 Solution

ID: past-exam-of-the-mathematical-mechanical-biology-course-of-the-university-of-oxford/2023/1/a/i/solution

d_{i}^{'} = u \times d_{i} .

u_{1} = 0

u_{2} = \overset{u}{^}

, and

u_{3} = 0

, since all three stiffnesses are positive. Thus

u = \overset{u}{^} d_{2}

d_{2}

vanishes, so

d_{2}^{'} = \overset{u}{^} d_{2} \times d_{2} = 0 .

Consequently this director is a fixed vector in space, even though the other two directors rotate along the elastic rod.

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