# Kleppner - Example 1.18: Representation of Position Vector

1. Nov 24, 2016

### StudentOfScience

1. The problem statement, all variables and given/known data
constant speed $u$ along spoke
it starts at center at $t=0$
angular position is given by $\theta=\omega t$, where $\omega$ is a constant

2. Relevant equations
$\frac{d\hat r}{dt} = \dot \theta \hat \theta$ (1)
$\frac{d\hat \theta}{dt} = -\dot \theta \hat r$ (2)
$\vec r = | \vec r| \hat r$ (3)
Note: velocity and acceleration in polar coordinates can be obtained by differentiating the above equation with resptect to time and using equations (1) and (2)

3. The attempt at a solution
I'm more or less confused as to why $\vec r (t)$ doesn't take the form
$\vec r(t) = r_r \hat r + r_\theta \hat \theta$ (4).
They derived the velocity and acceleration equations by using a general vector in the plane given by
$\vec r = |\vec r|\hat r$ (5)
and differentiating it with respect to time. I realize that (5) is just a vector that is not time dependent; we would write $\vec r(t)$ if it were a function of time. Here is the approach that I took (I'm not exactly sure what is wrong with it, but here we go):

Define $\vec r(t)=r_r \hat r + r_\theta \hat \theta$ (6). We are given radial velocity:
$v_r (t)=u$ (7)
We also know that
$\mathbf r_0 = \vec 0$ (8)
since the bead starts at the pole. Hence,
$r_r =\int_0^t udt' = ut$. (9)
In the problem statement, we are given angular position:
$r_\theta=\omega t$ (10)
Hence, the position of the bead is given by
$\vec r(t) = ut\hat r + \omega t \hat \theta$ (11).
I will go ahead and confirm that this is an Archimedean spiral:
Because
$\hat r = \cos\theta \hat i + \sin\theta \hat j, \hat \theta = -\sin\theta \hat i + \cos\theta \hat j$, (12)
it follows that
$\vec r(t) = (ut\cos\theta - \omega t \sin \theta) \hat i + (ut\sin\theta + \omega t \cos\theta)\hat j$ (13)
And we defined $\theta$ earlier as
$\theta = \omega t$ (14). Ok, here is a question I have: was it incorrect for me to assume that $r_\theta$ and $\theta$ in (13) were the same ($=\omega t$)? Perhaps I misinterpreted angular position as $r_\theta$? We can check this though on a graphing calculator in parametric mode and, by inspection, see that it is an Archimedean spiral (ok, I was too lazy to graph it by hand)

