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The minimum radius of a plane's circular path

  1. Oct 4, 2009 #1
    1. The problem statement, all variables and given/known data
    An 82 kg pilot flying a stunt airplane pulls out of a dive at a constant speed of 540 km/h.
    a) What is the minimum radius of the plane's circular path if the pilot's acceleration at the lowest point is not to exceed 7.0g.

    2. Relevant equations
    Fg = mg
    Fnet = mv^2/r
    Fnet = ma

    3. The attempt at a solution
    At the lowest point I said that there were only 2 forces acting on the plane: Fg and I said Fn for the upward lift on the plane.

    m=82 kg
    v=540 km/h=150 m/s
    a=7.0g

    Fnet = mv^2/r=ma
    Fnet = Fn - Fg = ma
    Fnet = (82 kg)(7 x 9.8 N/kg) - (82 kg)(9.8 N/kg)
    = 4821.6 N = mv^2/r
    4821.6 N = [(82 kg)(150 m/s)^2]/r
    r=[(82 kg)(150 m/s)^2]/4821.6 N
    = 382.65 m
    = 3.8 x 10^2 m

    The answer is actually 3.3 x 10^2 m... The book is sometimes wrong though.

    What did I do wrong?

    Thanks.
     
  2. jcsd
  3. Oct 4, 2009 #2
    I'm not sure what you did (your calculations are a bit messy), but here are the two equations I derived:

    [tex]n= m(g+a_{y}) = 6435.36 N[/tex]

    [tex] r = \frac{mv^{2}}{n-mg} = \frac{(82)(150^{2})}{6435.36-(82)(9.81)} = 3.27 \cdot 10^{2} \, m[/tex]
     
  4. Oct 4, 2009 #3
    That's essentially what I did, but I thought that the net force was Fn - Fg... How come it is Fn + Fg?

    Attached is what I thought the FBD should look like.
     

    Attached Files:

  5. Oct 4, 2009 #4
    [tex] F_{net} = n - mg = ma \iff n = mg + ma = m(g+a)[/tex]
     
  6. Oct 4, 2009 #5
    What's n? I think we use different symbols where I take this course...
     
  7. Oct 4, 2009 #6
    n is the normal force of the pilot.
     
  8. Oct 4, 2009 #7
    Ok, that's what I thought... Thanks.
     
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