Lagrangian approach for the inclined plane

[Rick16]

TL;DR Summary

getting the sign right for the potential energy in the Lagrangian

I want to use the Lagrangian approach to find the equation of motion for a mass sliding down a frictionless inclined plane. I call the length of the incline a and the angle that the incline makes with the horizontal b. Then the mass has kinetic energy 1/2m(da/dt)² and the potential energy should be -- as I see it -- mgh = mgasin(b). So the Lagrangian is 1/2m(da/dt)² - mgasin(b), which leads to the equation of motion
md²s/dt² = -mgsin(b). I am bothered by the minus sign in front of the right-hand term. The Newtonian method leads to md²s/dt² = +mgsin(b).

I searched the internet and looked at 3 solutions to this problem, none of which was helpful. The first internet expert sets up the Lagrangian the way I do, he also gets the same equation of motion, and he does not see anything wrong with it. The second expert also sets up the Lagrangian the way I do, but he makes a mistake solving the Lagrange equation and ends up with md²s/dt² = +mgsin(b). The third expert simply writes down the potential energy as -mgh, without any further comment. With a negative potential energy he naturally gets the solution md²s/dt² = +mgsin(b), but I can't see how the potential energy could be negative.

Could someone help me see clearer? Thanks ahead.

 

[Frabjous]

As “a“ increases, the potential energy increases. What does this say about the direction of the slope of the plane?

 

[BvU]

Hi,

Please clarify your symbols and don't jump around from ##a## to ##s##.
You have ##\mathcal L = T - V##. Write it out and write out the Euler- Lagrange equation.

(and learn some ##\LaTeX## ...)

##\ ##

 

[Rick16]

Sorry about the mixup with the variables. I used "s" in my handwritten sketch, but changed it to "a" because "ssin" looks strange on the screen. As for Latex, I actually did learn it, but I never use it because writing equations is much easier with MS Word, and I have forgotten all my Latex.

Anyway, here is my Lagrangian again (awkwardly written):

L = T - U = 1/2m(da/dt)² - mgasin(b). And the solution of the Euler-Lagrange equation is md²a/dt² = -mgsin(b).

About the direction of the slope: I don't know what to make of this. As the mass slides down the incline, its potential energy decreases, from a more positive value to a less positive value. So on any one point on the slope the potential energy is positive. Why would someone write U = -mgh?

 

[erobz]

Sorry about the mixup with the variables. I used "s" in my handwritten sketch, but changed it to "a" because "ssin" looks strange on the screen. As for Latex, I actually did learn it, but I never use it because writing equations is much easier with MS Word, and I have forgotten all my Latex.

It doesn't look strange if you write it in Latex. It takes all of 15 min to become proficient enough for what you wish to express here. Click on this link Learn LaTeX it explains how latex is used on this site.

 

[Rick16]

Okay, if you insist. I now rename my variables to ##s## for the displacement of the mass and ##\theta## for the angle that the incline makes with the horizontal.

Then the Lagrangian is: $$\mathcal {L}=\frac 1 2 m \dot s^2 - mgs \sin \theta$$

And the resulting equation of motion is $$m \ddot s = -mg \sin \theta$$

This looks certainly better, but it is still wrong, or so I believe.

 

[BvU]

The potential ! Does it increase or decrease with ##s## ?

##\ ##

 

[erobz]

Okay, if you insist. I now rename my variables to ##s## for the displacement of the mass and ##\theta## for the angle that the incline makes with the horizontal.

Then the Lagrangian is: $$\mathcal {L}=\frac 1 2 m \dot s^2 - mgs \sin \theta$$

And the resulting equation of motion is $$m \ddot s = -mg \sin \theta$$

This looks certainly better, but it is still wrong, or so I believe.

Ok, ##s## is down the slope. You apparently chose your PE = 0 datum at the top of the incline, but take the change in potential energy as positive. Thats where the problem is.

If the box is below the PE=0 datum it has negative gravitational potential as ##s## increases.

