 Solution

ID: past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2018/paper-1/section-c/c10/physical-interpretation/solution

Both the magnitude

J

of angular momentum and rotational kinetic energy are conserved after the impulse. At fixed

J

, eliminate

J_{3}^{2} = J^{2} - J_{1}^{2} - J_{2}^{2}

to write

T = \frac{J ^{2}}{2 I _{3}} + \frac{J _{1}^{2}}{2} (\frac{1}{I _{1}} - \frac{1}{I _{3}}) + \frac{J _{2}^{2}}{2} (\frac{1}{I _{2}} - \frac{1}{I _{3}}) .

The fixed-angular-momentum energy test for principal-axis rotation makes the stability geometric. The largest moment gives a constrained energy minimum, the smallest a maximum, and the intermediate moment a saddle. Conservation of energy keeps a nearby motion near either isolated extremum, but a saddle permits departure in one direction while compensating in another. Thus both extremes are stable without dissipation; it would be incorrect to demand an energy minimum alone in this conservative problem.

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