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Centripetal Acceleration and period of rotation

  1. Sep 26, 2007 #1
    1. The problem statement, all variables and given/known data

    In a test of a “g -suit,” a volunteer is rotated in a horizontal circle of radius 7.7m.

    What must the period of rotation be so that the centripetal acceleration has a magnitude of 3.7g ?

    What must the period of rotation be so that the centripetal acceleration has a magnitude of 10g?



    2. Relevant equations
    F=ma
    w=ma
    4pi^2r/T

    3. The attempt at a solution

    Ok I started by converting 3.7g, by doing this 9.8*3.7 = 36.26

    4pi^2(7.7)/ 36.26

    but it is wrong, how would I solve this?
     
  2. jcsd
  3. Sep 26, 2007 #2

    hage567

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    This isn't right, check the units. It's supposed to be acceleration, right? Did you derive it yourself?
     
  4. Sep 26, 2007 #3
    I've been told it has something to do with the equation, although I don't think I am using properly.

    what would you do in the following situation?
     
  5. Sep 26, 2007 #4

    hage567

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    Well, start with the equation for centripetal acceleration, a = v^2/r. You need to relate it to the period. You can do that by the velocity term. What's a way to express the velocity of an object moving in a circle?
     
  6. Sep 26, 2007 #5
    velocity would be distance/time.

    Circumference = 2*pi*Radius

    v=2pir/t ?
     
  7. Sep 26, 2007 #6

    hage567

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    That's right. Now put that together with the a=v^2/r and you should come up with the equation you need.
     
  8. Sep 26, 2007 #7
    a = 4pi^2r / t^2

    ?
     
  9. Sep 26, 2007 #8

    hage567

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    Yes. So solve that for t and use it to find the period.
     
  10. Sep 26, 2007 #9
    ok got it. thank you
     
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