 

[Rick16]

I think I am getting closer. If PE = 0 at the top of the incline, then the potential energy would be negative throughout the displacement. But I actually chose PE = 0 at the bottom of the incline, and therefore my potential energy seems to be positive. It should not matter where I set the potential energy equal to zero, I should always come to the same conclusion. Intuitively I would set PE = 0 at the bottom of the incline, but I don't see what reasoning I should use in this case to arrive at a negative potential energy for the mass.

 

[BvU]

The value doesn't matter. Zeros for potentials can be chosen arbitrarily.
The sign of the change in value is what counts !

 

[erobz]

I think I am getting closer. If PE = 0 at the top of the incline, then the potential energy would be negative throughout the displacement. But I actually chose PE = 0 at the bottom of the incline, and therefore my potential energy seems to be positive. It should not matter where I set the potential energy equal to zero, I should always come to the same conclusion. Intuitively I would set PE = 0 at the bottom of the incline, but I don't see what reasoning I should use in this case to arrive at a negative potential energy for the mass.

## mgs \sin \theta## is not the potential for choosing PE=0 at the bottom of the ramp. Try that part again.

 

[Rick16]

## mgs \sin \theta## is not the potential for choosing PE=0 at the bottom of the ramp. Try that part again.

I am sorry, but I don't understand this. If ##h## is the height of the mass above the ground, then ##U=mgh=mgs\sin\theta=0## when ##h=0##, i.e. at the bottom of the ramp.

 

[kuruman]

You have to be careful when you use constraints in Lagrangians. A pitfall-free method for beginners is to start with Cartesian coordinates, write the Lagrangian then apply the constraints. In this case, I choose conventional Cartesian coordinates: ##x## increases to the right, ##y## increases away from the center of the Earth. Then the kinetic energy is
$$T=\frac{1}{2}m (\dot x^2+\dot y^2).$$
The potential energy function is the negative of the work done by gravity. If we displace the mass from the origin, where we take the potential energy to be zero, to some positive ##y## (watch the negative signs),
$$U(y)=-W_g=-\int_0^y \vec F_ g\cdot dy~\hat y=-\int_0^y m g(-\hat y)\cdot dy~\hat y=+mgy.$$Note that the potential energy increases as the mass moves away from the center of the Earth and that is at it should be.

Now for the constraint that the mass stay on the incline. The Cartesian coordinates are be replaced by a single coordinate ##s## that increases up the incline, in which case ##x=s~\cos\!\theta## and ##y=s~\sin\!\theta.## The Lagrangian is $$\mathcal{L} =T-U=\frac{1}{2}m\dot s^2 -mgs\sin\!\theta.$$Then $$\frac{d}{dt}\frac{\partial \mathcal{L}}{\partial \dot s}=m\ddot s~;~~\frac{\partial \mathcal{L}}{\partial s}=-mg\sin\!\theta$$and the EOM is $$m\ddot s+g\sin\!\theta=0 \implies m\ddot s=-g\sin\!\theta $$The minus sign indicates that the acceleration of the block is negative, i.e. down the incline and this is as it should be.

The lesson to be learned here is that if you want to be formal and use the Lagrangian, then be formal and don't cut corners.

 

[erobz]

I am sorry, but I don't understand this. If ##h## is the height of the mass above the ground, then ##U=mgh=mgs\sin\theta=0## when ##h=0##, i.e. at the bottom of the ramp.

What is the height of the mass above the ground? The coordinate ##s## is originating at the top of the incline. Select an arbitrary initial height ##h##, you must write the potential in terms of ##h## and ##s##.

 

[kuruman]

I am sorry, but I don't understand this. If ##h## is the height of the mass above the ground, then ##U=mgh=mgs\sin\theta=0## when ##h=0##, i.e. at the bottom of the ramp.

You are looking for the potential energy function ##U## which, by definition, is zero at the origin of whatever coordinate you are using to describe it. See post #13 for its derivation.

 

[erobz]

Here is what you were apparently considering first.

