Lagrangian of a block on an inclined plane

[member 731016]

Homework Statement

Please see below

Relevant Equations

$F = mg$

For this problem,

I am confused where they get these two terms from

Can someone please explain? I get all the other terms expect those.

Thanks alot!

 

[haruspex]

Homework Statement: Please see below
Relevant Equations: $F = mg$

For this problem,
View attachment 343565
I am confused where they get these two terms from
View attachment 343566
Can someone please explain? I get all the other terms expect those.

Thanks alot!

Note that three of the four KE terms involving $m_1$ have factors that naturally combine to form a perfect square.

 

[member 731016]

Note that three of the four KE terms involving $m_1$ have factors that naturally combine to form a perfect square.

Thanks for your reply @haruspex!

Sorry what do you mean? Are you suggesting I rewrite the the highlighted bit in terms of cotangent?

Thanks!

 

[kuruman]

Also note that generalized coordinates $x_2$ and $y_1$ are, respectively, the horizontal position of the wedge and the vertical position of the block. How do you think you can get the horizontal position of the block $x_1(t)$ if you have successfully solved the equations of motion to get $x_2(t)$ and $y_1(t)$?

In other words, how have you handled the constraint that the block maintain contact with the wedge which accelerates to the right while the block accelerates down the incline? It would be easier for us to explain where you went wrong if you show us your work. We are not very good at guessing.

 

[haruspex]

Thanks for your reply @haruspex!

Sorry what do you mean? Are you suggesting I rewrite the the highlighted bit in terms of cotangent?

Thanks!

No, ignore your highlighting.
There are four $m_1$ terms contributing to the KE. Which three of them can be combined into an expression of the form $\frac 12m_1(…)^2$?

 

[member 731016]

Thank you both for your replies @kuruman and @haruspex!

Also note that generalized coordinates $x_2$ and $y_1$ are, respectively, the horizontal position of the wedge and the vertical position of the block. How do you think you can get the horizontal position of the block $x_1(t)$ if you have successfully solved the equations of motion to get $x_2(t)$ and $y_1(t)$?

In other words, how have you handled the constraint that the block maintain contact with the wedge which accelerates to the right while the block accelerates down the incline? It would be easier for us to explain where you went wrong if you show us your work. We are not very good at guessing.

I think that is part of my confusion. I am unsure how the wedge accelerates to the right and why it does, since as far as I know, there is no horizontal force acting on it since all contact in the system is frictionless. Maybe it is from the weight of m_1 that causes m_2 to accelerate?

No, ignore your highlighting.
There are four $m_1$ terms contributing to the KE. Which three of them can be combined into an expression of the form $\frac 12m_1(…)^2$?

Sorry, I am not sure how there are three $m_1$ terms for KE that can be combined into a expression of the form $\frac 12m_1(…)^2$. I only get two, which are $\frac{1}{2}m_1([\frac{dx_2}{dt}]^2 + [\frac{dy_1}{dt}\sinβ]^2)$ since the $m_1$ moves at the same speed as the block in the positive $x_2$ direction which moves at $\frac{dx_2}{dt}$ and $-\frac{dy_1}{dt}\sinβ$ from block moving down the incline since only a component is parallel to the wedge.

Thanks!

 

[Orodruin]

there is no horizontal force acting on it since all contact in the system is frictionless.

What about the force from the block on the wedge?

Either way, the beauty of Lagrangians is (among other things) that you don’t need to think about forces.

Maybe it is from the weight of m_1 that causes m_2 to accelerate?

The weight on m1 is vertical and acts only on m1. The contact force between block and wedge though …

Sorry, I am not sure how there are three m1 terms for KE that can be combined into a expression of the form 12m1(…)2. I only get two, which are 12m1([dx2dt]2+[dy1dtsin⁡β]2) since the m1 moves at the same speed as the block which moves at dx2dt and [dy1dtsin⁡β]2 from block moving down the incline.

Please write out your full derivation as suggested by @kuruman earlier. In particular, can you write the horizontal position $x_1$ as a function of $x_2$ and $y_1$?

 

[haruspex]

I only get two

$\frac 12m_1\dot x_2^2$, $\frac 12m_1\dot y_1^2$, $\frac 12m_1\frac 1{\tan^2(\beta)}\dot y_1^2$, $m_1\frac 1{\tan(\beta)}\dot y_1\dot x_2$.

 

[member 731016]

Thank you for your replies @Orodruin and @haruspex !

What about the force from the block on the wedge?

Either way, the beauty of Lagrangians is (among other things) that you don’t need to think about forces.


The weight on m1 is vertical and acts only on m1. The contact force between block and wedge though …


Please write out your full derivation as suggested by @kuruman earlier. In particular, can you write the horizontal position $x_1$ as a function of $x_2$ and $y_1$?

