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Centripetal acceleration centrifuge

  1. Oct 17, 2009 #1
    A centrifuge is a device in which a small container of material is rotated at a high speed on a circular path. Such a device is used in medical laboratories, for instance, to cause the more dense red blood cells to settle through the less dense blood serum and collect at the bottom of the container. Suppose the centripetal acceleration of the sample is 5.92 x 10^3 times as large as the acceleration due to gravity. How many revolutions per minute is the sample making, if it is located at a radius of 6.37 cm from the axis of rotation?

    Acceleration due to gravity is 9.80 and I converted 6.37cm to .0637m.
    So this is how I solved it: Ac=V^2/r so set it up where V^2=Ac*r=sqrt(9.80*5920)*.0637= 60.79160
    Then V=(2pi*r)/T so T=(2pi*r)/V: (2pi*.0637)/60.79160=.006584

    Is this right??
     
  2. jcsd
  3. Oct 17, 2009 #2

    Nabeshin

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    I didn't double check your numbers, but you solved for period and the question asks for rpm, just remember to do the appropriate conversion.
     
  4. Oct 17, 2009 #3
    I'm not sure on how to go from period to RPM
     
  5. Oct 17, 2009 #4

    Nabeshin

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    Well period is seconds per revolution, while rpm is revolutions per minute. Recall 1/T=F, the frequency, which has units revolutions per second.
     
  6. Oct 17, 2009 #5
    So would it be .006584*60=.39504??
     
  7. Oct 17, 2009 #6

    Nabeshin

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    No... check your units and make sure they come out alright. What you have is this:
    [tex].0065 \frac{\textrm{seconds}}{\textrm{revolution}} \times 60\frac{\textrm{seconds}}{\textrm{minute}} = .395 \frac{\textrm{seconds}^2}{\textrm{revolution minute}}[/tex]

    Which is obviously not right. Remember where you want to go, which is revolutions/minute.
     
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