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Magnitude of avg acc. of train moving in a circle?

  1. Mar 26, 2016 #1
    1. The problem statement, all variables and given/known data
    "Assume that a train is traveling at 65 cm/s, and the track has radius 20cm. What is the magnitude of the train's average acceleration as the train goes from the tree marked A to the next tree B, located exactly 1/4 of the way around the track from tree A? [Hint: This question is not asking for the train's centripetal acceleration.]"

    2. Relevant equations
    Equations of motion of object moving in circle, using ω,∝, θ.
    aavg=(vf-vi)/Δt


    3. The attempt at a solution
    The answer is given to be 190 cm/s2 but why wouldn't it be zero? Maybe I'm misunderstanding the question, but if it doesn't want centripetal acceleration, it must mean tangential acceleration. And I thought that only the direction of tangential velocity changes, not magnitude. I also included a picture. 20160326_130018.jpg
     
  2. jcsd
  3. Mar 26, 2016 #2

    gneill

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    Staff: Mentor

    Consider that velocity is a vector. Make a sketch showing the two velocity vectors of interest.
     
  4. Mar 26, 2016 #3

    mfb

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    2016 Award

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    If the answer would be zero, the average acceleration would be zero, so initial and final velocity would be the same.

    The answer is not the centripetal acceleration directly (and you don't have to calculate it), but this acceleration does contribute to the answer.
     
  5. Mar 26, 2016 #4
    Would vi=<0,65> and vf=<65,0>?
     

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  6. Mar 26, 2016 #5

    gneill

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    Yes, those are two candidates for vi and vf (the particular vectors depend upon what you choose as the orientation of the track in your frame of reference). They will work for determining the average acceleration over the interval.
     
  7. Mar 27, 2016 #6
    So if I subtract vi from vf, I get <65,-65>. To find time I used the formula acentripetal=v2/r=ω2r, and solved for ω by plugging in 65 cm/s as v and 20 cm as r and found ω to be 3.25rad/s. I converted this to rev/s and multiplied by 1/.25rev (since 1/4 the way around track) and got time to be 2.07s. Did I go wrong somewhere?
     
  8. Mar 27, 2016 #7

    gneill

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    Your time is incorrect. You really went the long way around to try to calculate it.

    Your angular velocity ω looks good. What is the angular distance of a quarter circle? Distance over velocity yields time...
     
  9. Mar 27, 2016 #8
    Ah, θ=90°=1.57rad. t=Δθ/ω. t=.48s. Then √(652+652) divided by .48s gives the 190cm/s2. Thank you!
     
  10. Mar 28, 2016 #9

    mfb

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    Note that you do not need ω. You can calculate the length of 1/4 of a circle with radius 20 cm, and divide that by the speed of the train. It gives the same result in a more direct way.
     
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