Why is Izz different from Ixx and Iyy for a cube?

[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 referred 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 ?

 

[berkeman]

Good question. Looks wrong to me too.

 

[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>

 

[Fermat]

Gottit!

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

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

Ixx = Ix_cx_c + 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.

 

[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!

 

[hawknosed]

In that same page on engineering.com, (http://www.engineering.com/Library/...138/PageID/336/Geometry-of-Common-Solids.aspx) the mass moment of inertia of the block about the z1 axis is marked with a "-". Isn't is easily calculable? Isn't it M/3(a^2+b^2)?

Thank you.