 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ii/paper-1/39c/b/iii/solution

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

σ_{r θ} = μ [r \frac{\partial}{\partial r} (\frac{u _{θ}}{r}) + \frac{1}{r} \frac{\partial u _{r}}{\partial θ}] .

r = a

, no slip holds for every

θ

, so

u_{θ} = 0

and

\partial_{θ} u_{r} = 0

. Therefore

σ_{r θ} (a, θ) = μ \partial_{r} u_{θ} (a, θ) = - μ ψ_{rr} (a, θ) .

Differentiating the solution from part (ii) gives

ψ_{rr} (a, θ) = γ - 2 γ cos 2 θ,

and hence

σ_{r θ} (a, θ) = - μ γ + 2 μ γ cos 2 θ .

The moment arm and line element are both

a

, so the torque exerted by the fluid on the disk per unit axial length is

T_{z} = a^{2} \int_{0}^{2 π} σ_{r θ} (a, θ) d θ = - 2 π μ γ a^{2} .

For

γ > 0

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