What is the purpose of radius of gyration

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Discussion Overview

The discussion revolves around the purpose and application of the radius of gyration in the context of a mass rolling down an incline. Participants explore its relationship to the moment of inertia and the implications for writing the Lagrangian for the motion of the object.

Discussion Character

  • Technical explanation, Conceptual clarification, Debate/contested

Main Points Raised

  • One participant expresses confusion about the necessity of both the physical radius 'r' and the radius of gyration 'Rg' for calculating the moment of inertia.
  • Another participant explains that the radius of gyration relates to the moment of inertia by stating that it allows for the equivalence of the object to a point mass at a distance of 'Rg' from the center of mass, leading to the formula I = m R_g^2.
  • A different participant points out the discrepancy in using 'Rg' versus 'r' for the moment of inertia, noting that using 'r' yields I = 1/2 * m r^2 for a cylinder.
  • Another participant questions the assumption that I = 1/2 * m r^2 is valid, suggesting that the problem may not specify the object as a solid disk or cylinder, and raises the possibility that R_g could be equal to r/√2.

Areas of Agreement / Disagreement

Participants do not reach a consensus, as there are competing views regarding the assumptions about the object's shape and the appropriate use of 'r' and 'Rg' in calculating the moment of inertia.

Contextual Notes

There are unresolved assumptions regarding the shape of the object in question and the specific definitions of 'r' and 'Rg', which may affect the calculations and interpretations presented.

TheBlueDot
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Hello,

I'm given a problem of a mass rolling down an incline with mass 'm', radius 'r', and radius of gyration Rg, and I need to write the Lagrangian for the motion. I'm confused on why both r and Rg are given. Don't I just need one of the two for the moment of inertia?
Thanks!
 
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The radius of gyration gives the moment of inertia via the equivalence of the given object and an object with all its mass concentrated at a distance of the radius of gyration from the center of mass. So I = m R_g^2.

You can treat the object as a massless cylinder of the given physical radius with a hollow cylinder of mass m and radius R_g embedded within it.
 
Hi jambaugh,

Thank you for your response. The part I'm confused on is if I use Rg the moment of inertia (assume a cylinder) would be I= mRg^2, but if I use 'r', then I=1/2 *mr^2. I guess I don't understand where the question is leading...
 
Why would I = \frac{m}{2} r^2? Is R_g given to be equal to r/\sqrt{2}? Did the problem specifically say the object was a solid disk or cylinder? Maybe the object is not what you are assuming with the \frac{m}{2}r^2 formula.
 

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