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Simple circular motion conceptual question

  1. Feb 17, 2015 #1

    RJLiberator

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    1. The problem statement, all variables and given/known data

    If the speed was doubled, what would the radius and period be to keep the same acceleration?

    2. Relevant equations

    a = v^2/R
    T = 2piR/t


    3. The attempt at a solution

    So with my understanding it is as follows:
    (2v)^2/R
    This makes it 4v^2/x = a
    since a=v^2/R

    Setting this into a proportion
    we get
    4v^2*R=v^2*x
    so 4R=x

    Radius must be 4 times larger to support a doubling in speed.

    T = 8piR/t

    Correct?
     
  2. jcsd
  3. Feb 17, 2015 #2
    Looks like it's correct to me.
     
  4. Feb 17, 2015 #3

    RJLiberator

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    Yahoooo!!! I did extremely well on this physics exam, many thanks to people like you on this forum.

    Cheers.
     
  5. Feb 18, 2015 #4

    haruspex

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    It depends what speed is referred to here. Is it the linear speed at the periphery or the angular speed? Maybe you could post the whole question, word for word.
     
  6. Feb 18, 2015 #5

    RJLiberator

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    Unfortunately, I don't have the question as it was an exam question.

    However, the "car" was going at constant speed on the outside (atan=0?) and the speed was doubled according to arad from my memory.
     
  7. Feb 18, 2015 #6

    BvU

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    You also have a T = 2piR/t
    and you think it changes to T = 8piR/t

    But forgot to mention what it T is or what t is. Dimensionally, this looks suspicious.
     
  8. Feb 18, 2015 #7

    RJLiberator

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    Hm.
    T = period.
    period = 2pir/time right?

    When the r becomes 4R it becomes 8pi*R/t
    the only part i am unsure about there is the denominator, t for time, if it is needed.
     
  9. Feb 18, 2015 #8

    NascentOxygen

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    The calculation of period will involve speed.
     
  10. Feb 18, 2015 #9

    RJLiberator

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    Yes, wouldn't the units be 2piR = meters and t = seconds
    so m/s ?
     
  11. Feb 18, 2015 #10

    NascentOxygen

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    Period is measured in seconds.
     
  12. Feb 18, 2015 #11

    RJLiberator

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    Ah.... I see my error, I was under the impression that the definition of period was T = 2piR/t however, that is incorrect. The definition is actually
    V = 2piR/T which becomes 2piR/V = T and in my case it would be 8piR/2v which simplifies to 4piR/v and the units becomes m/m/s which is just seconds.

    Ah, that is unfortunate, I won't score 100% because i had the wrong definition in my head :p.

    Thank you.
     
  13. Feb 18, 2015 #12

    BvU

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    So if the acceleration stays the same, and the speed while making a turn doubles, the radius has to become four times as big. And the time needed to complete the turn doubles.
     
  14. Feb 18, 2015 #13

    RJLiberator

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    Indeed. Thank you for that understanding. :)
     
  15. Feb 19, 2015 #14
    trying to help here but it seems most questions I know are quickly answered, from where I'm standing the only algebra was sqauring your two and moving your centripittal acceleration equation around, then you just plugged those values in for T, if you happen to confuse T equation and did everything else right I doubt your teacher will dock you too much
     
  16. Feb 19, 2015 #15

    RJLiberator

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    Thanks Jediknight. Yeah, I'm not too bothered, I'm very happy with how the exam went :D
     
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