Expression for effective potential energy

In summary, the central potential energy is negative for particles of identical mass when their distance from the center is smaller than a certain value. The constants involved in the expression are the moment of inertia and the mass. If the angular momentum is nonzero, then the critical value of the central potential energy can be expressed in terms of the reduced mass and the angular momentum.
  • #1
Thorscira
4
1
<< Mentor Note -- Poster has been reminded to use the Template when starting new schoolwork threads >>

Two particles of identical mass m interact with each other via central potential energy

Vcentral(r) = -V0(1-|r|/a), if 0 <= |r| <= a
0, if a < |r|

Constants are V0 > 0 and a > 0

What's the expression for the effective potential energy and what are the constants in your expression?

My attempt:

Veff = -V0(1-|r|/a) + M/2mr^2

M is the moment of inertia which is constant/conserved in any system relative to the centre.

/*I'm not sure about this any and all help would be very much appreciated! Thank you in advance :) */
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
Hello Thorscira ##\qquad## :welcome: ##\qquad## !

Please don't erase the template -- guidelines

I don't agree with
Thorscira said:
M is the moment of inertia which is constant/conserved in any system relative to the centre.
there is something else (involving M) that is conserved

The expression (Please use the sub- and superscript buttons)
Veff = -V0(1-|r|/a) + M/2mr2
can not be right: check the dimensions !
 
Last edited:
  • Like
Likes berkeman
  • #3
BvU said:
The expression (Please use the sub- and superscript buttons)
I'm so sorry! It's my first time posting on here and I wasn't sure how to define vectors and things. r is a vector describing two dimensions.
 
  • #4
can you fix the dimensions problem I mentioned ?
 
  • #5
Veffective(r) = -V0(1-|r|/α) + M/2mr2

I hope this is okay, I underlined the vectors.
 
  • #6
second term is length squared
 
  • #7
This is what I've got so far:

Veffective(r) = -V0(1-|r|/α) + M/2μr2

In this case the angular momentum would be constant and μ, right?

If we keep the distance α fixed but V0 can be varied and the angular momentum is nonzero. Is there a way one can express the critical value of V0 as a function of the reduced mass, α, and the angular momentum?
 
  • #8
you can not add length squared to energy
is it clear what mu is?
 

1. What is the expression for effective potential energy?

The expression for effective potential energy is a mathematical representation of the potential energy of a system that takes into account the effects of both conservative and non-conservative forces. It is typically represented as Ueff(r), where r is the distance between two interacting bodies.

2. How is the effective potential energy related to the total energy of a system?

The effective potential energy is a component of the total energy of a system. It represents the potential energy that can be converted into kinetic energy by a conservative force. The total energy of a system is the sum of its kinetic energy and potential energy, including the effective potential energy.

3. What factors affect the expression for effective potential energy?

The expression for effective potential energy is affected by factors such as the distance between interacting bodies, the masses of the bodies, and the strength and type of forces acting on the system. It can also be affected by external factors such as gravitational or electromagnetic fields.

4. Can the expression for effective potential energy be used to predict the motion of a system?

Yes, the expression for effective potential energy can be used to determine the behavior and motion of a system. By analyzing the shape and properties of the effective potential energy curve, we can determine the equilibrium points, stable and unstable orbits, and other important characteristics of the system.

5. How is the expression for effective potential energy derived?

The expression for effective potential energy is derived using principles from classical mechanics, such as the conservation of energy and Newton's laws of motion. It also takes into account the specific forces and interactions present in the system. The exact derivation process may vary depending on the specific scenario and forces at play.

Similar threads

  • Introductory Physics Homework Help
Replies
15
Views
272
  • Introductory Physics Homework Help
2
Replies
55
Views
2K
Replies
10
Views
438
  • Introductory Physics Homework Help
Replies
7
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
734
  • Introductory Physics Homework Help
Replies
8
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
671
  • Introductory Physics Homework Help
Replies
1
Views
846
  • Introductory Physics Homework Help
Replies
5
Views
1K
Back
Top