Revs per min from metres per second

  1. okay so here is the problem and here is wat I did!
    I am just tryin to study up for exams now... and this is one of the problems!

    A sample of blood is placed in a centrifuge of radius 17.5 cm. The mass of a red corpuscle is 3.00×10-16 kg, and the magnitude of the force required to make it settle out of the plasma is 4.08×10-11 N. At how many revolutions per second should the centrifuge be operated?

    Basically wat I did is found the velocity in metres per second first which I find is 154m/s which I found by usin F=(mv^2)/r
    But here is the simple problem that I am confused about... how do I change this to revs/sec. Is it somethin to do with C=2PIr, thats all I can think of!
    Thanks for your help!
     
  2. jcsd
  3. Hootenanny

    Hootenanny 9,677
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    You could do it that way, however it is easier to remember that if we resolve newton's second law radially we achieve;

    [tex]F = m\alpha[/tex]

    Where alpha is centripetal acceleration and [itex]\alpha = r\omega^2[/itex] where omega is angular velocity (rads/s). Thus;

    [tex]\fbox{F = mr\omega^2}[/tex]

    You method is completely valid, but is a bit long winded :biggrin:. You can convert radians per second into revolutions per second by dividing by [itex]2\pi[/itex].
     
    Last edited: Jun 6, 2006
  4. okay I did that and now I am so far off... I did 123686.127rev/sec.
    so wat I did was w^2=f/mr
    but then I got that in rads per second =7.77E5 which is bigger then I expected.
    So I then divided by 2*PI... then it was wrong... the answer is meant to be 140.3
     
  5. Hootenanny

    Hootenanny 9,677
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    You forgot to square root the 7.77x105. If you square root that value then divide by [itex]2\pi[/itex] you sould obtain the correct answer.
     
  6. thanks a lot I appreciate it, I thought I did sqaure root the answer but now I got the right answer.! ;P
     
  7. Hootenanny

    Hootenanny 9,677
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    No problem :smile:
     
  8. Andrew Mason

    Andrew Mason 6,880
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    I am sure you meant to say that the quantity [itex]r\omega^2 = v^2/r[/itex] is the centripetal acceleration not angular acceleration.

    AM
     
  9. Hootenanny

    Hootenanny 9,677
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    Thank-you andrew, duly corrected.
     
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