Rotational Inertia concept help

[Zlex]

Alright, I absolutly do not understand this concept. I missed the lecture and the textbook does not seem to explain it very well.

Here is a sample question:


Two particles, each with mass m = 1.10 kg, are fastened to each other, and to a rotation axis at O, by two thin rods, each with length d = 0.670 m and mass M = 0.252 kg. The combination rotates around the rotation axis with angular speed w = 0.417 rad/s. Measured about O, what are the combination's (a) rotational inertia and (b) kinetic energy?

http://www.intercomrealestate.com/FileSystem/Fig10_35.gif [Broken]



I know how to get $$E_k$$ from the rotational intertia, pretty much just plug and chug, but as far as finding the rotational intertia...

When I opened my textbook I just saw a lot of integrals and paniced. Well, actually they give you the common integals for all the shapes, the one for a rod is $$\frac{1}{12}ML^2$$. But I'm not sure what I'm supposed to do with that.

I have tried to do the question, but with no avail. But I will attempt to give some working that I've done.

$$I = \sum m_i*r_i^2$$
$$I = \frac{1}{12}ML^2$$ (For rod around central axis)
$$I = I_com + Mh^2$$ (Parallel axis theorm)

Applying all that I got:

$$md^2 + m(2d)^2 + \frac{1}{12}M(2d)^2 + Md^2$$


Its not right, and I'm not 100% sure why.

Edit: Oh wait, there are two rods. Ah, well, I'll try to figure it out from there. Any help is still appreciated

Edit2: $$md^2 + m(2d)^2 + \frac{1}{12}(2M)(2d)^2 + (2M)d^2$$

Heh; figured it out. Wish I could delete this thread.

 

[Nate810]

For the rotational inertia, you need to use the Parallel Axis Theorem. This states that the rotational inertia of a system is equal to the rotational inertia of its center of mass plus the mass times the square of the distance between the center of mass and the point at which the angular momentum is measured. So for this problem, you will need to add together the rotational inertia of each rod around its own center of mass (\frac{1}{12}ML^2), plus the mass of each rod times the square of the length of the rod (Md^2). You will need to do this twice since there are two rods. The kinetic energy can then be found using the equation E_k = \frac{1}{2}Iw^2.