Determining parametrization of curve from its acceleration

Question

I am doing a project in which I have an object experiencing acceleration in a direction changing with time. I know the along-track and transverse acceleration components $a_{||}(t)$ and $a_{\perp}(t)$ in the comoving frame of the object, but not in an inertial reference frame.

As time passes $a_{\perp}(t)$ increases while $a_{||}(t)$ decreases, so I expect the motion of the object to be a spiral in an inertial reference frame, and it is a parametrization of this spiral I am looking for.

I have attempted to describe the curvature by $\kappa(t) = \frac{v(t)^{2}}{a_{\perp}(t)}$, where $v(t) = \sqrt{a_{\perp}(t)^{2} + a_{||}(t)^{2}} \ t$, and as expected I find the curvature to be increasing in time.

Now how do I find a parametrization for the resulting motion in an inertial frame? I have looked into frenet-serret basis but the only thing resembling a solution I have not been able to find a solution. The closes I have found is this parametrization for an arc-length parametrized curvature:

$\eta(t) = \int_0^t\cos\left( \int_0^t \kappa(s)ds\right)ds$

But I get curve that spirals outwards in time from origo, not inwards.

Any help is appreciated, am I going about this the wrong way?

Answer 1

I'm going to try out an idea here where we start with

$$z(s)=\int e^{i\int \kappa(s)ds}ds$$

and $|\ddot z(s)|=\kappa(s)$. Now, allow that $v=\frac{ds}{dt}$ is the velocity parallel to the curve, so that

$$ z(t)=\int e^{i\int \kappa(t)vdt}vdt\\ \dot z=v(t)e^{i\int \kappa(t)vdt}\\ \ddot z=\left[\frac{dv}{dt}+iv^2(t)\kappa(t) \right]e^{i\int \kappa(t)vdt} $$ The terms in the bracket for $\ddot z$ represent the parallel acceleration and, perpendicular to it, the centrifugal acceleration. Notice that everything is dimensionally correct. Therefore, I think we can say

$$|\ddot z|=\sqrt{a_{||}(t)^{2} + a_{\perp}(t)^{2}}=\sqrt{\left(\frac{dv}{dt}\right)^2+[v^2(t)\kappa(t)]^2}$$

and solve for $\kappa$ as follows

$$\kappa(t)=\frac{a_{\perp}}{v^2}$$

since $\frac{dv}{dt}=a_{||}$. This is the equation we should have known in the first place, since $a_{\perp}=v^2/\rho=\kappa v^2$

We can then return to the equation for $z(t)$ to determine the curve (spiral) itself.