Solving projectile motion problems using Lagrangian mechanics

[Melkor77]

TL;DR Summary: Lagrangian for projectile motion in an inclined plane with negative slope.

I am a bit unsure on how to find the Lagrangian for projectile motion in an inclined plane with negative slope. I can solve it using Newton Mechanics, but am a bit new to lagrangian mechanics. Also could someone tell me about some free resources to learn more about the topic

 

[anuttarasammyak]

I am not sure of projectile motion IN a inclined plane you say. Do You mean 2D motion constrained in a slope plane ?

 

[Melkor77]

yes, 2d motion constrained in a slope plane

 

[anuttarasammyak]

Thanks. Then how about it for xy slope plane,
L=\frac{1}{2}m(\dot{x}^2+\dot{y}^2) - mg \sin \theta \ y
where plus y corresponds to upward slope of angle $\theta$ >0 ?

 

[PeroK]

TL;DR Summary: Lagrangian for projectile motion in an inclined plane with negative slope.

I am a bit unsure on how to find the Lagrangian for projectile motion in an inclined plane with negative slope. I can solve it using Newton Mechanics, but am a bit new to lagrangian mechanics. Also could someone tell me about some free resources to learn more about the topic

The Lagrangian does not depend on the slope. It depends only on the kinetic and potential energy.

Note that the shape of the paths that a projectile can take are the same in both cases. The only difference is the valid endpoints.

 

[Orodruin]

The Lagrangian does not depend on the slope. It depends only on the kinetic and potential energy.

Note that the shape of the paths that a projectile can take are the same in both cases. The only difference is the valid endpoints.

This is not correct for the situation the OP is describing:

yes, 2d motion constrained in a slope plane

 

[Orodruin]

However, you can of course still get the Lagrangian from taking the difference between the kinetic and potential energies.

In general for this type of problems:

1. Introduce your generalized coordinates.
2. Write down the typical standard 3D coordinates as a function of the generalized coordinates.
3. Take the time derivative of the 3D coordinates and insert into $m\dot{\vec x}^2/2$ to get the kinetic energy.
4. Insert the expression for the 3D coordinates in terms of your generalized coordinates into your expression for the potential to get potential as a function of the generalized coordinates.
5. Take the difference between kinetic and potential energy and you have the Lagrangian.

 

[anuttarasammyak]

In general for this type of problems:

　6. Express the constraint condition with method of Lagrange multiplier and add the $\lambda$ term to Lagrangean.

This is what I have gusessed from OP.

 

[Orodruin]

　6. Express the constraint condition with method of Lagrange multiplier and add the $\lambda$ term to Lagrangean.

This is what I have gusessed from OP.

The thing is, you don’t need to do this if you follow the procedure I described. This is one of the beauties of generalized coordinates.

 

[anuttarasammyak]

The thing is, you don’t need to do this if you follow the procedure I described. This is one of the beauties of generalized coordinates.

I assume my y in post#4 which is not y of xyz and has no constraints would be a kind of it.