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Homework Help: Revolutions per sec (centripital acceleration)

  1. Oct 23, 2007 #1

    ~christina~

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    [SOLVED] Revolutions per sec (centripital acceleration)

    1. The problem statement, all variables and given/known data
    As their booster rockets seperate, space shuttle astronauts typically feel acclerations up to 3g where g= 9.80ms^2. In their training astronauts ride in a device in which they experience such accelrations as centripital accleration. Specifically, the astronaut is fastened securely at the end of a mechanical arm and that turns at constant speed in a horizontal circle.

    determine the rotation rate in revolutions per sec required to give a astronaut a centripital accelration of 3.00kg while in a circular motion with the radius of 9.45m


    2. Relevant equations
    [tex]a_c = v^2/r [/tex]

    T= 2pi*r/ v


    3. The attempt at a solution

    all I know is that the

    [tex]a_c[/tex] = 3.00kg
    and that

    r= 9.45m

    I'm not sure what I do with this though since wouldn't the accelration usually be in m/s^2 but in this case it is in kg...how can this be?

    well I also don't have the velocity either...

    I was thinking of substituting the centripital accelration equation into the period T equation to get rid of v however I do know that speed is constant but velocity magnitude doesn't change since speed doesn't change but the magnitude I know does..

    Can someone help me out with this...

    Thanks :smile:
     
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  3. Oct 23, 2007 #2

    Kurdt

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    I think the question is supposed to say "centripetal acceleration of 3g" not 3.00 kg. Just seems like a typo. Your initial thoughts are quite correct. You can eliminate v to work out what T will be.
     
  4. Oct 23, 2007 #3

    ~christina~

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    thanks Kurdt
     
    Last edited: Oct 23, 2007
  5. Oct 23, 2007 #4

    Kurdt

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    I don't think so. You will never get an acceleration reported in units of mass unless there is a typo or mistake. :smile:
     
  6. Oct 23, 2007 #5

    ~christina~

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    wait so how can it be in g then ??

    you did say that it was in g not kg but now you say it can't be in units of mass...I'm confused..
     
  7. Oct 23, 2007 #6

    Kurdt

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    g in this context is the acceleration due to gravity. So when somebody says there was an acceleration of 3g, they mean that the acceleration is 3 times the magnitude of the acceleration due to gravity. g was given in your question as g=9.80ms-2.
     
  8. Oct 23, 2007 #7

    ~christina~

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    Oh I get it now... I thought it meant grams..
    Thanks Kurdt :smile:
     
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