Answer :
Answer:
The radius of curvature of the curved path of the airplane is 23784.356 meters (23.784 kilometers).
Explanation:
We assume that airplane can be represented as a particle. The free body diagram of the vehicle is presented below as attachment, whose variables are:
[tex]W[/tex] - Weight of the airplane, measured in newtons.
[tex]F[/tex] - Lift, measured in newtons.
[tex]\theta[/tex] - Banking angle, measured in sexagesimal degrees.
The equations of equilibrium associated with the airplane are, respectively:
[tex]\Sigma F_{r} = F\cdot \sin \theta = m\cdot \frac{v^{2}}{R}[/tex] (Eq. 1)
[tex]\Sigma F_{z} = F\cdot \cos \theta - W = 0[/tex] (Eq. 2)
From (Eq. 2):
[tex]F = \frac{W}{\cos \theta}[/tex]
In (Eq. 1):
[tex]W\cdot \tan \theta = m\cdot \frac{v^{2}}{R}[/tex]
By using the definition of weight, we eliminate the mass of the airplane:
[tex]g\cdot \tan \theta = \frac{v^{2}}{R}[/tex]
Where:
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]v[/tex] - Speed, measured in meters per second.
[tex]R[/tex] - Radius of curvature, measured in meters.
Lastly, we clear the radius of curvature with the expression:
[tex]R = \frac{v^{2}}{g\cdot \tan \theta}[/tex]
If we know that [tex]v = 250\,\frac{m}{s}[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]\theta = 15^{\circ}[/tex], the radius of curvature is:
[tex]R = \frac{\left(250\,\frac{m}{s} \right)^{2}}{\left(9.807\,\frac{m}{s^{2}} \right)\cdot \tan 15^{\circ}}[/tex]
[tex]R = 23784.356\,m[/tex]
The radius of curvature of the curved path of the airplane is 23784.356 meters (23.784 kilometers).
