# Water Pressure in Centrifuge

1. Aug 26, 2014

### bphys348

1. The problem statement, all variables and given/known data
A test tube filled with water is being spun around in an ultra centrifuge with angular velocity ω. The test tube is lying along a radius, and the free surface of the water is at a radius r0.

(a) Show that the pressure at radius r within the test tube is P = (1/2)rho*ω2(r2-r02). Ignore gravity and atmospheric pressure

2. Relevant equations
dP= -ρgdz

3. The attempt at a solution
Since the test tube is undergoing circular motion, and the water within the test the test tube lies at a varying radius from center, the acceleration (g) is given by a=(v2)/r = ω2r

So dP = -ρgdz = -ρω2rdr

P= -(1/2)ρω2r2 from r to r0

I think that much is fine, and this gives me the correct answer... But, here is where my understanding is hazy. If I define r to be increasing radially from center, then it is my understanding that my lower bound should be r and my upper to be r0.

Here's my question: If I define an axis to be increasing in one direction, then should I always integrate from the lower value to higher?

If I use this same reasoning with a simple pressure vs. water depth example then I run into trouble. If I want to find the pressure at a depth of 30m and I define my z axis to be increasing upwards then it seems to me the integral should look like this:

dP = ∫-ρgdz from -30 to 0 = -ρg(0 - -30) = -ρg*30

But the answer should be positive!

Any help would be much appreciated
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Aug 26, 2014

### Orodruin

Staff Emeritus
In the formula you have used, the gravitational acceleration is pointing in the negative z direction. If you change the direction of the axis, then you must also change the direction of the field.