Kinetic energy in polar coordinates

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Homework Help Overview

The discussion revolves around the concept of kinetic energy in a system described using polar coordinates. Participants are exploring whether an additional term for rotational kinetic energy is necessary when velocity is expressed in terms of polar coordinates, specifically focusing on the components related to radial and angular motion.

Discussion Character

  • Conceptual clarification, Assumption checking, Mixed

Approaches and Questions Raised

  • Participants are questioning the need for a separate term for rotational kinetic energy in polar coordinates, with some suggesting that the existing terms may already account for both translational and rotational motion. Others are asking for clarification on the nature of the motion and the relationships between the particles involved.

Discussion Status

The discussion is active, with participants providing varying perspectives on the necessity of additional kinetic energy terms. Some have suggested writing the kinetic energy in rectangular coordinates before converting to polar coordinates as a potential approach to clarify the situation. There is no explicit consensus, but several lines of reasoning are being explored.

Contextual Notes

Participants have noted the importance of understanding the specific motion of the system, including whether it involves simple spinning or more complex interactions between multiple particles. There are indications that assumptions about the system's constraints and relationships may need to be examined further.

teclo
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if i have a system that I'm describing using polar coordinates, do i need to have an additional term for rotational kinetic energy? it would seem like this is covered since my velocity is in terms of the r and theta basis vectors. (i.e. i will have a term that covers the rotational movement ala theta). I'm not really sure about this, though. thanks if anyone is able to help.
 
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More information would be helpful, because in a general way you're wrong (example, if your velocity just describes linear motion). What's it rotating about? Is it just spinning, etc.
 
Office_Shredder said:
More information would be helpful, because in a general way you're wrong (example, if your velocity just describes linear motion). What's it rotating about? Is it just spinning, etc.

two particles constrained to each other, free to rotate. each one is described by the position vector a(rhat) + b(theta hat). my velocity would have two components, r and theta. the velocity for each one would be in terms of r and theta, which would (i would think) would cover translational and rotational movement. the last sentence sums up my question fairly well. thanks for your input!-
 
I assume you are trying to write a lagrangian though I may be incorrect. You will have two kinetic energy terms in your lagrangian in this case, one proportional to the time derivative of your r position, and one proportional to the time derivative of the theta direction.
 
teclo said:
if i have a system that I'm describing using polar coordinates, do i need to have an additional term for rotational kinetic energy?

YES

marlon
 
marlon said:
YES

marlon

could you elaborate, please?
 
Write the kinetic energy of each point-particle in rectangular coordinates, then convert to polar coordinates. If there are relations among your point-particles [e.g. extended rigid bodies], you may be able to group terms and reinterpret.
 
robphy said:
Write the kinetic energy of each point-particle in rectangular coordinates, then convert to polar coordinates. If there are relations among your point-particles [e.g. extended rigid bodies], you may be able to group terms and reinterpret.

yes, this is definitely the best way to see what is going on if you are confused. Start with just a 1/2mv^2 term for each direction and then sub in the substitutions into polar coordinates, yuo should much of it will cancel out and you will be left with a simple expression with a kinetic energy term for each of your coordinates, r and theta.
 
http://web.me.com/dmwilliams/photo.jpg
Here is the derivation.
 

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