# Finding the radius of curvature of trajectory

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1. Dec 6, 2016

### doktorwho

1. The problem statement, all variables and given/known data
The functions are given:
$r(t)=pe^{kt}$
$\theta (t)=kt$
$v(r)=\sqrt2kr$
$a(t)=2k^2r$
Find the radius of the curvature of the trajectory in the function of $r$
2. Relevant equations
$$R=\frac{(\dot x^2 + \dot y^2)^{3/2}}{(\dot x\ddot y - \ddot x\dot y)}$$
There is also a second equation:
$$R=\frac{(1- y'^2)^{3/2}}{y''}$$
3. The attempt at a solution
I tried using the first one to get the dependance of $t$ and then transforming to the dependance of $r$ but i get stuck. Here:
$R=\frac{(\dot x^2 + \dot y^2)^{3/2}}{(\dot x\ddot y - \ddot x\dot y)}$
I did not know what exactly are $x, y$ in my problem statement so i supposed that they are the radial and angle component of the velocity vector.
$v(t)=pke^{kt}\vec e_r + pke^{kt}\vec e_{\theta}$
So the $(\dot x^2 + \dot y^2)^{3/2}=(2p^2k^2e^{2kt})^{3/2}$
$=\sqrt2pke^{kt}$
I have continued like this and used the acceleration in the polar coordinates for the below part but fail to get anything. Is my thinking from the start wrong? Could i have used an easier way? Perhaps finding the dependance on $r$ immediately from the result insted of first from $t$?

2. Dec 6, 2016

### Staff: Mentor

If you calculated the acceleration, would you be able to resolve the acceleration into tangential and centripetal components?

3. Dec 6, 2016

### doktorwho

$a=(\ddot r-\dot \theta^2)\vec e_r +(r\ddot \theta +2\dot r\dot \theta)\vec e_\theta$
From here my centripetal component is the radial one here, right? and the tangential is the angle one? Or is there another formula that connects the two?

4. Dec 6, 2016

### Staff: Mentor

No. The tangential component of acceleration is what you get if you dot the acceleration vector with a unit vector in the direction of the velocity vector. If this component is subtracted from the acceleration vector, what remains is the centripetal acceleration.

5. Dec 6, 2016

### doktorwho

So
$a_t=(a_rv_r+a_\theta v_\theta)$
And
$a_c=\sqrt{a^2-a^2_t}$
Where $a=\sqrt{a^2_r+a^2_\theta}$?
How does this relate to the problem? Is my work on the above of the fraction part correct?

6. Dec 6, 2016

### Staff: Mentor

No. The unit vector tangent to the trajectory is given by $$\vec{i}_t=\frac{\vec{v}}{|v|}$$The tangential component of the acceleration is given by:
$$\vec{a}_t=(\vec{a}\centerdot \vec{i}_t)\vec{i}_t=\left(\frac{\vec{a}\centerdot \vec{v}}{|v|}\right)\frac{\vec{v}}{|v|}$$
The component of the acceleration normal to the trajectory is given by:$$\vec{a}_n=\vec{a}-\vec{a}_t=\vec{a}-\left(\frac{\vec{a}\centerdot \vec{v}}{|v|}\right)\frac{\vec{v}}{|v|}$$The square of the normal component of acceleration is given by:
$$|\vec{a}_n|^2=\vec{a}\centerdot {\vec{a}}-\left(\frac{\vec{a}\centerdot \vec{v}}{|v|}\right)^2$$
This is related to the radius of curvature R of the trajectory by $$\frac{v^2}{R}=|\vec{a}_n|$$
Try this approach with Cartesian coordinates to verify that it gives the correct result for the radius of curvature.

7. Dec 7, 2016

### doktorwho

Before i go on to calculate the tangential acceleration lets get something straight
$a_t=\frac{(a_rv_r+a_\theta v_theta)}{\sqrt{v^r_r+v^2_\theta}}$ this is the fomula for the tangetial acceleration right?

8. Dec 7, 2016

Yes.