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Dynamics relative acceleration question

  1. Apr 4, 2015 #1
    1. The problem statement, all variables and given/known data
    Aircraft A is flying NorthWest at 800km/h. Aircraft B is flying in a circular, anticlockwise holding pattern, radius of 5km, at 400km/h.


    Show that the acceleration of A relative to B is

    A(A/B)= 2.469cosθBi+2.469sinθBj m/s2

    2. Relevant equations


    3. The attempt at a solution
    I know that aA/B = aA - ab and that a=v2/r

    I changed 400km/h into 2.47m/s (multiplied by 1000 and divided by 360) and used it with the 5000m radius to get an acceleration of 2.47m/s but I’m not sure if this is correct or not? I don’t know how to get the acceleration of A.

    I know that aA = |aA|cosθAi + |aA|sinθAj and I know the same equation applies for aB and for aA/B

    If anyone can tell me if the acceleration I have worked out is correct or not that would be great and also how to work out the acceleration of A that would be very helpful, thank you:)
     
  2. jcsd
  3. Apr 4, 2015 #2

    Matternot

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    2.469 is the correct answer as provided in the question. Make sure you don't have a rounding error with your 2.47.

    What you say you "used it with the 5000m radius", how did you use it.

    I presume you mean divided by 3600 rather than 360
     
  4. Apr 4, 2015 #3
    Thank you for responding! Yeah I meant I divided by 3600 rather than 360. What happens to the acceleration of A? As it is the acceleration of A relative to B so I thought I would need to use the equation AA/B = Aa - Ab?
     
  5. Apr 4, 2015 #4

    Matternot

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    But A is not accelerating?
     
  6. Apr 4, 2015 #5

    SteamKing

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    A speed of 400 km / hr is not equivalent to 2.47 m/s. When you convert 400 km / hr to the correct speed in m/s and use it with a radius of 5000 m, you get an acceleration of 2.47 m/s2 [note the units].
     
  7. Apr 4, 2015 #6
    The m/s was a typing error, I meant m/s2 :) As A is not accelerating then using the equation aA/B = aa - ab = 0cos135i + 0sin135j -2.469cosθBi-2.469sinθBj
    = 0 + 0 -2.469cosθBi-2.469sinθBj
    =-2.469cosθBi-2.469sinθBj m/s2
    Which is negative of the answer I'm trying to get? Is the acceleration negative due to B flying in an anticlockwise direction?
     
  8. Apr 4, 2015 #7

    Matternot

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    why is ab necessarily 2.469cosθi+2.469sinθj ? Draw a diagram of B and take a look at the direction the acceleration is pointing.
     
  9. Apr 4, 2015 #8
    Snapshot.jpg The acceleration is going anticlockwise, making it negative and so ab = -2.469cosθi - 2.469sinθj ?
     
  10. Apr 4, 2015 #9

    Matternot

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    You've drawn on the path it takes. Draw on the acceleration.
     
  11. Apr 4, 2015 #10
    Snapshot2.jpg Is that what you mean? If not, then I'm not really sure what else it could be
     
  12. Apr 4, 2015 #11

    Matternot

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    Aha. There's the hole in the understanding!

    If something is moving in circular motion (e.g. I'm whizzing round a ball on a string) In which direction do you think the acceleration will be in? (Think about the forces on the ball)
     
  13. Apr 4, 2015 #12

    Matternot

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    Or think about the forces on the Earth orbiting the Sun
     
  14. Apr 4, 2015 #13
    The earth accelerates towards the sun which results in a centripetal force on the earth?
     
  15. Apr 4, 2015 #14

    Matternot

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    Exactly! for circular motion, the acceleration is always towards the centre of rotation.

    So now you can redraw the diagram with acceleration. Think about how you could parametrise the acceleration in terms of θ
     
  16. Apr 5, 2015 #15
    Snapshot3.jpg Is this the diagram you meant? Because the acceleration of B is into the centre of rotation, does this make the acceleration negative and so when I use the equation aA/B = aa - ab this gives me the correct answer?
     
  17. Apr 5, 2015 #16

    Matternot

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    So if you wanted to write the position vector of B in terms of r and θ, how would you do this?
     
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