Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 352 / 3 / b / i

Table of contents
- Solution i

Solution

Let

G = \frac{Δ P}{L} > 0.

Axial momentum balance gives shear-stress magnitude

∣ τ_{rz} ∣ = G r /2

. The wall stress must exceed the pressure-dependent yield stress, so motion requires

Δ P > Δ P_{m i n} = \frac{2 L μ p _{s}}{R} .

When this holds, define the plug radius

a = \frac{2 μ p _{s}}{G} < R .

The no-slip solid velocity is

w_{s} (r) = ⎩ ⎨ ⎧ \frac{G}{4 η} (R - a)^{2}, \frac{G}{4 η} (R^{2} - r^{2}) - \frac{μ p _{s}}{η} (R - r), 0 \leq r \leq a, a < r \leq R .

In the sheared annulus,

∣ w_{s}^{'} ∣ = \frac{G ( r - a )}{2 η}, I_{v} = \frac{G ( r - a )}{2 p _{s}},

so

ϕ (r) = ⎩ ⎨ ⎧ ϕ_{m}, \frac{ϕ _{m}}{1 + G ( r - a ) / ( 2 p _{s} )}, 0 \leq r \leq a, a < r \leq R .

 Ancestors (11)

 View article source

 Discussion (0)

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version