How Does Radar Track a Plane's Motion in Different Coordinate Systems?

[Santilopez10]

Homework Statement

A plane flies along a flat trajectory with constant velocity $v$, at height h= 8km. It is followed by a radar placed exactly under the trajectory at an origin O.
Find:
a) Velocity at which the plane flies away from the radar as a function of time. What would be the equation as a function of the angle?
b) The cross component of the velocity vector (with O as fixed pole) as a function of time. What would be the equation as a function of the angle?
c) The time variation of the radial component of the velocity vector as a function of time. Find the equation as a function of angle aswell.
d) The angular acceleration as a function of time, and angle.

Relevant Equations

Kinematics equations in Polar and cartesian coordinates

I tried to workout the problem but I find motion in different coordinates systems a bit weird at the moment, so only thing I could do is realize that the x component of $\vec r(t)$ is: $$vt +x_0$$ but for simplicity we will use the initial condition $x_0=0$ so that $t_0$ is the moment the plane is 90 degrees with the radar. so: $$x=vt$$ and $$y= 8000 meters$$ so $$\vec r(t)=\begin{pmatrix} vt \\ 8000 \end{pmatrix} $$
It is obvious that the velocity at which the plane flies away from the radar is then $v$. That is all I could do, any help would be appreciated.

 

[FactChecker]

What is the straight-line distance between the airplane and O? The rate of change of that distance would be the velocity they are looking for.

 

[Santilopez10]

What is the straight-line distance between the airplane and O? The rate of change of that distance would be the velocity they are looking for.

So I should use the fact that $\vec r = |\vec r(t)|e_r$ then $\theta = \arctan{\frac {y(t)}{x(t)}}e_{\theta}$ and use the equation $\vec v = \dot r e_r + r \dot \theta e_{\theta}$?

 

[tnich]

The question uses the term velocity to mean speed and the term velocity vector to indicate the vector $\vec v$. This makes figuring out what is actually being asked for a bit difficult. I think the way to approach it is to think of parts a) and b) as hints for how to do parts c) and d). If you can solve parts c) and d), then you can go back and fill in something for a) and b). So how do you solve c) and d)?

As @FactChecker pointed out, you can find the radial velocity component by first writing an equation for $|\vec r|$ and then differentiating with respect to $t$. You could use that result and the chain rule to get an expression for the radial velocity as a function of $\theta$.

 

[Santilopez10]

So I used the fact that $$\dot r= \frac{{v_0}^2t}{r}$$ and $$\dot{\theta}= \frac{-8000v_0}{r^2}$$ so then $$\vec v= \frac{{v_0}^2t}{r}e_r -\frac{8000v_0}{r} e_{\theta}$$ (these would be my velocity components as a function of time right?) but $t=\frac{x}{v_o}=\frac{r \cos{\theta}}{v_0}$ and $8000=r\sin{\theta}$. Plugging these in the velocity formula I get: $$\vec v=v_0 \cos{\theta}e_r - v_0 \sin{\theta}e_{\theta}$$
Which I guess is the velocity as a function of the angle.
Then I use the fact that the angular acceleration $\ddot{\theta} = \frac {d\dot{\theta(t)}}{dt} = \frac{16000{v_0}^3 t}{r^4}$ then I used same relations as before but I get that $$\ddot{\theta}=\frac{{v_0}^2 \sin{2\theta}}{r^2}$$ and I sense there is a mistake somewhere because I can not express the angular acceleration as a function of the angle because of that r^2. What do you think? thanks!

 

[tnich]

Then I use the fact that the angular acceleration $\ddot{\theta} = \frac {d\dot{\theta(t)}}{dt} = \frac{16000{v_0}^3 t}{r^4}$ then I used same relations as before but I get that $$\ddot{\theta}=\frac{{v_0}^2 \sin{2\theta}}{r^2}$$ and I sense there is a mistake somewhere because I can not express the angular acceleration as a function of the angle because of that r^2. What do you think? thanks!

You have done well to this point.

You know $r(x)$ and $x(\theta)$. What prevents you from expressing $r^2$ as a function of $\theta$?

Once you get that, I think you still need to figure out $\dot r(\theta)$.

 

[Santilopez10]

You have done well to this point.

You know $r(x)$ and $x(\theta)$. What prevents you from expressing $r^2$ as a function of $\theta$?

Once you get that, I think you still need to figure out $\dot r(\theta)$.

What I did was $r=\frac{8000}{\sin{\theta}}$ then I plugged it in in the angular acceleration formula to get $$\ddot{\theta}(\theta)=\frac{{v_0}^2 \sin^{2}{\theta} \sin{2\theta}}{8000^2}$$
Is it correct? For the $\dot r$ part I know already that it is equal to $v_0 \cos{\theta}$

 

[tnich]

What I did was $r=\frac{8000}{\sin{\theta}}$ then I plugged it in in the angular acceleration formula to get $$\ddot{\theta}(\theta)=\frac{{v_0}^2 \sin^{2}{\theta} \sin{2\theta}}{8000^2}$$
Is it correct? For the $\dot r$ part I know already that it is equal to $v_0 \cos{\theta}$

That works.