91Ó°ÊÓ

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

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,

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

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

The relation of horizontal forces is given by,

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

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

On dividing the above two relations.

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

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