Posted by **chelsea** on Wednesday, May 27, 2009 at 9:39pm.

an airplane is flying in a horizontal circle at a speed of 430 km/h if its 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.

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**drwls**, Wednesday, May 27, 2009 at 11:10pm
The horizontal component of the lift force equals the centripetal force and the vertical component equals the weight, M g.

C*A (1/2)(rho)V^2 sin 40 = M V^2/R

C*A (1/2)(rho)V^2 cos 40 = M g

C is the lift coefficient and A is the wing area. "rho" is the air density.

Divide one equation by the other to eliminate a lot of unknowns.

tan 40 = V^2/(gR)= 0.839

R = 1.192 V^2/g

Be sure to convert V = 430 km/h to m/s before using the formula. g = 9.8 m/s^2

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