 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-329/1/b/solution

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

At leading order the particle is a sphere. A pure applied couple produces no translation, while torque balance with the rotating sphere in Stokes flow gives

U_{0} = 0, Ω_{0} = \frac{G}{8 π μ a ^{3}} .

With

ma t hb f n

directed from the particle into the fluid, the exact conditions on the true surface

S

are

u = U + Ω \times x, \int_{S} σ n d S = 0, \int_{S} x \times (σ n) d S = - G .

Evaluate no slip at

ma t hb f x_{s} = (a + ε f) n

and expand about

ma t hb f x = a n

. The order-

ε

terms give

u_{1} = U_{1} + Ω_{1} \times x + \frac{f}{a} Ω_{0} \times x - f \frac{\partial u _{0}}{\partial r} (r = a) .

The divergence of the Newtonian stress vanishes in the surrounding fluid, and its symmetry makes the divergence of angular-momentum flux vanish as well. The total force and torque may therefore be evaluated on any homologous enclosing surface, in particular the fixed reference sphere. Since the applied force is zero and the applied couple is fixed independently of

ε

\int_{r = a} σ_{1} n d S = 0, \int_{r = a} x \times (σ_{1} n) d S = 0 .

Using a fixed enclosing sphere is also why no separate terms involving the shape

f

and leading stress

σ_{0}

appear.

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