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Centrifuge acceleration expressed in terms of gravity

  1. Nov 28, 2006 #1
    A Biophysicist wants to separate sub-cellular particles with an analytic ultra-centrifuge. The biophysicist must determine the magnitude of the centripetal acceleration provided by the centrifuge at various speeds and radii.

    Calculate the magnitude of the centripetal acceleration at 8.4 cm from the centre of the centrifuge when it is spinning at 6.0*10^4 rpm. Express your answer in terms of g (acceleration due to Earth's gravity).


    Given:

    r = 0.084m
    f = 60000 / 60s

    Required: ac

    Analaysis:

    ac = 4pie^2 * r * f^2

    Solution:

    ac = 4(3.14)^2 * 0.084m * (60000 / 60s)^2

    = 3312825.6 m / s^2

    Now I am a bit unclear as to how to express this in terms of g (Earth's gravity).

    I could factor in 9.8 m / s^2 but that only changes the direction of the ac force by .001 m / s2

    /F/ = /ac^2/ + /Fg^2/

    I think I am approaching this from the wrong angle. Any suggestions?
     
  2. jcsd
  3. Nov 28, 2006 #2

    berkeman

    User Avatar

    Staff: Mentor

    Use the equation 1g = 9.8m/s^2 (g = accel due to gravity, not gram).

    Multiply your answer by 1 ( = 1g/(9.8m/s^2) )....
     
  4. Nov 28, 2006 #3
    Doh! So easy. Thanks
     
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