I Hamiltonian of the bead rotating on a horizontal stick

AI Thread Summary
The discussion centers on the derivation of the Hamiltonian for a bead rotating on a horizontal stick with constant angular speed, as presented in David Morin's "Introduction to Classical Mechanics." The Lagrangian does not include the derivative over angular speed because the angular speed is constant, leading to the conclusion that the only generalized coordinate is the bead's radial position, r. This results in the Hamiltonian reflecting only the kinetic energy associated with the radial motion. The participants clarify that the stick's position is a function of time and does not vary with initial conditions, confirming the system has only one degree of freedom. The conversation highlights the importance of recognizing fixed parameters in dynamical systems.
Michael Korobov
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Why Lagrangian derivative over angle isn't included in the Hamiltonian calculation?
Hi,
In David Morin's "Introduction to classical mechanics", Problem 6.8, when deriving Hamiltonian of the bead rotating on a horizontal stick with constant angular speed, the Lagrangian derivative over angular speed isn't included.
Why is that?
Specifically, the Lagrangian takes form $$L=T=\frac{m {\dot r}^2}{2}+\frac{mr^2\omega^2}{2}=\frac{m {\dot r}^2}{2}+\frac{mr^2{\dot \theta}^2}{2}$$
The Hamiltonian is then calculated as
$$H=\left(\sum_{i=1}^N \frac{\partial L}{\partial {\dot q_i}}{\dot q_i} \right)-L=\frac{m {\dot r}^2}{2}-\frac{mr^2\omega^2}{2}$$
This implies either $$\frac{\partial L}{\partial {\dot \theta}}{\dot \theta}=0$$
or that the only generalized coordinate is ##r## and therefore ##N=1##
Why is that? I understand this relates to ##{\dot \theta}=const## but didn't manage to understand why the partial derivate should be 0.

Later on, in the "Analytical Mechanics" by Hand and Finch, in Question 14 on p.22, there is identical question and then they ask how Hamiltonian looks like if ##\omega=\omega (t) \neq const## which I presume should include Lagrangian dependency on ##\dot \theta##

Thanks a lot,
Michael
 
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In the second equation the second term of RHS has plus sign. In the third equation LHS is nonzero.
 
anuttarasammyak said:
In the second equation the second term of RHS has plus sign. In the third equation LHS is nonzero.
This is the exact solution from Morin.
The equation 6.142 corresponds to the second equation in my question
If we presume ##\omega=\dot \theta## then we have to take partial derivative over it into account, But this doesn't happen.
Looks like I'm missing something obvious...

bead_on_stick.png
 
Looks like I understood the problem.
The position of the stick is not the dynamical variable as it's a given function of time not depending on initial conditions therefore shouldn't be considered.
So, the system effectively has only one degree of freedom - bead's position on the wire.
Thanks!
 
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