1. ## Kinematics problem

- correct problem appears in the next post

2. To begin with I'll just make this question more readable and fill in the bits that have obviously been removed in the copy and paste. Also I've corrected any equations that I think haven't been written correctly. Let me know if I guess the missing bits incorrectly. I've highlighted the bits of text I added in red:

1. A pilot attempts to fly with constant speed $v$ from the point $P = (D, 0)$ on the $x$-axis to the origin $O = (0; 0)$. A wind blows with speed $w$ in the positive $y$ direction. The pilot is not familiar with vector addition and thinks the shortest path to $O$ is achieved by flying his plane so that it always points directly towards $O$.

(a) Show that the actual flight path of the plane (in Cartesian coordinates) is given by

$y(x) = f(x) \sinh [g(x)]$ ; (1)

where $f(x)$ and $g(x)$ are scalar functions of $x$ that are to be determined.

(b) Consider the three cases $w > v$, $w = v$, and $w < v$ separately, and determine in which cases the pilot actually reaches $O$.

(c) Show that the flight path of the plane (in polar coordinates) is given by the polar equation

$r = r(\theta) = D ( \cos \theta )^{c_1} (1 + \sin \theta )^{c_2}$ ; (2)

where $c_1$ and $c_2$ are constants (that depend on $v$ and $w$) that are to be determined.

(d) Show that the radial component of the acceleration $a_r = 0$ and determine the transverse component $a$ in terms of the polar coordinates $r$ and $\theta$ (at time $t$).

(e) Determine the flight path of the plane (in Cartesian coordinates) if the wind is blowing with a velocity $w$ in a direction that makes an angle with the vertical (i.e., the $y$ axis).
I'll put up a solution in a little while.

3. It's late in the UK right now so will do so tomorrow now.

4. Hi!

I have been struggling with this problem for quite some time so would also appreciate it if you would be able to post an answer ASAP or at least some clarification!

5. hey, yes the corrections are right. not sure why it didn't include them when i pasted the question. thanks for changing them

6. Here are the solutions (a) to (d):

(a)

Spoiler:

The pilot's apparent (to himself) velocity is the relative velocity, $\mathbf{v'}$ , given by

$\mathbf{v'} = \mathbf{v} - \mathbf{w}$

where $\mathbf{v}$ is the actual velocity and

$\mathbf{w} = w \hat{\mathbf{j}}$

is the wind velocity. Now, since he is always flying towards $O$ we have that

$\mathbf{v'} = -v \hat{\mathbf{e}}_r = -\frac{v}{\sqrt{x^2 +y^2}} (x \hat{\mathbf{i}} + y\hat{\mathbf{j}})$

so

$\mathbf{v} = \mathbf{v'} + \mathbf{w}$
$= -v \hat{\mathbf{e}}_r +w \hat{\mathbf{j}}$
$= -\frac{v x}{\sqrt{x^2 +y^2}} \hat{\mathbf{i}} + \left(w -\frac{vy}{\sqrt{x^2 +y^2}} \right)\hat{\mathbf{j}}$.

Now, note that $v_x = \dot{x}$ and $v_y = \dot{y}$ so

$\frac{dy}{dx} = \frac{v_y}{v_x}$

$\Leftrightarrow \frac{dy}{dx} = \frac{y}{x} - \frac{w}{v} \sqrt{1+ \left(\frac{y}{x}\right)^2}$.

Now let $y = z x$ so

$\frac{dy}{dx} = x \frac{dz}{dx} + z$ so our equation becomes

$x \frac{dz}{dx} + z = z - \frac{w}{v} \sqrt{1 + z^2}$

$\Leftrightarrow x \frac{dz}{dx} = - \frac{w}{v} \sqrt{1 + z^2}$

Integrating and using the fact that at $x=D$, $y = 0$ we have

$\int_{0}^{z} \frac{1}{\sqrt{1+{z'}^2}} \, \mathrm{d}z' = -\frac{w}{v} \int_{D}^{x} \frac{1}{x'}\, \mathrm{d}x'$

giving us

$\sinh^{-1} z = \frac{w}{v} \ln \left(\frac{D}{x} \right)$,

$\Leftrightarrow z = \sinh \left[ \frac{w}{v} \ln \left(\frac{D}{x} \right) \right]$,

$\boxed{ \Leftrightarrow y = x \sinh \left[ \frac{w}{v} \ln \left(\frac{D}{x} \right) \right]}$.

