Trajectory of a particle when its position vector changes

In summary: I think that the derivative would be ##2 \vec r## if I use the product rule.With respect to time? ##\vec r.\vec r## would be 0. And ##\vec r^2## would be 0 too,... but it should be different, right? I think that the derivative would be ##2 \vec r## if I use the product rule.Yes, so what can you say about ##\vec r^2##?Yes, so what can you say about ##\vec r^2##?In summary, the position vector of a particle changes only by its module or direction. The trajectory of the particle's movement can be a line or a circle
  • #36
Davidllerenav said:
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.
 
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  • #37
haruspex said:
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?
 
  • #38
Davidllerenav said:
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.
 
  • #39
haruspex said:
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##?
 
  • #40
Davidllerenav said:
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 ...?
 
  • #41
haruspex said:
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
 
  • #42
Davidllerenav said:
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.
 
  • #43
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.
 
  • #44
haruspex said:
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?
 
  • #45
Davidllerenav said:
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.
 
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  • #46
Davidllerenav said:
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.
 
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  • #47
kuruman said:
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.
 

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