Density and circumference relationship? From Example in book

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SUMMARY

The discussion focuses on the equilibrium position of a mass m placed in the plane of a thin uniform circular ring with radius a and mass M, as described in Example 5.3 of the fifth edition of The Marion Thornton book on Classical Dynamics. The confusion arises regarding the linear mass density, represented as ρ = M/(2πa), which is derived from the mass of the ring divided by its circumference rather than its volume. Additionally, the differential mass element dM is expressed in terms of the angular displacement dφ, leading to the equation dΦ = -G(dM/b) = (-Gaρ/b)dφ, where G is the gravitational constant. Clarifying these concepts is essential for solving the equilibrium problem presented in the example.

PREREQUISITES
  • Understanding of linear mass density and its calculation.
  • Familiarity with gravitational potential and its differential form.
  • Basic knowledge of circular motion and equilibrium conditions.
  • Proficiency in calculus, particularly in dealing with infinitesimal quantities.
NEXT STEPS
  • Study the derivation of linear mass density in circular objects.
  • Learn about gravitational potential energy and its applications in dynamics.
  • Explore the concept of stability in equilibrium positions for various shapes.
  • Review calculus techniques for handling infinitesimal changes in angular coordinates.
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Homework Statement


Example 5.3 from The Marion Thornton book (fifth edition) of Classical Dynamics states the following problem:

Consider a thin uniform circular ring of radius a and mass M. A mass m is placed in the plane of the ring. Find a position of equilibrium and determine whether it is stable. I'm following the example in the book, and there's two things I don't understand. First is why /rho = \frac{M}{2*/pi * a}. I always thought that the density was the mass over the volume, so I don't see why this is mass over circumference.

Second, in the next piece, it says that d\Phi = -G \frac{dM}{b} = \frac{-Ga\rho}{b}d\phi

I'm failing to see two things. 1) why the dM element ends up with a small phi, and 2) why rho is as it is. From here, I believe I can find the rest of the example, but why are these two things true?
 
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The textbook is apparently using the symbol ##\rho ## to denote the linear mass density.

An infinitesimal arc length can be written in terms of ##d \phi##.
 

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