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Is it possible to determine the position of a particle undergoing circular motion, in x-y coordinates, at any given time and velocity?

I'm thinking it has something to do with $\pi$

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And what is the question? –  Marek Jun 4 '11 at 21:28
    
it's in the title. –  Dan the Man Jun 4 '11 at 21:29
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what you have in the title is an indicative sentence followed by a question mark. But even if it were grammatically correct, I am not sure what you want. Do you want express particle's trajectory in cartesian coordinates? –  Marek Jun 4 '11 at 21:42
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well, maybe. We can guess at what you meant but it's hard to be sure unless you are very clear about your question. It's easy to miscommunicate in physics if you're not careful, so we place a lot of emphasis on clarity in writing questions and answers here. I did you the favor of editing your question to make it more clear. –  David Z Jun 4 '11 at 22:23
    
ok. i will keep that in mind. thanks –  Dan the Man Jun 4 '11 at 22:39
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2 Answers 2

up vote 1 down vote accepted

Yes, unless you want to be very picky. The coordinates are described by the trig functions sine and cosine.

Suppose something is moving on a circle with radius $R$, so its $x$ and $y$ coordinates obey

$$x^2 + y^2 = R^2$$

If it moves at a uniform velocity of $v$, then the angle its path subtends is linear in time (by dimensional analysis), so we can define the angle $\theta$ as

$$\theta = vt/R$$

One full rotation turns out to be an angle of about 6.28, but it's really a transcendental number, $2\pi$.

If you draw a circle, then draw the angle from the equation above, you find a unique intersection point that tells you where the object is. If you project that point onto the $x$ and $y$ axes, you've graphically determined the coordinates.

enter image description here

The coordinate you obtain, for a given angle, is linear in $R$ (again by dimensional analysis, coupled with a feature of Euclidean geometry - that it has no innate length scale), so if we divide the coordinates by $R$ we get two functions that map $[0,2\pi) \to [-1,1]$ because $-1$ and $1$ are the minimum and maximum values, since $-R$ and $R$ are the minimum and maximum values of the coordinates.

These functions are give the names "cosine" and "sine", and we say

$$x/R = \cos\theta$$

$$y/R = \sin\theta$$

The sine and cosine functions can be represented as infinite series,

$$\cos\theta = 1 - \theta^2/2 + \theta^4/24 - \ldots$$

$$\sin\theta = \theta - \theta^3/6 + \theta^5/120 - \ldots$$

We know many of the properties of these functions, and they are used extensively in mathematics and physics. You can calculate their values to high precision with a computer. However, we cannot simply write down exactly what $\sin(1)$ is in any simple way that you can explain to an elementary-school student. If you want to know what it is, you can simply use a calculator, like this.

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I see we are not very far away from explaining how addition works too on this site :) In any case, good job figuring out what OP wanted (if he wanted this)... –  Marek Jun 4 '11 at 22:04
    
@Marek Yeah, it's definitely an elementary question, but at least it's not "plzz du my homework!! my test iz in one huor!!" Also, the OP is 16 years old according to his profile, so he might never have been taught this material, or not taught it well. –  Mark Eichenlaub Jun 4 '11 at 22:06
    
WOW! thank you very much! i needed those equations. –  Dan the Man Jun 4 '11 at 22:08
    
I'm teaching myself lol –  Dan the Man Jun 4 '11 at 22:08
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Of course. How do you represent a position in 2d space? as a couple of numbers, right? Just make them variables, functions in respect to time. The following functions describe circular motion (with radius a):

$x(t)=a \sin(t)$

$y(t)=a \cos(t)$

Here, go to this site: http://fooplot.com , change the style to parametric (they use $s$ instead of $t$) and use the above functions. Try different functions :)

I'm not sure if I answered your question.. What level are you?

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good answer! ty for the site. very cool. –  Dan the Man Jun 4 '11 at 22:09
    
I'm in high school –  Dan the Man Jun 4 '11 at 22:18
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