Equations of motion of a point sliding on a line of arbitrary shape

[Atomillo]

Homework Statement

Let f(x) be the shape of a line. A point is located in some initial position of this line. What are the equations of motion of this point?

Relevant Equations

Conservation of Energy.

Hi!

First of all, mention that this is not a "homework" problem in the sense that no teacher ever gave it to me or that I have the obligation to do it. It is a question that came to mind when repasing the theory done in class and though interesting. I still post it here because I suppose that what I'm missing is something basic, but in case it didn't belong to this category please tell me so that I can delete it and comply with the rules. Thanks in advance.

So, I'm currently in first semester of Electronic Engineering after doing HL Physics in IB. In class we studied the motion of a point in an inclined plane. I wonder however how could we find the equations of motion for a point in a line (2D, to keep things as simple as possible) that is not straight, i.e that is defined by some function f(x).

I naturally came to the conservation of energy theorem. After some algebraic manipulation to find the velocity, I got:

v(h) = √(2*(E-mgh)/m)
​

Where E is the total mechanical energy at the starting point and mgh is the potential energy at a height h. However, if the shape of the plane is expressed in the function f(x) as stated previously, then the height h at a point a is simply f(a). Re writing:

v(x) = √(2*(E-mgf(x))/m)
​

Here I came to my first problem: I was looking for v(t) and got v(x). Perusing the internet, I found this very helpful thread which I followed: Time dependence of velocity from position dependence of velocity. After doing as stated in the forum, I got:

t + C = ∫1/(v(x)) dx
​

After solving the integral, the only thing left to do would be to solve for x. In order to check that this result is correct, I then try to derive from this general case the equations of motion for an inclined plane of 45 degrees. Now, this plane is characterized by the equation f(x) = -x. Replacing, I get:

v(x) = √(2*(E+mgx)/m)
​

Using an HP Prime I solved the integral to obtain:

t + C = √2 * √(gm^2x+mE)/gm
​

Assuming C = 0, x is then:

x(t) = (g^2mt^2-2E)/2gm
​

Now, assuming the point starts at x = 0, that the initial velocity is cero, then E = 0 (since E is the initial mechanic energy, and starting at x=0 that means h=0 and there is no kinetic energy). After all this, I get:

x(t) = (g*t^2)/2
​

This doesn't make any sense! This equation is describing the motion of free fall i.e an angle of 90 degrees! How could this happen? Does this prove that the third equation is incorrect, or did I screw something up in the way to the result?

This is my first post, so if I'm doing something incorrectly, posting on the wrong place, etc... please tell me so that I can correct it as soon as possible.

Thanks!

EDIT: I said that this was my first post. This is not true! It has been so long since I posted (lurking only) that I completely forgot. My apologies.

​

 

[PeroK]

t + C = ∫1/(v(x)) dx
​

What does this equation mean?

 

[Atomillo]

What does this equation mean?

I'm following the process stated in the forum thread linked to convert v(x) to v(t). That equation is supposed to give (when solved) the velocity of a point in a time t in any line f(x).
EDIT: Sorry, I floped. The equation when solved does NOT give v(t) but x(t). My bad

 

[PeroK]

I'm following the process stated in the forum thread linked to convert v(x) to v(t). That equation is supposed to give (when solved) the velocity of a point in a time t in any line f(x).
EDIT: Sorry, I floped. The equation when solved does NOT give v(t) but x(t). My bad

At least one confusion appears to be that if you define a curve by $(x, f(x))$, then $\frac{dx}{dt}$ is not the speed of the particle along the curve.

Fundamentally, you have a 2D motion with $\vec{v}(t) = (v_x(t), v_y(t))$, where there is a relationship between $x$ and $y$ as long as the particle stays on the curve. That said, unless there are constraining forces, the particle may fly off the curve.

I'm not sure anything of what you've done makes much sense.

 

[Atomillo]

At least one confusion appears to be that if you define a curve by $(x, f(x))$, then $\frac{dx}{dt}$ is not the speed of the particle along the curve.

Fundamentally, you have a 2D motion with $\vec{v}(t) = (v_x(t), v_y(t))$, where there is a relationship between $x$ and $y$ as long as the particle stays on the curve. That said, unless there are constraining forces, the particle may fly off the curve.

I'm not sure anything of what you've done makes much sense.

I'm trying to replicate the equations of motion that appear in an inclined plane when the plane is not simply a straight line.
Being honest I never even considered that when the plane isn't simply straight you cannot treat it as a one dimensional problem with one velocity but instead need one for each axis. I simply tried extending the same concept. That means that everything discussed above is simply pointless, right?

