Block sliding down a movable wedge - solution using Lagrangian mechanics

Here is a block sliding down a movable wedge without friction.
Here is my introduction to Lagrangian
• Introduction to Lagran...
Here is the same problem solved using work-energy and momentum.
• What are the final spe...

[@nukelewman]

Great video man!

[@BjarturMortensen]

3:04 "I put negative because I do matter" yes you do 🤗

[@daltonz]

Great review problem for catching up on classical mechanics thanks!

[@DotPhysics]

Glad it was useful!

[@not_vinkami]

11:53 Lagrangian mechanics is useful especially when you want to generate wrong equations

[@mariomuysensual]

?

[@General12th]

I am also not understanding. Are you saying that Lagrangian mechanics is easy to mess up?

[@christianjourneytv1003]

Thank you so much for the clear explanation.

[@DotPhysics]

You are welcome!

[@virabhadra2]

Thank you for the nice explanation! I'm refreshing my knowledge from the past with your videos.
I have a question. Do you have any video, how to consider the friction forces using the Lagrange mechanics?
I mean the sliding friction (as k*R) and viscous friction in gas/fluid (as C*v²)

[@DotPhysics]

Including friction (sliding) with Lagrangian is not simple since friction is a non-conservative force.

[@sajaalkhader3227]

Thank you so much

[@chilledvibes99]

with the dot above the variable is that always the derivative with respect to time? For example s dot is actually ds/dt?

[@moustachescarz]

yes those expressions are equivalent

[@LipeKleiz]

Shouldn't the potential energy be defined as U=m1*g*(Ol -s*sin(θ)) where Ol is the opposite leg to the θ angle?
If we define U=-m1*g*s*sin(θ), that way U has maximum magnitude in the lowest point (while still having lower potential energy at the lowest point).

[@vytah]

The magnitude does not matter at all, all matters is the differences of potential energy being correct. The Ol is just a unnecessary constant, so it can be ignored.

[@christianjourneytv1003]

How do we usually know that theta won't change. I think theta only changes in pendulum right

[@DotPhysics]

theta is the angle of the wedge - the wedge doesn't change in this case

[@sivasakthisaravanan4850]

x1 is not independent of s; so why use it at all? x and s completely describe the system.

[@DotPhysics]

The awesome thing about Lagrangian mechanics is that you can use non-independent variables.

[@milanrai6988]

better express s in terms of x and y coz s is not even a coordinate

[@vx7482]

6:10 isn't U supposed to be positive? Since mg is going down so it's negative, and the box is going down as well so (s sin Ɵ) is also negative?

[@vytah]

s becomes positive when going down, the direction of the s coordinate is diagonally downwards. U is negative, since it's zero at the highest point.

[@wynautvideos4263]

It would be if the coordinates were centered below the objects