Question

An aircraft executes a horizontal loop at a speed of 720 km/h with its wings banked at 15°. What is the radius of the loop?

Answer 1

Data:
s (speed) = 720 km/h
g (gravity) ≈ 10 m/s²
$F_{c} = Centripetal\:Force$
$W = Weight$
√3 ≈ 1.7
"â" is the angle ($\alpha$) of inclination of the wings relative to the horizontal plane = 15º


$\overrightarrow{N}=\overrightarrow{ F_{s}}$ Is the holding force applied by the air and which is perpendicular to the wings - $\overrightarrow{W}$ is the weight.

$Tg\:\^{a} = \frac{opposite\:leg}{adjacent\:leg}$
$Tg\:\alpha = \frac{ F_{c} }{W}$
$tg\alpha= \frac{\diagup\!\!\!\!\!m* \frac{s^2}{R} }{\diagup\!\!\!\!\!m*g}$
$tg\alpha= \frac{\frac{s^2}{R} }{g}$
$tg\alpha= \frac{s^2}{R} * \frac{1}{g}$
multiply
$s^2 = R*g*tg \alpha$
$s = \sqrt{R*g*tg \alpha }$
$720 = \sqrt{R*10*tg15^0}$
$720 = \sqrt{R*10*2- \sqrt{3} }$
$(720)^2 = (\sqrt{R*10*2- \sqrt{3} })^2$
$518400 = R*10*2- \sqrt{3} }$
$518400 = 20R- \sqrt{3} }$
$518400 = 20R - 1.7$
$20R = 518400 + 1.7$
$20R = 518401.7$
$R = \frac{518401.7}{20}$
$\boxed{\boxed{R = 25920.085\:m}}\end{array}}\qquad\quad\checkmark$
For the sake of clarity, see the annex:

Answer 2

Explanation-

# Given-

v = 720 km/h = 200 m/s

θ = 15°

# Solution-

For circular motion with inclination θ, we have

tanθ = v^2 / rg

Putting values,

tan15 = (200)^2 / (r×10)

r = 4000/0.268

r = 14925 m

r = 15 km

Radius of the loop is 15 km.