Hamiltonian of a Point particle on a frictionless plane

In summary, the conversation discusses a problem involving Hamiltonian expressions and circular orbits. The person is struggling with questions e and f and is unsure how to apply Hamilton's equation and use Latex. They receive advice on how to solve the problem and are reminded to use Latex in the future.
  • #1
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1
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Homework Statement
Find the Hamiltonian of a Point particle on a frictionless plane with a given potential (kr^2)/2
Relevant Equations
L = T - V

dot rep time derivative

L = (m/2) * ( rdot^2 + r ^2 θdot^2) - V (r,θ)
Lagrange eq
d/dt (∂L/∂xdot) = ∂L/∂x
H = Σ pi qdoti - L
9efPZ.png
I am stuck on Question e and then how to proceed to f. I cannot seem to show this using the steps in the prior questions. My answers are:

a)
1673167910977.png


b)
1673167927405.png

1673167962364.png


c)

1673167979252.png


c) continued - and d) at the bottom of the page
d)
1673168014929.png
I am not sure where I have gone wrong, as I am not sure how to apply the relevant Hamilton's eq to the Hamiltonian in e)
I can sub in p_r into the Hamiltonian in e) however, I cannot use the p_theta 'dot' expression.
and then also

for f) how to show the energy is equal to kr_0.
 
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  • #2
Most people here won't have the patience to wade through that ugly mess.
You'll need to learn quickly how to use Latex on this forum. :oldfrown:
(Do a search for "latex" to find instructions.)
 
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  • #3
I agree with your Hamiltonian Expression, to complete the problem (the last few parts) realize that ##\dot{r} = 0## if we're talking about circular orbit ##r = r_0##. Which from the EL equation involving ##r## should tell you ##p_r = 0## (Which in turn tells you ##\dot{p_r} = 0##) (THIS IS THE KEY PART)

This cancels out a term in your Hamiltonian leaving.

## H = \frac{p_\theta^2}{2mr^2} + \frac{kr^2}{2}##

Take the expression for ##H## and find ##\dot{p_r} = - \frac{\partial H}{\partial R} ## set it equal to ##0## and solve for ##\frac{p_\theta^2}{2mr^2}## and plug it back into ##H## your answer should pop out.

Your assignment is likely turned in by now but you may find this helpful anyway.

But yes, use Latex next time, It is awesome.
 

FAQ: Hamiltonian of a Point particle on a frictionless plane

What is the Hamiltonian of a point particle on a frictionless plane?

The Hamiltonian of a point particle on a frictionless plane is given by the expression \( H = \frac{p_x^2}{2m} + \frac{p_y^2}{2m} \), where \( p_x \) and \( p_y \) are the momenta in the x and y directions, respectively, and \( m \) is the mass of the particle.

How is the Hamiltonian related to the total energy of the system?

The Hamiltonian represents the total energy of the system, which in this case is purely kinetic energy since the plane is frictionless and there is no potential energy involved. Therefore, \( H = T \), where \( T \) is the kinetic energy.

What are the equations of motion derived from this Hamiltonian?

The equations of motion can be derived using Hamilton's equations: \( \dot{q_i} = \frac{\partial H}{\partial p_i} \) and \( \dot{p_i} = -\frac{\partial H}{\partial q_i} \). For a point particle on a frictionless plane, these yield \( \dot{x} = \frac{p_x}{m} \), \( \dot{y} = \frac{p_y}{m} \), \( \dot{p_x} = 0 \), and \( \dot{p_y} = 0 \), indicating that the particle moves with constant velocity.

What assumptions are made in this Hamiltonian formulation?

The key assumptions are that the plane is perfectly frictionless, the particle is point-like (i.e., it has no internal structure or rotational degrees of freedom), and there are no external forces acting on the particle. Additionally, the system is assumed to be isolated and non-relativistic.

How does the Hamiltonian change if the plane is not frictionless?

If the plane is not frictionless, the Hamiltonian would need to account for the potential energy associated with friction. This could involve adding a term that represents the work done against friction forces. However, friction is typically a non-conservative force, and incorporating it into the Hamiltonian formalism can be complex and may require a different approach, such as using a Rayleigh dissipation function.

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