0
$\begingroup$

I'm studying polar coordinates, and can't wrap my head around a few things about a few things (and believe me I've tried). I'll get to the point quickly, and I'm not sure if I'm allowed to ask two kind of unrelated questions at once (?). Either way, I'll try. If not, I apologize.

Question 1: First off, if an object has the position-vector $\vec{r} = r\hat{r}$ from the origin in a reference frame, with the angle between $\hat{r}$ and the x-axis being $\theta$, then in terms of polar coordinates, you can refer to its position as $(r, \theta)$. Now, assume this object has a velocity $\vec{v}$. I did an assignment, and when asked to express the velocity vector $\vec{v}$ in polar coordinates, my book expressed the position in the "position coordinate frame".

Basically, they used the old $\hat{r}$ and $\hat{\theta}$, which were used for the object's position, and then expressed the velocity vector in terms of them, such as $v_r \hat{r} + v_{\theta} \hat{\theta} = (v_r, v_\theta)$. I on the other hand moved the $\vec{v}$-vector so that it pointed from the origin, and expressed it as $(|\vec{v}|, \phi)$, where $\phi$ is the angle between $\vec{v}$ and the x-axis. I'm already confused about this enough as it is already, and can't figure out which one is right. Can anyone give me some guidance in this? If a velocity should be expressed like the book wants, is it really "expressed in polar coordinates"? To me, it seems like it's not, but considering how awfully confused I am about all of this, I may be wrong.

Question 2: The acceleration can be expressed in polar coordinates such as:

$$\vec{a} = \left( \frac{d^2|\vec{r}|}{dt^2} -\omega \vec{r}^2 \right)\hat{r} + ... $$ If $a \cdot \hat{r}$ is known somehow (as a function), and I want to find the radial velocity, I thought I would integrate that function (or if it's not only a function of time, solve the differential equation). However, several times, I see that it's $\frac{d^2|\vec{r}|}{dt^2}$ being integrated instead. If $\vec{a} \cdot \hat{r}$ is the actual radial component of the acceleration, why isn't it the one being integrated? What's the actual difference between integrating either one? (I know I should know the answer to the latter question, but confusion, etc. etc. makes it difficult.)

Thanks in advance. I'm tired, so there may be errors in my reasoning. Either case, I hope it's comprehensible.

$\endgroup$
  • $\begingroup$ Position can be thought as a function from time to coords, ex $x=x(t)=r(t)\partial_r + \theta(t)\partial_\theta$, I think. $\endgroup$ – Emil Jan 19 '17 at 0:07
2
$\begingroup$

Polar coordinates are a coordinate system spanned by a set of orthonormal time dependent unit vectors $\left\{ \hat r, \hat \theta \right\},$ where relative to the familiar rectangular Cartesian coordinates, $\hat r = \cos \theta \hat x + \sin \theta \hat y $ and $\hat \theta = -\sin \theta \hat x + \cos \theta \hat y$. The notation $(v_r, v_{\theta})$ is not synonymous with $(r,\theta)$ in that the former are components of vectors while the latter are not. Any point in the polar coordinate plane can be described by the tuplet $(r,\theta)$ or via $\mathbf r = r \hat r(\theta)$. This is different to the cartesian coordinate description, where the point $(x,y)$ are the components of the vector in the Cartesian basis $\left\{\hat x, \hat y \right\}$, with $\mathbf r = x \hat x + y \hat y$.

More explicitly, the position of the object in polar coordinate representation is $\mathbf r = r \hat r(\theta) $ and then the velocity vector is $$\mathbf{ \dot r} = \dot r \hat{r} + r \dot{\hat r} = \dot r \hat r + r \dot \theta \hat \theta,$$ noting the time dependence of the unit vectors and using the explicit expression for $\hat r(\theta)$ as given above. So, here $v_r = \dot r$ and $v_{\theta} = r \dot \theta$. Performing another time derivative, we obtain $\ddot{\mathbf r} = (\ddot r - r \dot \theta^2) \hat r + (2 \dot r \dot \theta + r \ddot \theta) \hat \theta,$ where $\ddot{\mathbf r} \equiv \mathbf a = a_r \hat r + a_{\theta} \hat \theta$ and so $$\sum F_r = m a_r.$$The physical interpretation of the extra terms come from considering the fictitious forces present in the non-inertial co-rotating frame. The equation as written is valid to an inertial frame.

$\endgroup$
1
$\begingroup$

Answer 1:

$v_r \hat{r} + v_{\theta} \hat{\theta} = (v_r, v_\theta)$ can also be thought of as a vector centered on the origin, expressed in cartesian co-ordinates with one axis radial and one circumfrential to the polar frame of reference. I think you misunderstood what it represents - are you sure the book actually says that is a polar representation?
It is also possible to translate this into a polar representation just as you have done, with $\phi=\theta+sin^{-1}(v_\theta/v_r)$ (at least for $\theta<180, otherwise it will be subtraction, not addition).

Answer 2:

Is that the radial acceleration you are after? The equation you quoted doesn't seem to be purely derrived from expression in polar co-ordinates. If it's not a free particle to answer I need to know the geometry or other constraints.

If it's a free particle then integrating the acceleration to a velocity and projecting it onto $\hat r$ will be the same as projecting it onto $\hat r$ and then integrating it.

(maybe return when not tired)

$\endgroup$
  • $\begingroup$ Thank you for the response! I will read it tomorrow when I can focus :) $\endgroup$ – Max Jan 18 '17 at 23:36
  • $\begingroup$ Okay, so now I'm all awake, and I've read everything. Regarding Q1, you made me realize that what I was talking about yesterday is, as you're saying, a normal-tangential coordinate system, rather than a polar coordinate system. The book clearly said express in polar and gave that normal-tangential representation. Does that mean it's wrong? $\endgroup$ – Max Jan 19 '17 at 7:57
  • $\begingroup$ Regarding Q2: Yes, it is the radial acceleration I'm after. I only gave half the equation, but if we assume that acceleration is written as $\vec{a} = a_r \hat{r} + a_\theta \hat{\theta}$, then I assumed that the radial velocity would be obtained by integrating $a_r$. I now realize that this may not be the case, since the $\hat{r}$-vector is time-dependent. Is this the case? Because the $-r\omega^2$-term shouldn't really contribute to any radial velocity, since it only changes the direction of the velocity vector. Although I can't see the bigger picture in this. Is it correct? $\endgroup$ – Max Jan 19 '17 at 8:01
  • $\begingroup$ Last few notes: It's a free particle. Als0, didn't I already project it onto $\hat{r}$ and integrate it? $\endgroup$ – Max Jan 19 '17 at 8:02
  • $\begingroup$ Last comment: It's not a normal-tangential coordinate system. Ignore all the times I use that term :P $\endgroup$ – Max Jan 19 '17 at 8:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.