advanced26 min studyLesson 155 in Physics

Torque-free precession of a symmetric body

Solve the symmetric free rotor and distinguish body-component circulation from precession of the symmetry axis around fixed angular momentum.

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What you’ll understand

  • Solve the body-frame transverse angular-velocity equations for a symmetric rotor.
  • Separate signed body circulation from inertial symmetry-axis precession.
  • Check conservation of tilt, angular momentum, and energy.

A released symmetric rotor can wobble without an external torque. Its angular momentum remains fixed in space, but its symmetry axis need not coincide with that momentum. Several rates can then be called precession, and they are not interchangeable. We will define each rate by the object that moves and the frame in which its motion is observed.

Solve the body-component motion

Let the first two principal moments equal A and the third equal C. The third Euler equation immediately shows that the third angular-velocity component is constant without torque. The remaining two equations become a pair of first-order linear equations. Combining them into a complex transverse component turns the pair into one phase-rotation equation, whose signed rate depends on the inertia difference.

Define a positive body circulation as motion from the first toward the second body axis. If C exceeds A and the third angular velocity is positive, the transverse angular-velocity components circulate in that positive sense. If C is smaller, the circulation reverses. The transverse amplitude stays constant. Its period uses the absolute value of the signed rate; when the rate vanishes, the component vector does not circulate.

ω1+iω2=ω⊥ei(Ωbt+α),Ωb=C−AAω3,ω˙3=0\omega_1+i\omega_2=\omega_\perp e^{i(\Omega_b t+\alpha)},\qquad\Omega_b=\frac{C-A}{A}\omega_3,\qquad\dot\omega_3=0
Ωb is the signed circulation rate of transverse angular-velocity components in the principal body frame. It is not the inertial azimuth rate of the symmetry axis.↔ Scroll wide equations horizontally.

Watch the symmetry axis from space

Choose the fixed space z direction along the conserved angular momentum, whose magnitude is L. The body symmetry axis maintains a constant angle theta to that direction because its projection on angular momentum is C times the constant third angular velocity. This produces a cone in space. For nonzero tilt, its azimuth increases at L divided by A in the right-handed convention about positive angular momentum.

This inertial rate is positive by the choice of positive space axis. The final Euler spin-coordinate rate is instead negative Omega_b in the declared active Z-X-Z convention. Neither should be mislabeled as the physical third component of angular velocity: that component also includes the projection of azimuthal precession. At exact alignment the cone collapses, and an azimuth rate is a redundant coordinate description rather than an observable sweep.

cos⁡θ=Cω3L,ϕ˙=LA,ψ˙=ω3−ϕ˙cos⁡θ=−Ωb\cos\theta=\frac{C\omega_3}{L},\qquad\dot\phi=\frac{L}{A},\qquad\dot\psi=\omega_3-\dot\phi\cos\theta=-\Omega_b
These Euler-angle relations use the fixed space z axis along positive angular momentum and R = Rz(φ) Rx(θ) Rz(ψ). The separate azimuth interpretation assumes nonzero tilt.↔ Scroll wide equations horizontally.

Distinguish geometry from energy loss

Torque-free precession does not require friction, energy input, or a force continually pushing the symmetry axis sideways. The geometry follows from the relationship between angular momentum and angular velocity in an anisotropic body. The ideal model keeps both the rotational energy and angular-momentum magnitude constant. Internal dissipation would invalidate that energy conservation while possibly leaving total angular momentum approximately fixed.

The vector lab can test the geometry of a projection: predict that increasing an angle from thirty to sixty degrees decreases its axial cosine projection while preserving the chosen magnitude, then inspect the components. Connect that observation to the tilt formula on paper. This single planar vector does not animate either precession cone, integrate Euler’s equations, or demonstrate the different body and space rates.

Two different rates for the same free rotor

A symmetric rotor has A = 2 and C = 3 kilogram square meters. Its third angular velocity is 4 radians per second and its transverse amplitude is 3. Find the signed body circulation rate, the angular-momentum magnitude, and the inertial precession rate.

  1. Compute the signed body rate from the inertia difference: (3 - 2) times 4 divided by 2 gives positive 2 radians per second. The transverse components circulate from body axis one toward body axis two.

  2. The transverse angular momentum has magnitude A times 3, or 6, while the axial momentum is C times 4, or 12. Their perpendicular combination gives L = square root of 180, approximately 13.42 kilogram square meters per second.

  3. Divide L by A to obtain the inertial symmetry-axis azimuth rate, approximately 6.71 radians per second. Its tilt satisfies cosine theta = 12 divided by square root of 180, approximately 0.894.

The result

The body circulation rate is 2, whereas the inertial azimuth rate is about 6.71 radians per second. The Euler spin-coordinate rate is -2, and the physical third angular-velocity component remains +4.

Explore the model

Interactive lab

Interactive lab · vectors

Check the fixed-magnitude projection geometry

Predict which component decreases when the angle changes from 30 to 60 degrees at fixed magnitude. Verify with the angle control, then apply the cosine projection to the tilt relation.

Check the fixed-magnitude projection geometryPlayback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Current readings: x component: 1.393; y component: 0.975; Magnitude: 1.7. All parameters are adjustable with the labeled controls below.−7.46−3.7303.737.46−3.5−1.7501.753.5xyvvₓ = 1.39
Vectorx / y projectionsTrace marker
x component
1.393
y component
0.975
Magnitude
1.7
t = 0 s

Playback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. 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 lab shows a planar projection only. It does not simulate body-cone or space-cone precession, whose distinct rates are calculated in the lesson.

Check your understanding

A symmetric rotor has C smaller than A, positive omega_3, and nonzero tilt. With space z along its angular momentum, which statement is correct?

Try it yourself

Take A = 4, C = 2, omega_3 = 6, and transverse amplitude 2 in consistent SI units. Find the signed body circulation rate and the inertial symmetry-axis azimuth rate.

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 transverse momentum as A times the transverse angular velocity and axial momentum as C omega_3. Their magnitudes combine in quadrature.

Compare with the solution

The signed body circulation rate is (2 - 4) times 6 divided by 4, or -3 radians per second. Angular momentum has perpendicular components of magnitudes 8 and 12, hence magnitude square root of 208, about 14.42. The inertial azimuth rate is square root of 208 divided by 4, about 3.61 radians per second. Reversing the body circulation sign does not reverse the azimuth around the chosen positive angular-momentum axis.

Take it with you

  • Name the moving direction and the observing frame whenever reporting a precession rate.
  • A symmetric free rotor keeps its axial angular velocity and tilt to angular momentum constant.
  • Signed body circulation and positive inertial symmetry-axis precession are different observables.

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