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A question on centrifugal artifical gravity

  1. Jul 31, 2013 #1
    How fast would a hollow cylindrical object seven miles in diameter need to spin to maintain earth gravity on the interior surface?
     
  2. jcsd
  3. Jul 31, 2013 #2

    berkeman

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    Welcome to the PF.

    What is the context of the question? Is this for schoolwork? What do you know already about centriptal forces?
     
  4. Jul 31, 2013 #3
    the question does not relate to schoolwork, I'm working on a science fiction novel and my math skills are inferior so I can't crack a book and easily solve the equation myself.
    I need to know the rpm the cylinder would need to spin at to maintain normal earth gravity (would it be a four minute revolution, a half hour revolution?).
     
  5. Aug 1, 2013 #4
    Do you recognise the equation ω2r for calculating centripetal acceleration
    Also, when you come to do the calculation I would suggest that you give the radius in metres rather than miles.
     
  6. Aug 1, 2013 #5

    jfizzix

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    If the space station is 7 miles in diameter, it is 3.5 miles in radius.

    If we assume that we want to feel Earth-level artificial gravity at this distance from the center, we want the centripedal acceleration of a point on this cylinder to be the same as the acceleration due to gravity. In short,

    [itex]a_{edge} = g[/itex]
    but
    [itex]a_{edge} = R \omega^{2}[/itex]
    where R is the radius of the station (3.5 miles or 5607 meters) and omega is the angular velocity of the space station in radians per second.

    Then we solve for [itex]\omega[/itex], finding that
    [itex]\omega=\sqrt{\frac{g}{R}}[/itex]
    so that [itex]\omega[/itex] is about 6.64thousandths of a revolution per second or about 0.40 revolutions per minute.
     
  7. Aug 1, 2013 #6

    jfizzix

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    hope that helps:)
     
  8. Aug 1, 2013 #7

    jfizzix

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    4/10 revolutions per minute means 10/4 minutes per revolution or about 2 1/2 minutes per revolution
     
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