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Rotating space station

  1. Oct 14, 2004 #1
    Another problem (oy!):

    ------A projected space station consists of a circular tube that is set rotating about its center (like a tubular bicycle tire). The circle formed by the tube has a diameter of about D = 1.02km. What must be the rotation speed (in revolutions per day) if an effect equal to gravity at the surface of the Earth (1g) is to be felt? Do not enter units. ---------

    Ok, so first I found the velocity of the station by doing v= sqrt r x g . I got 70.69 m/s. And then I used the formula T= 2PiR/V ....and I got 45.33 rps....or at least I think it's revolutions per second. It asks for revolutions per day, so I multiply 3600s (per hour) by 24 hours (in a day) and get 86400...so I multiplied that by 45.33 to get 3.9e6....but apparently I'm wrong. If anyone could be so kind as to enlighten me as to what I'm doing wrong I'd greatly appreciate it. Thanks! (again).
  2. jcsd
  3. Oct 14, 2004 #2
    Looks to me like the problem lies in the units again.

    or at least I think it's revolutions per second. --> don't be too sure.
  4. Oct 14, 2004 #3
    lol i had a similar problem last time and never figured it out. i can't seem to figure that one out...you have meters on the top and m/s on the bottom...so wouldn't the answer be in seconds?
  5. Oct 14, 2004 #4
    Yes, the period would be measured in seconds. However, it is actually in radians per second (2pi radians being 1 revolution). I believe (and i'm not sure as i only did this 3 weeks ago in class and am not an expert) that this is because of how it is defined. It all comes from the rotational motion formula s = r*theta - which is a radian measure. So as you go through it all, the final answer will be radians per second. Anyone care to expand on this?
  6. Oct 14, 2004 #5
    ohh...that would change my answer. lol. so i can't use that equation then? or if i could, how would i convert?
  7. Oct 14, 2004 #6
    1 revolution = 2pi radians
  8. Oct 14, 2004 #7
    wait so the r in 2piR is radians? in my physics book it says it's the radius...sorry if i sound stupid...they just didn't teach this to us, so i'm having difficulty understanding
  9. Oct 14, 2004 #8
    I understand your confusion (heck im confused too), but lets see how i can explain this. R is the radius yes. However, the answer is a radian measure (not revolution). Actually, I'm not sure i quite understand this either. Is there anyone that can clarify? Because in my textbook they say the period is a "radian measure" but what do they mean by that?
  10. Oct 14, 2004 #9


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    Radian is the distance in a unit circle. An unit circle has a radius 1 so therefore it has a 2pi circunference, you measure around this circle a distance x. Of course we define radian as angle, but a radian is exactly the same as the distance in a unit circle.

    Also Radian is also considered an artificial unit, you can see this because

    Arc Distance

    [tex] s = r\theta [/tex]

    s is in meters and r is in meters too, so when you divide to get theta you get 1, rather than an unit.
    Last edited: Oct 14, 2004
  11. Oct 14, 2004 #10
    alright...but how would i use that in this problem? it seems that if i divided my answer by the angular speed i could get the answer in seconds...but there aren't any angles involved in this problem...that's a few chapters ahead of where we are. there must be something simpler....but what...
  12. Oct 14, 2004 #11


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    Show me the [tex] v = \sqrt{rg} [/tex] calculation
  13. Oct 14, 2004 #12
    v= sqrt rg....sqrt (510m x 9.8 m/s^2) = 70.69 m/s
  14. Oct 14, 2004 #13


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    Check the complete work again, i'll say.
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