ISEGORIA / MATH ENCYCLOPEDIA
Rigid-body dynamics: spinning, tumbling, precessing
The tennis-racket flip, the gyroscope that refuses to fall, and the inertia tensor behind both.
Before you begin: Classical mechanics and linear algebra
Predict, manipulate, then check your reasoning against the example and question. Graphs illustrate the mathematics; they do not replace a proof.
1. The tennis-racket theorem
A box with three different moments of inertia spins with no torque acting on it. About the long axis or the short axis the spin is stable. About the middle axis the smallest disturbance grows, and the body flips over again and again, as the cosmonaut Vladimir Dzhanibekov saw with a wing nut in orbit in 1985. Conservation of energy and of angular momentum forces the angular-momentum point to move along the curves where an ellipsoid meets a sphere.
Worked example. Linearising about the middle axis at spin rate \(\omega\) gives growth like \(e^{\lambda t}\) with \(\lambda=\omega\sqrt{(I_2-I_1)(I_3-I_2)/(I_1I_3)}\). For this box \(I_1:I_2:I_3\approx0.363:0.780:1.083\), so \(\lambda\approx0.567\,\omega\).
Watch out. No torque causes the flip: energy and angular momentum are conserved throughout, and the readout shows the numerical drift. With internal friction a real body ends up spinning about its largest axis, which is not modelled here.
Why can a small disturbance about the long axis never grow large?
Near that axis the polhodes are small closed loops. Conservation of energy and of \(|\mathbf L|\) confines the motion to one of them, so the angular momentum can only circle nearby.
2. The gyroscope: precession and nutation
A heavy wheel spins on an axle whose end rests on a pivot. Gravity pulls the wheel down, yet the axle swings sideways instead: the torque changes the direction of the angular momentum, not its size. Released from rest, the tip bobs as it goes round (nutation) and draws cusps. Launched at the right sideways speed, it precesses perfectly steadily.
Worked example. For a fast top the mean precession rate is \(\Omega\approx mgl/(I_3\omega_3)\), so doubling the spin halves the precession. With \(I_3=0.8\), \(mgl=4\) and \(\omega_3=10\) this is 0.5 radians per unit time.
Watch out. Here \(\theta\) is measured from the upward vertical and the pivot is frictionless, with \(I_1=1\). Friction at a real pivot slowly drains the spin, so the precession speeds up as the wheel slows.
Where does the tip stop moving completely?
At the top of each cusp when the wheel is released from rest. There \(\dot\theta=0\), and \(\dot\phi=0\) because \(p_\phi-I_3\omega_3\cos\theta\) vanishes exactly when \(\theta\) is back at its starting value.
3. The inertia tensor and principal axes
Drag four point masses in a plane. The moment of inertia about an axis through the reference point depends on the axis direction only through a symmetric matrix, so I(θ) is a sinusoid in 2θ. Its extremes, always 90° apart, are the principal axes, and the momental ellipse shows them at a glance. Switch the reference point to the origin to see the parallel-axis theorem at work.
Worked example. Unit masses at \((\pm1,0)\) and \((0,\pm2)\) give \(I_{xx}=8\), \(I_{yy}=2\) and \(I_{xy}=0\): the coordinate axes are already principal, and \(I_z=10\).
Watch out. The rule \(I_z=I_{xx}+I_{yy}\) holds only for a flat body. A solid body needs the full \(3\times3\) tensor, which, being symmetric, also has three perpendicular principal axes.
Why are the principal axes always perpendicular?
The inertia tensor is a real symmetric matrix, and eigenvectors of a real symmetric matrix belonging to different eigenvalues are orthogonal.