 

[PeroK]

I'm trying to replicate the equations of motion that appear in an inclined plane when the plane is not simply a straight line.
Being honest I never even considered that when the plane isn't simply straight you cannot treat it as a one dimensional problem with one velocity but instead need one for each axis. I simply tried extending the same concept. That means that everything discussed above is simply pointless, right?

You can study motion on a stright line curve as 1D by defining your variable along the curve.

For other problems, the energy will be in terms of the y-coordinate. That will give you the speed in terms of the y-coordinate.

You can relate $\frac{dy}{dt}$ to $\frac{dx}{dt}$, as long as the particle stays on the curve. In simple cases you should be able to get a formula for $v(t)$ where $v$ is the speed of the particle.

I'd definitely start again, but you may need the calculus of variations for one thing in genuine 2D problems, like the brachistrochrone:

https://en.wikipedia.org/wiki/Brachistochrone_curve

 

[Atomillo]

It's the first time I heard about such a curve. It might be a good idea to wait a little bit and ask my calculus/physics teacher if we are going to study it in class.

"For other problems, the energy will be in terms of the y-coordinate. That will give you the speed in terms of the y-coordinate." Is that why I got the equations of free fall? If I understood correctly, that would mean that I'm only calculating the velocity in the y direction.

For simplicity I'm always assuming the particle stays on the curve.

¿Could you please elaborate on relating the velocities of the two axis into one expression?

EDIT: Is v(t) the celerity i.e the modulus of v? In that case, wouldn't it be possible to use conservation of energy to get such a function?
I'm grateful for the help provided: I think I'm starting to see the problems.

 

[haruspex]

If I understood correctly, that would mean that I'm only calculating the velocity in the y direction.

No, both x direction speed and y direction speed contribute to the kinetic energy, but the value of the kinetic energy relates to the y coordinate by conservation of mechanical energy. If it is a plane at angle θ to the horizontal, x is the horizontal coordinate and y is the up plane coordinate then $\frac 12(\dot x^2+\dot y^2)+gy\sin(\theta)=E$.
If the curve is y=f(x), what is the relationship between f', $\dot x$ and $\dot y$?

 

[Atomillo]

I'm really sorry, but I'm not completely following you. When you say: "but the value of the kinetic energy relates to the y coordinate by conservation of mechanical energy" I do not understand it. In the same line, I don't understand the second term in the calculation of E (assuming E is the kinetic energy discussed so far and not the mechanical). I'm probably missing something very obvious and very basic, I apologize for being slow in understanding. Up until this point in my education when the kinetic energy of a sistem in 2d had to be calculed we splitted the coordinates into x and y and calculated the kinetic energies of each one separately. Is there any resource (a web, PDF, etc...) I could read to understand this?

 

[hutchphd]

The constraint forces do no work. Hence the speed is all produced by gravity (change in PE).

 

[Atomillo]

Oh, so I can simply do (assuming rest at the beggining):

Kinetic energy = Potential Energy
1/2(dx/dt^2 + dy/dt^2)m = mgxsin(θ)
1/2(dx/dt^2 + dy/dt^2) - gxsin(θ) = 0
​

However, in the expression given by haruspex there is y instead of x. There is also addition, no substraction Furthermore, isn't he calculating kinetic energy and not mechanical energy? Furthermore, for this to work doesn't the particle have to be at h=0 when we calculate it's velocity so that all the potential energy is converted into kinetic energy? I'm pretty sure this is not what haruspex wrote

​

 

[hutchphd]

What exactly do you wish to know? The speed of the particle at any position is determined simply using energy conservation and its direction is tangent to the curve of constraint. It is probably useful to be able to solve this using more direct (and complicated) methods for pedagogic reasons. This was mentioned by @PeroK .

 

[Atomillo]

The speed of the particle at any position is just what I want to know, and in fact energy conservation is what I tried to use in order to find it. The reason I made this post is that the result that I have obtained is not correct.
More specifically, when I let f(x) = -x (a simple inclined plane with angle = 45 degrees) I obtain the equations of motion of free fall, not those of the corresponding plane. I'm asking for help in spotting the mistakes in the process I described above (that is, why I don't obtain the expected equation for this particular case).

 

[PeroK]

The speed of the particle at any position is just what I want to know, and I fact energy conservation is what I tried to use in order to find it. The reason I made this post is that the result that I have obtained is not correct.
More specifically, when I let f(x) = -x (a simple inclined plane with angle = 45 degrees) I obtain the equations of motion of free fall, not those of the corresponding plane. I'm asking for help in spotting the mistakes in the process I described above (that is, why I don't obtain the expected equation for this particular case).

