Kinematics, curvilinear motion (a couple of questions)

[srnixo]

Homework Statement

Our teacher taught us about curvilinear motion. The lesson was divided into two parts. First, we studied it using [normal and tangential components]. Then, we studied it using [Cartesian components ] . And so, I have a couple of questions because he couldn't explain things well to me. And I did some research as well , but I am unsure if the information I found is correct. HELP!


The questions are not related to any school assignment (homework). They are elements from the lesson presented to us and I want to understand them clearly and without any ambiguity. Regarding the first question, I only need the answer and not any proof.

Relevant Equations

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First question:​

Does the position vector exist in the tangential and normal coordinate system in curvilinear motion?

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Regarding my teacher's response, he confirmed the existence of the position vector formula but was unable to provide it. Upon returning home, I attempted to find its formula.
However, I found three different answers from different sources, and I don't know which one is correct.

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On the other hand, based on some sources from google , I found that :
No, the position vector does not exist in the tangential and normal coordinate system in curvilinear motion. because Normal-tangential (n-t) coordinates are attached to, and move with, a particle. Therefore there is no position vector in n-t coordinates. A typical n-t problem will either give the exact location of the particle on a path, or it will give kinematics information from which the position can be determined.

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Second question:​

Why do we sometimes use [tangential and normal coordinates] , and other times use [Cartesian coordinates] to study curvilinear motion?

According to my teacher:
[the tangential and normal coordinate system should be used when the nature of the path is known, such as when it is curved but will be circular in the end , or curved and then straight, like a train track etc..... ]
On the other hand, the Cartesian coordinate system should be used when the type of path is unknown, which means a general case.

but according to the research conducted:
It has been found that (n-t) coordinates are used for precise descriptions of motion along curved paths, while Cartesian coordinates are used for approximating curved paths with small straight segments.
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When dealing with curvilinear motion, it is recommended to use (n-t) coordinates instead of Cartesian coordinates as they align better with the natural motion of an object along a curved path.

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Third question:​

Is there really a difference between curvilinear and curved motion, or are they just different names for the same motion? In other words, is saying 'curved' or 'curvilinear' essentially the same thing?

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Thank you all for your efforts and willingness to assist me. I appreciate it.

 

[haruspex]

There may be some confusion between curvilinear motion and curvilinear coordinates.
The former just means the motion is not in straight lines, and is independent of the choice of coordinate system. That said, in some cases of curvilinear motion it may be more convenient to use curvilinear coordinates.

Does the position vector exist in the tangential and normal coordinate system in curvilinear motion?

No. https://web.mst.edu/~reflori/be150/FloriNotes/ntCoordLectureNotes1.htm
Please provide links for the answers you found. I cannot make sense of (1) and (3). (2) is ok if it defines $r_T$ as the displacement from the origin in the direction of (current) $\hat t$ and $r_n$ as the displacement from the origin in the direction of (current) $\hat n$, but that is trivially true and not at all helpful.

Why do we sometimes use [tangential and normal coordinates] , and other times use [Cartesian coordinates] to study curvilinear motion?

I don't like either of the answers you quote. I suggest curvilinear coordinates are useful when they correspond to the known facts, i.e. if the equations you can write down from physical considerations relate to the parameters those coordinates use: distance along a path, normal and tangential velocities, normal and tangential accelerations.

Another link: https://www.kpu.ca/sites/default/files/Faculty of Science & Horticulture/Physics/CH12 - 5 - NormalTangential.pdf

 

[srnixo]

Mr @haruspex :

You said it's useful to use curvilinear coordinates but, As I mentioned previously, this lesson was presented by the professor, So I can't change coordinates like i want.
He chose ( n-t coordinates and Cartesian coordinates).

In addition, for the first question about position vector, The link you provided is the same one from which I extracted the information, and since you are more knowledgeable, I think the final answer will be no ( there is no position vector).

 

[Delta2]

The first two formulas you provide for the position vector are correct (and the third too given that the normal component is always zero) BUT:

The position vector needs an origin in order to be defined and those formulas give the position vector with respect to the origin of another coordinate system that has a fixed origin. With respect to the origin of the n-t system the position vector is always ZERO!

So ultimately I ll have to agree with this quote of yours:

No, the position vector does not exist in the tangential and normal coordinate system in curvilinear motion. because Normal-tangential (n-t) coordinates are attached to, and move with, a particle. Therefore there is no position vector in n-t coordinates. A typical n-t problem will either give the exact location of the particle on a path, or it will give kinematics information from which the position can be determined.

