# Mass Moment of inertia

1. Dec 7, 2005

### Fermat

Can someone explain something to me ?

I refer you to http://www.engineering.com/content/ContentDisplay?contentId=41005050".

If we consider the rectangular block to be a cube, Then I would expect that we would get the mass moments of inertia all equal, Ixx = Iyy = Izz.
But, in the page I refered you to, if we let a = b = L, then we get Ixx = Iyy, but Izz is different.

Can anyone explain why Izz is different from Ixx and Iyy ?

Last edited by a moderator: Apr 21, 2017
2. Dec 7, 2005

### Staff: Mentor

Good question. Looks wrong to me too.

3. Dec 7, 2005

### Fermat

Thanks. Now I don't feel so bad.

Someone had asked me for help in calculatring MOI for a rectangular block.

I did some working and got the Izz format in that page. And posted my working on the forum I was at.

Then I decided to check it out on the web.

Imagine my consternation when I found that web page !!

I hope some more people tell me it is wrong <hopes mightily>

4. Dec 7, 2005

### Fermat

Gottit!

The x-axis and the y-axis don't pass through the COM of the block, but the z-axis does.

Ixcxc is of the same form as Izz, but Ixx needs the parallel axes theorem giving,

Ixx = Ixcxc + M(L/2)²

which gives,

Ixx = (1/12)Ma² + (1/3)ML²

So, my working was right after all

Oh, yes. And that web page is also correct.

5. Dec 7, 2005

### Cyrus

Are you sure you looked at the right fomulas?

You had to observe x_C, y_c,z_c, NOT X,y,Z.

x_c,y_c,z_c ARE equal.

(You are right in your observation abotu x and y not being in the center.)

Be aware that the COM is locationally dependent!

Last edited: Dec 7, 2005
6. Nov 23, 2011