Why Does the Potential Energy of the Wedge Appear in Lagrangian Mechanics?

[Another]

Homework Statement

Find the Lagrangian of system.
In the question say the wedge can move without friction on a smooth surface.
Why does the potential energy of the wedge appear in Lagrangian?

Relevant Equations

$\frac{d}{dt}\frac{\partial L}{\partial \dot{q}} - \frac{\partial L}{\partial q}$

In Solution https://www.slader.com/textbook/9780201657029-classical-mechanics-3rd-edition/67/derivations-and-exercises/20/

In the question say the wedge can move without friction on a smooth surface.

Why does the potential energy of the wedge appear in Lagrangian?

(You can see the Larangian of this system at below.)

 

[etotheipi]

The potential energy of the wedge is constant anyway, so isn't going to affect the equation of motion.

But in any case the expression $U = Mgy_M$ isn't correct, because the centre of mass of the wedge is not the coordinate $y_M$ (that is the coordinate of the left corner). In any case they will still stumble upon the correct answer, because the mistake happens to be constant and drops out.

 

[PeroK]

It looks like strange thing to do. I wouldn't have the terms involving $y_M$ and $\dot y_M$ as the wedge is constrained to move in the x-direction only.

 

[Steve4Physics]

They are just following a systematic approach and plugging-in all values of potential energy, even the constant one which is going to disappear. If you do this for all problems, it helps you to establish a standard approach and makes it less likely you'll forget some term(s) in more complex problems. Not essential but probably a useful practice for some students.

As etotheipi points out, they've done it incorrectly, which is a good example of irony.

 

[wrobel]

Moreover, the Lagrangian is defined up to an additive function
$$\dot f(t,q)=\frac{\partial f}{\partial t}+\frac{\partial f}{\partial q^i}\dot q^i$$
that is the Lagrangians $L$ and $L'=L+\dot f(t,q)$ generate the same equations

 

[vela]

They are just following a systematic approach and plugging-in all values of potential energy, even the constant one which is going to disappear. If you do this for all problems, it helps you to establish a standard approach and makes it less likely you'll forget some term(s) in more complex problems. Not essential but probably a useful practice for some students.

I don't think this was the motivation here. The problem asks for the forces of constraint as well, so you don't want to impose the constraint right from the start. It's not entirely clear to me from the problem statement, however, if it was asking for just the forces of constraint on just the particle or for all of the forces of constraint within the system.