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Uniform Circular Motion Problem

  1. Jan 31, 2010 #1
    1. The fast French train known as the TGV (Train à Grande Vitesse) has a scheduled average speed of 216 km/h. (a) If the train goes around a curve at that speed and the magnitude of the acceleration experienced by the passengers is to be limited to 0.050g, what is the smallest radius of curvature for the track that can be tolerated? (in km) (b) At what speed must the train go around a curve with a 1.08 km radius to be at the acceleration limit? (in km/h)

    2. a=v2/r and T=2pir/v

    3. I have attempted both part a and part b but I have no success in getting the correct answer. Here are m attempts
    a) a=0.05g
    r=v2/a
    r=2162/0.05= 933120


    b) r=1.08km a=0.05g
    v=sqrtar
    v=sqrt0.05x1.08= 0.232
     
  2. jcsd
  3. Jan 31, 2010 #2

    hage567

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    Your approach is fine, but you need to remember to make sure your units are consistent. Look carefully at them. You can't mix km/h, m, and s like you've done. Also, don't forget that g has a value, you can't just drop it out of your calculation.
     
  4. Jan 31, 2010 #3
    Well I really was confused about g. I couldn't find out a conversion factor for it anywhere. I don't know how to use it.
     
  5. Jan 31, 2010 #4

    hage567

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    I would just leave g as 9.81 m/s2, and convert all the other quantities so they are in meters and seconds. Then do the calculations, then convert the final answer back into km/h.

    But if you wanted to convert g to km/h2, you don't need to "find" a conversion factor. Just do it dimensionally using the same relations you would use to convert anything else (1 km = 1000 m, 1 hr = 3600 s, etc).
     
  6. Feb 1, 2010 #5
    Oh okay but I'm still confused on g. Would I multiply 0.05X9.81 to get the correct acceleration?
     
  7. Feb 1, 2010 #6
    1 G = 9.81 m/s^2

    so

    0.05 G = ?


    yes, u only need to multiply
     
  8. Feb 1, 2010 #7
    OKay thank you I got the right answer
     
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