# Expression for the moment of inertia

1. May 20, 2014

### Firben

1. The problem statement, all variables and given/known data
The thin, homogeneous bent rod has the mass m and the total length of 4b. It rotates with the angular speed of ω = ω0(24i + 12j - 6.0k) (only rotation)

Determine the expression for the moment of inertia with consideration of the center of mass of the rod

The figure:
http://s716.photobucket.com/user/Pitoraq/media/Mek4_zps8205c021.png.html

2. Relevant equations

H = (Ixxωx-Ixyωy-Ixzωz)i+(-Iyzωx+Iyyωy-Iyzωz)j+(-Izxωx-Izyωy+Izzωz)k

m = ρb

3. The attempt at a solution¨

I plotted the figure in yx-axis and could see that

Iyz=Iyx=Iyz=Izy=Ixz=Izx=0 is symmetric

But i'm not sure about Ixx, Iyy and Izz

2. May 20, 2014

Can you post your graph?

3. May 20, 2014

### Firben

4. May 20, 2014

### SteamKing

Staff Emeritus
You might have to, you know, calculate Ixx, Iyy, and Izz. It's a shocking suggestion, I'm sure.

5. May 27, 2014

### unscientific

The centre of mass is easy: it is at the origin of y-z axis.

There are three parts to this problem: the rotating center bit, and the two sides. By symmetry, the MOI of the left and right side are equal.

Calculate the MOI for rotation in the $i$ and $j$ and $k$ axes. Use parallel axes theorem to calculate the MOI of the sides.