Inclined plane with atypical axes/ Lagrangian

In summary, the author is having difficulty solving an intro level box on an inclined plane problem and has been struggling with the axis choice. He/she eventually solves it using the Lagrangian method, which makes sense because the acceleration in the y direction is not simply g. However, this method requires solving for the acceleration along the incline which is a separate question.
  • #1
david13579
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Homework Statement


I'm asked to solve the typical intro level box on an inclined plane problem but I need to do it using the lagrangian.

My difficulty with it is that the axis I am required to use are not the typical axes used when solving this using Newtonian mechanics. Instead of the x(or y) axis being along the incline with the other perpendicular to it, the axes here are the x-axis going from the high end of the incline to the low end and the y-axis from the top to the bottom (so x is horizontal facing left and y is vertical facing down)

I am then asked to compare the answer using lagragian to the answer using Newtonian stuff.
http://i.imgur.com/OAosTDI.png

Homework Equations





The Attempt at a Solution


I've tried a billion different ways and always and always end up with the acceleration in the y direction being g(sin(a))^2 and in the x direction g(sina)(cosa).

On one hand it makes no sense since obviously the acceleration in the y direction should be just g but on the other hand it makes sense because if we find the acceleration along the incline it is clearly gsin(a) as doing with a the typical coordinates of x along the incline would show. Then, if the acceleration is gsina along the incline, breaking it into components along the x and y axes would multiply it further by another sina in the y direction and cosa in the x direction which shows the same thing I got using the Lagrangian method.

So I have been going in circles with something that makes sense yet doesn't make sense.
 
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  • #2
If you use those coordinate axes, you have to add a contraint equation that forces the motion to be along the incline, and use the method of undetermined multipliers. Have you done that?

When you have solved the accelerations in x and y directions, make a transformation to a coordinate system that has been rotated by angle a in the clockwise direction (you know rotation matrices, don't you?) and see whether the accelerations along the new axes are 0 and gsin(a).
 
  • #3
hilbert2 said:
If you use those coordinate axes, you have to add a contraint equation that forces the motion to be along the incline, and use the method of undetermined multipliers. Have you done that?

When you have solved the accelerations in x and y directions, make a transformation to a coordinate system that has been rotated by angle a in the clockwise direction (you know rotation matrices, don't you?) and see whether the accelerations along the new axes are 0 and gsin(a).

Doing the same question with constraints is a separate question so there has to be a way without constraints. The first question is to do it and the second is to do it again using a constraint (which I think should be f(x,y) = y-xtan(theta)=0

And no, I don't know rotation matrices.
 
  • #4
Here are 2 of my attempts at it:
without constraints http://i.imgur.com/wQ9RKSA.png
With constraints http://i.imgur.com/7zvNv6w.png
both give me the same result. And like I said before, if I draw on paper and try it as a problem of breaking vectors into components, I get the same answer but then again it also contradicts that if y is vertical then logically only g should be the components and nothing else.
 
  • #5
Your answer is correct. The acceleration in y direction is not g because both gravity and the support force of the inclined plane act on the object, not only gravity.
 
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  • #6
And now, after further thinking I even see that the constraint is what prevents g from being the acceleration in the y direction. mg-λ=mgsin2(a) which would be the force in the y direction with the acceleration I found.

:) Thanks a lot.
 

Related to Inclined plane with atypical axes/ Lagrangian

1. What is an inclined plane with atypical axes?

An inclined plane refers to a flat surface that is tilted at an angle, rather than being completely horizontal. Atypical axes refer to the orientation of the plane, which may not align with the traditional x and y axes. This can make it more challenging to analyze and understand the movement of objects on the plane.

2. What is the significance of using Lagrangian mechanics in analyzing an inclined plane with atypical axes?

Lagrangian mechanics is a mathematical approach to solving problems in classical mechanics. It is particularly useful when dealing with complex systems, such as an inclined plane with atypical axes, as it allows for a more comprehensive and accurate analysis of the system's behavior.

3. How is the Lagrangian defined in the context of an inclined plane with atypical axes?

The Lagrangian, denoted by L, is a function that describes the kinetic and potential energies of a system. In the case of an inclined plane with atypical axes, the Lagrangian takes into account the orientation and shape of the plane, as well as the mass and velocity of the object on the plane.

4. Can Lagrangian mechanics be used to analyze real-world situations involving inclined planes with atypical axes?

Yes, Lagrangian mechanics can be applied to real-world scenarios involving inclined planes with atypical axes. It is commonly used in physics and engineering to model and understand the motion of objects on inclined planes, which have various applications in everyday life, such as ramps and hills.

5. What are the advantages of using Lagrangian mechanics over other methods in studying an inclined plane with atypical axes?

One of the main advantages of using Lagrangian mechanics is its ability to handle complex systems with multiple variables, such as an inclined plane with atypical axes. It also provides a more general and elegant approach to solving problems in classical mechanics, making it a valuable tool for scientists and engineers.

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