Trajectory of a particle when its position vector changes

[Davidllerenav]

Right. And what do those two perpendicular vectors represent in terms of the motion?

I think that they represent the components of the vector, is it right?

 

[haruspex]

I think that they represent the components of the vector, is it right?

No, they are two different vectors, $\vec r$ and $\dot{\vec r}$. What does each represent in terms the motion?

 

[Davidllerenav]

No, they are two different vectors, $\vec r$ and $\dot{\vec r}$. What does each represent in terms the motion?

The position and the velocity?

 

[haruspex]

The position and the velocity?

Right. So we deduce that in this motion the position vector is perpendicular, always, to the velocity vector.

 

[Davidllerenav]

Right. So we deduce that in this motion the position vector is perpendicular, always, to the velocity vector.

But what does that mean?

 

[haruspex]

But what does that mean?

It means thar if you draw a line from the origin to the point, the particle moves at right angles to that line - neither towards nor away from the origin.

 

[Davidllerenav]

It means thar if you draw a line from the origin to the point, the particle moves at right angles to that line - neither towards nor away from the origin.

So it only changes its directio, right? And how would it be when it only changes its magnitude?

 

[haruspex]

So it only changes its directio, right? And how would it be when it only changes its magnitude?

Well, you had the answer to that in post#1, but I suppose you are looking for an analytical approach.
The only thing I can think of is to write the vector in coordinates. I don't see a neat vector approach.

 

[Davidllerenav]

Well, you had the answer to that in post#1, but I suppose you are looking for an analytical approach.
The only thing I can think of is to write the vector in coordinates. I don't see a neat vector approach.

Somethinh like $\vec r = r cos \theta + r sin \theta$?

 

[haruspex]

Somethinh like $\vec r = r cos \theta + r sin \theta$?

Nearly. The right hand side needs to be a vector too.
And if the direction doesn't change then ...?

 

[Davidllerenav]

Nearly. The right hand side needs to be a vector too.
And if the direction doesn't change then ...?

Well if the direction doesn't change it is a one dimesional movement, so de y coordinate is 0

 

[haruspex]

Well if the direction doesn't change it is a one dimesional movement, so de y coordinate is 0

Coordinate directions are arbitrary. You can choose coordinates such that the x coordinate is the direction of movement, but it does not have to be that way.

 

[Delta2]

I think both questions can be answered by considering the position vector expressed as

$\vec{r}=|\vec{r}|\hat{r}$ where $\hat {r}$ is the unit position vector.

So at least the first question becomes easy to answer since it is given that only $|\vec{r}|$ changes while $\hat{r}$ remains constant.

 

[Davidllerenav]

Nearly. The right hand side needs to be a vector too.
And if the direction doesn't change then ...?

Then it would be $}\vec r = \vec r cos \theta + \ vec r sin \theta$?
If the direction doesn't change, it only moves in an stright line, thus the magnitude changes, right?

 

[haruspex]

Then it would be $\vec r = \vec r \cos \theta + \vec r \sin \theta$?
If the direction doesn't change, it only moves in an stright line, thus the magnitude changes, right?

No, that can't be right either. You could rewrite that as $\vec r = \vec r (\cos \theta + \sin \theta)$, i.e $(\cos \theta + \sin \theta)=1$.
Use @Delta2 's formulation in post #43.

 

[kuruman]

Somethinh like $\vec r = r cos \theta + r sin \theta$?

You almost got there with this post but not quite. Look at #40 by @haruspex and write the right hand side of the equation as a vector. This means find expressions for $r_x$ and $r_y$ such that $\vec r=r_x~\hat i+r_y~\hat j.$ Note that once you choose the coordinate system, the Cartesian axes are fixed and so are the unit vectors $\hat i~$ and $\hat j~$. Therefore in general when vector $\vec r$ changes, only $r_x$ and $r_y$ change and not the unit vectors. The question before you then is "how should $r_x$ and $r_y$ change in particular so that the direction of $\vec r$ remains constant?" You can find an analytical expression describing that either in terms of $r_x$ and $r_y$ or in terms of $\theta$ and $r$, whichever you think is easier.

What I suggest is not really different from the suggestion by @Delta2 in #43 which implies writing $\hat r~$ in terms of Cartesian unit vectors. The point is that the direction of $\hat r$ depends on $\theta$ so if you want to express a fixed direction analytically, you need to express it in terms of fixed (Cartesian) unit vectors.

 

[Davidllerenav]

You almost got there with this post but not quite. Look at #40 by @haruspex and write the right hand side of the equation as a vector. This means find expressions for $r_x$ and $r_y$ such that $\vec r=r_x~\hat i+r_y~\hat j.$ Note that once you choose the coordinate system, the Cartesian axes are fixed and so are the unit vectors $\hat i~$ and $\hat j~$. Therefore in general when vector $\vec r$ changes, only $r_x$ and $r_y$ change and not the unit vectors. The question before you then is "how should $r_x$ and $r_y$ change in particular so that the direction of $\vec r$ remains constant?" You can find an analytical expression describing that either in terms of $r_x$ and $r_y$ or in terms of $\theta$ and $r$, whichever you think is easier.

What I suggest is not really different from the suggestion by @Delta2 in #43 which implies writing $\hat r~$ in terms of Cartesian unit vectors. The point is that the direction of $\hat r$ depends on $\theta$ so if you want to express a fixed direction analytically, you need to express it in terms of fixed (Cartesian) unit vectors.

Oh, I see thanks, I will try to describe $r_x$ or $r_y$ on terms of $\theta$ and $r$, I'll post when I do it.