Rest one end of a spinning gyroscope's axle on a stand and let go. Instead of tipping over, it swings slowly around in a horizontal circle. Angular Momentum of a Particle gave $d\vec{L}/dt = \vec{\tau}$; here that vector equation does something surprising.
Why it works
The spin gives the wheel a large angular momentum $\vec{L}$ along its axle, which is horizontal. Gravity pulls down on the wheel's center, a distance $r$ from the support, so its torque about the support is $\vec{\tau} = \vec{r}\times M\vec{g}$: horizontal and perpendicular to $\vec{L}$. In a short time $dt$, $\vec{L}$ changes by $d\vec{L} = \vec{\tau}\,dt$, sideways. A sideways nudge to a vector turns it without changing its length, through an angle $d\phi = \dfrac{dL}{L} = \dfrac{\tau\,dt}{L}$. So the axle sweeps around at $$\Omega = \dfrac{d\phi}{dt} = \dfrac{\tau}{L} = \dfrac{Mgr}{I\omega_s}.$$
Key idea
A fast-spinning gyroscope supported at one end precesses at $\Omega = \dfrac{\tau}{L} = \dfrac{Mgr}{I\omega_s}$: the faster the spin, the slower the precession. The torque changes the direction of $\vec{L}$, not its size. (This holds when the spin is fast, $\Omega \ll \omega_s$.)
Worked example: A toy gyroscope
A gyroscope's wheel is a uniform disk ($M = 0.50\ \text{kg}$, $R = 0.050\ \text{m}$) spinning at $600\ \text{rad/s}$, with its center $0.080\ \text{m}$ from the support. Find the precession rate and period.
- $$L = \tfrac{1}{2}MR^2\omega_s = \tfrac{1}{2}(0.5)(0.05)^2(600) = 0.375\ \text{kg·m}^2\text{/s}$$
- $\tau = Mgr = 0.5(9.8)(0.08) = 0.392\ \text{N·m}$.
- $\Omega = 0.392/0.375 = 1.05\ \text{rad/s}$; one lap takes $2\pi/1.05 = 6.0\ \text{s}$.
Answer: $\Omega \approx 1.05\ \text{rad/s}$; period about $6.0\ \text{s}$.
Worked example: Spin it faster
If the same wheel spins twice as fast, what happens to the precession?
- $L$ doubles while $\tau = Mgr$ stays the same.
- $\Omega = \tau/L$ is halved: the gyroscope precesses half as fast.
Answer: Half the precession rate.
Common mistake
Expecting the spinning top or gyroscope to fall over like a non-spinning one. Without spin, $L$ starts at zero, and the torque builds angular momentum in its own direction: the wheel tips down. With a large spin $\vec{L}$, the same torque adds only a small sideways change to a big vector, which turns it. Check the directions: $d\vec{L}$ points along $\vec{\tau}$, and $\vec{\tau}$ is horizontal, so the axle moves sideways, not down. As friction slows the spin, $\Omega$ grows and the motion eventually becomes a fall.
Earth itself precesses: the Sun's and Moon's pulls on its equatorial bulge make its spin axis trace a circle about once every 26,000 years, slowly changing which star sits near the north celestial pole.