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Homework Help: Centripetal acceleration and airplane lift

  1. Oct 3, 2009 #1
    1. The problem statement, all variables and given/known data

    An airplane is flying in a horizontal circle at a speed of 460 km/h (Fig. 6-42). If its wings are tilted at angle θ = 37° 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.


    2. Relevant equations

    Newton's second law, for centripetal motion:

    [tex]F_{net} = m*\left( \frac{v^2}{r} \right)[/tex]

    3. The attempt at a solution

    First let us identify the forces acting on the plane. There are exactly 3: The force of gravity, [tex]F_g[/tex], the force of the lift, causing it to fly, [tex]F_L[/tex], and the centripetal force caused by the rotation, [tex] m*\left( \frac{v^2}{r} \right)[/tex]

    We know that there must be some lift on the plane, keeping it in air. Because there is no vertical motion, we know that [tex]F_L - F_g*\cos(\theta) = 0[/tex]. So [tex]F_L = F_g*\cos(\theta)[/tex]

    Next, we also know that the plane must be pulled in by the centripetal force. So the horizontal component of the lift must be caused by this force, and we have

    [tex]m*\left( \frac{v^2}{r} \right) = F_L_x = F_L*\sin(\theta) = F_g*\cos(\theta)*\sin(\theta) [/tex]

    Solving for the radius r

    [tex]\frac{m*v^2}{F_g*\cos(\theta)*\sin(\theta)} = \frac{v^2}{g*\cos(\theta)*\sin(\theta)} = r[/tex]

    Subbing the values

    [tex]\frac{460^2}{9.8*\cos(37)*\sin(37)} = 44923 = r[/tex]

    The computer system for my homework has me entering this as meters instead of kilometers, which is the units of the shown result. However, it's not lining up. I suspect that there might be something wrong with my lifting force, but I'm not sure.
  2. jcsd
  3. Oct 3, 2009 #2


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    Homework Helper

    I tried it, interesting problem. My approach was to draw the wing going like the "/" character, Fg down, Fc to the left toward the center and FL up and to the left 53 degrees above horizontal. I reasoned that the vertical component of FL must cancel Fg.
    So I got FL*sin(53) = Fg.

    I ended up with a much smaller radius than you did.
  4. Oct 4, 2009 #3

    This is the free body diagram I used for the problem. The blue line represents the airplane. The solid green line pointing downward represents the force of gravity, and the dashed green line is the perpendicular component of that force. The solid green line is the force of lift on the plane, and the dashed green line is the horizontal component of the lift force.

    Oh yea, I tried your answer.

    [tex]F_L*\sin(90 - \theta) = F_L*\cos(\theta) = F_g[/tex]


    [tex]\frac{F_g}{\cos(\theta)} = F_L[/tex]

    Plugging this into the centripetal motion equation

    [tex]m*\left( \frac{v^2}{r} \right) = F_L_x = F_L*\sin(\theta) = \frac{F_g}{\cos(\theta)}*\sin(\theta) = F_g*\tan(\theta) [/tex]

    Then for radius:

    [tex]\frac{m*v^2}{F_g*\tan(\theta)} = \frac{v^2}{g*\tan(\theta)} = r[/tex]

    [tex]\frac{460^2}{9.8*\tan(37)} = 28653 = r[/tex]

    Suffice to say, still not correct :(

    Attached Files:

  5. Oct 4, 2009 #4

    Doc Al

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    Staff: Mentor

    Careful with units. What's the speed in m/s?
  6. Oct 4, 2009 #5


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    Homework Helper

    Thanks, Doc!
    knight, I agree with all that right down to your last line. Using the velocity in m/s I've got R just over 2200 m.
  7. Oct 4, 2009 #6
    Okay, I see what I was doing wrong.. My units was wrong but my model was wrong also. I was treating [tex]F_L[/tex] as a normal force, therefore [tex]F_L - F_g*\cos(\theta) = 0[/tex]. But this is different. A lifting force counteracts gravity, so instead you'd have [tex]F_L_y - F_g = 0[/tex], which is why tangent turns up later..

    Thanks for the help, guys..
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