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Finding revolution time at surface

  1. Feb 3, 2015 #1
    1. The problem statement, all variables and given/known data
    For a 0.90km radius cylinder, find the time for one revolution if "gravity" at the surface is to be 9.8 m/s2.

    2. Relevant equations
    rω^2=a

    3. The attempt at a solution
    i tried solving for omega but i couldnt find a solution that i only had 1 variable in it.
     
  2. jcsd
  3. Feb 3, 2015 #2

    haruspex

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    Which other variable do you think you have no value for?
     
  4. Feb 3, 2015 #3
    i dont know the velocity for solving for omega, is this what you meant?
     
  5. Feb 3, 2015 #4

    gneill

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    ω is a velocity: an angular velocity. How is ω related to the frequency and the period of one "cycle" (one revolution in this case)?
     
  6. Feb 3, 2015 #5

    haruspex

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    I don't know where you are stuck. The first step is to calculate ω - can you do that from the formula you quoted? Next, as gneill says, you can find the time of one revolution from ω. Is that part your problem?
     
  7. Feb 4, 2015 #6
    i know that ω = v/r but i do not know v. I guess i am confused about what to do in order to solve for ω.
     
  8. Feb 4, 2015 #7

    haruspex

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    In the OP you quoted the right equation to be using. That has three variables and you are given the values of two of them. Forget v for now.
     
  9. Feb 4, 2015 #8
    wow can't believe i missed that. so then 450ω^2=9.8, ω=.133
    i converted .9 km to 900 m
     
    Last edited: Feb 4, 2015
  10. Feb 4, 2015 #9

    haruspex

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    Ok, but why did you halve it to 450? Either way, I don't get .133.
     
  11. Feb 4, 2015 #10
    the cylinder's diameter is 900m and i need the radius. 900/2=450. then i did 9.8/450=ω^2=.022, ω=.147
    (i had a math error in the previous post, this is what i actually got)
     
  12. Feb 4, 2015 #11

    haruspex

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    It says radius.
     
  13. Feb 4, 2015 #12
    my bad

    then i get .104
     
  14. Feb 4, 2015 #13

    haruspex

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    Sounds right.
     
  15. Feb 4, 2015 #14
    so after this do i just plug it into the equation? the answer i get seems high
     
  16. Feb 4, 2015 #15

    haruspex

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    You know what I'm going to say, surely? Please post your working and answer.
     
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