A spinning gyroscope can precess instead of simply falling under gravity. The essential idea is that gravity
produces a torque about the pivot, and that torque changes the direction of the spin angular momentum.
\[
\vec{\tau}_{\mathrm{net}}=\frac{d\vec{L}}{dt}.
\]
1. Torque due to gravity
If the centre of mass is a perpendicular distance \(r_\perp\) from the pivot, the gravitational torque magnitude is
\[
\tau=M g r_\perp.
\]
Here, \(M\) is the mass of the gyroscope and \(g\) is the gravitational field.
If the distance from the pivot to the centre of mass along the gyroscope axis is \(\ell\), and the axis makes angle
\(\theta\) from the vertical, then the perpendicular torque arm is
\[
r_\perp=\ell\sin\theta.
\]
2. Spin angular momentum
For a rotor spinning about its symmetry axis,
\[
L=I_0\omega_s.
\]
Here, \(I_0\) is the moment of inertia about the spin axis and \(\omega_s\) is the spin angular speed.
3. Steady precession formula
In steady precession, the magnitude of \(\vec L\) is nearly constant while its direction rotates around the
vertical. The torque changes the direction of \(\vec L\), so
\[
\tau=L\Omega_p.
\]
Therefore,
\[
\Omega_p=\frac{\tau}{L}.
\]
Substituting \(\tau=M g r_\perp\) and \(L=I_0\omega_s\):
\[
\boxed{
\Omega_p=
\frac{M g r_\perp}{I_0\omega_s}
}.
\]
4. Precession period
The time for one complete precession cycle is
\[
T_p=\frac{2\pi}{\Omega_p}.
\]
A smaller \(\Omega_p\) means a longer precession period.
5. Why faster spin gives slower precession
Since
\[
\Omega_p=\frac{M g r_\perp}{I_0\omega_s},
\]
increasing the spin speed \(\omega_s\) increases the spin angular momentum
\[
L=I_0\omega_s.
\]
For the same gravitational torque, a larger angular momentum vector changes direction more slowly. Therefore, a
faster-spinning gyroscope precesses more slowly.
6. High-spin approximation
The simple precession formula works best when the precession rate is much smaller than the spin rate:
\[
\Omega_p\ll\omega_s.
\]
If \(\Omega_p\) becomes comparable to \(\omega_s\), the motion can include wobbling, called nutation, and the simple
steady-precession model becomes less accurate.
7. Solving inverse problems
The same formula can be rearranged to solve other quantities.
Required spin speed:
\[
\omega_s=\frac{M g r_\perp}{I_0\Omega_p}.
\]
Required moment of inertia:
\[
I_0=\frac{M g r_\perp}{\Omega_p\omega_s}.
\]
Required mass:
\[
M=\frac{\Omega_p I_0\omega_s}{g r_\perp}.
\]
Required torque arm:
\[
r_\perp=\frac{\Omega_p I_0\omega_s}{M g}.
\]
8. Worked example
Suppose
\[
M=2.0\ \mathrm{kg},
\qquad
r_\perp=0.12\ \mathrm{m},
\qquad
I_0=0.004\ \mathrm{kg\,m^2},
\qquad
\omega_s=200\ \mathrm{rad/s}.
\]
Take
\[
g=9.81\ \mathrm{m/s^2}.
\]
First compute the torque:
\[
\tau=M g r_\perp.
\]
\[
\tau=(2.0)(9.81)(0.12)=2.3544\ \mathrm{N\,m}.
\]
Then compute the spin angular momentum:
\[
L=I_0\omega_s.
\]
\[
L=(0.004)(200)=0.800\ \mathrm{kg\,m^2/s}.
\]
Now compute the precession rate:
\[
\Omega_p=\frac{\tau}{L}.
\]
\[
\Omega_p=\frac{2.3544}{0.800}=2.943\ \mathrm{rad/s}.
\]
Finally, the precession period is
\[
T_p=\frac{2\pi}{\Omega_p}.
\]
\[
T_p=\frac{2\pi}{2.943}\approx2.14\ \mathrm{s}.
\]
The gyroscope precesses once every about \(2.14\ \mathrm{s}\) while spinning much faster about its own axis.
Key idea: torque does not simply make the gyroscope fall; it changes the direction of \(\vec L\), producing
precession.
