= Solution
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 $d\mathbf L/dt=\boldsymbol\tau$; 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, $d\mathbf L/dt=\Omega\hat{\mathbf z}\times\mathbf L$, giving
$$
\boxed{\Omega\simeq\frac{Mgl}{I_3\omega_3}.}
$$
The 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.<University Physics Volume 1 by William Moebs and others/11.4 Precession of a Gyroscope>[Equation11.12 and torque/precession geometry]{ref}
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