 

[srnixo]

Thank you so much Mr @Delta2 . Your explanation is excellent and precise. Now I understand well about this matter. Thank you again.

 

[haruspex]

The first two formulas you provide for the position vector are correct (and the third too given that the normal component is always zero)

I cannot see that the first makes sense at all. What is #R(t)##, and what has the radius of curvature to do with it?
I gave an interpretation in which the second could be correct (but unhelpful). Are you using the same interpretation? If not, what?

Each of the equations could happen to be true in some specific context, such as a straight line trajectory for the third one, but surely what the OP is asking for is a general formula.

 

[kuruman]

Is there really a difference between curvilinear and curved motion, or are they just different names for the same motion? In other words, is saying 'curved' or 'curvilinear' essentially the same thing?

I am not sure I understand what you mean by "curved" motion. I know that curvilinear motion is one-dimensional motion along a curved path. By one-dimensional I mean that there is a single variable $s$ that describes the position of the moving object along the curved path relative to a fixed point on the path. Associated with this variable is a velocity and an acceleration also along the path.

A good example of an object exhibiting curvilinear motion is a car. Its odometer shows $s$ up to that point as the car followed an incredibly complex curvilinear path since it rolled out the manufacturing plant. Furthermore, its speedometer shows the instantaneous rate of change of $s$ with respect to time. Both $s$ and $\frac{ds}{dt}$ are one-dimensional in the sense that there is no reference to a system of coordinate axes, e.g. North-South, East-West or "in the same direction as the motion" $\mathbf{\hat t}$ and "perpendicular to the direction of motion" $\mathbf{\hat n}$.

By contrast, rectilinear motion is one-dimensional motion along a straight line. An object undergoing rectilinear motion at constant speed necessarily has zero acceleration. An object undergoing curvilinear (but not rectilinear) motion at constant speed necessarily has non-zero acceleration. Do you see why?

 

[haruspex]

curvilinear motion is one-dimensional motion along a curved path. By one-dimensional I mean that there is a single variable s that describes the position

Isn't that confusing the motion with the coordinates used to express it?

From Wikipedia,

"The motion of an object moving in a curved path is called curvilinear motion.[1] Example: A stone thrown into the air at an angle."

 

[Delta2]

I cannot see that the first makes sense at all. What is #R(t)##, and what has the radius of curvature to do with it?

It is correct but the math are not so obvious $R(t)$ is the total arc length (or the length of the trajectory curve).. You can get this expression by considering the vector integral $$\vec{r}=\int \vec{v}dt=\int v(t)\hat e_t dt$$ where $v(t)$ the speed of the particle and doing integration by factors for $v(t)=\frac{dR(t)}{dt}$. The radius of curvature comes into play by considering the $$\int R(t)\frac{d\hat e_t}{dt} dt$$ which is not always zero.

EDIT: Unlucky I overloaded symbol t, it is both time and the symbol for the tangential component.

 

[haruspex]

The radius of curvature comes into play by considering the $$\int R(t)\frac{d\hat e_t}{dt} dt$$ which is not always zero.

Consider a path which has so far reached some point $\vec r$. It then continues straight for a short distance. $\rho$ is infinite and $\hat n$ exists.

 

[Delta2]

Consider a path which has so far reached some point $\vec r$. It then continues straight for a short distance. $\rho$ is infinite and $\hat n$ exists.

yes ok I guess some smoothness conditions are needed such as for example that the radius of curvature and perhaps $\frac{d\hat e_t}{dt}$ are continuous functions along the path.

 

[haruspex]

yes ok I guess some smoothness conditions are needed such as for example that the radius of curvature and perhaps $\frac{d\hat e_t}{dt}$ are continuous functions along the path.

You can make the transition from a curved path to a straight one as smooth as you like. E.g. $y=x^n, x>0; y=0, x<0$. The higher you make n, the more derivatives are zero both sides of the origin.
I get that $\hat n\dot R=\rho\dot e_t$.
How about you verify your equation for some simple case?

 

[Delta2]

You can make the transition from a curved path to a straight one as smooth as you like. E.g. $y=x^n, x>0; y=0, x<0$. The higher you make n, the more derivatives are zero both sides of the origin.
I get that $\hat n\dot R=\rho\dot e_t$.
How about you verify your equation for some simple case?

It's not my equation e hehe , I checked hastily the tangential part which seem correct to me so I thought the normal part would also be correct, but yeah it might need some *slight* correction regarding the normal part, hold on while I check it in more detail

 

[haruspex]

I thought the normal part would also be correct

It was always that part that struck me as ludicrous. How could the instantaneous radius of curvature at the endpoint be relevant?
@srnixo , where did you find (1)?

