threshold. The familiar upright-limit critical magnitude is $2\sqrt{I_{1}mgh}$; strict inequality gives gyroscopic stabilization of an inverted symmetric top. This limiting stability condition should not be confused with a positive minimum over every possible inclination. As $\theta$ approaches $\pi /2$ from below, the fixed-inclination threshold approaches zero. At the horizontal angle $\Omega =mgh/J_3$ for $J_3\ne 0$; below the horizontal the quadratic admits real solutions even at $J_3=0$. The stated minimum therefore needs the inclination convention made explicit.

For slow gyroscopic precession, the gravitational torque turns the spin angular momentum: $Mg\ell \sin \theta =\Omega I_3\omega \sin \theta$, with $I_3=M{R}^2/2$. ThusThe inclination cancels at this order. More precisely, uniform precession of a heavy symmetric top satisfiesIts slow root gives $T_p\simeq 1.57\mathrm{s}$. About $1.6\mathrm{s}$ is the slow-precession answer; arbitrary initial conditions can also produce nutation rather than exact steady precession.

Model the assembly as a heavy symmetric top with a fixed frictionless support, a uniform gravitational field, and centre of mass on its symmetry axis. Assume constant spin, negligible damping, and initial conditions prepared for uniform precession of a heavy symmetric top. Large spin means the transverse angular momentum from precession is small: $I_1|\Omega |\sin \theta \ll |I_3{\omega }_3|$. Also require $|\Omega |\ll |{\omega }_3|$. A large spin rate alone does not select steady motion after an arbitrary release.

For an original illustrative calculation, stipulateThe transverse principal moments of inertia $I_1,I_2$ are about the pivot, including the parallel axis theorem contribution. These inputs giveThe ratio ${\Omega }_0/{\omega }_3=0.00368$ is small; the more direct angular momentum ratio $I_1{\Omega }_0\sin 60^{\circ }/L_s=0.00358$ is small too. These are model calculations, not experimental results.

Check the approximation against the existing exact uniform precession of a heavy symmetric top condition. Its total angular momentum isAt constant $\theta ,\Omega ,L_s$, differentiating this expression and equating it to the gravitational torque givesFor $L_s>0$ the slow solution can be evaluated without subtracting two nearly equal numbers:It exceeds ${\Omega }_0$ by about $0.208\%$. The other solution is about $177.409rad/s$ and violates the assumed slow precession; the quadratic's two roots must not be conflated. Its discriminant must be nonnegative for these steady solutions to exist.

The original curves hold $I_1,I_3,\theta$ fixed and vary spin and the imposed torque coefficient $K$ in ${\mathbf{\tau }}_O=K\hat{\mathbf{z}}\times \hat{\mathbf{e}}$. For gravity alone, $K=Mgd$; the other curves are hypothetical changes in the transverse torque, not observations of this wheel. Solid curves use the slow quadratic root; dashed curves use $K/(I_3{\omega }_3)$. The second panel displays their relative difference. Increasing spin improves this approximation for the parameters shown. It does not remove nutation caused by inappropriate initial conditions.

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.