The radar station at O is tracking the meteor P as it moves through the atmosphere. At the instant shown, the station measures the following data for the motion of the meteor: r = 21,000 ft, θ = 40° , ṙ= -2,440 ft/s, and θ = 2.935 rad/s. Use Eq. (2.62) to determine the magnitude and direction

Question

The radar station at O is tracking the meteor P as it moves through the atmosphere. At the instant shown, the station measures the following data for the motion of the meteor: r = 21,000 ft, θ = 40° , ṙ= -2,440 ft/s, and [latex]\dot{ heta}[/latex] = 2.935 rad/s. Use Eq. (2.62) to determine the magnitude and direction (relative to the xy coordinate system shown) of the velocity vector at this instan.

Accepted Answer

We define a coordinate system as shown in the figure. Then Eq. (2.62) become

[latex]\begin{aligned}\dot{\vec{r}} & =\dot{r} \hat{u}_r+\vec{\omega}_r imes \vec{r} \& =\dot{r} \hat{u}_r+r \omega_r \hat{u}_ heta . \quad \quad ext{(1)}\end{aligned}[/latex]

The [latex]\left(\hat{u}_r, \hat{u}_ heta ight)[/latex] coordinate system is expressed in the (x,y) coordinate system

[latex]\hat{u}_r=\cos heta \hat{\imath}+\sin heta \hat{\jmath}, \quad \hat{u}_ heta=-\sin heta \hat{\imath}+\cos heta \hat{\jmath} .[/latex] (2)

Now we replace [latex]\omega_r ext { with } \dot{ heta}, \dot{\vec{r}} ext { with } \vec{v}[/latex] , and plug Eq. (2) into Eq. (1) to

[latex]\begin{aligned}\vec{v} & =\dot{r}(\cos heta \hat{\imath}+\sin heta \hat{\jmath})+r \dot{ heta}(-\sin heta \hat{\imath}+\cos heta \hat{\jmath}) ,\& =(\dot{r} \cos heta-r \dot{ heta} \sin heta) \hat{\imath}+(\dot{r} \sin heta+r \dot{ heta} \cos heta) \hat{\jmath}. \quad \quad ext {(3)}\end{aligned}[/latex]

Recalling that r = 21,000 ft, [latex] heta[/latex] = 40°, ṙ = 22.440 ft/s, and [latex]\dot{ heta}[/latex] = 2.935 rad/s, we can evaluate the terms in Eq. (3) to

[latex]\vec{v}=(22,430 \hat{\imath}-61,640 \hat{\jmath}) ft / s[/latex] (4)

The magnitude of the velocity is

[latex]|\vec{v}|=\sqrt{v_x^2+x_y^2}[/latex] (5)

Hence, since from Eq. (4) we have that [latex]v_x[/latex] = 22,430 ft=s and [latex]v_y[/latex] = - 61,640 ft/s, evaluating Eq. (5)

[latex]|\vec{v}|=65,600 \ ft / s .[/latex]

Referring to the figure on the right, let [latex]\phi[/latex] denote the angle formed by the velocity vector and the x axis such that [latex] an \phi=v_y / v_x[/latex] . Then the we can obtain [latex]\phi[/latex] as [latex] an ^{-1}\left(v_y / v_x ight)[/latex]. Hence, recalling again that from Eq. (4) we have [latex]v_x=22,430 \ ft / s[/latex] and [latex]v_y=-61,640 \ ft / s[/latex].

[latex]\phi= an ^{-1}\left(\frac{v_y}{v_x} ight)=-70.0^{\circ}[/latex] (6)

Question 2.112: The radar station at O is tracking the meteor P as it moves ...

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