(b)

Spoiler:

If you recall the definition of the $\sinh$ function (in terms of exponentials) we can rewrite this as

$y = \frac{x}{2} \left(\exp\left[\frac{w}{v} \ln \left(\frac{D}{x} \right) \right] - \exp\left[-\frac{w}{v} \ln \left(\frac{D}{x} \right) \right] \right)$

$= \frac{x}{2} \left( \left(\frac{D}{x}\right)^{\frac{w}{v}} - \left( \frac{D}{x}\right)^{-\frac{w}{v}} \right)$

$= \frac{1}{2} \left( D^{\frac{w}{v}} x^{1-\frac{w}{v}} - D^{-\frac{w}{v}} x^{1+ \frac{w}{v}} \right)$

Now what we're interested in is the behaviour of $y$ as $x \to 0$ since if $y \to 0$ then the pilot will arrive at $O$.

When $w > v$ then $1- \frac{w}{v} < 0$ so then as $x \to 0$, $y \to \infty$ so the pilot never arrives at O. This should be obvious since even if he were to head directly into the wind he would always be moving vertically away from the $x$-axis!

When $w < v$ then $1- \frac{w}{v} > 0$ so as $x \to 0$, $y \to 0$ also. Hence the pilot does infact arrive at $O$ in this case.

When $w = v$ then $1- \frac{w}{v} = 0$ so we have that

$y = \frac{1}{2} \left(D - D^{-1} x^2 \right)$

so at $x=0$ we have $y = \frac{D}{2}$ so again the pilot does not reach $O$.

(c)

Spoiler:

Continuing from the previous part you can substitute for $x$ and $y$ using $x = r \cos \theta$ and $y = r \sin \theta$ so the equation from (b) becomes:

$2 r \sin \theta = D^{\frac{w}{v}} (r \cos \theta)^{1-\frac{w}{v}} - D^{-\frac{w}{v}} (r \cos \theta)^{1+ \frac{w}{v}}$

then division by $r \cos \theta$ gives

$2 \tan \theta = D^{\frac{w}{v}} (r \cos \theta)^{-\frac{w}{v}} - D^{-\frac{w}{v}} (r \cos \theta)^{\frac{w}{v}}$

and if we let

$u = \left(\frac{r\cos \theta}{D}\right)^{\frac{w}{v}}$ we get the quadratic

$u^2 + 2 \tan \theta \, u - 1 = 0$,

which is easily solved as

$u = - \tan \theta \pm \sec \theta$

but since $r = D$ when $\theta = 0$ then

$u = \sec \theta -\tan \theta = \frac{1}{\cos \theta} (1- \sin \theta)$

but note that

$1- \sin^2 \theta \equiv \cos^2 \theta$

$\Leftrightarrow (1-\sin \theta) (1+ \sin \theta) \equiv \cos ^2 \theta$
$\Leftrightarrow 1- \sin \theta = \cos^2 \theta (1+ \sin \theta)^{-1}$

so

$u = \cos \theta (1+ \sin \theta)^{-1}$

and putting back in terms of $r$ we get

$\boxed{r = D \left( \cos \theta \right)^{\frac{v}{w} -1} \left(1+ \sin \theta \right)^{-\frac{v}{w}}}$.

Note that had I used the substitution $u = \left(\frac{r\cos \theta}{D}\right)^{-\frac{w}{v}}$, solving the quadratic would have directly yielded an expression in terms of $(1 + \sin \theta)$ rather than having to use the identity. I did it this way because I felt people would naturally choose the positive exponent so wanted to show how it could be done.

An alternative method would be to convert the velocity in terms of polar coordinate vectors and relate to the general expression for velocity in polar coordinates to get a differential equation in terms of $r$ and $\theta$ and then solve.

