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Homework Help: Need help with centripetal acceleration problem.

  1. Dec 16, 2007 #1
    A biophysicist wants to separate sub-cellular particles with an analytic ultracentrifuge. 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 8.4cm from the centre of the centrifuge when it is spinning at 6.0 x 10^4rpm. Express your answer in terms of g (acceleration due to Earth's gravity).

    r = 0.84m
    f = 6.0 x 10^4rpm = 1000s

    ac = 4pi^2rf^2
    = 4pi^2(0.84m)(1000s)^2
    = 3.32 x 10^7 m/s^2

    So now that I have the centripetal acceleration I'm unsure of how to express my answer in terms of g as per the second part of the question. What I did was the following:

    ac / g
    3.32 x 10^7 m/s^2 / 9.8 m/s^2 = 3.39 x 10^6 m/s^2

    but I am second guessing this because I really have no idea why I dividedt he acceleration by the acceleration of earths gravity. Could somebody shed some light on this and if I am wrong?
  2. jcsd
  3. Dec 17, 2007 #2


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    ac / g
    3.32 x 10^7 m/s^2 / 9.8 m/s^2 = 3.39 x 10^6 m/s^2

    You have to wright the answer as
    (ac / g)*g
    3.32 x 10^7 m/s^2 / 9.8 m/s^2 = (3.39 x 10^6 )g. That is whar required.
  4. Dec 20, 2007 #3

    Shooting Star

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    Some light has already been shed, but I'm just asking you this simple question: if ac was equal to, say, 4.1*9.8 m/s^2, how much would it be in terms of g?
  5. Dec 21, 2008 #4
    Bro...The QUestion Stated that r=8.4cm, so r=0.084m not r=0.84m...Correct that and so u r answer would be one decimal off....right answer would be ac= 3.32*10^6...in terms of g = 3.39*10^5 no unit.
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