advanced24 min studyLesson 156 in Physics

Why the intermediate axis is unstable

Linearize torque-free Euler equations and distinguish oscillatory stability from exponential growth without invoking a torque.

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What you’ll understand

  • Derive the perturbation growth rate about the middle principal axis.
  • Check the result against conserved energy and angular momentum.
  • Distinguish an instability from a sustained change in spin magnitude.

A book tossed while spinning about one carefully chosen axis can turn over unexpectedly, even when air resistance and external torques are negligible. The surprising motion is not a failure of angular-momentum conservation. The conserved vector lives in inertial space, while the book changes its orientation relative to it. We will examine small departures from steady rotation, keep the first terms that can grow, and identify which principal moment makes the difference. This gives a quantitative explanation of the intermediate-axis instability rather than a rule learned from a demonstration.

Perturb a steady principal-axis spin

Order three distinct positive principal moments as I₁ < I₂ < I₃, all about the center of mass of a freely moving rigid body. Exact rotation with body angular velocity (0, Ω, 0) solves the torque-free Euler equations. Introduce small body components u along axis one and v along axis three. The change in the second component is second order because its Euler equation contains the product uv. At linear order, hold that component equal to Ω while solving for the two transverse perturbations.

The resulting first-order equations couple u and v with coefficients of the same sign when Ω is positive. Differentiate once to eliminate v. The coefficient multiplying u in its second derivative is positive, so the solutions include an exponentially increasing term. A negative coefficient would instead give an oscillation. Notice that this is a statement about body components near a particular steady solution; it does not say that the magnitude of the full angular momentum grows, or that the rigid body gains kinetic energy.

u˙=I2−I3I1Ωv,v˙=I1−I2I3Ωu,u¨=λ2u\dot u=\frac{I_2-I_3}{I_1}\Omega v,\qquad \dot v=\frac{I_1-I_2}{I_3}\Omega u,\qquad \ddot u=\lambda^2u
Both coupling coefficients are negative for positive Ω, so their product is positive. These are first-order perturbation equations about spin along axis two.↔ Scroll wide equations horizontally.

Read the growth rate and its limits

The positive growth rate depends on the gaps between the middle moment and the two extreme moments. Doubling Ω halves the local growth time. Making either gap small weakens this particular linear instability. If two principal moments coincide, this distinct-axis calculation reaches a degenerate limit and cannot be used to claim exponential growth with a finite rate. Near the smallest or largest principal axis, repeating the same calculation gives a negative second-derivative coefficient and bounded linear oscillations of the transverse components.

Linear stability has a local meaning. The approximation predicts an exponential departure only while u and v remain small compared with Ω. Once the body tumbles substantially, neglected terms matter and the full Euler equations must be used. The finite body cannot sustain unlimited exponential growth of its angular speed with fixed energy. In the ideal conservative problem, nearby extreme-axis motions remain close in the appropriate stability sense, but they do not relax back to the exact axis because there is no damping.

λ=∣Ω∣(I3−I2)(I2−I1)I1I3,T=12∑aIaωa2,L2=∑aIa2ωa2\lambda=|\Omega|\sqrt{\frac{(I_3-I_2)(I_2-I_1)}{I_1I_3}},\qquad T=\frac12\sum_a I_a\omega_a^2,\qquad L^2=\sum_a I_a^2\omega_a^2
The first formula is the local middle-axis growth rate. The last two quantities remain constant for exact torque-free rigid-body motion.↔ Scroll wide equations horizontally.

A saddle on a conserved surface

Express the rotational energy in terms of body angular-momentum components and restrict attention to the sphere of fixed angular-momentum magnitude. Rotation about the largest moment minimizes that energy on the sphere, rotation about the smallest moment maximizes it, and rotation about the middle moment is a saddle. Nearby intersections of the energy ellipsoid with the momentum sphere therefore have different geometry near the three axes. Conservation restricts the trajectory but does not require every allowed steady motion to be stable.

Use the quadratic-function lab as a local-curvature illustration. Predict that changing coefficient a from positive to negative reverses the sign of the tangent slope at the same positive position x, while the stationary point stays at zero. Check that prediction using the displayed slope 2ax. This single parabola illustrates a curvature sign, not both directions of a saddle and not the Euler equations. No control in this lab specifies principal moments or predicts a three-dimensional flip; the growth rate must come from the mechanical calculation above.

A middle-axis growth time

A free body has principal moments (2, 3, 4) kg m² and spins nearly about the middle axis at Ω = 2 rad/s. Find the linear growth rate and the growth factor after one e-folding time.

  1. Insert the moment gaps: (I₃−I₂)(I₂−I₁)/(I₁I₃) = (1×1)/(2×4) = 1/8. The square root is 1/(2√2).

  2. Multiply by |Ω| = 2 s⁻¹ to obtain λ = 1/√2 s⁻¹. Thus the e-folding time is 1/λ = √2 s, approximately 1.41 s.

  3. A perturbation prepared in the growing eigenmode multiplies by exp(λt), so it grows by e after √2 s. This estimate applies only until the perturbation ceases to be small; a general perturbation also contains a decaying component.

λ=21⋅12⋅4=12 s−1\lambda=2\sqrt{\frac{1\cdot1}{2\cdot4}}=\frac1{\sqrt2}\ \mathrm{s}^{-1}
Radians are dimensionless, and the inertia factors enter only through a dimensionless ratio.↔ Scroll wide equations horizontally.
The result

The local e-folding time is √2 s. Energy and angular-momentum magnitude remain constant while the orientation departs from middle-axis spin.

Explore the model

Interactive lab

Interactive lab · function

Inspect a local curvature sign

At x = 1, compare coefficient a = 1 and a = −1. Predict the tangent slope before changing the coefficient. This quadratic is a curvature analogy, not a tumbling-body simulation.

Inspect a local curvature signThis lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Current readings: y at chosen x: 0.608; Tangent slope · 2ax: 1.35; Signed area · ax³/3: 0.182. All parameters are adjustable with the labeled controls below.−2.4−1.201.22.4−9−4.504.59xy(x, ax²)
y = ax²TangentSigned integral
y at chosen x
0.608
Tangent slope · 2ax
1.35
Signed area · ax³/3
0.182
t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forThe slope changes from 2 to −2, but a complete saddle requires two perturbation directions and the rigid-body constraints.

Check your understanding

A torque-free body initially near middle-axis spin begins tumbling. Which statement is consistent with the ideal rigid-body model?

Try it yourself

For moments (1, 2, 3) kg m² and middle-axis spin Ω = 3 rad/s, find λ. What happens to λ if the spin rate doubles?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Compute the moment-gap factor first, then multiply its square root by |Ω|.

Compare with the solution

The factor is (3−2)(2−1)/(1×3) = 1/3. Therefore λ = √3 s⁻¹. Doubling Ω doubles λ and halves the e-folding time, while leaving the distinction between middle-axis and extreme-axis spin unchanged.

Take it with you

  • Instability concerns nearby orientations, not failure of torque-free conservation.
  • The intermediate-axis perturbation has a positive second-derivative coefficient.
  • Exponential growth is a local approximation and stops describing large tumbles.

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