Question

An airplane is flying in a horizontal circle at a speed of 364km/h. If the wings of the plane are tilted 38.6° 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

here we shall focus on the following assumption;

Vertical component of normal forces will balance the weight.

$\therefore$ $N*\cos\theta =mg$ ...eqn1

Horizontal component of normal forces will provide the necessary centripetal force

$N*\sin\theta=F\scriptscriptstyle L$ $=\tfrac{mv^2}{r}$ ..eqn2

dividing eqn2 by eqn 1,we get

$\tfrac{\sin \theta}{cos \theta} =tan\theta =\tfrac{v^2}{g*r}$

v=364km/hr=101.111m/s

g=9.81m/s², $\theta = 38.6\degree$

$r=\tfrac{V^2}{g*tan \theta}$ =$\tfrac{101.111^2}{9.81*tan(38.6 \degree)}$

=$\tfrac{10223.2321}{7.83122}$

=1305.445633

$\approx 1305.446 m$