Hamiltonian of the bead rotating on a horizontal stick

  • #1
Michael Korobov
6
0
TL;DR Summary
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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  • #2
In the second equation the second term of RHS has plus sign. In the third equation LHS is nonzero.
 
  • #3
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
 
  • #4
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!
 

Related to Hamiltonian of the bead rotating on a horizontal stick

What is the Hamiltonian of a bead rotating on a horizontal stick?

The Hamiltonian of a bead rotating on a horizontal stick is the total energy of the system, which includes both the kinetic and potential energy. For a bead of mass \( m \) rotating on a stick of length \( L \), the Hamiltonian can be expressed as \( H = \frac{p^2}{2m} + V(\theta) \), where \( p \) is the conjugate momentum associated with the angular coordinate \( \theta \), and \( V(\theta) \) is the potential energy, which is usually zero if we consider a frictionless, horizontal stick.

How do you derive the kinetic energy term in the Hamiltonian for a bead on a stick?

The kinetic energy \( T \) of a bead rotating on a horizontal stick is given by \( T = \frac{1}{2} m v^2 \). Since the bead is constrained to move in a circle of radius \( L \), its velocity \( v \) is \( L \frac{d\theta}{dt} \). Therefore, the kinetic energy becomes \( T = \frac{1}{2} m (L \frac{d\theta}{dt})^2 \). In terms of the conjugate momentum \( p = m L^2 \frac{d\theta}{dt} \), the kinetic energy can be written as \( T = \frac{p^2}{2 m L^2} \).

What is the role of the potential energy in the Hamiltonian of a bead on a horizontal stick?

In the idealized case of a bead rotating on a frictionless, horizontal stick, the potential energy \( V(\theta) \) is typically considered to be zero because there is no change in height and no other forces acting on the bead. Therefore, the Hamiltonian simplifies to \( H = \frac{p^2}{2 m L^2} \). If there were external forces or constraints, these would be included in \( V(\theta) \).

How does the Hamiltonian change if the stick is not horizontal but inclined?

If the stick is inclined at an angle \( \alpha \) with respect to the horizontal, the potential energy \( V(\theta) \) needs to account for the gravitational potential energy. For a bead at an angle \( \theta \) around the stick, the height \( h \) relative to the horizontal is \( L \sin(\alpha) \sin(\theta) \). Therefore, the potential energy becomes \( V(\theta) = m g L \sin(\alpha) \sin(\theta) \). The Hamiltonian then becomes \( H = \frac{p^2}{2 m L^2} + m g L

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