Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 1 / 39D / c / iii / Solution

Pressure acts normally to the circular cylinder and so has no moment about its axis. To leading lubrication order, the tangential shear at the cylinder is

τ_{c} = μ \frac{\partial u}{\partial y}_{y = h} = \frac{μ ( a Ω - V )}{h} + \frac{h}{2} \frac{d p}{d x} .

Substitution of the pressure gradient gives

τ_{c} = μ [\frac{4 a Ω + 2 V}{h} - \frac{6 Q}{h ^{2}}] .

A torque-free cylinder requires

\int_{- \infty}^{\infty} τ_{c} d x = 0

. Also,

\int h^{- 1} d x = π 2 a h_{0} h_{0}^{- 1}, \int h^{- 2} d x = \frac{π}{2} 2 a h_{0} h_{0}^{- 2} .

Using

Q = 2 h_{0} (V + a Ω) /3

, the torque condition reduces to

(4 a Ω + 2 V) - 2 (V + a Ω) = 2 a Ω = 0.

Hence

Ω = 0.

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