Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 329 / 3 / a

Table of contents
- Solution a

Solution

Near the closest point, the cylinder–plane gap is

h (x) = \frac{a}{2} ε + \frac{x ^{2}}{2 a} = \frac{a}{2} (ε + ξ^{2}), ξ = \frac{x}{a} .

(1)

For translation parallel to the cylinder axis, the leading flow is Couette flow. Its shear traction is

μU / h

, so

\frac{F _{y}}{U} = μL \int_{- \infty}^{\infty} \frac{d x}{h ( x )} = 2 μL \int_{- \infty}^{\infty} \frac{d ξ}{ε + ξ ^{2}} = 2 π ε^{- 1/2} μL .

(2)

Hence

A_{yy} = 2 π ε^{- 1/2} μL

For transverse translation, the local Couette-Poiseuille flow in a thin gap and mass conservation give the Reynolds lubrication equation. Writing

x = a ε X

and using pressure recovery at both ends determines its integration constant. The pressure and viscous contributions to the horizontal traction reduce to

\frac{F _{x}}{U} = \frac{μL}{ε} (16 I_{2} - 16 I_{3}) = \frac{μL}{ε} (8 π - 6 π) + \frac{2 π μL}{ε},

(3)

where the final term is the direct Couette shear contribution. Therefore

A_{xx} = 4 π ε^{- 1/2} μL .

(4)

 Ancestors (10)

 View article source

 Discussion (0)

New discussion

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

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 329 / 3 / a

Solution

 Ancestors (10)

 Discussion (0)  Subscribe (1)

 Articles by others on the same topic (0)

 Discussion (0)