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Centripetal? Airplane problem

  1. Jun 1, 2004 #1
    In this problem I need to figure out how far I am from the airport.

    The pilot tells us that we have to circle the aiport before we can land. We will maintain a speed of 450mph at an altitude of 20,000ft. while traveling in a horizontal circle around the airport. I notice that the pilot banks the plane so that the wings are oriented at 10deg to the horizontal. An article in the in-flight magazine says that planes can fly because the air exerts a force "lift" on the wings which is perpendicular to the wing surface.

    V=660 ft/s
    h=20,000 ft
    bank = 10deg to the horizontal

    So I drew a FBD with my normal force perpendicular to my wings, which were banked to the horizontal.

    Sum Fx=Wx=ma where a=v^2/r
    sin10 (mg)=mv^2/r (mass drops out)
    sin10 (32.2ft/s^2)=(660ft/s)/r
    solving for r=118.03ft

    add that to the current altitude of 20,000ft gives a distance of 20,118ft from the airport?

    It doesn't seem like this problem could be this "easy". I feel like I am totally missing a point here.

  2. jcsd
  3. Jun 1, 2004 #2
    oops! I forgot to square my velocity. That being said, I have a new r=77,904ft.

    If you figure 77,904 as the x value of a right triangle, and 20,000 as the y value, then the hypotenuse or distance from the airport would be 80,431ft or approx 15miles. This sounds more reasonable?
  4. Jun 1, 2004 #3
    Why have you taken "mg" as your lift force?
    Take another look!
  5. Jun 1, 2004 #4
    It seems from what you wrote above, the force F that is giving the plane lift has magnitude of mg. This is clearly not true.
    Using F = ma in the vertical direction gives
    [tex]0 = Fcos(\theta) - mg \rightarrow F = mg/cos(\theta)[/tex]​
    where [tex]\theta[/tex] is the bank angle.
    Plug this F back into what you have above, solve for r, use the pythagorean theorem and you're done.

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