# Moment of Inertia of ring

1. May 14, 2012

### catmando

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

Find the moment of inertia of a circular thin cylindrical surface which ranges from -α/2 to α/2.

So looks like - )

The dash being the origin.

It basically looks like one fifth of a circular ring.

2. Relevant equations

I = mr²

3. The attempt at a solution

area of ring = ar . dr

I = ∫ (area of ring) (radius of ring)²

I = ∫ (ar . dr) (r)²

integrated this between 0 and r to get

I = ar^4 / 4

2. May 14, 2012

### tiny-tim

welcome to pf!

hi catmando! welcome to pf!

if it's about the axis of the cylindcr, why are you using area? why not just m ?

3. May 16, 2012

### catmando

Re: welcome to pf!

thanks

about the axis of the cylinder, im really not sure

i know how to find the moment of inertia about a complete cylinder but i really dont understand how to find the moment of inertia when its only a section of that cylinder, and the cylinder has a very small thickness. we are told that the object has a mass m kg/m along the arc.

4. May 16, 2012

### tiny-tim

hi catmando!
only the distance (r) matters …

the same mass at distance r (from a particular axis) has the same moment of inertia whether it's on a cylindrical shell or all at one point

(that's why, when you have to integrate, you always choose to integrate over r, with cylindrical shells )

5. May 16, 2012

### catmando

hello :D

yeah i understand that, but when you have only a section of that cylindrical shell, if you integrate over r, wont that include the whole cylinder shell, not just a section?

http://s13.postimage.org/90249nij9/picture.jpg

if i integrated over r, wouldn't it include the rest of the circle?

6. May 18, 2012

### catmando

bumpity

7. May 18, 2012

### tiny-tim

hello catmando!
i'm very sorry, i didn't see that post until today

no, when you integrate a body over r, you divide the body into cylindrical shells,

each shell looking like the shell in your picture

you then say "the area of this shell is the thickness, dr, times the angle, θ(r), ie θ(r) dr"

and finally you multiply that area by ρr2 (or whatever) … ∫ θ(r)ρr2 dr

in this case, you only have the one shell, so no integration is needed!

8. May 23, 2012

### catmando

thanks i ended up getting I = mr^2

then using the parallel axis theorem I = Io + md^2

the moment of inertia about the centre of gravity is Io = mr^2 - md^2

Last edited: May 23, 2012
9. May 23, 2012

### tiny-tim

hi catmando!

(try using the X2 button just above the Reply box )
that's correct …

the whole body is distance r from the axis, so it's just mr2
(does the question ask for that? )

correct

10. May 23, 2012

### catmando

yeah the question asked for that. thanks for you help :)