(d)

Spoiler:

Recall that we can express $\mathbf{v}$ as

$\mathbf{v} = -v \hat{\mathbf{e}}_r +w \hat{\mathbf{j}}$

(don't worry about the mixed geometry just yet as it makes life easier) so then the acceleration is given by the derivative wrt $t$ of this so we have

$\mathbf{a} = \frac{d}{dt} \mathbf{v} = -v \dot{\theta} \hat{\mathbf{e}}_{\theta}$ since $\frac{d}{d \theta} \hat{\mathbf{e}}_{r} = \hat{\mathbf{e}}_{\theta}$,

so $\mathbf{a} \cdot \hat{\mathbf{e}}_r = 0$ hence $a_r = 0$.

Now, from the above, we know that the transverse acceleration, $a_t$ , is given by

$a_t = -v \dot{\theta}$.

Now note that in polar coordinates

$\mathbf{v} = \dot{r} \, \hat{\mathbf{e}}_{r} + r \dot{\theta} \hat{\mathbf{e}}_{\theta}$

so if we express our $\mathbf{v}$ in polar coordinates it will give us $\dot{\theta}$ at time $t$.

So to convert to polar coordinates recall that

$\hat{\mathbf{j}} = \sin \theta \, \hat{\mathbf{e}}_{r} + \cos \theta \, \hat{\mathbf{e}}_{\theta}$

$\Rightarrow \mathbf{v} = \left(w \sin \theta - v \right) \, \hat{\mathbf{e}}_{r} + w \cos \theta \,\hat{\mathbf{e}}_{\theta}$,

hence by comparison we have

$r \dot{\theta} = w \cos \theta$ so

$a_t = -v \dot{\theta} = -\frac{v w}{r} \cos \theta$.

$\boxed{a_t = -\frac{v w}{r} \cos \theta }$.

Will put up solution to (e) in a little while (going to have my supper!).

If anyone else would like to put up the solution to (e) in the meantime they are welcome to!

7. I love ya~~~!!!

8. Thanks, I ended up figuring out a,b and c but your solution to d helped alot.

for e however, i'm still clueless. midterm soon but i feel much more prepared now.

9. ## Concluding part!

(e)

Spoiler:

First let $\alpha$ be the angle that $\mathbf{w}$ makes with the $y$-axis (positive in clockwise sense).

The thing to notice about this problem is that if you were to rotate the $x$- $y$ coordinate system clockwise through an angle of $\alpha$, to the new $x'$- $y'$ coordinate system, the wind is then again parallel to the $y'$-axis, so we arrive back at our original problem with the only difference being that our initial starting point for the trajectory is now at ( $D \cos \alpha$ , $D \sin \alpha$ ).

So we can treat this problem exactly the same as part (a) up to the part where the initial conditions come in i.e. the limits of the integral in terms of $z$.

So at the starting point in the $x'$- $y'$ frame coordinate system we have $x' = D \cos \alpha$ and $z = y' / x' = \tan \alpha$ so the integral becomes

$\int_{\tan \alpha}^{z} \frac{1}{\sqrt{1+{z'}^2}} \, \mathrm{d}z' = -\frac{w}{v} \int_{D \cos \alpha}^{x'} \frac{1}{x''}\, \mathrm{d}x''$

where $z'$ and $x''$ are the dummy variables of integration in place of $z$ and $x'$ respectively.

From the integral we get:

$\sinh^{-1} z - \sinh^{-1} \left(\tan \alpha \right) = -\frac{w}{v} \ln \left(\frac{x'}{D \cos \alpha} \right)$.

Note this solution can be tidied up in various ways but I'll leave that to you.

Finally, the solution is in terms of the rotated coordinates $x'$ and $y'$ so it needs to be put back in terms of $x$ and $y$ using the transformation equations

$x' = x \cos \alpha - y \sin \alpha$,
$y' = x \sin \alpha + y \cos \alpha$ and

$z = \frac{y'}{x'} = \frac{x \cos \alpha - y \sin \alpha}{x \sin \alpha + y \cos \alpha}$.

The final result is then a big messy implicit equation in terms of $x$ and $y$ which I can't be bothered to late $\chi$ so I'll leave to you to work out from the above!