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A train rounding a Curve

  1. Feb 27, 2009 #1
    1. The problem statement, all variables and given/known data

    A train traveling at a constant speed rounds a curve of radius 218 m. A lamp suspended from the ceiling swings out to an angle of 16.6° throughout the curve. What is the speed of the train?


    2. Relevant equations

    mv^2/r


    3. The attempt at a solution

    I drew a diagram and attempted to calculate v by setting mv^2/r = -cos16.6 mg
    I don't really know how to approach this?
     
  2. jcsd
  3. Feb 27, 2009 #2

    LowlyPion

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    I think the angle is with the vertical, which makes the deflection in x given as Sin16.6 not Cos 16.6.

    V2/r = Sin16.6*g

    V = (r*sin16.6*g)1/2
     
  4. Feb 27, 2009 #3
    I tried and got 24.705 but that was wrong. Are there any other ways to approach the problem?
     
  5. Feb 27, 2009 #4

    LowlyPion

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    What units do they want the answer in? m/s or km/h?

    24.705 m/s = 88.9 km/h
     
  6. Feb 27, 2009 #5
    I tried both ways but neither options were correct. I think it is supposed to be in m/s and I think my answer is wrong in general, is there anything else I can do to get another answer?
     
  7. Feb 27, 2009 #6

    LowlyPion

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    Ooops. Sorry. I did a sketch and realized vertical is g and that means then that

    V2 = tan16.6*g*r
     
  8. Feb 27, 2009 #7
    Ok, I got it! Thank you so much.
     
  9. Feb 27, 2009 #8

    LowlyPion

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    It's important you understand why.

    Draw the acceleration vectors. The acceleration vectors add to some resultant a that forms the angle. The vertical component is g which means the Resultant acceleration on the lamp is given by

    g = ay = a*cosθ

    So a = g/cosθ

    For the x component, that means that ax = a*sinθ = g*sinθ/cosθ = g*tanθ

    and that is what equals the centripetal acceleration.
     
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