How do I calculate the moment of inertia of a torus with given parameters?

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Homework Help Overview

The discussion revolves around calculating the moment of inertia of a torus, characterized by its mass, density, cross-sectional radius, and overall radius. Participants are exploring the mathematical formulation and integration techniques relevant to this problem.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the use of cylindrical slices and the integration of volume elements to derive the moment of inertia. Questions arise regarding the definition of the radius and the axis of rotation, as well as the correct formulation of the volume of cylindrical shells.

Discussion Status

The discussion is active, with participants providing hints and clarifications about the geometry involved. There is an ongoing exploration of different interpretations of the problem setup, particularly concerning the dimensions and integration methods.

Contextual Notes

Some participants question the definitions of the internal radius and the average radius, as well as the assumptions about the axis of rotation. There is also uncertainty regarding the volume calculations for cylindrical shells.

rock.freak667
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Homework Statement



Find the moment of inertia of a torus if mass is m and density [itex]\rho[/itex] is constant.
The cross-sectional radius is 'a' and the radius is R.

Homework Equations



[tex]I= \int r^2 dm[/tex]

The Attempt at a Solution



Well I looked up the answer to be

[tex]I_z= m(R^2 + \frac{3}{4}a^2)[/tex]

But I am not sure how to start. Can someone just point me in the right direction?
 
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rock.freak667 said:
Find the moment of inertia of a torus if mass is m and density [itex]\rho[/itex] is constant.
The cross-sectional radius is 'a' and the radius is R.

Homework Equations



[tex]I= \int r^2 dm[/tex]

Hi rock.freak667! :smile:

(have a rho: ρ :wink:)

Do you mean the moment of inertia about its axis of rotational symmetry? And is R the internal radius, or the average radius?

Hint: divide the torus into cylindrical slices of thickness dr, and integrate between R ± a :wink:
 
Yes R is the internal radius.


So if I am considering cylindrical shells of thickness dr.

if I draw it in 2d, it makes a circle such that [itex]x^2+z^2=a^2[/itex]


the volume of a cylindrical shell is

[tex]dV= \pi z^2 dr[/tex]

dV=pi z2 dr

so the moment of inertia of the small cylindrical element is

[tex]dI_c = \frac{1}{2} (\rho \pi z^2 dr) z^2[/tex]

dIc= (1/2) (p*pi*z2 dr)z2

But this would give me the inertia not about the z-axis right but the axis perpendicular to the cylindrical shell. Which is not about the z-axis.

and also I would be integrating z w.r.t. r
 
Last edited:
Hi rock.freak667! :smile:
rock.freak667 said:
the volume of a cylindrical shell is

dV=pi z2 dr

I'm not sure what your slices are, but that looks like the volume of a cylinder.

A cylindrical shell is the (thickened) surface of a cylinder. :wink:
 
tiny-tim said:
Hi rock.freak667! :smile:


I'm not sure what your slices are, but that looks like the volume of a cylinder.

A cylindrical shell is the (thickened) surface of a cylinder. :wink:


so dV= (2pi*z)x dr ? Not sure on the surface area of cylindrical shell that I'm considering
 
(have a pi: π :wink:)
rock.freak667 said:
so dV= (2pi*z)x dr ? Not sure on the surface area of cylindrical shell that I'm considering

(2π*z)x dr ? :confused:

what are z and x ?

slice it with a cookie cutter of radius r, then slice it again with a cookie cutter of radius r + dr :smile:
 

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