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Moment of inertia of a torus

  • #1
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?
 

Answers and Replies

  • #2
tiny-tim
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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:
 
  • #3
rock.freak667
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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:
  • #4
tiny-tim
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Hi rock.freak667! :smile:
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:
 
  • #5
rock.freak667
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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
 
  • #6
tiny-tim
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(have a pi: π :wink:)
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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