Past exam of the physics B course of the University of Cambridge / 2017 / Paper 2 / Section B / B8 / c

Take axes fixed in a rigid body, chosen along its principal axes, with moments of inertia

I_{1}, I_{2}, I_{3}

. The angular momentum components are

H_{i} = I_{i} ω_{i}

. Newton's rotational equation is

τ = (d H / d t)_{inertial}

, while the rotating-frame transport rule gives

(\frac{d H}{d t})_{inertial} = (\frac{d H}{d t})_{body} + ω \times H .

I_{1} \overset{ω}{˙}_{1} + (I_{3} - I_{2}) ω_{2} ω_{3} I_{2} \overset{ω}{˙}_{2} + (I_{1} - I_{3}) ω_{3} ω_{1} I_{3} \overset{ω}{˙}_{3} + (I_{2} - I_{1}) ω_{1} ω_{2} = τ_{1}, = τ_{2}, = τ_{3} .

The nonlinear cross terms arise because the principal axes rotate. Even with no external torque, the body-frame components of angular momentum can vary while the angular momentum vector stays fixed in inertial space.

With zero torque, the equations conserve

E = \frac{1}{2} \sum_{i} I_{i} ω_{i}^{2}, H^{2} = \sum_{i} I_{i}^{2} ω_{i}^{2} .

Uniform rotation about any principal axis is a torque-free solution. If

I_{1} < I_{2} < I_{3}

, linearization about the intermediate-axis solution

ω_{2} = Ω

gives

δ \overset{ω}{¨}_{1} = Ω^{2} \frac{( I _{2} - I _{1} ) ( I _{3} - I _{2} )}{I _{1} I _{3}} δ ω_{1} .

The positive coefficient gives exponential growth: intermediate-axis rotation is unstable, whereas rotation about the largest or smallest moment is linearly stable. For a symmetric body, Euler's equations also describe the precession of the transverse angular velocity around its symmetry axis. Torques from gravity or an applied couple enter through the right-hand sides rather than being absorbed into the inertial cross terms.

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