Hamilton equation for a block on an inclined plane

AI Thread Summary
To find the Hamilton equations for a block on an inclined plane without friction, it is essential to define the orientation of the x and y-axes clearly. The momentum must be treated separately for both components, resulting in two differential equations for the x and y components of momentum. The potential energy expression should include the y component, and care must be taken to avoid sign errors in the kinetic energy part of the Lagrangian. The equations of motion can be derived by taking partial derivatives of the Hamiltonian, focusing on the two momentum components. Theta is treated as a fixed parameter rather than a variable in this context.
Cocoleia
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Homework Statement


I am asked to find the Hamilton equations for a block on an inclined plane (no friction)

Homework Equations

The Attempt at a Solution


Please ignore the fact that my steps are written in French (sorry!)
upload_2016-11-30_16-50-45.png

I am no longer sure of what I'm doing when it comes to finding the momentum. I here chose y' and just ignored the x', because that seemed logical to me at the time. How can I correct my solution if it isn't right?
 
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You should make clear the orientation of the x and y-axes.

Treat the ##\dot{x}## term similarly to how you treated the ##\dot{y}## term.
 
TSny said:
You should make clear the orientation of the x and y-axes.

Treat the ##\dot{x}## term similarly to how you treated the ##\dot{y}## term.
Will I have two different equations for momentum then? And then add them together for a total momentum?
 
Cocoleia said:
Will I have two different equations for momentum then? And then add them together for a total momentum?
You will have a differential equation for the x-component of momentum and a separate differential equation for the y-component of momentum.
Since you didn't state the orientation of your axes, I am having to guess the orientations based on your equations.
 
TSny said:
You will have a differential equation for the x-component of momentum and a separate differential equation for the y-component of momentum.
Since you didn't state the orientation of your axes, I am having to guess the orientations based on your equations.
x measured horizontally across the slope, and y measured down the slope
 
Cocoleia said:
x measured horizontally across the slope, and y measured down the slope
OK. Thank you.
 
TSny said:
OK. Thank you.
I think I understand, I will have to use the definition:
upload_2016-11-30_18-1-41.png
 
Cocoleia said:
I think I understand, I will have to use the definition:
View attachment 109724
Yes. But ##p_i## should not be "dotted" (time derivative) in ## H = \sum p_i \dot{q}_i - L ##.
 
TSny said:
Yes. But ##p_i## should not be "dotted" (time derivative) in ## H = \sum p_i \dot{q}_i - L ##.
This is my final answer:
upload_2016-11-30_19-34-15.png
 
  • #10
In your second equation for ##L##, there is a sign error in the kinetic energy part. But you corrected it later.

In your expressions for ##H## you've dropped ##y## from the potential energy part.

Otherwise, looks OK.

Are you also supposed to write out the equations of motion?
 
  • #11
TSny said:
In your second equation for ##L##, there is a sign error in the kinetic energy part. But you corrected it later.

In your expressions for ##H## you've dropped ##y## from the potential energy part.

Otherwise, looks OK.

Are you also supposed to write out the equations of motion?
Yes. How do we go about doing this?
 
  • #12
Cocoleia said:
Yes. How do we go about doing this?
You showed at the end of your first post how to get the equations of motion by taking partial derivatives of H. But you have two components of p: px and py.
 
  • #13
TSny said:
You showed at the end of your first post how to get the equations of motion by taking partial derivatives of H. But you have two components of p: px and py.
So will I derive 3 times, once for px, once for py, and then for theta?
 
  • #14
No, just twice. Theta is not a variable. You could call theta a parameter. Theta is assumed to have a fixed value as the block slides on the incline.
 
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