# Moment of inertia about tangent of a hoop

LCKurtz
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Gold Member
In a thread this lengthy, it is not clear to which formulas you are referring.

The general formulas for moment of inertia given by Zondrina in Post #20 would be the correct ones to use.
No, they are not appropriate to this problem. In post #12 the OP makes it clear he is talking about a [thin] wire. This is not a 3D problem. There is no volume and no ##dV##. ##\rho## would be a linear density.

SteamKing
Staff Emeritus
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No, they are not appropriate to this problem. In post #12 the OP makes it clear he is talking about a [thin] wire. This is not a 3D problem. There is no volume and no ##dV##. ##\rho## would be a linear density.
I think they are, in general. This problem with the wire is a limiting case of the much more general one involving a figure, like a torus.

The wire would be assumed to have a constant cross-sectional area, dA, so that dV = dA * dl, and dm = ρ dV = ρ * dA * dl.

The dl can be expressed using polar coordinates in place of the cartesian coordinate in which these formulas are expressed.

The quantity ρ * dA is a constant, with units of M / L, as desired.

LCKurtz
Homework Helper
Gold Member
I think they are, in general. This problem with the wire is a limiting case of the much more general one involving a figure, like a torus.

The wire would be assumed to have a constant cross-sectional area, dA, so that dV = dA * dl, and dm = ρ dV = ρ * dA * dl.

The dl can be expressed using polar coordinates in place of the cartesian coordinate in which these formulas are expressed.

The quantity ρ * dA is a constant, with units of M / L, as desired.
In other words, if you take a 3d problem, put enough assumptions on ##\rho## and use an approximation to account for the "thinness" of the wire, and carefully take the limit, you will eventually wind up with the integral$$\int x^2\rho~ds$$which is the straightforward way to work the problem in the first place. I do not concede your point. This is a 1d, one parameter, problem.

vela and SammyS
SteamKing
Staff Emeritus
Homework Helper
In other words, if you take a 3d problem, put enough assumptions on ##\rho## and use an approximation to account for the "thinness" of the wire, and carefully take the limit, you will eventually wind up with the integral$$\int x^2\rho~ds$$which is the straightforward way to work the problem in the first place. I do not concede your point. This is a 1d, one parameter, problem.
Except, it's not.

You challenged the validity of the limiting assumptions I made, but you did not show that these were erroneous or mathematically invalid in any way.

The MMOI of the wire doesn't depend on just x2, it depends on the distance, (x2 + y2), from the axis of rotation of each differential element of mass, dm, according to the definition of the MMOI, which Zondrina kindly posted above in Post #20.

In other words, dMOI = (x2 + y2) * dm

Thus, the MOI = ∫ (x2 + y2) * dm, taken around the entire length of the hoop.

It just so happens that a convenient substitution for (x2 + y2) makes it easy to use polar coordinates for this problem, along with a suitable equivalent expression for dm.

This article shows the MMOI results for many different kinds of figures (including, I might add, the circular hoop made from a thin wire):

https://en.wikipedia.org/wiki/List_of_moments_of_inertia

Calculus is based on making limiting assumptions at some point, otherwise it would be unusable for analyzing many problems.

LCKurtz