An airplane is flying in a horizontal circle at a speed of $\mathbf{480}\mathbf{}\mathbf{km}\mathbf{/}\mathbf{h}$(Fig. 6-41). If its wings are tilted at angle$\mathbf{\theta }\mathbf{=}\mathbf{}\mathbf{40}\mathbf{°}$to the horizontal, what is the radius of the circle in which the plane is flying? Assume that the required force is provided entirely by an “aerodynamic lift” that is perpendicular to the wing surface.

The passenger of mass$m$ is riding around a horizontal circle of radius$R$at speed$v$.

The problem deals with Newton’s laws of motion which describe the relations between the forces F, acting on a body and the motion of the body. Also, it deals with centripetal acceleration. This is the acceleration, a of a body traversing a circular path of radius R.

Formula:

According to Newton’s laws of motion,

$\mathrm{Fsin\theta }=\frac{{\mathrm{mv}}^2}{\mathrm{R}}$

$\mathrm{Fcos\theta }=\mathrm{mg}$

Centripetal acceleration is given by,

$\mathrm{a}=\frac{{\mathrm{v}}^2}{\mathrm{R}}$

From the free body diagram, note that $\mathrm{F}$is the force of aerodynamic lift and$\mathrm{a}$is towards the right.

Centripetal acceleration,

$\mathrm{a}=\frac{{\mathrm{v}}^2}{\mathrm{R}}$ (i)

Applying Newton’s law,

$\mathrm{Fsin\theta }=\frac{{\mathrm{mv}}^2}{\mathrm{R}}$ (ii)

$\mathrm{Fcos\theta }=\mathrm{mg}$ (iii)

Eliminating mass from equations (ii) and (iii) as:

$\mathrm{tan\theta }=\frac{{\mathrm{v}}^2}{\mathrm{gR}}$

We have,

$\begin{array}{c}\mathrm{v}=480\text{\hspace{0.17em}km/h}\\ =133\text{\hspace{0.17em}m/s}\end{array}$

and $\mathrm{\theta }={49}^{\mathrm{o}}$

Then,

$\begin{array}{c}\mathrm{R}=\frac{{\mathrm{v}}^2}{\mathrm{gtan\theta }}\\ =\frac{{133}^2}{9.8\times \tan {40}^{\text{o}}}\\ =2151\text{\hspace{0.17em}m}\end{array}$

Thus, radius of the circle is ,$\mathrm{R}=2.2\times {10}^3\text{\hspace{0.17em}m}$.