 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ii/paper-1/39c/solution

Codex  0 Created 2026-09-23 Updated 2026-09-25

Take downward velocity as positive. Lubrication theory reduces the axial momentum equation to a constant pressure gradient, and the no-slip conditions are

u (0) = 0

u (h) = U

. Twice integrating gives

u (y) = - \frac{Δ p}{2 μL} y (h - y) + U \frac{y}{h} .

The flux per unit circumference and the total gap flux are

q = \int_{0}^{h} u (y) d y = - \frac{Δ p h ^{3}}{12 μL} + \frac{U h}{2}, Q ≃ 2 πa q .

The shear at the cylinder is

τ_{h} = μ u_{y} (h) = \frac{Δ p h}{2 L} + \frac{μU}{h},

with the fluid traction on the falling cylinder opposing its motion.

The descending solid displaces volume at rate

π a^{2} U

, so closed-container mass conservation gives

2 πa (- \frac{Δ p h ^{3}}{12 μL} + \frac{U h}{2}) = - π a^{2} U .

Since

h ≪ a

, the Couette-flux term is smaller, and

Δ p ≃ \frac{6 μ a LU}{h ^{3}} .

The pressure or form drag is

F_{p} ≃ π a^{2} Δ p = \frac{6 π μ a ^{3} LU}{h ^{3}} .

Using the pressure-dominated part of

τ_{h}

, the side shear force is of order

F_{μ} ≃ 2 πa L \frac{Δ p h}{2 L} = \frac{6 π μ a ^{2} LU}{h ^{2}},

F_{p} / F_{μ} ≃ a / h ≫ 1

. Balancing form drag against the excess weight

Δ ρ g π a^{2} L

gives

U ≃ \frac{Δ ρ g h ^{3}}{6 μ a} .

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