For velocity, I differentiated (11) to get
$$\vec v(t)$= (u-\omega t \frac{d\theta}{dt})\hat r + (\omega+ut\frac{d\theta}{dt})\hat\theta (14) [and LaTeX apparently is not being so nice right now] I didn't use \dot \theta = \omega (15) (maybe this is another mistake?); instead, since we have defined \vec r(t) = r_r \hat r + r_\theta \hat \theta , (16) we can differentiate this to obtain \vec v(t) = (\dot r_r -r_\theta \dot \theta)\hat r + (r_r\dot \theta +\dot r_\theta)\hat \theta (17) By equality of components for (14) and (17), we have v_r= u-\omega t \dot \theta = \dot r_r - r_\theta \dot\theta implies \dot \theta = \frac{u-\dot r_r}{\omega t-r_\theta}=\frac{0}{0}, which is indeterminate. I tried equating the other component -$v_\theta $- but, to no avail, I also got an indeterminate form. And thus a road block. The solution in Kleppner, though, is much cleaner: They derived earlier that \vec a(t)= (\ddot r - r(\dot \theta)^2)\hat r + (r\ddot \theta+2\dot r \dot \theta)\hat \theta (18) and they essentially used$\theta=\omega t$and$r=ut$and differentiated when appropriate. If I take this approach, how can I find position? They started with the general vector$\vec r=r\hat r $, which, as I said earlier, is not a function of time. So would I have to convert to Cartesian coordinates, integrate, and then go back to polar coordinates, or is there a nicer way to find position (finding position wasn't part of the question, but I ask out of curiosity)? It seems as though (11) is incorrect for position. Thank you all for your answers. Please let me know if you would like me to clarify something. I appreciate your time! 2. Nov 24, 2016 ### haruspex Please define rr and rθ. 3. Nov 24, 2016 ### StudentOfScience$r_r$is position in the radial (r) direction and$r_\theta$is position in the tangential direction (that is, the angle vector r makes with the x axis where theta is measured counterclockwise). In component form, the vector-valued function for position may be written as \vec r(t)=r_r \hat r + r_\theta \hat\theta where the$r_r$,$r_\theta$are functions of time. A general vector$\vec A$in polar coordinates can be written as \vec A=|\vec A|\hat r$\vec A $does depend on theta, but since it is not time-dependent, it is simply written as that above (according to my understanding; a correction would be appreciated) Again, I want to reiterate the above is based on my understanding, and corrections to this would be greatly appreciated. 4. Nov 24, 2016 ### haruspex There is only one position. If you mean rr is its distance in the radial direction (i.e. |r|) then$r_r\hat r$is its position.$\hat r$is the unit vector pointing towards the position$\vec r$. When you have gone distance |r| in that direction from the origin you have reached$\vec r$. Position of$\vec r$in the tangential direction means nothing to me. You could have some other position$\vec s$at distance a in the tangential direction from$\vec r$, in which case$\vec s=\vec r+|\vec s-\vec r|\hat\theta=\vec r+a\hat\theta$(or minus instead of plus). 5. Nov 24, 2016 ### StudentOfScience So, in essence, my problem boiled down to not recognizing that position is really \vec r=|\vec r|\hat r Of course, since the unit vecotr$\hat r$is dependent on the angle with the axis, the vector r is consequently dependent on theta. I suppose that differentiating the equation above with respect to time makes velocity and acceleration functions of time. Thus, the remaining problem is to find a vector valued function, in polar coordinates, that represents the trajectory of the bead. And hence, let us see what to do: \vec v(t)= \dot |\vec r| \hat r + |\vec r|\dot \theta \hat \theta By definition, \frac {d\vec r}{dt} = \vec v(t) Solving this differential equation yields \vec r(t)= \mathbf r_0 + \int_{t_0=0}^t \vec v(u)du We must be careful, though; integration with the unit vectors r and theta are different from i and j; that is, we cannot say that \vec A(t) - \vec A(0)= \int_0^t \vec a(u)\hat r du= A(u)|_0^t , where \frac{d\vec A}{dt}=\vec a(t) Thus, one way to proceed is to rewrite$\hat r$and$\hat \theta$in terms of$\hat i$and$\hat j$and$\theta$, in terms of time (in our problem, we had$\theta=\omega t$), and then integrate term by term and rewrite in polar coordinates. Is there perhaps a slicker way of doing this without the conversion? 6. Nov 25, 2016 ### haruspex Yes, but it will be clearer if we write r for$|\vec r|$:$\vec v=\dot r\hat r+r\dot \theta\hat\theta$. You know that (r, θ)=(ut, ωt). To show it is a particular type of spiral, you need a definition for that type. What definition is acceptable? 7. Nov 25, 2016 ### StudentOfScience Ok, let me get something straight: our position function is \vec r(t)=ut\hat r and since$\hat r$is dependent on$\theta$, which we know to be$\omega t$, we already have our theta/time dependence. (the unit vector r is defined to be dependent on theta, anyways). So, let's start with an intuitive definition of the Archimedean Spiral in a physical context. For polar coordinates, perhaps we can say as |\theta| \rightarrow \infty (I put the absolute value so that we can have a "backwards" spiral as well.) r(\theta) \rightarrow \infty That is,$r(\theta)$is "monotonically increasing" as a function of$\theta$. Where r: \mathbb R \rightarrow \mathbb R which maps$\theta$to$r$. However, note that this is not a vector valued function, so let's consider that case. I'll try to make this a bit more general: Define \vec A: \mathbb R \rightarrow \mathbb R^n Now suppose we have unit vectors$\hat r$and$\hat \theta$, which can be defined in terms of$\hat i$and$\hat j$. Thus, \vec r: \mathbb R \rightarrow \mathbb R^2, \vec r(t)=r\hat r=r(\cos\theta \hat i +\sin\theta \hat j) Where$r$and$\theta$are functions of a single, real variable that maps a real number to a real number (in our case, this variable is$t$). As before, if we let$|\theta (t)|$increase without bound, and if this causes$r$to increase without bound, then let's call$\vec r(t)$the trajectory of an Archimedean spiral. In our case, obviously this is true since both$\theta$and$r$are defined as linear functions of$t##. I probably made multiple mistakes with the notation and such, so please do correct me. If something is unclear, just let me know. Thank you 8. Nov 25, 2016 ### haruspex That's all fine, except that an Archimedean spiral is rather more specific than that. A reasonable definition is that r=aθ for some constant a. 9. Nov 25, 2016 ### StudentOfScience Yes, thank you for the correction. Using our equations for r and theta in terms of t:$$r=ut, \theta=\omega t$$and eliminating t from both equations yields$$r=\frac{u}{\omega}\theta which is the Archimedean spiral we were looking for.

10. Nov 25, 2016

Right.