 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-2/39d/iv/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

For a Newtonian fluid,

σ_{ij} = - p δ_{ij} + μ (\frac{\partial u _{i}}{\partial x _{j}} + \frac{\partial u _{j}}{\partial x _{i}}) .

r = a

, with

n = x / a

A^{'} = - \frac{3}{2 a}, B = 0, B^{'} = \frac{3}{2 a ^{3}}, p = \frac{3 μ}{2 a} (U \cdot n) .

\frac{\partial u _{j}}{\partial x _{i}} = \frac{3}{2 a} n_{i} ((U \cdot n) n_{j} - U_{j}) .

Contracting the stress with

n_{j}

, the normal part of the viscous stress cancels the pressure contribution, leaving the uniform traction

(σ \cdot n)_{i} = - \frac{3 μ}{2 a} U_{i} .

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