Question

An airplane is flying in a horizontal circle at a speed of 480 km/h (). If its wings are tilted at angle =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.

Answer 1

Answer:

$R = 2162 m$

Explanation:

When wings of the airplane makes an angle of 40 degree with the horizontal so here we can say that force due to air is having two components

$F_y = mg$

$F_x = \frac{mv^2}{R}$

now we know that

$F_y = F cos40$

$F_x = F sin40$

also we know that

$v = 480 km/h$

$v = 133.3 m/s$

now plug in all data in above equations

$tan 40 = \frac{v^2}{Rg}$

$R = \frac{v^2}{g tan40}$

$R = \frac{133.3^2}{9.8 tan40}$

$R = 2162 m$