Question

An airplane is traveling at 250 m/s in level flight. If the airplane is to make a change in direction, it must travel is a horizontal curved path. To fly in the curved path, the pilot banks the airplane at an angle such that the lift has a horizontal component that provides the horizontal centripetal acceleration to move in a horizontal circular path. If the airplane is banked at an angle of 15.0 degrees, then the radius of curvature of the curved path of the airplane is

Answer 1

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:

$W$ - Weight of the airplane, measured in newtons.

$F$ - Lift, measured in newtons.

$\theta$ - Banking angle, measured in sexagesimal degrees.

The equations of equilibrium associated with the airplane are, respectively:

$\Sigma F_{r} = F\cdot \sin \theta = m\cdot \frac{v^{2}}{R}$ (Eq. 1)

$\Sigma F_{z} = F\cdot \cos \theta - W = 0$ (Eq. 2)

From (Eq. 2):

$F = \frac{W}{\cos \theta}$

In (Eq. 1):

$W\cdot \tan \theta = m\cdot \frac{v^{2}}{R}$

By using the definition of weight, we eliminate the mass of the airplane:

$g\cdot \tan \theta = \frac{v^{2}}{R}$

Where:

$g$ - Gravitational acceleration, measured in meters per square second.

$v$ - Speed, measured in meters per second.

$R$ - Radius of curvature, measured in meters.

Lastly, we clear the radius of curvature with the expression:

$R = \frac{v^{2}}{g\cdot \tan \theta}$

If we know that $v = 250\,\frac{m}{s}$, $g = 9.807\,\frac{m}{s^{2}}$ and $\theta = 15^{\circ}$, the radius of curvature is:

$R = \frac{\left(250\,\frac{m}{s} \right)^{2}}{\left(9.807\,\frac{m}{s^{2}} \right)\cdot \tan 15^{\circ}}$

$R = 23784.356\,m$

The radius of curvature of the curved path of the airplane is 23784.356 meters (23.784 kilometers).