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Consider a particle with tangential and normal accelerations $\vec{a_T}(t)$ and $\vec{a_N}(t)$ respectively ($t$ is time). If the initial velocity and position vectors are both $\vec{0}$, how can the trajectory $y = f(x)$ (or $(x(t), y(t))$ in parametric form) of this particle be found?

In my attempts, I get stuck because I cannot figure out what the angle that the net acceleration ($= \vec{a_N} + \vec{a_T}$) makes with the axes is, even though the angle it makes with the velocity is simply $\tan^{-1}\Big(\frac{a_N}{a_T}\Big)$. If the angle were known, it would be a matter of integrating the acceleration vector to get the velocity, and hence the position vector as a function of time.

Is there any way to find the trajectory $y = f(x)$, and is any extra information necessary for this task? Any hints would be appreciated.

Aadi Prasad
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  • Perhaps I'm missing something, but if the total acceleration is $\vec{a} = \vec{a}_r + \vec{a}_t$ then why can't you just integrate $\vec{a}$ to get $\vec{v}(t)$ and then $\vec{r}(t)$? – Michael Seifert Sep 24 '22 at 22:23
  • @MichaelSeifert because of the the fact that the direction of $\vec{a}$ is not that easily obtained in standard Cartesian coordinates. – Aadi Prasad Sep 26 '22 at 07:57
  • Ah, so you don't have $\vec{a}_t$ and $\vec{a}_r$, you have $|\vec{a}_t|$ and $|\vec{a}_r|$ and you're trying to infer their directions from the motion. That makes more sense. – Michael Seifert Sep 26 '22 at 11:49
  • Related https://physics.stackexchange.com/a/787106/226902 – Quillo Nov 04 '23 at 10:20

1 Answers1

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the position of the trajectory is

$$\mathbf R= \left[ \begin {array}{c} x \left( s \right) \\ y \left( s \right) \end {array} \right] \tag 1$$

where $~s~$ is the line element hence

$$v_s=\frac{ds}{dt}\quad ,a_s=\frac{d^2s}{dt^2}$$

from equation (1) you obtain $$\mathbf v=\frac{d\mathbf R}{ds}\dot s\\ \mathbf a=\frac{d\mathbf R}{ds}\ddot s +\frac{d\mathbf v}{ds}\dot s$$

the tangential vector $~\mathbf t~$ is $$\mathbf t=\frac{\frac{d\mathbf R}{ds}}{\left|\frac{d\mathbf R}{ds}\right|}\quad, \mathbf t\cdot\mathbf t=1$$

the normal vector $~\mathbf n$ is

$$\mathbf n=\begin{bmatrix} \mathbf{t}_2\\ -\mathbf{t}_1 \\ \end{bmatrix}$$

hence

$$ a_t(s)=\mathbf a\cdot \mathbf t\tag 2$$ $$ a_n(s)=\mathbf a\cdot \mathbf n\tag 3$$

where $~a_t(s)~,a_n(s)~$ are the given trajectory tangential and normal accelerations

from equation (2) and (3) you obtain two second order differential equations $~\left(\frac{d^2}{ds^2}x(s)=...~,\frac{d^2}{ds^2}y(s)=...\right)~$ which you can solve to obtain $~x(s)~,y(s)$

$${a_t(s)}=\sqrt { \left( {\frac {d}{ds}}x \left( s \right) \right) ^{ 2}+ \left( {\frac {d}{ds}}y \left( s \right) \right) ^{2}}a_{{s}}+{ \frac { \left( \left( {\frac {d}{ds}}x \left( s \right) \right) { \frac {d^{2}}{d{s}^{2}}}x \left( s \right) + \left( {\frac {d}{ds}}y \left( s \right) \right) {\frac {d^{2}}{d{s}^{2}}}y \left( s \right) \right) {v_{{s}}}^{2}}{\sqrt { \left( {\frac {d}{ds}}x \left( s \right) \right) ^{2}+ \left( {\frac {d}{ds}}y \left( s \right) \right) ^{2}}}} $$

$$a_n(s)=-{\frac {{v_{{s}}}^{2} \left( - \left( {\frac {d}{ds}}y \left( s \right) \right) {\frac {d^{2}}{d{s}^{2}}}x \left( s \right) + \left( {\frac {d}{ds}}x \left( s \right) \right) {\frac {d^{2}}{d{s} ^{2}}}y \left( s \right) \right) }{\sqrt { \left( {\frac {d}{ds}}x \left( s \right) \right) ^{2}+ \left( {\frac {d}{ds}}y \left( s \right) \right) ^{2}}}} $$

Eli
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