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Homework Help: Artificial Gravity and a Rotating Space Station

  1. May 31, 2009 #1
    Last edited: May 31, 2009
  2. jcsd
  3. May 31, 2009 #2


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    This can be modeled with the equations [/itex]x= 500 cos(\omega t)[/itex], [itex]y= 500 sin(\omega t)[/itex] where [itex]\omega[/itex] is the angular velocity. Differentiating to get the velocity, [itex]v_x= -500\omega sin(\omega t)[/itex], [itex]v_y= 500\omega sin(\omega t)[/itex]. The acceleration is the derivative of that, [itex]a_x= -500 \omega^2 cos(\omega t)[/itex], [itex]a_y= -500 \omega^2 sin(\omega t)[/itex]. The magnitude of the acceleration is [itex]\sqrt{(-500 \omega^2 cos^2(\omega t))^2+ (500 \omega^2sin^2(\omega t)}[/itex][itex]= 500\omega^2[/itex] and that must be equal to the gravity of earth, 9.81 m/s2. Set them equal and solve for [itex]\omega[/itex]. To find the diameter at which to set the inner ring, go back to the original equation with that value for
    [itex]\omega[/itex], a variable, r, in place of the 500, do the same calculations, set equal to the gravitational acceleration on mars and solve for r.

  4. May 31, 2009 #3


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    I woudl make a free body diagram of a person standing on the space station. There is only one force acting on the person. for a person to feel as if they were on Earth, that force must be equal to mg. Now use Newton's second law, so you get mg = ma.
    For the acceleration, use the formula for centripetal acceleration. You will then find the speed easily
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