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Acceleration of centrifuge - SOLVED

  1. Aug 31, 2012 #1
    1. The problem statement, all variables and given/known data
    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)
    3-figure-34-alt.gif
    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.



    2. Relevant equations
    a = [itex]\frac{v^2}{r}[/itex]

    v = [itex]\frac{2\pi r}{t}[/itex]


    3. The attempt at a solution

    part a)
    a = [itex]\frac{v^2}{r}[/itex]

    v = 2[itex]\pi[/itex]r(f)

    r = [itex]\frac{2000g}{(2\pi f)^2}[/itex]

    r = 14.27cm

    part b)
    This is what gets me. Let's call rA the top of the tube and rB the bottom. Would rA not be r-(7.1cm*cos(45)) and rB = 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.
     
    Last edited: Aug 31, 2012
  2. jcsd
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