Fluid Mechanics: Cone-Plate viscometer

1. Jan 22, 2015

Feodalherren

1. The problem statement, all variables and given/known data

2. Relevant equations
Fluid Mechanics

τ =μ (du/dy)
3. The attempt at a solution

I got far enough to write down

dM = μ (Ωr/tanθ) dA

from just substitutions, easy enough.

I get confused when I'm solving for the differential surface area. I somehow need the dA for a cone.
For a circle it's easy enough. A=πr^2 so then dA = 2πr dr.

But how do I go about getting this for a cone? the book lists it as (2πr/cosθ) dr, without any explanation, of course.

2. Jan 22, 2015

erisedk

Take a strip element 2πxdr, where x is the radius of circle formed by that strip element. Try relating x to r using similarity of triangles.

3. Jan 22, 2015

Nathanael

Well you should first understand how to find the surface area of a "truncated cone" (ignoring the faces)
It is the average circumference of the cone multiplied by the side length (not the height)

So the surface area of a truncated cone (ignoring the faces) is $2\pi r_{avg}L$ or you could say $2\pi L\frac{r_1+r_2}{2}$
(Side note, this also applies to normal cones; just treat the tip as r2=0)

So with this understanding of truncated cones, look at the following picture I made:

The differential surface area would be $2\pi r_{avg}L$ but as you can see from the picture, $L=\frac{dr}{\cos\theta}$ therefore the differential surface area is $\frac{2\pi R_{avg}}{\cos\theta}dr$

The two radii are r and (r+dr) but since dr is obviously infinitesimal, it suffices to say $r_{avg}=r$

Last edited: Jan 22, 2015
4. Jan 22, 2015

erisedk

Does it really require all that? Isn't using similarity to relate x and R, considering a string element of radius x, just enough?

5. Jan 22, 2015

Nathanael

No that is incorrect. The differential area element is $\frac{2\pi x}{\cos\theta}dx$

Me and the OP were using r not to represent the radius of the base of the cone, but to represent what you called "x"

6. Jan 22, 2015

erisedk

Oh ok.

7. Jan 22, 2015

Feodalherren

Excellent explanation. Thank you very much, Sir.