13. Rotational Inertia & Energy

Parallel Axis Theorem

# Parallel Axis Theorem

Patrick Ford

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Hey, guys, in this video, we're gonna talk about something called the Parallel Axis. The're, um, which is something that we're gonna use to find what I call nonstandard moments of inertia. And it's going to be critical to solving a lot of rotational dynamics problems. All right, let's get to it. Now. The moment of inertia is really, really important to know and to be able to find because it's the rotational analog. It acts like mass okay and rotational equations. Most rotational equations look exactly like linear equations. You simply just replace the linear variables with the angular or rotational variables. For instance, Newton's second law looks like F equals M A for a linear system. For a rotational system, it would look like the rotational, um, analog to force, which would be torque the rotational analog to mass. Which would be moment of inertia, like I just reminded us off and the rotational analog toe acceleration, which would be angular acceleration. Every single equation looks like this. If you know the linear one, you know, the rotational one. You just replace the variables with the rotational versions of those variables. Okay, now the problem with the moment of inertia is that it's not a fixed quantity like the masses. If you know the mass who always know the mass, the moment of inertia is a relative quantity, and it depends upon the rotation about which the motion is occurring. Okay, typically, there will be a select few moments of inertia given on things like homework or your exam, which I would call typical or standard moments of inertia. Usually, they're given about the center of mass of the object. Typically, they're given for regular geometric shapes like spheres, discs, rings and typically they are given for those objects when they're uniformly distributed, meaning homogeneous as in the masses. Sorry, the density is the same throughout the object. Okay, the masses evenly distributed throughout the whole objects. Okay, those were very, very typical. Moments of inertia is, and you're typically going to be given them for things like solid spheres, thin spheres, discs, rods, etcetera. Okay, but how do you solve a problem when you have a non typical rotation and you need a non typical moment of inertia? Okay, this is where the peril access serum comes in. OK? If, for instance, we want to know the moment of inertia for disc, rotating about its rim. Okay, what if we want to know if it was rotating about its rim? First we would start with the disk rotating about an axis through its center, which hopefully you guys remember this. It's a very, very common one. The moment of inertia is one half M times the radius squared. But what if now I wanted to choose a parallel axis. So obviously parallel access serum, these axes must be parallel to one another. Okay, but the second access is on the rim, which means it's a distance of the radius away. This is where we need to use the parallel axis. The're, um, to find the new moment of inertia. Okay, The parallel access there, um, is going to be the moment of inertia through the center of mass, which, in this case, we know because we know that through the center, which is the center of mass for a uniformed disk, the moment of inertia is one half m r squared. OK, this is going to be plus m d squared. Where D is the distance from the center of mass axis to the new axis. Let me minimize myself, which, as the name implies, must be parallel. So, like I said, for a disk, this is the center of mass. Here is some access. Let's say we're rotating about this access. The moment of inertia is gonna be one half m r squared. And if we chose a parallel axis at the rim, then the moment of inertia is going to be the moment of inertia of the center of mass plus the mass of the disc times the distance squared where the distance is the distance between the the center of mass and the new parallel axis. You can see that these two axes are definitely parallel, okay? And they need to be parallel. Otherwise, you cannot use the parallel accessed era, right, as the name implies, only works for parallel axes. Okay, so if we have a disc of mass M and Radius R, what is the moment of inertia about an axis perpendicular to the surface of the disk at the rim of the disk? So the same hypothetical problem I've been giving, okay? All we need to do is apply the peril access there. Um, it's the center of mass plus M D squared, where the distance to the rim is the radius. And by the way, I told the Mass capital him in this problem. So I'm just going to stick with the notation. And as I said earlier, the moment of inertia through the center of a disc is one half m r squared plus M r squared. This is going to be three halves m R squared. OK, but that's only one half of the problem. The problem says, What about a parable? Access halfway to the rim. Okay, well, same exact idea. Right? Here's our disk. Here's our