What is ##U## given this datum, and coordinate ##s##?

 

[Rick16]

What is the height of the mass above the ground? The coordinate ##s## is originating at the top of the incline. Select an arbitrary initial height ##h##, you must write the potential in terms of ##h## and ##s##.

Thank you!!! This comment finally did it. ##s## is originating at the top of the incline. I made the mistake to think that ##s=0## at the bottom of the incline. Actually, I did not really think this, I rather did not think about ##s## at all. I just looked at the right triangle and wrote down ##h=s\sin\theta##, which is wrong.

If I use ##h_0## as the initial height of the mass, then ##h_0-h(t)=s(t)\sin\theta##. Then ##h(t)=h_0-s(t)\sin\theta##. So the potential energy is ##mg(h_0-s(t)\sin\theta)## and the correct Lagrangian is

$$\mathcal {L}=\frac 1 2 m \dot s^2-mg(h_0-s(t)\sin\theta)$$

Then ##\frac {\partial \mathcal {L}} {\partial s}=mg\sin\theta## and ##\frac {d}{dt} \frac {\partial \mathcal {L}} {\partial \dot s}=m\ddot s##, and the equation of motion is correctly ##m\ddot s = mg\sin\theta##.

Thank you everybody for your answers. They have all been helpful, but what I needed most was this one sentence: ##s## is originating at the top of the incline.

 

[kuruman]

I just looked at the right triangle and wrote down ##h=s\sin\theta##, which is wrong.

If I use ##h_0## as the initial height of the mass, then ##h_0-h(t)=s(t)\sin\theta##. Then ##h(t)=h_0-s(t)\sin\theta##. So the potential energy is ##mg(h_0-s(t)\sin\theta)## and the correct Lagrangian is

$$\mathcal {L}=\frac 1 2 m \dot s^2-mg(h_0-s(t)\sin\theta)$$

Then ##\frac {\partial \mathcal {L}} {\partial s}=mg\sin\theta## and ##\frac {d}{dt} \frac {\partial \mathcal {L}} {\partial \dot s}=m\ddot s##, and the equation of motion is correctly ##m\ddot s = mg\sin\theta##.

Thank you everybody for your answers. They have all been helpful, but what I needed most was this one sentence: ##s## is originating at the top of the incline.

Just to clarify things some more. What matters is how the generalized coordinate is defined as increasing because however one writes the (gravitational) potential energy function ##U##, it has to be consistent with Nature's demand that it increase when the mass is moved farther away from the center of the Earth. Thus,

• if ##s## originates at the top of the incline and increases downhill as shown in the figure in post #16, ##U=-mgs\sin\!\theta.##
• if ##s## originates at the bottom of the incline and increases uphill as I assumed in post #13, ##U=+mgs\sin\!\theta.##

Either choice leads to the same equation of motion. Your mistake was that you treated ##s## as a distance, not as a coordinate.

 

[vanhees71]

Okay, if you insist. I now rename my variables to ##s## for the displacement of the mass and ##\theta## for the angle that the incline makes with the horizontal.

Then the Lagrangian is: $$\mathcal {L}=\frac 1 2 m \dot s^2 - mgs \sin \theta$$

And the resulting equation of motion is $$m \ddot s = -mg \sin \theta$$

This looks certainly better, but it is still wrong, or so I believe.

Why should it be wrong? I suggest you provide a drawing, where you clearly define your coordinate(s). From your equation obviously ##s## is a coordinate wrt. a basis vector which is a tangent vector on the plane pointing upwards, i.e., with increasing ##s## the mass gets "higher" in the gravitational field of the Earth.

 

[BvU]

Why should it be wrong? I suggest you provide a drawing, where you clearly define your coordinate(s). From your equation obviously ##s## is a coordinate wrt. a basis vector which is a tangent vector on the plane pointing upwards, i.e., with increasing ##s## the mass gets "higher" in the gravitational field of the Earth.

See #16 for 'downwards' ...

