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Inertial reference frame

  1. Sep 12, 2007 #1
    I am standing (yes, the question actually goes like this!) on a level floor at the origin of an inertial frame S and kick a frictionless puck due north across the floor.

    a.) Write down the x and y coordinates of the puck as functions of time as seen from my inertial frame. (use x and y axes pointing east and north respectively). now consider two more observers, the first at rest in a frame S' that travels with constant velocity v due east releative to S, the second at rest in a frame S'' that travels with constant acceleration due east relative to S. (all three frames coincide at the moment i kick the puck and S'' is at rest relative to S at that same moment.

    b.) find the coordinates x',y' of the puck and describe the puck's path as seen from S'.

    C.) do the same for S''.
    Which of the frames is inertial?



    Ok, well I know the puck is obviously not travelling east at all, so the x coordinate will stay at 0 from 'my reference frame'. I'm not sure where to start for a formula for the y-coordinate from his reference frame for a.)

    For b.) Since S' is not accelerating, the puck will appear to just move west.

    c.) S'' which is not inertial, will make S appear to move south west? How do i start finding the coordinates?

    thanks!
     
  2. jcsd
  3. Sep 12, 2007 #2

    learningphysics

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    For a, remember the puck is frictionless... no force acting on it horizontally... so no accleration...

    b) careful... remember the puck is moving north in S's frame of reference.

    c) how do you get south west?

    One thing that the question doesn't state... the velocity of frame S'' with respect to frame S when the 3 frames coincide... that will affect the answer to part c). Did they give this velocity in the question? Are we to assume that the velocity of S'' is 0 when the 3 frames coincide?

    The position of the puck seen by an observer is the position of the puck - position of the observer... (measuring all the coordinates in the S frame)
     
  4. Sep 12, 2007 #3
    How does this look for a.)?
    x(t) = y + vt+ (F/2m)t^2
     
  5. Sep 12, 2007 #4
    and no velocity is given at all...
    I'm not sure how to get ACTUAL coordinates without the values...
     
    Last edited: Sep 12, 2007
  6. Sep 12, 2007 #5

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    I don't think they want you to consider the time spent hitting the puck which is negligible... I think they only care once the puck has left your foot...

    You know x(t)=0... since it's only going north. Once the puck has left your foot... assume the puck has velocity vp... what is y(t)

    Do they give information like the mass of the puck?
     
  7. Sep 12, 2007 #6
    no mass is given either..

    So, its y-coordinate will just be vxt
     
  8. Sep 12, 2007 #7

    learningphysics

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    Yeah, y(t) = vp*t (using a different variable from v, to not get it mixed up with the frame S' velocity v)
     
  9. Sep 12, 2007 #8
    introducing momentum?
     
  10. Sep 12, 2007 #9

    learningphysics

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    no sorry... by vp I just meant velocity of the puck.
     
  11. Sep 12, 2007 #10
    ok ok i get it!
     
  12. Sep 12, 2007 #11
    do you think they want actual values for the coordinates?
     
  13. Sep 12, 2007 #12

    learningphysics

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    No, I suspect they just want expressions.
     
  14. Sep 12, 2007 #13
    and the expression will need to express the sort of parabolic motion the puck will appear to make?
     
  15. Sep 12, 2007 #14

    learningphysics

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    yes... get the x and y coordinates in terms of time... in the 3 frames...

    then you can then express y in terms of x... that equation will give you the path... for example:

    if x = t and y = t^2... you can substitute t from the first equation into the second... giving

    y = x^2.... that means that the object follows a parabolic path.
     
  16. Sep 12, 2007 #15

    learningphysics

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    Find the coordinates of the two other observers... as seen by you.

    Then the coordinates of the puck - coordinates of the observer... that's the coordinates that particular observer sees.
     
  17. Sep 12, 2007 #16
    maybe I'm missing something , but how do you get coordinates of frames; where all the info about such frames is that they are moving in a certain direction, either constant or accelerating?
     
  18. Sep 12, 2007 #17

    learningphysics

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    Yeah, just use variables (they gave v as the velocity of S' relative to S)... for example the coordinates of frame S' relative to S are...

    x(t) = v*t
    y(t) = 0

    For the coordinates of S'', just use a for acceleration... maybe v0 for initial velocity...
     
  19. Sep 12, 2007 #18
    just thinking about it in my head, from S', wont S appear to move both in the y and x
    directions? instead of just x?
     
  20. Sep 12, 2007 #19

    learningphysics

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    why? S' is just moving east. so S will appear to move west... but the puck will appear to move in both directions...
     
  21. Sep 12, 2007 #20
    I was thinking the puck. My bad.
     
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