Position of Particle in a Circular Path: Solution

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In summary, the conversation discusses a parametrization for a particle's position in a circular motion and how it can be shown that the position is given by $x(t)^2 + y(t)^2 = R^2$. There is a mention of unit vectors, but it is clarified that the notation used is not familiar and that they are indeed unit vectors.
  • #1
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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  • #2
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.
 
  • #3
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?
 
  • #4
re: Particles position

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

$\displaystyle \vec{r}(t)=R\langle \cos(\omega t),\sin(\omega t) \rangle$
 
  • #5
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.
 
  • #6
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.
 
  • #7
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. ;)
 

FAQ: Position of Particle in a Circular Path: Solution

What is the formula for calculating the position of a particle in a circular path?

The formula for calculating the position of a particle in a circular path is x = r * cos(θ) and y = r * sin(θ), where r is the radius of the circle and θ is the angle of rotation from the center of the circle.

How do you determine the direction of motion of a particle in a circular path?

The direction of motion of a particle in a circular path can be determined by the direction of the velocity vector, which is always tangent to the circle at any given point. If the velocity vector is pointing counterclockwise, the particle is moving in a counterclockwise direction, and vice versa.

Can the position of a particle in a circular path be negative?

No, the position of a particle in a circular path cannot be negative. The position is measured from the center of the circle, so it can only be positive or zero. If the particle moves outside of the circle, the position can still be represented as a positive value using the formula x = r * cos(θ) and y = r * sin(θ).

How does the velocity of a particle in a circular path change as it moves around the circle?

The velocity of a particle in a circular path is constantly changing, as it is always directed tangent to the circle. As the particle moves along the circle, its velocity vector changes direction, but its magnitude remains constant. This means that the speed of the particle remains constant, but its direction changes.

Is the position of a particle in a circular path affected by the mass of the particle?

No, the position of a particle in a circular path is not affected by the mass of the particle. The position is determined solely by the radius of the circle and the angle of rotation from the center of the circle. The mass of the particle only affects the force required to keep it in circular motion, not its position.

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