 

[Rick16]

Just to clarify things some more. What matters is how the generalized coordinate is defined as increasing because however one writes the (gravitational) potential energy function ##U##, it has to be consistent with Nature's demand that it increase when the mass is moved farther away from the center of the Earth. Thus,

• if ##s## originates at the top of the incline and increases downhill as shown in the figure in post #16, ##U=-mgs\sin\!\theta.##
• if ##s## originates at the bottom of the incline and increases uphill as I assumed in post #13, ##U=+mgs\sin\!\theta.##

Either choice leads to the same equation of motion. Your mistake was that you treated ##s## as a distance, not as a coordinate.

This is exactly true. I saw a static picture with fixed distances, instead of seeing ##U## as a function of a moving coordinate. I now also realize that I don't need to include ##h_0## in the expression for the potential energy. ##h_0## is an arbitrary constant, and I can as well set it equal to zero. Then I finally get the result that I have been looking for from the start: ##U(s) = -mgs\sin\theta##. Of course, in this case the ramp no longer sits on top of the x-axis, but below it. I saw this setup elsewhere before, with the x-axis pointing to the right and the y-axis pointing down, but I did not understand why the axes were chosen this way. Now I know. I have learned a lot from this seemingly simple problem. Thanks a lot for your help.

 

[vanhees71]

Ok, let's take #16. Then of course ##s## is in the other direction. To derive the Lagrangian, you start with the calculation of the kinetic energy
$$T=\frac{m}{2} \dot{\vec{x}}^2$$
and the potential of the force acting on the particle
$$\vec{F}=m \vec{g} =-\vec{\nabla} V \; \Rightarrow \; V=-m \vec{g} \cdot \vec{x}$$
in terms of the coordinates.

Here we introduce first a new Cartesian basis. Let the plane shown in #16 be the (12)-plane with ##\vec{e}_1## pointing to the right ##\vec{e}_2## pointing up and ##\vec{e}_3## pointing out of the plane. In the corresponding Cartesian coordinates you have
$$\underline{g}=\begin{pmatrix}0 \\ -g \\ 0 \end{pmatrix}, \quad g>0.$$
Then we introduce
$$\underline{e}_s=\begin{pmatrix}\cos \alpha \\ -\sin \alpha \\0 \end{pmatrix},$$
where ##\alpha## is the angle of the red triangle on the edge on the 1-axis.

As the other basis vector in the plane shown in #16 we take
$$\underline{e}_n=\begin{pmatrix} \sin \alpha \\ \cos \alpha \\0 \end{pmatrix}.$$
As the third basis vector we can keep ##\vec{e}_3=\vec{e}_s \times \vec{e}_n##.

The point should now be restricted to the inclined plane, i.e.,
$$\vec{x}=s \vec{e}_s + z \vec{e}_3.$$
The kinetic energy is
$$T=\frac{m}{2} \dot{\vec{x}}^2=\frac{m}{2} (\dot{s}^2+\dot{z}^2)$$
and the potential is
$$V=-m \vec{g} \cdot \vec{x}=-m \begin{pmatrix}0\\ -g \\ 0 \end{pmatrix} \cdot \begin{pmatrix}s \cos \alpha \\ -s \sin \alpha \\ z \end{pmatrix}=-mg s \sin \alpha.$$
The Lagrangian thus is
$$L=\frac{m}{2} (\dot{s}^2+\dot{z}^2) +m g s \sin \alpha,$$
and the Euler-Lagrange equations read
$$\mathrm{d}_t \frac{\partial L}{\partial \dot{s}}=m \ddot{s}=\frac{\partial L}{\partial s}=mg$$
and
$$\mathrm{d}_t \frac{\partial L}{\partial \dot{z}}=m \ddot{z} = \frac{\partial L}{\partial z}=0.$$
The solutions are
$$s=s_0 + v_{s0} t +\frac{g}{2} \sin \alpha t^2, \quad z=z_0 + v_{30} t,$$
as expected.