It would be better to describe the curve as $y = -x$. We've already said that you are mixing up 1D motion with a single parameter $x$ and 2D motion with $x, y$ parameters.

You can get the KE as described above - from simple conservation of energy. This gives you the velocity as a function of position. Velocity as function of time is often much harder to get, because you need to calculate the length of the curve, which is often difficult.

 

[Atomillo]

I think I see the problem. Because there are two parameters (x, y) velocity is not simply dx/dt but dx/dt, dy/dt (as stated previously by yourself ).

Thus, the method I was using to convert v(x) to v(t) (https://math.stackexchange.com/ques...-velocity-from-position-dependece-of-velocity) is invalid. Thus, I have to find another method. (¡Which is just also what you mentioned with the Brachistochrone curve!)

For some reason it seems my brain is not able to work as well with two parameters as with one; lack of practice I suppose. It also seems unable to properly understand what's reading, taking into account the problem was mentioned explicitly in the fifth answer...

Assuming the particle does not leave the track, is there any metod similar to the described in the above link (for example, taking two partial derivatives, one for vx and another for vy), or do I have to use tools like the Brachistochrone curve (with which I have no experience whatsoever)? When you say "because you need to calculate the length of the curve, which is often difficult." to what are you referring exactly?

Again, thanks all for the help given despite my slowness.

 

[PeroK]

I think I see the problem. Because there are two parameters (x, y) velocity is not simply dx/dt but dx/dt, dy/dt (as stated previously by yourself ).

Thus, the method I was using to convert v(x) to v(t) (https://math.stackexchange.com/ques...-velocity-from-position-dependece-of-velocity) is invalid. Thus, I have to find another method. (¡Which is just also what you mentioned with the Brachistochrone curve!)

For some reason it seems my brain is not able to work as well with two parameters as with one; lack of practice I suppose. It also seems unable to properly understand what's reading, taking into account the problem was mentioned explicitly in the fifth answer...

Assuming the particle does not leave the track, is there any metod similar to the described in the above link (for example, taking two partial derivatives, one for vx and another for vy), or do I have to use tools like the Brachistochrone curve (with which I have no experience whatsoever)? When you say "because you need to calculate the length of the curve, which is often difficult." to what are you referring exactly?

Again, thanks all for the help given despite my slowness.

The Brachistrochrone is just a good example of what you can do in 2D problems.

It's not too hard to write down the equations of motion for an obect on a 2D curve. But, solving those equations can be hard.

You can start with $E = mgy + \frac 1 2 mv^2$

What happens if you differentiate that?

 

[Atomillo]

With respect to what should I differentiate it?

 

[PeroK]

With respect to what should I differentiate it?

Time!

 

[Atomillo]

The result is zero, and that is basically the principle of conservation of energy i.e energy is invariant and thus it's derivative must be zero (a constant). Correct?

 

[PeroK]

The result is zero, and that is basically the principle of conservation of energy i.e energy is invariant and thus it's derivative must be zero (a constant). Correct?

There are two sides to every equation.

 

[TSny]

You can start with $E = mgy + \frac 1 2 mv^2$

What happens if you differentiate that?

I wouldn't bother to differentiate the expression. Note that if you can find $x$ as a function of $t$ then you will automatically have $y$ as a function of $t$ via $y(t) = f[x(t)]$.

Write $$E = mgy + \frac 1 2 mv^2$$ as $$E = mgf(x) + \frac 1 2 m \left( \dot x^2 + \dot y^2 \right)$$

Then follow @haruspex 's suggestion of writing $\dot y$ in terms of $\dot x$ by making use of $y = f(x)$. You should end up with an equation involving just $x$ and $\dot x$ and for which you can separate variables and integrate (in principle) to find $t$ as a function of $x$. If you're lucky, you'll then be able to invert this to get $x(t)$.

 

[Atomillo]

This is where my knowledge of calculus is just unable to keep up. Just to check if I have structured the equation correctly:
f(x) = x^2 (for example)
E = mgx^2 + 1/2m((dx/dt)^2+(x^2dx/dt)^2)
And I would put this expression in some online solver.
Thanks for all the help. The conclusion is that I really look forward to differential calculus classes next semester, as well as more advanced calculus in general (my level is very very basic, since I only took SL Math).

 

[PeroK]

I wouldn't bother to differentiate the expression. Note that if you can find $x$ as a function of $t$ then you will automatically have $y$ as a function of $t$ via $y(t) = f[x(t)]$.

