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Centripical Acceleration!

  1. Sep 3, 2009 #1
    The fast French train known as the TGV (Train a Grande Vitesse) has a scheduled average speed of 216 km/h. If the train goes around the a corner and the maximum acceleration experienced by the passengers is limited to 0.06 g, what is the smallest radius of curvature the track can have? 6.12Km

    b) What is the maximum speed the train can go around a curve of radius 0.9 km if train is not to exceed this acceleration limit?

    I dont know what I'm doing wrong for Part B. For part A i did the following:

    Vo/ Ac= radius

    216/ (.06*g*60)= 6.12 km

    now I am trying to find just the speed or Vo which should be like the above but just altered to find the speed:

    Vo= Ac*radius

    Vo= (.06*g*60)*.9= 31.752 Km/hr...but thats not right.

    I then tried to put it into a ratio: 216/6.12= Vo/.9 and solved for Vo which i got as 31.764...very close to my orignal answer.

    PLEASE HELP!
     
  2. jcsd
  3. Sep 3, 2009 #2
    I also tried to square root 31.752 but it was wrong.
     
  4. Sep 3, 2009 #3

    rl.bhat

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    Vo/ Ac= radius

    216/ (.06*g*60)= 6.12 km

    Can you explain this calculation? What is 60? what is the value of g you have taken?
     
  5. Sep 3, 2009 #4
    well the 60sec came from sec=> hours which is 3600sec= 1 hour. the equation is this:
    Ac= Vo^2 (squared)/R

    So all i did was take the square root of 3600=> 60 (just pretty much eliminating the square root

    the problem wants it in km/hr...not m/s
     
  6. Sep 3, 2009 #5

    rl.bhat

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    You have written Ac = Vo^2/R.
    So to calculate R, where is the equation of taking square root arises?
     
  7. Sep 3, 2009 #6
    i have no idea what you mean...

    I solved for A already...that answer i gave you is correct 6.12

    I am solving for B which involves solving for Velocity...not radius.

    the radius is .9Km
    the acceleration is .06g which is in m/s^2

    i think my conversions are wrong..i dont know what im doing wrong with my conversions
     
  8. Sep 3, 2009 #7
    its 6.12 Km..so i was able to convert correctly
    the velocity they want is Km/hr

    and the first problem A..the 216 velocity is in Km/hr

    so doing .06*g*60 is the conversion for acceleration in Km/hr
     
  9. Sep 3, 2009 #8

    rl.bhat

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    0.06g* (1/1000)/(1/3600)^2 = 0.06*g*3.6*3600
     
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