 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 $d\mathbf{L}/dt=\mathbf{\tau }$; 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, $d\mathbf{L}/dt=\Omega \hat{\mathbf{z}}\times \mathbf{L}$, givingThe 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.^[ref]