original access through the center of Mass. And here is our new axis. Okay, where this distance is one half of the radius, right half way to the rim means that the distance is half of the radius. So what is that new center of Mass? Will that new Sorry? What's that new moment of inertia? The new moment of inertia is still the moment of inertia through the center of mass plus capital limb D squared, where now the distance is one half of the radius. Okay, And remember that it's the distance squared. So the one half is squared as well. So this is one half M r squared, which is simply the moment of inertia through the center mass plus 1/4 m r. Squared. Right, Because this one half also gets squared. Okay, one half plus a quarter is three quarters. So this is three quarters M R. Squared. Okay, so you see that the moment of inertia half way to the rim happens to be one half off the moment of inertia at the rim. Okay, Now, let's do another example right here. The moment of inertia of a thin rod. So we have a rod of some length l in some mass m The moment of inertia about an axis perpendicular to the rod at the end of the rod is one third. So this moment of inertia is one third. This should be a capital in Just because I call this the capital them, that will be corrected. And you guys pdf before it reaches you. Okay, so I'll go with capital M. L squared. Okay. What we want to know is, what is the moment of inertia? Halfway between the center of the rod and the edge of the rod. Okay, so This is one quarter of the length of the rod because, okay, the distance to the center of mass is clearly one half of the length. Right? That's clearly true. Now, should we simply say that this moment of inertia is one third m l squared plus 1/4? Sorry plus M 1/4 l squared. Should we say that that's true? No, we shouldn't say that. That's true. And the reason is because you have to remember that the moment Sorry. The parallax is here. Um um in order to use it, you need to know the moment of inertia about the center of mass, which we don't know in this case we know. What about the edge? So first we need to sort of use the moment of the parallax Ethereum in reverse toe. Find the moment of inertia at the center of Mass. Then we can find it at the point we're interested in. Okay, so here is the disk. Sorry. The rod right here is the center of mass. This moment of inertia, we don't know, but we want to find Here's the moment of inertia at the edge, which we do know is one third m l squared And this guy is one half elf. So using the peril access there, um, here we know that I is I times the center of mass plus m times one half l squared. OK, we know I thistles clearly a constant and we want to solve for this. So this is I minus 1/4 M l squared. Right? Let me minimize myself 1/4 because this one half just get squared and I we know is one third m l squared minus 1/4 m l squared If we multiply this guy by 3/3. This guy Sorry by 4/4, the least common denominator is 12. This guy by 3/3. Then what we're going to get is I'm running out of room here, so hold on. Sorry about this. Guys. This is 4/12 m l squared minus 3/12 m. L squared. So this whole thing right is just four minus 3/ m l squared or it's 1. 12 m. L squared. So now that we know the moment of inertia about the center of mass, now we can apply the parallel axis there. Um, so this distance is what we're interested in, which is still one forth. L Okay, but just to be technical, we're moving it from the radius from the access through the center of Mass. We're doing that from there too. This new axis. Okay, so this is the center of mass Moment of inertia. Who knows? 1/12 M l squared plus m times 1/4. Sorry. This is all capital IMS. I keep writing them is lower case Sims. Guys, that's my bad. It's just sloppy notation. All these air capital, IMS Capital in capital him. Capital limb capital him. L squared. OK, so this is gonna be 1/12 m l squared plus 1/16 m l squared now to find the least common denominator. How I did this is 12 is four times 3. 16 is four times four. They both share that common factor of four. So if I multiply the 12 times four and the 16 times three, I should get a common denominator. So this is gonna be 4/48 m. L squared plus 3/48 m. L squared. And it turns out I do get a common denominator. And four plus three is just seven. So 7/48 m. L squared. Okay, so that is the moment of inertia off axis, halfway between the center of the rod and the edge of the rod. Okay, remember that the moment the parallax s there and requires you to know the moment of inertia at the center of mass. So you cannot do this right here and try to move the access from the edge to halfway to the edge. Because the moment of inertia at the edge is not the center of mass. You need to use the parallel access. The're, um, like this toe find at the moment of inertia at the center of mass. Then you can move that access to the edge. All right, guys, Thanks so much for watching. I'll see you guys in another video.

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