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Artificial Gravity

  1. Jun 30, 2007 #1
    "Artificial Gravity"

    1. The problem statement, all variables and given/known data

    A space station in the form of a large wheel, 333m in diameter, rotates to provide an "artificial gravity" of 9m/s^2 for people located on the outer rim. Find the rotational frequency of the wheel that will produce this effect. Answer in units of rpm [revolutions per minute]

    2. Relevant equations

    w = 2pi/T , Frequency = 1/T

    3. The attempt at a solution

    I first equated the normal force with the centripetal force, which is mg = mv^2/r. Cancelling out the m, I got gr = v^2. After I solved for V, I plugged it into the w = 2pi/T equation to find T, then 1/T. To convert the answer to rpm, I multiplied what I got for F by 60 and divided by 2pi. However I kept getting the wrong answer. Can someone please help? Thank you!
  2. jcsd
  3. Jun 30, 2007 #2


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    How did you relate v to w when finding T in this step?

    Be sure to keep track of the units to make sure things make sense.

    What answer did you get?
  4. Jun 30, 2007 #3
    I got 3.487429162 rpm
  5. Jun 30, 2007 #4
    I also know that w = 2 x pi x frequency
  6. Jun 30, 2007 #5


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    Yes, but remember that v is linear velocity, and w is angular velocity. They are not interchangeable. Do you know the relation between those two quantities?
  7. Jun 30, 2007 #6
    I know that w = v/r
  8. Jun 30, 2007 #7


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    So what do you get if you make use of that? You found what v is, and ultimately what you want to find is w (in rev/min).

    You're initial approach is OK, it's just the long way around. But you still need to use v=rw.
  9. Jun 30, 2007 #8
    oh! I got it now! Thank you so much!
  10. Jun 30, 2007 #9


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    You're welcome. :smile:
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