## A Sticky Physics Problem

As I often do when I’m too busy to write anything strenuous I thought I’d post something from my back catalogue of physics problems. I don’t remember where this one comes from but I think you’ll find it interesting…

Oil of viscosity η and density ρ flows downhill in a flat shallow channel of width w which is sloped at an angle θ. The oil is everywhere of the same depth, d, where d<<w. The effect of viscosity on the side walls can be assumed to be negligible.

If x is a coordinate that represents the vertical position within the flow (i.e. x=0 at the bottom and x=d at the top), write down a differential equation for the velocity within the flow  v(x) as a function of x. Use physical arguments to derive appropriate boundary conditions at x=0 and x=d and use these to solve the equation, thereby determining an explicit form for v(x). Hence determine the volume flow rate in terms of η, ρ, θ, d and w as well as the acceleration due to gravity, g.

### 7 Responses to “A Sticky Physics Problem”

1. Way over my head. More interesting to me: Say Mars was 36 million miles from earth. How long would it take to travel there on a spaceship accelerating at 1g and then turning around at at the half-way point and decelerating also at 1g?

2. Is this happening on Jupiter?

• telescoper Says:

I doubt if oil would flow on Jupiter…

• Themos Says:

I must have completely missed the last bit “as well as the acceleration due to gravity, g”, sorry! Isn’t this a case where a dimensional argument gives you the answer?

• telescoper Says:

Only up to dimensionless factors (like, e.g., 1/3)…

3. Chris Brunt Says:

I have a million things to do, therefore I am doing this instead, though I can’t do maths symbols or Greek symbols.

Using big D derivative, with del dot v = 0, assuming x is perpendicular to the inclined plane and nu=constant –

0 = g sin(theta) + nu d^2v/dx^2

v = – (g sin(theta) / 2nu) x^2 + c_1 x + c_0

Assuming the flow is stuck on the bottom (v = 0 at x = 0) and there is no shear at the air boundary (dv/dx = 0 at x = d)

v = (g sin(theta) / 2nu) ( 2xd – x^2 )

(fastest at the top)

The flow rate is integral of this over x, times width w

= (g sin(theta) / 3 nu) d^3 w

This looks right dimensionally (m^3 s^-1)

• Chris Brunt Says:

What I’ve called nu is your eta over rho, unless you meant kinematic viscosity, in which case my nu is your eta.