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I am aware that there exists duplicates to the title and have gone through the answers but it still doesn't answer my issue with a statement in the last image.

These two similar situations with slight variations are being introduced, I am not able to understand the reasoning provided as to why the quantity $E$ is equal to total energy in one and not in the other.

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I am not able to understand how the minus sign before $r^2ω^2$ implies that the conserved quantity is not total energy as stated in the last slide.

Qmechanic
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Anonymousstriker38596
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    OP's example is essentially a duplicate of https://physics.stackexchange.com/q/427144/2451 – Qmechanic Sep 13 '23 at 18:39
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    This is a dupe of a lot of different question asking when the Hamiltonian is equal to the total mechanical energy. – hft Sep 13 '23 at 18:48
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    Does this answer your question? What is the difference between total energy and the Lagrangian energy function? – hft Sep 13 '23 at 18:49
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    In short, assuming you don't care about velocity dependent potentials, you need to have $\dot q_i \frac{\partial T}{\partial \dot q_i} = 2T$. – hft Sep 13 '23 at 18:51
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    If you do care about velocity dependent potentials then the condition is: $\dot q_i \frac{\partial T}{\partial \dot q_i} - \dot q_i\frac{\partial V}{\partial \dot q_i}= 2T$ – hft Sep 13 '23 at 18:52
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    As you can see your $T$ in the last part (the one you are interested in) does not satisfy the requirement that it is a homogeneous function of order two in the velocity (because there is a piece of $T$ that depends on $r$). – hft Sep 13 '23 at 18:58
  • @hft The condition you formulated for 2$T$ is clear to me now but I didn't get your last comment "because there is a piece of T that depends on r", the same is true for even the first case but $E$ equals to total energy over there. – Anonymousstriker38596 Sep 14 '23 at 07:26
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    In the first case, the piece of $T$ that depends on $r$ is multiplied by a time derivative (the $\dot \theta^2$ factor) so that you still have $\sum_i\dot q_i\frac{\partial T}{\partial \dot q_i} = 2T$, but in the second case there is no generalized velocity variable multiplying $r$ (there is a constant $\omega^2$ instead) so $\sum_i\dot q_i\frac{\partial T}{\partial \dot q_i} = 2T - mr^2\omega^2$. – hft Sep 14 '23 at 15:01

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