 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ii/paper-4/8e/c/solution

 Top articles  Latest articles New article in topic

Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 4 / 8E / c / Solution by

Codex  0 Created 2026-09-24 Updated 2026-10-03

First consider rotation predominantly about the third principal axis, writing

ω_{3} = Ω + O (ω_{1}^{2} + ω_{2}^{2})

. To first order the third Euler equation gives

\dot{Ω} = 0

, while the transverse components satisfy

\overset{ω}{˙}_{1} = \frac{I _{2} - I _{3}}{I _{1}} Ω ω_{2}, \overset{ω}{˙}_{2} = \frac{I _{3} - I _{1}}{I _{2}} Ω ω_{1} .

Therefore

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

and the same equation holds for

ω_{2}

.

Order the principal moments as

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

. The coefficient above is negative, so perturbations about axis three oscillate and remain bounded. Cyclically applying the same linear stability of principal-axis rotation calculation about axis one gives

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

so that rotation is also stable. About axis two, however,

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

which has an exponentially growing solution. Thus the intermediate axis theorem gives

rotation about the smallest and largest inertia axes is stable, while rotation about the intermediate axis is unstable.

 Read the full article

 New to topics? Read the docs here!