= Slow steady gyroscope precession
{title2=$\Omega_{\mathrm{slow}}\simeq Mgd/(I_3\omega_3)$}
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, stipulate
$$
M=2.0\,\mathrm{kg},\quad d=0.15\,\mathrm m,\quad g=9.81\,\mathrm{m/s^2},\quad
I_3=0.080\,\mathrm{kg\,m^2},\quad I_1=I_2=0.090\,\mathrm{kg\,m^2},\quad
\omega_3=100\,\mathrm{rad/s},\quad\theta=60^\circ.
$$
The transverse <principal moments of inertia> $I_1,I_2$ are about the pivot, including the <parallel axis theorem> contribution. These inputs give
$$
L_s=8.0\,\mathrm{kg\,m^2/s},\quad K=Mgd=2.943\,\mathrm{N\,m},\quad
|\boldsymbol\tau_O|=K\sin60^\circ=2.5487\,\mathrm{N\,m},
$$
$$
\Omega_0=K/L_s=0.367875\,\mathrm{rad/s},\qquad T_0=2\pi/\Omega_0\simeq17.080\,\mathrm s.
$$
The ratio $\Omega_0/\omega_3=0.00368$ is small; the more direct <angular momentum> ratio $I_1\Omega_0\sin60^\circ/L_s=0.00358$ 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> is
$$
\mathbf L_O=I_1\Omega(\widehat{\mathbf z}-\cos\theta\widehat{\mathbf e})+L_s\widehat{\mathbf e}.
$$
At constant $\theta,\Omega,L_s$, differentiating this expression and equating it to the gravitational <torque> gives
$$
I_1\cos\theta\,\Omega^2-L_s\Omega+K=0.
$$
For $L_s>0$ the slow solution can be evaluated without subtracting two nearly equal numbers:
$$
\Omega_{\mathrm{slow}}=\frac{2K}{L_s+\sqrt{L_s^2-4I_1K\cos\theta}}
\simeq0.368639\,\mathrm{rad/s}.
$$
It exceeds $\Omega_0$ by about $0.208\%$. The other solution is about $177.409\,\mathrm{rad/s}$ 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.
\Image[/gyroscope-precession-spin-torque-sweep.png]
{title=Synthetic steady-precession calculations: larger spin slows the drift and larger transverse torque speeds it up}
The original curves hold $I_1,I_3,\theta$ fixed and vary spin and the imposed <torque> coefficient $K$ in $\boldsymbol\tau_O=K\widehat{\mathbf z}\times\widehat{\mathbf e}$. 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.
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