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Circular motion(Normal-Tangential)

  1. Apr 24, 2008 #1
  2. jcsd
  3. Apr 24, 2008 #2

    nrqed

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    Consider the situation when the train is just about to "fly off" and leave the tracks. In that case, the normal force on the train is zero. This should alllow you to solve for rho.

    EDIT: Question b does not make sense to me...What do they mean by "bottom of a hill"? Maybe we are to assume that we have now a track curved upward with the same radius?
     
  4. Apr 24, 2008 #3
    ok i got

    A = N-g

    A = 0 - 9.8m/s^2

    so

    r = 106^2/-9.8 = -1148 m

    does that make sense for the first question?
     
  5. Apr 24, 2008 #4

    nrqed

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    Well a radius can't be negative.

    Be careful, you are using

    [tex] \sum F_y = m a_y [/tex]

    so your a_y should be [tex] - v^2/ r [/tex] not v^2/r.
    Other than that it looks fine
     
  6. Apr 24, 2008 #5
    how is Velocity negative when gravity is acting downward in the y direction?
     
  7. Apr 24, 2008 #6
    when you solve A-sub-n you get -9.8m/s^2 however A-sub-n in the FBD is acting upward. So how do you know which sign to use?
     
  8. Apr 24, 2008 #7

    nrqed

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    I never said that a velocity is negative. I am talking about the y component of the acceleration. The radial (or centripetal) acceleration a_rad always points to ward the center of the circle. In your example, the center of the circle is below the train. Therefore

    [tex] a_y = - a_{Rad} = - \frac{v^2}{R} [/tex]
     
  9. Apr 24, 2008 #8
    ok I see what your saying, than my FBD would be wrong? I have the A-rad pointing upward.

    In terms of Normal and Tangential accerlation, when I draw the FBD if the normal Accerleration is upward, does that mean the Normal Force is also upwards?
     
  10. Apr 24, 2008 #9

    nrqed

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    I am not sure what you mean by "normal acceleration". If we have circular motion, there is a radial accceleration and there is a tangential acceleration. The radial acceleration is toward the center of the circle. So downward in your diagram. The normal force is always perpendicualr and away from the surface so it's upward.
     
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