# Moment of inertia of a torus

1. Apr 17, 2009

### rock.freak667

1. The problem statement, all variables and given/known data

Find the moment of inertia of a torus if mass is m and density $\rho$ is constant.

2. Relevant equations

$$I= \int r^2 dm$$

3. The attempt at a solution

Well I looked up the answer to be

$$I_z= m(R^2 + \frac{3}{4}a^2)$$

But I am not sure how to start. Can someone just point me in the right direction?

2. Apr 19, 2009

### tiny-tim

Hi rock.freak667!

(have a rho: ρ )

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

3. Apr 21, 2009

### rock.freak667

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 $x^2+z^2=a^2$

the volume of a cylindrical shell is

$$dV= \pi z^2 dr$$

dV=pi z2 dr

so the moment of inertia of the small cylindrical element is

$$dI_c = \frac{1}{2} (\rho \pi z^2 dr) z^2$$

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: Apr 21, 2009
4. Apr 21, 2009

### tiny-tim

Hi rock.freak667!
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.

5. Apr 21, 2009

### rock.freak667

so dV= (2pi*z)x dr ? Not sure on the surface area of cylindrical shell that I'm considering

6. Apr 21, 2009

### tiny-tim

(have a pi: π )
(2π*z)x dr ?

what are z and x ?