Banking turn is a basic motion of an airplane used to change the direction the airplane is headed
(a) The time needed to reverse is approximately 54.65 s
(b) An additional force the passengers experience is the lift force which is is approximately 12.45 times the mass of the airplane
The reason the above values are arrived at are as follows:
The given parameter are;
Velocity of the airplane, v = 480 km/h = 400/3 m/s
Path of motion of the airplane = Semicircle, to reverse course
Banking angle of the wings, θ = 38°
Question hint as obtained online; Take the lift force, L due to aerodynamics as acting perpendicular to the airplane
(a) The time required to reverse course is found as follows;
Let L represent the lift force acting perpendicular tot he wings, we have;
L·cos(θ) = m·g
[tex]L \cdot sin(\theta) = \dfrac{m \cdot v^2}{r}[/tex]
Where;
r = The radius of the circular path as the airplane turns
m = The mass of the airplane
Therefore;
[tex]tan(\theta) = \dfrac{L \cdot sin(\theta) }{L \cdot cos(\theta)} = \dfrac{\dfrac{m \cdot v^2}{r} }{m \cdot g} = \dfrac{v^2}{r \cdot g}[/tex]
[tex]r = \dfrac{v^2}{g \cdot tan(\theta) }[/tex]
Which gives;
[tex]r = \dfrac{\left(\dfrac{400}{3} \right)^2}{9.81 \times tan(38^{\circ}) } \approx 2,319.5[/tex]
The radius of the reverse path, r ≈ 2,319.5 m
[tex]Time, \ t = \dfrac{Distance}{Speed}[/tex]
The distance of the reverse path = The length of the semicircular path = π·r
[tex]t = \mathbf{\dfrac{2,319.5 \times \pi}{\dfrac{400}{3} }} \approx 54.65[/tex]
The time needed to reverse, t ≈ 54.65 s
(b) The additional force experienced during the turn is the lift force, L, which is given as follows;
L·cos(θ) = m·g
[tex]L \approx \dfrac{m \times 9.81}{cos(38^{\circ})} \approx 12.45 \cdot m[/tex]
The additional force, L ≈ 12.45·m
Learn more about circular motion of an airplane here:
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