# Thread: from (centripetal) acceleration a(x,y) to x(t) and y(t)

1. ## from (centripetal) acceleration a(x,y) to x(t) and y(t)

Hi, all! I am trying to predict the position of an object O under a certain control policy, which defines what acceleration should be applied to O for every point in space. All I have is the start conditions (velocity $\displaystyle v_0$ and position $\displaystyle p_0$) and acceleration $\displaystyle a(x,y)$. What makes things complicated is that $\displaystyle a(x,y)$ is a function not only of $\displaystyle (x,y)$ but also of the instantaneous velocity $\displaystyle v(t)$. What's worse, each component of $\displaystyle a(t)$ depends on both components of $\displaystyle v(t)$. Any resemblance to centripetal acceleration is not accidental, and if there are methods to "derive" $\displaystyle x(t)$/$\displaystyle y(t)$ from $\displaystyle a(t)$ for Uniform Circular Motion, they might came in handy here.

Already in a ODE-friendly form, the two components of $\displaystyle a(t)$ can be expressed thus ($\displaystyle r$ is a given constant, and $\displaystyle x'(t)$ and $\displaystyle y'(t)$ are the components of $\displaystyle v(t)$):

$\displaystyle a_x(t) = \frac{2 y'(t) \left(\sqrt{r^2-y(t)^2} y'(t)-y(t) x'(t)\right)}{r^2}$

$\displaystyle a_y(t) = \frac{2 x'(t) \left(y(t) x'(t)-\sqrt{r^2-y(t)^2} y'(t)\right)}{r^2}$

So the problem lends itself to a direct formulation as an ODE system:

$\displaystyle x''(t) - a_x(t) = 0$
$\displaystyle y''(t) - a_y(t) = 0$

I am no Mathematica wizard, but using a direct application of DSolve I was able to get the following replacement rule:

$\displaystyle x'(t) \rightarrow e^{-\frac{y(t)^2}{r^2}} \int_1^t \frac{2 e^{\frac{y(K[1])^2}{r^2}} \sqrt{r^2-y(K[1])^2} y'(K[1])^2}{r^2} \, dK[1]+c_1 e^{-\frac{y(t)^2}{r^2}}$,

whose mere application to the equation for $\displaystyle y(t)$ yielded an equation that DSolve wasn't able to treat.

Thanks in advance for any pointers on all this. Cheers,

Jorge.

2. Is

$\displaystyle r^2 = x^2 +y^2$

for this problem. If so then I can solve this easily.

Also are there any initial conditions etc?

3. ## almost that

hi, the_doc! Thank you for your attention to this problem, and sorry for my failing to post the invariants right from the start. You almost got this right, but two alternative invariants that are acceptable in my case are a bit different, namely either:

(i) $\displaystyle c^2=x'(t)^2+y'(t)^2$
or
(ii) $\displaystyle c_1^2=f(t)^2+g(t)^2\land x'(t)=c_2+f(t)\land y'(t)=c_3+g(t)$.

Preferrably (ii) .

The start conditions are: $\displaystyle x[0]=c_4\land y[0]=c_5\land x'[0]=c_6\land y'[0]=c_7$

Thanks in advance for any pointers. Cheers,

4. I was not stating a solution. I was simply asking if by writing $\displaystyle r$ you were referring to the $\displaystyle r$ in polar coordinates. I didn't think it was that simple but just wanted to eliminate the possibility before attempting to solve a more difficult problem.

As for your invariants (i) must always be true for these accelerations since

$\displaystyle x' a_x + y' a_y = 0$ so it is an invariant for this problem!

However (ii) is not necessarily true so when you say invariant do you actually mean to say constraint here?

So assuming (ii) is a constraint this means that

$\displaystyle c_1^2 = (x'-c_2)^2 +(y'-c_3)^2$

so differentiating wrt $\displaystyle t$ gives us

$\displaystyle 0 = (x'-c_2)x'' +(y'-c_3)y''$

$\displaystyle \Leftrightarrow 0 = (x' a_x + y' a_y ) - (c_2 a_x + c_3 a_y)$

but since $\displaystyle x' a_x + y' a_y = 0$ we have that ($\displaystyle c_2,c_3 \neq 0$)

$\displaystyle a_x = -\frac{c_3}{c_2} a_y$

Now let $\displaystyle \alpha = \frac{c_3}{c_2}$ and substituting for $\displaystyle a_x$ and $\displaystyle a_y$ we get

$\displaystyle (y'-\alpha x')(\sqrt{r^2-y^2} \, y'-y x') = 0$ .

So either $\displaystyle y' = \alpha x'$, which leads to the trivial solution that $\displaystyle x'$ and $\displaystyle y'$ are constants (independent of $\displaystyle t$) and so the accelerations are zero and the motion is linear, or

$\displaystyle \sqrt{r^2-y^2} \, y'-y x' = 0$ (*)

i.e.

$\displaystyle \int \, \mathrm{d}x = \int \frac{\sqrt{r^2 - y^2}}{y} \, \mathrm{d} y$.

So there is your relationship between $\displaystyle x$ and $\displaystyle y$ which I'll leave for you to integrate (as it is simple enough) and then determine the constant of integration by your initial conditions.

Furthermore using (*) to eliminate $\displaystyle x'$ from the original constraint (ii) will give you a differential equation in terms of $\displaystyle y$, $\displaystyle y'$ and the constants $\displaystyle c_1$, $\displaystyle c_2$ and $\displaystyle c_3$ which hence will give you $\displaystyle y(t)$.

So this gives the solution to your problem subject to constraint (ii). This is essentially a trajectory for which the accelerations are zero. This doesn't make any physical sense but that's a consequence of the constraint which is not physically meaningful.

If I've misunderstood what you wrote let me know.

5. ## bad constraint

The_doc, thanks again for your reply and sorry! My bad: constraint (ii) simply does not hold. Actually, there are two versions of this problem, the simplest of which is already expressed here (ignoring the erroneous constraint). Can it be solved as is?

Anyway, the hardest and most interesting version (which I precipitately thought (ii) would be enough to express) involves an "external" source of velocity (e.g. wind) the acceleration cannot act on. In this case:

$\displaystyle a_x = -\frac{2 y' \left(\gamma w_x+\sqrt{r^2-\gamma ^2} w_y+\gamma x'+\sqrt{r^2-\gamma ^2} y'\right) \sqrt{k_w+2 w_x x'+2 w_y y'}}{r^2 v_a}$
$\displaystyle a_y = \frac{2 x' \left(\gamma w_x+\sqrt{r^2-\gamma ^2} w_y+\gamma x'+\sqrt{r^2-\gamma ^2} y'\right) \sqrt{k_w+2 w_x x'+2 w_y y'}}{r^2 v_a}$

where

$\displaystyle \gamma (t)=t w_y+y(t)$
$\displaystyle k_w=v_a^2+w_x^2+w_y^2 \text{ (constant)}$
$\displaystyle v_a^2=x'^2 + y'^2 \text{ (constant)}$

and all other symbols ($\displaystyle w_x$, $\displaystyle r$, etc.) are constants.

Thanks in advance. Cheers,

Jorge.