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Uniform circular motion - airplane

  1. Oct 7, 2003 #1
    I have been having trouble getting key components of this problem:

    A plan is flying in a horizontal circle at a speed of 480 km/h. The wings are tilted 40 degrees to the horizontal. What is the radius of the circle in which the plane is flying? Assume that the required force is provided entirely by an "aerodynamic lift" that is perpendicular to the wing surface.

    okay, so I have:

    V = 480 km/h
    @ = 40 deg
    Find: R

    That required force would look much like the Normal force of a box on an incline plane, right?

    So, could i use:

    Fy = N - mgcos@ = ma ?

    with a = V^2/R

    but, I don't have a mass to use. So I'm unsure as how to approach this. Would I also need to find Fx?

    Fx = mgcos@ = ma, right?
  2. jcsd
  3. Oct 8, 2003 #2


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    You don't need to know the mass because you don't really have to find the force. You only need to know the acceleration toward the center of the circle. That together with the speed around the circle will give you the radius. Either calculate N/m and then cancel m later or just assume m=1.

    Well, I don't know. What are Fx and Fy? How do you have your coordinate system set up?
  4. Oct 8, 2003 #3
    +y is in the direction of N
    +x is along the wing of the plane that is up 40degrees from the horizontal

    Since the plane is flying in a horizontal direction, I was thinking I just had to use:

    Fx = mgsin@ = ma

    Then, I can see how the mass doesn't matter:

    gsin@ = a

    and since a = v^2/R:

    gsin@ = V^2/R


    R = V^2/gsin@


    V = 480 km/h = 133.3333 m/s
    @ = 40

    I got R = 2822.18 m = 2.8 km

    But, I checked my book and the answer is supposed to be 2.2 km.

    I was having trouble at first because I forgot to change the unit of measurement, but now .. I still don't get the right answer
  5. Oct 8, 2003 #4


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    Then there is no net force in either the x or y directions- the plane isn't moving thos directions!
    But you just said that Fx is NOT horizontal! There is a "lift" in the "y" direction (normal to the airplane): you need to break that into vertical and horizontal components. Calling the normal force N, I get that the vertical component is N cos(θ) and the horizontal component is N sin(θ). Since the airplane is going neither up nor down, the vertical component must of set gravity: N cos(&theta)= mg which tells us that N= mg/cos(θ). (I think that's the part you missed.)
    It is the horizontal component that causes the airplane to turn.
    From "f= ma", since the horizontal force, f, is N sin(θ), we must have N sin(&theta)= ma. (this is the part you have!)

    Since N= mg/cos(θ), ma= (mg/cos(θ))sin(θ) so
    a= g tan(θ)= v2/R. Solving for R,
    R= (v2/g)cot(θ).

    I see that you did correctly convert the speed to m/s (I almost missed that myself!). If you put v= 133.3 m/s, θ= 40 degrees and g= 9.8 m/s2, you should get the right answer.
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