Is Angular Momentum Conserved in a Vertical Plane with Additional Force?

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Homework Help Overview

The problem involves a particle moving in a vertical plane under the influence of gravity and an additional force directed toward the origin. The discussion centers on determining whether angular momentum about the origin is conserved in this context.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the choice of generalized coordinates and the formulation of the Lagrangian equations of motion. There is a focus on the potential and kinetic energy expressions, as well as the implications of the derived equations on angular momentum conservation.

Discussion Status

The discussion is ongoing, with participants providing feedback on the formulation of the Lagrangian and questioning specific assumptions made in the setup. Some participants express confidence in the material while others seek clarification on the implications of their findings.

Contextual Notes

There are indications of potential misunderstandings in the initial setup, particularly regarding the kinetic energy term and the implications for angular momentum. The original poster acknowledges possible errors in their calculations.

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Homework Statement


Consider a vertical plane in a constant gravitational field. Let the origin of a coordinate system be located at some point in this plane. A particle of mass m moves in the vertical plane under the influence of gravity and under the influence of an aditional force f = -Ar^(a-1) directed toward the origin (r is the distance from the origin; A and a [does not = 0 or 1] are constants). Choose appropriate generalized coordinates, and find the Langrangian equations of motion. Is the angular momentum about thet origin conserved?

Homework Equations


L = T - U
\frac{\partial L}{\partial q_i} - \frac{d}{dt}\frac{\partial L}{\partial \dot q_i} = 0

The Attempt at a Solution


Choose the coordinate system (r', y') centered at the origin, so that the r' axis is in the direction of the origin to the particle of mass m and the y' axis is perpendicular to r'.

First, \arrow F = (-mg\cos{\theta}, -mg\sin{\theta}) + (-Ar^{a-1}, 0) = (-mg\cos{\theta} - Ar^{a-1}, -mg\sin{\theta}).

From here we can guess the potential:
U = mgr\cos{\theta} + \frac{A}{a}r^a + mgy\sin{\theta}.

(The preceding steps may be wrong, but the following is what I'm not sure about):
Notice that in our coordinate system y is always 0. So our potential is U = mgr\cos{\theta} + \frac{A}{a}r^a. For the kinetic energy, we get T = 1/2m(\dot r^2 + \dot \theta^2r^2) <ACCIDENT = 1/2m\dot \theta^2r^2>. So our Langrangian is L = 1/2m(\dot r^2 + \dot \theta^2r^2) - mgr\cos{\theta} - \frac{A}{a}r^a.

The rest is easy if I did the preceding correctly. You get that angular momentum is not conserved when you find \frac{\partial L}{\partial \theta} - \frac{d}{dt}\frac{\partial L}{\partial \dot \theta} = 0.

Thanks in advanced!
 
Last edited:
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Looks OK. You should get two equations of motion. What does this mean? Are you saying r-dot is zero?

<br /> T = 1/2m(\dot r^2 + \dot \theta^2r^2) = 1/2m\dot \theta^2r^2<br />
 
Last edited:
That was an accident. In the Lagrangian, you can see I did not carry that part through. I will put the incorrect part in brackets. Thank you! My confidence in this material has been elevated.
 
Last edited:
Just checking. Lagrangians are fun once you get used to them.
 

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