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



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 theorem)

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.

 

[member 553083]

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.

 

[wais]

I understand your frustration with the concept of rotational inertia. It can be a difficult concept to grasp, especially without attending the lecture. However, don't worry, with some practice and understanding of the main principles, you will be able to solve problems like the one provided.

First, let's start with the definition of rotational inertia. It is the measure of an object's resistance to changes in its rotational motion. In other words, it is the object's tendency to keep rotating at a constant speed or to resist changes in its rotational speed.

Now, let's break down the steps to solve this problem:

1. Identify the objects and their masses: In this problem, there are two particles with mass m = 1.10 kg each, and two thin rods with mass M = 0.252 kg each.

2. Determine the rotational axis: The rotation axis is located at point O, where the two particles and the two rods are fastened together.

3. Find the rotational inertia of each object: Since the rods are thin and have a length d = 0.670 m, we can use the formula for the rotational inertia of a rod around its central axis, which is given as I = \frac{1}{12}ML^2. Therefore, the rotational inertia of each rod is \frac{1}{12}(0.252)(0.670)^2 = 0.0094 kg*m^2.

For the two particles, we can use the formula for the rotational inertia of a point particle, which is given as I = mr^2, where r is the distance from the rotation axis to the particle. Since the particles are located at a distance d and 2d from the rotation axis, their rotational inertia will be md^2 and m(2d)^2, respectively.

4. Apply the parallel axis theorem: Since the particles and the rods are not rotating around their own center of mass, we need to use the parallel axis theorem to find their total rotational inertia about the rotation axis at O. The parallel axis theorem states that the rotational inertia of an object about an axis is equal to its rotational inertia about its center of mass plus the product of its mass and the square of the distance between the two axes. In this case, we need to apply this theorem to both the particles and the rods.

For the two particles, the distance between their center of mass and the rotation axis is d, so the additional rotational inertia