An airplane is flying in a horizontal circle at a speed of 480 km/h. If its wings are tilted 40° 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.

r = 2161.9 m

Explanation:

Aerodynamic lift(L) is perpendicular to the wing, which is tilted 40 degrees to the horizontal.

Since the plane is moving in a horizontal circle, the vertical component of the lift must cancel the weight W of the airplane, but the horizontal component is the centripetal force that keeps it in a circle.

L is perpendicular to wing at angle θ with respect to horizontal

Thus,

Vertical component of lift is:

L cosθ = W = mg

Thus, m = L cosθ / g - - - - (eq1)

Horizontal component of lift is:

L sinθ = centripetal force = mv² / r - - - - (eq2)

Combining equations 1 and 2,we have;

L sinθ = (L cosθ / g)(v² / r)

L cancels out on both sides to give;

tanθ = v²/ rg

r = v² / (g tanθ)

We are given;

velocity; v = 480 km/hr = 480 x 10/36 = 133.33 m/s

r = 133.33²/[(9.8) tan(40)] = 2161.9 m

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