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Rotation period of a space station

  1. Feb 3, 2015 #1
    1. The problem statement, all variables and given/known data
    To simulate the extreme accelerations during launch, astronauts train in a large centrifuge. If the centrifuge diameter is 13.5m , what should be its rotation period to produce a centripetal acceleration of
    If the centrifuge diameter is 13.5m , what should be its rotation period to produce a centripetal acceleration of 2 g? of 5g?

    2. Relevant equations
    T=2*pi*r/v

    3. The attempt at a solution
    i tried solving for v using a=v^2/r and then plugging it in but i got the wrong answer
     
  2. jcsd
  3. Feb 3, 2015 #2

    haruspex

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    Please post your working, or we can't tell if or where you went wrong.
    (Are you sure it asks for a centripetal acceleration of that magnitude, not a net g-force corresponding to it?)
     
  4. Feb 3, 2015 #3
    i did 9.8=v^2/6.75, v^2=(9.8)(6.75)=66.15 v=8.13
    then i plugged v into the equation i gave above
     
  5. Feb 4, 2015 #4

    haruspex

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    It says 2g and 5g, not 1g.
     
  6. Feb 4, 2015 #5
    You need to introduce T into the root equation.
    You have a = v ² / r
    But v = ( 2 * π * r ) / T
     
  7. Feb 4, 2015 #6
    then what equation should i use to solve for v since i dont know T?
     
  8. Feb 4, 2015 #7

    haruspex

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    I think dean barry is suggesting you eliminate v between the two equations so that you can go straight to finding T without having to calculate v. That's good advice generally, since it serves to reduce accumulation of rounding errors, but I don't think it matters here.
    Do you have a response to my post #4?
     
  9. Feb 4, 2015 #8
    i did accidently forget to do 2g instead of just 9.8.
     
  10. Feb 4, 2015 #9

    haruspex

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    So does that resolve your issue, or do you still have the wrong answer?
     
  11. Feb 4, 2015 #10
    that fixed it. thanks a ton
     
  12. Feb 4, 2015 #11
    so after solving for omega i plug it into T=2*pi*r/v? i get 407 when i do this and this seeems a little high
     
  13. Feb 4, 2015 #12

    haruspex

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    Yes, that's way too high. Please post all your steps.
     
  14. Feb 5, 2015 #13
    Take: a = v ² / r
    You know:
    v = ( 2 * π * r ) / T
    ( which introduces T into the game )
    You get: a = ( ( 2 * π * r ) / T ) ² ) / r
    Transpose for T
     
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