Question 5 of 6: Torque-Free Rigid-Body Rotation — Euler's Equations After an Impulsive Moment

Question 5: Torque-Free Rigid-Body Rotation — Euler's Equations After an Impulsive Moment (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A rigid body spinning about its own C.M., described in its principal-axis body frame by Euler's equations of torque-free rotation for $t>0$ (the moment is an instantaneous impulse at $t=0$ only): $$I_{xx}\dot\omega_x=(I_{yy}-I_{zz})\omega_y\omega_z,\quad I_{yy}\dot\omega_y=(I_{zz}-I_{xx})\omega_z\omega_x,\quad I_{zz}\dot\omega_z=(I_{xx}-I_{yy})\omega_x\omega_y.$$ The impulsive moment sets the initial angular velocity via $\Delta\vec L=\vec M_0=\int\vec M\,dt$, i.e.\ $\omega_i(0^+)=M_{0,i}/I_{ii}$.

Find. Qualitative behaviour of $\vec\omega(t)$ for $t>0$ in each of the two inertia cases.

Approach. Compute $\vec\omega(0^+)$ from the impulse, then examine whether that state is a fixed point (steady spin) of the torque-free Euler equations for each inertia distribution, using the two conserved quantities of torque-free motion — kinetic energy $2T=I_{xx}\omega_x^2+I_{yy}\omega_y^2+I_{zz}\omega_z^2$ and angular momentum magnitude $L^2$ — to characterize the resulting motion.

1. Part (a) — spherical inertia, $I_{xx}=I_{yy}=I_{zz}=I$. The impulse gives $\vec\omega(0^+)=\vec M_0/I=(a/I,b/I,c/I)$. Because all three principal moments are equal, every term $(I_{yy}-I_{zz})$, $(I_{zz}-I_{xx})$, $(I_{xx}-I_{yy})$ in Euler's equations vanishes identically — regardless of the direction of $\vec\omega$. The equations collapse to $I\dot{\vec\omega}=0$, i.e.\ $\vec\omega$ is exactly constant for all $t>0$. The body immediately settles into steady, unchanging rotation about the fixed axis along $\vec M_0=(a,b,c)$, with constant magnitude $|\vec M_0|/I$ — no wobble, no precession. This is because a spherically symmetric inertia tensor treats every axis identically: there is no gyroscopic coupling between components to drive any change, so whatever axis the impulse sets up is automatically a permanent axis of rotation (confirmed: integrating Euler's equations from this initial condition returns $\vec\omega(t)\equiv\vec\omega(0^+)$ to machine precision).
2. Part (b) — asymmetric (triaxial) inertia, $I_{xx}>I_{yy}>I_{zz}$. The impulse gives $\vec\omega(0^+)=(a/I_{xx},\,0,\,c/I_{zz})$ — zero component along the intermediate-inertia axis $y$, nonzero along both the maximum ($x$) and minimum ($z$) inertia axes. Check whether $\omega_y\equiv 0$ can persist: from the middle Euler equation, $$I_{yy}\dot\omega_y\big|_{t=0^+} = (I_{zz}-I_{xx})\,\omega_z\omega_x\Big|_{t=0^+} = (I_{zz}-I_{xx})\cdot\frac{c}{I_{zz}}\cdot\frac{a}{I_{xx}} \neq 0$$ because $I_{zz}\neq I_{xx}$ and $a,c>0$ are both nonzero by hypothesis. So $\omega_y=0$ is not maintained — the state $(\omega_x,0,\omega_z)$ with both $\omega_x,\omega_z\neq0$ is not an equilibrium of Euler's equations (only the two pure states, spin about $x$ alone or about $z$ alone, are steady). The $y$-component is therefore driven away from zero immediately, while kinetic energy $2T$ and $|\vec L|$ stay exactly conserved (no external torque for $t>0$) — confirmed: $\omega_y$ grows from 0 to a magnitude comparable to $\omega_x(0),\omega_z(0)$ within a few characteristic periods, while $2T$ and $|\vec L|$ drift by less than $10^{-6}$ over that time. The rotation does not settle into steady spin about a fixed axis; it develops a genuine, time-varying $y$-component and nutates/tumbles, tracing a closed (bounded) polhode curve on Poinsot's construction around whichever of the two stable axes ($x$ or $z$) the particular energy/momentum combination favours — qualitatively different from part (a)'s permanently steady rotation, precisely because the gyroscopic coupling terms in Euler's equations no longer vanish when the three principal moments are distinct.