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I dont understand how to find acceleration for the second part

  1. Feb 15, 2013 #1
    for the first two I used the equation a=v^2/r
    but for the second part I dont know what it is asking for or how to approach it. is there an equation that goes with it?
     

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  3. Feb 15, 2013 #2
    the 1st two are wrong when I use a=v/r it is also wrong, and I converted km/h to m/s
     
  4. Feb 15, 2013 #3

    tiny-tim

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    Hi Sneakatone! :smile:

    a = v2/r should be correct.

    How exactly did you get 74.3 and 197.7 ? :confused:
     
  5. Feb 15, 2013 #4

    HallsofIvy

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    [tex]a= v^2/r[/tex] will give the acceleration due to the motion around the circle. The actual force the pilot experiences will be those plus or minus the weight of the pilot.
     
  6. Feb 15, 2013 #5
    105.5^2/150=74.2 m/s
    172.22^2/150= 197.7 ms
     
  7. Feb 15, 2013 #6

    tiny-tim

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    erm :redface:

    it's not 150 ! :wink:
     
  8. Feb 15, 2013 #7
    the diameter is 300 so 1/2 of it is 150 , thats where I got it. and then I tried using the whole diameter ans that is not right
     
  9. Feb 15, 2013 #8

    tiny-tim

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    no, the diameter is 1000 :wink:
     
  10. Feb 15, 2013 #9
    well sonb , I cant believe I didnt see that thanks!
    now that we have that I still dont know how to do the second part.
     
  11. Feb 15, 2013 #10

    tiny-tim

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    you mean the centripetal acceleration at the bottom?

    same method :smile:
     
  12. Feb 15, 2013 #11
    would i use gravity and the magnitude?
     
  13. Feb 15, 2013 #12

    tiny-tim

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    do you mean the third part? :confused:

    yes, you have to take gravity into account
     
  14. Feb 15, 2013 #13
    Im thinking it would be something like 22.1-(1/2*9.82) because gravity is downward on a horizontal acceleration.?
     
  15. Feb 15, 2013 #14

    tiny-tim

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    you're doing it again!

    where do those figures come from? :cry:
     
  16. Feb 15, 2013 #15
    the way im thinkg of it is that the found acceleration which is 22.1 is being acted upon a down ward acceleration of gravity.

    how would you do this?
     
  17. Feb 15, 2013 #16

    tiny-tim

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    (i don't make it exactly 22.1)

    where did the 1/2 come from? :confused:

    has your professor taught you about rotating frames of reference, and centrifugal force? what do you know about them?

    (i'm off to bed now :zzz:)
     
  18. Feb 15, 2013 #17
    I believe not , i dont even know which equation I should appy
     
  19. Feb 15, 2013 #18
    I belive not , I ont even know which equaton to apply.
    it seems like w^2r is a possibility.
     
  20. Feb 16, 2013 #19

    tiny-tim

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    (just got up :zzz:)

    has your professor taught you about rotating frames of reference, and centrifugal force? what do you know about them?

    2r and v2/r are the same, since v = ωr, even when ω isn't constant)
     
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