 

[Delta2]

You got me in a bad time , its early morning, my sleep disease is at a peak today and did very bad sleep, it seems that that integral is equal $$\int \frac{R\dot R}{\rho} \hat e_n dt$$ but I see no way of proceeding further, maybe later if I manage to feel a bit better, if we do integration by factors again it seems to give a term $$\frac{R^2}{\rho}\hat e_n$$ but it gives another term too which I find difficulty to prove that it is zero, maybe later today when my mind becomes more clear.

 

[Delta2]

It was always that part that struck me as ludicrous. How could the instantaneous radius of curvature at the endpoint be relevant?

It seems to be relevant cause of that term at post#15 but for sure not in the way the OP claims.

@srnixo , where did you find (1)?

Maybe he did some sort of understanding error when he was reading the page, and that $\rho(t)$ is in fact something else.

 

[Orodruin]

On the other hand, based on some sources from google

This is not good as reference. We have no way of checking what exactly you have read. As such, we have no way of telling whether youf sources were incorrect or if you just drew incorrect conclusions.

I think the final answer will be no ( there is no position vector).

Wrong. The existence of a position vector is tied to whatever underlying space you have (ie, if it is affine or not), not on any particular choice of coordinates.

With respect to the origin of the n-t system the position vector is always ZERO!

The position vector of the object itself. This is an important distinction, there may be other objects.

 

[srnixo]

where did you find (1)?

I found it on a website in Arabic, as far as I remember. they said that: R(t) represents the radius of curvature in tangent direction and P(T) is the radius of curvature in the normal direction at that point.

so that the equation (1) provides a vector that accurately describes the position and curvature of the object at any point along the curved path.

and they said also that the equation (1) is similar to (2) but uses different notations.

 

[Delta2]

@srnixo Can you give us more detail on how they define $R$ the radius of curvature in the tangent direction. I got a hunch though I think it is defined so that $$v\hat e_t=R \frac{d \hat e_n}{dt}$$

 

[haruspex]

: R(t) represents the radius of curvature in tangent direction and P(T) is the radius of curvature in the normal direction at that point.

For a curve (not a surface) there is only one radius of curvature at a point. I have no idea what "radius of curvature in tangent direction" means.

In 3D it gets more interesting. I did wonder whether the Arabic link related to that.
I use s for distance along the curve. All derivatives are wrt s.
If the position vector from some arbitrary origin is $\vec r$ then the unit tangent vector is $\hat t=\vec r'$. If $\hat t'$ is nonzero at a point then the curvature there is $\kappa=|\hat t'|$. The radius of curvature is $\rho=1/\kappa$.
The principal normal unit vector is $\hat n=\hat t'\rho$.
The plane containing those two unit vectors is called the osculating plane. As we progress along the curve, the osculating plane can rotate about the tangent.
The "binormal" is normal to both the tangent and the principal normal, $\hat b=\hat t\times\hat n$, and so is normal to the osculating plane.
$\hat b'$ is necessarily parallel to $\hat n$, so we can define the rate of rotation, or "torsion", by $\hat b'=\tau\hat n$.
But this does not correspond to any radius that I can envision.

 

[haruspex]

@srnixo Can you give us more detail on how they define $R$ the radius of curvature in the tangent direction. I got a hunch though I think it is defined so that $$v\hat e_t=R \frac{d \hat e_n}{dt}$$

In 2D, the derivatives of $e_t$ and $e_n$ necessarily have the same magnitude.

 

[Delta2]

Hmmm This gets quite complex. But my own question here is:
Since the velocity vector can clearly be given in n-t coordinate what does the integral $$\int \vec{v} dt=\int v(t)\hat e_t dt$$ represents??? $v(t)$ the speed.

In my opinion it represents the position vector where we take as (fixed )origin the position of the particle at t=0.

 

[haruspex]

Hmmm This gets quite complex. But my own question here is:
Since the velocity vector can clearly be given in n-t coordinate what does the integral $$\int \vec{v} dt=\int v(t)\hat e_t dt$$ represents??? $v(t)$ the speed.

In my opinion it represents the position vector where we take as (fixed )origin the position of the particle at t=0.

Of course, since $\vec v=\frac{d\vec r}{dt}$.
But an integral form for the position vector is not what we are after.

 

[Delta2]

But an integral form for the position vector is not what we are after.

Hmm yes it is not clear if the integral of a vector in n-t coordinates can also be expressed as a vector in n-t coordinates.

