Physics Question pt. 2

Question

I'm really struggling with this problem. If anyone could help me out with these it would be greatly appreciated:

2.) An airplane is flying in a horizontal circle at a speed of 450 km/h (see the figure). If its wings are tilted at angle θ = 39° 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. Use g = 9.8 m/s2.

Thanks

Answer 1

L = aerodynamic lift

 

L is perpendicular to the wing, which is tilted 39 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 perpendicular to wing, wing at angle θ with respect to horizontal, then:

 

Vertical component of lift is:

 

L cosθ = W = mg ⇒ m = L cosθ / g

 

Horizontal component of lift is:

 

L sinθ = centripetal force = mv² / r

 

Combining these two equations,

 

L sinθ = (L cosθ / g)(v² / r)

 

Note L cancels out on both sides.

 

tanθ = v² / rg

 

r = v² / (g tanθ)

 

v = 450 km/hr = 450 x 10³ m / (3600 s) = 125 m/s

 

r = (125 m/s)² / [(9.8 m/s²) tan(39)] = 1968.9 m