A jet is flying straight and level at a speed v0 = 1100 km/h when it turns to change its course by 90° as shown. The turn is performed by decreasing the path’s radius of curvature uniformly as a function of the position s along the path while keeping the normal acceleration constant and equal to 8g

Question

A jet is flying straight and level at a speed [latex]v_{ 0 }[/latex] = 1100 km/h when it turns to change its course by 90° as shown. The turn is performed by decreasing the path’s radius of curvature uniformly as a function of the position s along the path while keeping the normal acceleration constant and equal to 8g, where g is the acceleration due to gravity. At the end of the turn, the speed of the plane is [latex]v_{ f }[/latex] = 800 km/h. Determine the radius of curvature [latex]{ \rho }_{ f }[/latex] at the end of the turn and the time [latex]{ t }_{ f }[/latex] that the plane takes to complete its change in course.

Accepted Answer

Recalling that we are given [latex]a_n[/latex] = 8g = 78.48 m/s² (since g = 9.81 m/s²), [latex]v_0[/latex] = 1100 km/h = 305.6 m/s, and [latex]v_f[/latex] = 800 km/s = 222.2 m/s, the radii of curvature at the beginning and end of the turn are

[latex] ho_0=\frac{v_0^2}{a_n}=1190 \ m \quad ext { and } \quad ho_f=\frac{v_f^2}{a_n}=629.1 \ m ext {. }[/latex] (1)

Expressing the value of [latex] ho_f[/latex] to 3 significant figures, we have

[latex] ho_f=629 \ m .[/latex]

Next, recalling that we are told that ρ decreases uniformly with the position s along the airplane’s path, i.e., that [latex]d ho / d s[/latex] = constant, denoting the constant in question by [latex]\gamma[/latex] , we have that the radius of curvature, expressed as a function of s must have the following form

[latex] ho(s)= ho_0+\gamma s,[/latex] (2)

where, referring to the figure on the right, we have set s = 0 to correspond to the location of the airplane along the path at the beginning of the turn, and where the nondimensional constant [latex]\gamma[/latex] will be determined later. Now, we recall that that, at every point along the airplane’s trajectory, we can define an osculating circle.

Let C(s) denote the center of the osculating circle corresponding to the position s along the path. Then, consider the radial segment of length ρ(s) going from C(s) to the airplane. In addition, let the orientation of the segment in question be described by the angle θ(s) that the segment forms with the line connecting C(0) and the airplane when s = 0. Based on how we have defined θ(s) we have [latex] heta(0)=0, heta\left(s_f ight)=(\pi / 2) rad[/latex], and we have that θ changes with time with an angular velocity we will denote by ω, i.e., [latex]\dot{ heta}=\omega[/latex]. Because of the properties of the osculating circle and because of how we have defined θ, we must have

[latex]\omega(s)=\frac{v(s)}{ ho(s)},[/latex] (3)

where [latex]v[/latex] is the speed of the plane. Then manipulating the above equation, we have

[latex]\omega=\frac{v}{ ho}=\frac{d heta}{d t}=\frac{d heta}{d s} \frac{d s}{d t}=v \frac{d heta}{d s} \quad \Rightarrow \quad \frac{d heta}{d s}=\frac{1}{ ho}.[/latex] (4)

Separating variables in the last of Eqs. (4) and integrating, we have

[latex]\int_0^{(\pi / 2) rad } d heta=\int_0^{s_f} \frac{d s}{ ho_0+\gamma s} \Rightarrow \frac{\pi}{2} \operatorname{rad}=\frac{1}{\gamma} \ln \left(\frac{ ho_0+\gamma s_f}{ ho_0} ight) \quad \Rightarrow \quad \frac{\pi}{2} \operatorname{rad}=\frac{1}{\gamma} \ln \left(\frac{ ho_f}{ ho_0} ight),[/latex] (5)

which can be solved for [latex]\gamma[/latex] to obtain

[latex]\gamma=\frac{2}{\pi} \ln \left(\frac{ ho_f}{ ho_0} ight)=-0.4058.[/latex] (6)

Knowing that represents the rate of change of curvature with respect to position [latex]\left(\gamma=\frac{d ho}{d s} ight)[/latex] we can take the derivative with respect to s of the equation [latex]a_n=\frac{v^2}{ ho}=8 g[/latex] to obtain

[latex]v^2=8 g ho \Rightarrow 2 v \frac{d v}{d s}=8 g \frac{d ho}{d s} \Rightarrow v \frac{d v}{d s}=4 g \gamma.[/latex] (7)

However, [latex]v \frac{d v}{d s}[/latex] represents the quantity [latex]\dot{v}[/latex]. Therefore, we have

[latex]\frac{d v}{d t}=4 g \gamma \quad \Rightarrow \quad \int_0^{t_f} d t=\int_{v_0}^{v_f} \frac{d v}{4 g \gamma} \Rightarrow t_f=\frac{1}{4 g \gamma}\left(v_f-v_0 ight).[/latex] (8)

Using the value of [latex]\gamma[/latex] computed in Eq. (6), and recalling that we have g = 9.81 m/s², [latex]v_0[/latex] = 1100 km/h = 305.6 m/s, and [latex]v_f[/latex] = 800 km/s = 222.2 m/s, we can evaluate the last of Eqs. (8) to obtain

[latex]t_f=5.24 \ s .[/latex]

Question 2.246: A jet is flying straight and level at a speed v0 = 1100 km/h...

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