 

[Delta2]

I can almost feel it that there must be some suitable definition of $R(t)$ and $\rho(t)$ such that the integral $$\int v(t) \hat e_t dt$$ equals $$ R(t) \hat e_t+\rho (t)\hat e_n$$ if not for all trajectories at least for some trajectories with some additional properties.
If the trajectory is straight line it holds for $$R(t)=\int v(t) dt$$ and $$\rho(t)=0$$ for example.

 

[srnixo]

With respect to the origin of the n-t system the position vector is always ZERO!

I've done some more research and found out that the tangential and normal coordinate systems do not have a fixed origin (the origin moves with the object). Therefore, the position vector of the object will constantly change to reflect its changing position and will not be equal to ZERO.

It has been found also that equations 1 and 2 are both correct, albeit with different notations. However, it is recommended to use equation (2) as it provides a more explicit and clear representation of the key parameter involved in describing the position vector in curvilinear motion.

So, in most cases, the position vector will not be zero unless the point is at the origin itself, which means that both the (rt) and the (rn) components would have to be 0 at the same time, which would imply that the point is at the origin of the n-t coordinate system.

However, as the object moves along the curve, the values of rt and rn will change, resulting in a changing position vector.

[ I hope that this information is accurate ]

 

[Delta2]

I've done some more research and found out that the tangential and normal coordinate systems do not have a fixed origin (the origin moves with the object). Therefore, the position vector of the object will constantly change to reflect its changing position and will not be equal to ZERO.

It has been found also that equations 1 and 2 are both correct, albeit with different notations. However, it is recommended to use equation (2) as it provides a more explicit and clear representation of the key parameter involved in describing the position vector in curvilinear motion.

So, in most cases, the position vector will not be zero unless the point is at the origin itself, which means that both the (rt) and the (rn) components would have to be 0 at the same time, which would imply that the point is at the origin of the n-t coordinate system.

However, as the object moves along the curve, the values of rt and rn will change, resulting in a changing position vector.

[ I hope that this information is accurate ]

Ehm the definition of the position vector is such that it requires a FIXED origin(sorry for the caps). In n-t coordinate system the origin is not fixed hence the position vector cannot even be defined for an n-t system.

However we can find an expression $$\vec{r}=R(t) \hat e_t+\rho(t) \hat e_n$$ that gives the position vector not in the n-t coordinate system (though it is expressed with the unit tangential and normal vectors $\hat e_t,\hat e_n$) but this expression is the position vector in any coordinate system where we take as FIXED origin the point where the particles lies at t=0. Of course the unit tangential and normal vectors have to be expressed as functions of the chosen coordinate's system unit vectors. If the chosen coordinate system is cartesian and we know the trajectory as $y=f(x)$ then there are well known expressions that give $\hat e_t$ and $\hat e_n$ as expressions that include the function f and the cartesian unit vectors i,j.

 

[srnixo]

Could you Mr @Delta2 clarify for me something!

Is this formula (equation1) which you provided only used when the object is at rest in the origin or moving along a curved path? In other words, It should not be used when the object is in motion?

 

[Delta2]

Could you Mr @Delta2 clarify for me something!

Is this formula (equation1) which you provided only used when the object is at rest in the origin or moving along a curved path? In other words, It should not be used when the object is in motion?

the formula $$\vec{r}=R(t)\hat e_t+\rho(t)\hat e_n$$ can be used for any t, that is even the object is moving, it is just that due to FIXED origin requirement it doesn't give us the position vector in n-t coordinate system, though the expression is such that it might trick us that it does so.

 

[srnixo]

Alright, Thank you so much for your efforts. This is so helpful.

 

[haruspex]

the tangential and normal coordinate systems do not have a fixed origin (the origin moves with the object). Therefore, the position vector of the object will constantly change

Ok, so in eqn 1, what does $\vec r$ mean? It is not the location of the object moving along the path s since that is by definition (0,0) in n-t. It must be the location of some other point expressed in n-t coordinates relative to the object, but what point?

$\rho \hat n$ is the location of the centre of curvature, but what is $R(t)$? I can find no online reference to "radius of curvature in tangent direction", and @Delta2's proposed interpretation in post #19 makes it the same as $\rho$.

Here's another interpretation: $R$ and $\rho$ (or perhaps it is $P$, not $\rho$) have nothing to do with radii of curvature. All the eqn is saying is that given some arbitrary point $\vec r$ and a point on a curve as origin, $\vec r$ can be expressed in curvilinear coordinates as $(R, P)$, meaning $R\hat t+P\hat n$.