What is the Moment of Inertia of a Circular Thin Cylindrical Surface?

[catmando]

Homework Statement

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.

Homework Equations

I = mr²

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

 

[tiny-tim]

welcome to pf!

hi catmando! welcome to pf!

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

about which axis?

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

 

[catmando]

hi catmando! welcome to pf! about which axis?

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

thanks

about the axis of the cylinder, I am really not sure

i know how to find the moment of inertia about a complete cylinder but i really don't 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.

 

[tiny-tim]

hi catmando!

i know how to find the moment of inertia about a complete cylinder but i really don't understand how to find the moment of inertia when its only a section of that cylinder.

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 )

 

[catmando]

hello :D

yeah i understand that, but when you have only a section of that cylindrical shell, if you integrate over r, won't 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?

 

[catmando]

bumpity

 

[tiny-tim]

hello catmando!

hello :D

yeah i understand that, but when you have only a section of that cylindrical shell, if you integrate over r, won't 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?

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! ​

 

[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

 

[tiny-tim]

hi catmando!

(try using the X² button just above the Reply box )

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

So looks like - )

thanks i ended up getting I = mr^2

that's correct …

the whole body is distance r from the axis, so it's just mr²

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

(does the question ask for that? )

correct

 

[catmando]

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