Question

Why do airplanes bank when they turn? How would you compute the banking angle given the airspeed and radius of the turn? [Hint: Assume an aerodynamic "lift" force acts perpendicular to the wings. See also Example 5 -7.]

Short answer

The Airplanes bank when they turn, because in order to turn, they must be a force that will be exerted towards the center of a circle. The banking angle is equal to $\theta ={\tan }^{-1}\left(\frac{v^2}{rg}\right)$.

Step-by-step solution

Step 1. Understanding the force exerting on an airplane in level flight

While traveling in level flight, the gravitational force on the airplane is directed downwards and this force is opposed by the lift force.

Step 2. Explaining the banking angle

The lift force acting on the airplane is perpendicular to the wings of the airplane. As the airplane banks, the lift force vector will be having vertical as well as horizontal components. The weight of the airplane is balanced by the vertical component, whereas the horizontal lift force provides the centripetal force. Hence, when the airplane turns they banks because the centripetal is exerted on the it.

The following is the free body diagram of the airplane.

The relation of vertical forces is given by,

$\begin{array}{c}{F}_{\mathrm{y}}=0\\ F_{\mathrm{L}}\cos \theta -mg=0\\ F_{\mathrm{L}}\cos \theta =mg\end{array}$

Here, $F_{\mathrm{L}}$is the lift force, $\theta$is the banking angle, m is the mass and g is the gravitational acceleration.

Step 3. Evaluating the banking angle of the airplane

The relation of horizontal forces is given by,

$\begin{array}{c}{F}_{\mathrm{x}}=ma\\ F_{\mathrm{L}}\sin \theta =\frac{m{v}^2}{r}\end{array}$

Here, a is the acceleration of the airplane, v is the velocity and r is the radius.

On dividing the above two relations.

$\begin{array}{c}\frac{F_{\mathrm{L}}\sin \theta }{F_{\mathrm{L}}\cos \theta }=\left(\frac{\frac{m{v}^2}{r}}{mg}\right)\\ \tan \theta =\frac{v^2}{rg}\\ \theta ={\tan }^{-1}\left(\frac{v^2}{rg}\right)\end{array}$

Thus, $\theta ={\tan }^{-1}\left(\frac{v^2}{rg}\right)$is the required angle.