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Circular Motion w/ gravity

  1. Nov 7, 2007 #1
    1. The problem statement, all variables and given/known data
    A train has a speed of v = 66.5 km/h. If the acceleration experienced by the passengers is to be less than 0.35 g, (g = 9.81 m/s2), find the smallest radius of curvature R acceptable for the track.

    2. Relevant equations
    A=v^2/r
    a=(mv^2)/r



    3. The attempt at a solution
    I tried putting 10.16 m/s^2 (added 9.81 + .35) for the acceleration and 18.4722 m/s for the velocity and putting it in the first equation to get 33.5848 m, however, this answer is close but not correct... I think the problem is, I don't understand what to do with the gravity? Is their another equation I am suppost to use or what? or am I only suppost to use .35 m/s/s as my acceleration? Please help, I cannot find any other simular examples in the text; the text only has examples involving tension and mass.
     
  2. jcsd
  3. Nov 7, 2007 #2

    D H

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    Staff Emeritus
    Science Advisor

    That 0.35 is a unitless scale factor. You can't add it to 9.81 m/s^2. The problem says "0.35 g" which means "0.35 * g", not "0.35 + g". Multiply!

    Carrying out this multiplication will give you the limit on the tolerable centripetal acceleration.
     
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