Sorry, I don't understand the technique of introducing a third variable to describe the two generalized variables $x_2$ and $y_1$. We have not been taught that. May you please explain why it is needed?

$\frac 12m_1\dot x_2^2$, $\frac 12m_1\dot y_1^2$, $\frac 12m_1\frac 1{\tan^2(\beta)}\dot y_1^2$, $m_1\frac 1{\tan(\beta)}\dot y_1\dot x_2$.

Sorry do you please know how you got the later three terms?

Thanks!

 

[Orodruin]

Sorry, I don't understand the technique of introducing a third variable to describe the two generalized variables x2 and y1. We have not been taught that. May you please explain why it is needed?

It is not a third variable as it can be expressed in terms of the other two. It is a help function that you will differentiate with respect to time to find the horizontal velocity of the block (which is necessary to write down its kinetic energy).

 

[Orodruin]

Again, it would be much more helpful if you would actually provide your work.

 

[haruspex]

Sorry do you please know how you got the later three terms?

Thanks!

Sorry, I should have made it clear that I am working backwards from the expression you are supposed to arrive at. (You had not posted your own equation, so that is all I could have been referring to.)
It may become clear then how to get that answer.

 

[Orodruin]

$\frac 12m_1\dot x_2^2$, $\frac 12m_1\dot y_1^2$, $\frac 12m_1\frac 1{\tan^2(\beta)}\dot y_1^2$, $m_1\frac 1{\tan(\beta)}\dot y_1\dot x_2$.

In addition, it is many times useful to consider an expression in some particular case. In this scenario, it could be instructive to consider the case where the wedge does not move, i.e., $\dot x_2 = 0$. In that case only two of these terms will be non-zero and it can be instructive to try to figure these out and why they appear.

 

[member 731016]

Thank you for your replies @haruspex and @Orodruin!

I've done so more thinking about the problem and don't think I physically understand it. Do you please know why the block moves to the right? This is because I also found a similar problem: https://farside.ph.utexas.edu/teaching/336k/Newton/node80.html that has the same block that moves to the right, expect, it is facing in the opposite direction. I would please like to know a general principle that can be used/assumption when determining what direction the wedge moves in and why.

Thanks!

 

[haruspex]

Do you please know why the block moves to the right?

It doesn't. Why do you think it does? The displacement, velocity and acceleration may be measured as positive to the right, but that just means the values will be negative.

Please try working backwards from the answer as I suggested in posts #8 and #12. Do you agree with the four terms I extracted from the target equation? Can you see which three can be combined into the form I gave in post #5?

 

[Orodruin]

It doesn't.

It can of course in general, but not if released from rest as in the specific problem.

I would please like to know a general principle that can be used/assumption when determining what direction the wedge moves in and why.

You don’t need to make any assumptions on how the block moves. All you need to do is to introduce a set of coordinates that describe the system fully. Then you need to figure out how the speed of the block and wedge can be expressed in those coordinates to find the kinetic energy and how the potential can be written in those same coordinates. That’s all there is to it. After that you apply the EL equations to find the equations of motion.

 

[member 731016]

It doesn't. Why do you think it does? The displacement, velocity and acceleration may be measured as positive to the right, but that just means the values will be negative.

Please try working backwards from the answer as I suggested in posts #8 and #12. Do you agree with the four terms I extracted from the target equation? Can you see which three can be combined into the form I gave in post #5?

It can of course in general, but not if released from rest as in the specific problem.


You don’t need to make any assumptions on how the block moves. All you need to do is to introduce a set of coordinates that describe the system fully. Then you need to figure out how the speed of the block and wedge can be expressed in those coordinates to find the kinetic energy and how the potential can be written in those same coordinates. That’s all there is to it. After that you apply the EL equations to find the equations of motion.

Thank you for your replies @haruspex and @Orodruin!

I have done so more thinking about the problem now if you could please check.

We know that the potential energy V and kinetic energy T are scalar potential functions.

This means that one of the ways they can be (most trivially) defined is explicitly in terms of scalar products of the physical variables in the formula. I show a example below. Thus this implies that since T and V are scalar potential functions, then by the Lagrangian which is a linear combination of T and V, $\mathcal{L} = eT + fV + g$, $e, f, g \in \mathbb{R}$ must be a scalar function.

Note that I have written the $\mathcal{L}$ in it most general form which I have not seen in most before. Using initial conditions used to define the Lagrangian, we get $e = 1, f = - 1, g \in \mathbb{R}$. This gives the most standard definition of Lagrangian $\mathcal{L} = T - V + g$, however, most textbooks set $g = 0$

$V = -m_1g \hat y_1 \cdot y_1 \hat y_1 = -m_1gy_1$

Since problem defines $x_2 > 0$ to the right, this implies $\hat x_2 > 0$ to the right where sloped block slides.

Is my working so far correct?

Thanks!