Position of Particle in a Circular Path: Solution

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Discussion Overview

The discussion centers around the mathematical representation of a particle's position moving in a circular path, specifically using parametric equations. Participants explore the implications of the notation used and the properties of the vectors involved.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents the position of the particle as a function of time using the equation $\mathbf{r}(t) = \hat{\mathbf{x}}R\cos(\omega t) + \hat{\mathbf{y}}R\sin(\omega t)$, noting that at $t = 0$, the particle is at $(R,0)$.
  • Another participant suggests considering the parametrization $x(t)=R\cos(\omega t)$ and $y(t)=R\sin(\omega t)$, and proposes squaring both equations and adding them to eliminate the parameter.
  • Some participants discuss the notation of unit vectors $\hat{\mathbf{x}}$ and $\hat{\mathbf{y}}$, questioning their interpretation and whether they are orthogonal unit vectors.
  • One participant expresses confusion over the notation and proposes an alternative representation using $\vec{r}(t)=R\langle \cos(\omega t),\sin(\omega t) \rangle$.
  • Another participant confirms that the vectors are indeed unit vectors, expressing frustration over the notation used in the context of the discussion.

Areas of Agreement / Disagreement

Participants exhibit some agreement on the mathematical representation of the particle's position, but there is disagreement regarding the interpretation of the notation and the properties of the vectors involved. The discussion remains unresolved on these points.

Contextual Notes

There is uncertainty regarding the notation of unit vectors and their properties, as well as the implications of the parametrization used. Some assumptions about the vectors being orthogonal are not explicitly confirmed.

Dustinsfl
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Relatively easy question but I haven't done anything like this in awhile.
$$
\mathbf{r}(t) = \hat{\mathbf{x}}R\cos(\omega t) + \hat{\mathbf{y}}R\sin(\omega t)
$$
The particle moves in a circle so I want to show that the position is given by the above.
I know at $t = 0$ the particle is on the x-axis.
Therefore, we know that at t = 0, $(R,0)$.
Obviously it has to be that but I guess there is a way to show this basically.
 
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re: Particles position

Consider the parametrization:

$\displaystyle x(t)=R\cos(\omega t)$

$\displaystyle y(t)=R\sin(\omega t)$

Square both and add to eliminate the parameter.
 
re: Particles position

MarkFL said:
Consider the parametrization:

$\displaystyle x(t)=R\cos(\omega t)$

$\displaystyle y(t)=R\sin(\omega t)$

Square both and add to eliminate the parameter.

That will just be $x(t)^2 + y(t)^2 = R^2$.
What about x and y hat?
 
re: Particles position

Aren't they orthogonal unit vectors? I'm sorry, I'm not familiar with that notation. I interpreted it as:

$\displaystyle \vec{r}(t)=R\langle \cos(\omega t),\sin(\omega t) \rangle$
 
re: Particles position

MarkFL said:
Aren't they orthogonal unit vectors? I'm sorry, I'm not familiar with that notation. I interpreted it as:

$\displaystyle \vec{r}(t)=R\langle \cos(\omega t),\sin(\omega t) \rangle$

They could be but it doesn't specify. All we know is that they are vectors.
 
re: Particles position

dwsmith said:
They could be but it doesn't specify. All we know is that they are vectors.

I just looked through the book. They are unit vectors. What a dumb notation.
 
re: Particles position

Good deal. I believe I have seen that notation a long time ago in a galaxy far away, but I wasn't sure. ;)
 

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