How Does Centrifuge Radius Affect Blood Component Separation?

[cowmoo32]

Homework Statement

Human blood contains plasma, platelets, and blood cells. To separate the plasma from other components, centrifugation is used. Effective centrifugation requires subjecting blood to an acceleration of 2000g or more. In this situation, assume that blood is contained in test tubes of length L = 14.2 cm that are full of blood. These tubes ride in the centrifuge tilted at an angle of 45.0° above the horizontal (see figure below)

What is the distance of a sample of blood from the rotation axis of a centrifuge rotating at a frequency f = 3540 rpm, if it has an acceleration of 2000g?

If the blood at the center of the tubes revolves around the rotation axis at the radius calculated in Part (a), calculate the accelerations experienced by the blood at each end of the test tube. Express all accelerations as multiples of g.

Homework Equations

a = $\frac{v^2}{r}$

v = $\frac{2\pi r}{t}$

The Attempt at a Solution

part a)
a = $\frac{v^2}{r}$

v = 2$\pi$r(f)

r = $\frac{2000g}{(2\pi f)^2}$

r = 14.27cm

part b)
This is what gets me. Let's call r_A the top of the tube and r_B the bottom. Would r_A not be r-(7.1cm*cos(45)) and r_B = r+(7.1cm*cos(45))? This seems logical to me but it isn't giving me the correct answer for the acceleration.

edit: I figured it out. I worked on this for a couple hours last night and kept using the same velocity. It just dawned on me that the velocity changes with the radius.

 

[jbsteele]

I apologize for the mental lapse. aA = \frac{(2\pi*(r-(7.1cm*cos(45))*f)^2}{r-(7.1cm*cos(45))} = 3.48gaB = \frac{(2\pi*(r+(7.1cm*cos(45))*f)^2}{r+(7.1cm*cos(45))} = 5.54g