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Centrifuge - find centripetal acceleration

  1. Feb 2, 2010 #1
    1. The problem statement, all variables and given/known data
    You are designing a centrifuge to spin at a rate of 15,300 rev/min.
    (a) Calculate the maximum centripetal acceleration that a test-tube sample held in the centrifuge arm 14.7 cm from the rotation axis must withstand.
    377361.36 m/s2

    (b) It takes 1 min, 16 s for the centrifuge to spin up to its maximum rate of revolution from rest. Calculate the magnitude of the tangential acceleration of the centrifuge while it is spinning up, assuming that the tangential acceleration is constant.

    2. Relevant equations
    Is the correct equation for the tangential acceleration at[SUB=r*a?
    If not, what is the correct formula(s)?

    [b]3. The attempt at a solution[/b]
    at=.147*377361.36=55472.12 m/s2
  2. jcsd
  3. Feb 2, 2010 #2


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    Homework Helper

    Re: Centrifuge

    I get the 377361.36 m/s2 answer. You are supposed to show how you got yours so we can find out where you went wrong . . .
    I started with v = 2πrn/T where n is the number of turns and T the time taken.
    Then the standard formula for centripetal acceleration can be used to finish it.
  4. Feb 2, 2010 #3
    Re: Centrifuge

  5. Feb 2, 2010 #4
    Re: Centrifuge

    Thank you. So let me make sure I got this correctly...for my problem:

    Vfinal=(2*pi*.147)/(1/255)=235.53 m/s
    Vinitial= 0
    Tfinal=76 seconds

    235.53/76=3.099 m/s2

    Is this correct?
  6. Feb 2, 2010 #5
    Re: Centrifuge

    Thank you very very much. I got the answer.
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