Hamiltonian does NOT equal energy?

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The discussion clarifies that in a system with a rod rotating at a constant angular speed and a movable ring, the binding forces are rheonomic, which are a type of non-holonomic constraint. This rheonomic nature of the forces is cited as the reason why the Hamiltonian (H) does not equal the total energy (E) of the system. The relationship between the Hamiltonian and energy is explored through the Lagrangian formalism. The participants confirm that the distinction between H and E arises from the specific constraints applied to the system. Understanding these concepts is essential for accurately analyzing such mechanical systems.
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Hello,

Just to be sure: is the following correct?

Imagine a long rod rotating at a constant angular speed (driven by a little motor). Now say there's a small ring on the rod that can move on the rod without friction. The ring is then held onto the rod by an ideal binding force (I don't know the correct term in English, a force that doesn't deliver virtual work). This ring, free to move along the rod, can be described using the Lagrangian formalism.

Now is it true that the binding forces are non-holonomic? And this is the reason why H \neq E? It seems like it anyway, but I remember reading a thread on here a while ago looking for simple systems for which H \neq E while all examples given were very mathematical and far-fetched in a physical sense, so this makes me a bit unsure about what I've written above, hence the double-check.
 
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Oops, important EDIT:

"Now is it true that the binding forces are non-holonomic?"

I meant non-SCLERONOMIC, i.e. rheonomic.
 
mr. vodka said:
Hello,

Just to be sure: is the following correct?

Now is it true that the binding forces are non-holonomic? And this is the reason why H \neq E?

Yes,the constraint forces are rheonomic and this is whry H\neq E=T+V
 
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