 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ii/paper-1/38a/b/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

Away from the rim's

O (h)

edge region, the geometry and uniform injection have no radial scale other than the factor required by continuity. It is therefore consistent to take

w = w (z)

. Regularity at

r = 0

then lets the continuity equation integrate to

u (r, z) = - \frac{r}{2} w^{'} (z) .

w (0) = V, w^{'} (0) = 0, w (h) = 0, w^{'} (h) = 0.

Since

p

is independent of

z

, the radial equation implies that

w^{'''}

is constant. The four boundary conditions give

w (z) = V [1 - 3 (\frac{z}{h})^{2} + 2 (\frac{z}{h})^{3}],

and hence

u (r, z) = \frac{3 V r}{h} \frac{z}{h} (1 - \frac{z}{h}) .

Using

w^{'''} = 12 V / h^{3}

p_{r} = μ u_{zz}

gives

p_{r} = - \frac{6 μ V r}{h ^{3}} .

Taking the pressure at the rim to be atmospheric,

p (R) = p_{0}

p (r) - p_{0} = \frac{3 μ V}{h ^{3}} (R^{2} - r^{2}) .

W = \int_{0}^{R} [p (r) - p_{0}], 2 π r d r = \frac{3 π μ V R ^{4}}{2 h ^{3}} .

Since

ε = h / R

W = \frac{3 π μ V R}{2 ε ^{3}} .

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