Write $$E = mgy + \frac 1 2 mv^2$$ as $$E = mgf(x) + \frac 1 2 m \left( \dot x^2 + \dot y^2 \right)$$

Then follow @haruspex 's suggestion of writing $\dot y$ in terms of $\dot x$ by making use of $y = f(x)$. You should end up with an equation involving just $x$ and $\dot x$ and for which you can separate variables and integrate (in principle) to find $t$ as a function of $x$. If you're lucky, you'll then be able to invert this to get $x(t)$.

Do you know any good examples for $f(x)$ other than a straight line?

 

[Atomillo]

Ok, so assuming the straight line y=x previously discussed and plugin the equation in Wolfram:

Solution (the energy E is N, because otherwise it interpreted it has the number e).

It seems as the constant c1 should play a role, since by setting c1 = 0 (E=0, thus assuming we start at the origin and zero initial velocity) we still don't get the equation for the plane, which for an angle of 45 should be √2/4 *g*t^2. How can I calculate the value of c1?

 

[TSny]

Do you know any good examples for $f(x)$ other than a straight line?

The only other curve that I have found in which the integration is elementary is the cycloid. The cycloid is usually given in parametric form. For a parametrically defined curve, $\left\{x(\tau), y(\tau)\right\}$, the energy equation can be written as $$E = mgy(\tau) +\frac{m}{2} \left(x'(\tau)^2 + y'(\tau)^2 \right) \left( \frac{d \tau}{dt} \right)^2$$ where the primes denote differentiation with respect to the parameter $\tau$.

For the case where the particle starts from rest at the origin, $E = 0$. If we also redefine the positive direction for $y$ to be downward instead of upward, the energy equation can be rearranged to $$\sqrt{2g} dt = \sqrt{\left[\frac{x'(\tau)^2+y'(\tau)^2}{y(\tau)}\right]}d\tau$$ Integrating both sides will yield $t$ as a function of $\tau$. So, if you pick some value of $\tau$ that corresponds to a certain point on the curve, you can find the time $t$ at which the particle reaches that point. Unfortunately, it doesn't seem to be easy to find curves for which the $\tau$ integration can be done by elementary means.

Also, this parametric form is easily applicable to the case where the curve is given in non-parametric form as $y=f(x)$ or $x=g(y)$. For example, $y= f(x)$ can be written in parametric form by letting $\tau = x$. Then the curve is represented by the parametric equations $x(\tau) = \tau$ and $y(\tau) = f(\tau)$.

For the case of a straight line sloping downward at 45^o from the origin, we have $x(\tau) = \tau$ and $y(\tau) = \tau$ (keeping in mind that the y-axis points downward). It is then easy to do the integral $$\int \sqrt{\left[\frac{x'(\tau)^2+y'(\tau)^2}{y(\tau)}\right]}d\tau$$ and you get the expected result.

 

[Atomillo]

How does the result emerge? When I take x=phi, so that y=x=phi:

t*√2*√g = 2*√2*√phi
√phi = (t*√g)/2
Since phi=x
x = (t^2*g)/4
​

Which is not the expected result. Where have I done wrong?

 

[PeroK]

[WQUOTE="Atomillo, post: 6252019, member: 623873"]
How does the result emerge? When I take x=phi, so that y=x=phi:

t*√2*√g = 2*√2*√phi
√phi = (t*√g)/2
Since phi=x
x = (t^2*g)/4
​

Which is not the expected result. Where have I done wrong?
[/QUOTE]
Why do you think that's wrong? What should it be?

 

[Atomillo]

When the angle is 45 degrees, isn't the effective force sin(45)mg, from which we find the acceleration to be (√2/2)*g, and subsequently v= (gt√2)/2 and then x = (gt^2√2)/4?
Shouldn't both expressions be the same?

 

[PeroK]

When the angle is 45 degrees, isn't the effective force sin(45)mg, from which we find the acceleration to be (√2/2)*g, and subsequently v= (gt√2)/2 and then x = (gt^2√2)/4?
Shouldn't both expressions be the same?

You're still confusing 2D and 1D motion. Those are the equations for the coordinate tangential to the slope. Not for the x- or y-coordinates.

To find $x$ you need to convert to Cartesian coordinates.

 

[Atomillo]

Oh. True. Same mistake twice. So how could the conversion occur? Multiplying the result by the sinus and cosinus (y and x) of the angle formed by the tangent line of the shape to the horizontal?

 

[PeroK]

Oh. True. Same mistake twice. So how could the conversion occur? Multiplying the result by the sinus and cosinus (y and x) of the angle formed by the tangent line of the shape to the horizontal?

A good exercise is to solve the motion on an inclined plane at angle $\theta$. The usual way by looking at tangential motion. Then, convert back to Cartesian coordinates.