Physics: Rotational Inertia of Solid Bodies

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The moment of inertia of a solid object around some particular axis of rotation is a physically important quantity. It is used when you're calculating kinetic energy of rotation; it's used in lots of situations where you have rotating objects. And the computation of the moment of inertia is simple when the object is just composed of a bunch of little chunks. If you have some object that's built-up out of a mass here, m[1], and a mass here, m[2], and here's your axis about which you're rotating, so you have to imagine that these are somehow rigidly connected together into a body and everybody's going to start rotating. The moment of inertia would just be m[1]r[1]^2 plus m[2]r[2]^2 squared, etc. You'd just add-up the moment of inertia, mr^2, for each little chunk. So it's easy.
If you have barbells, and I imagine that the mass of the barbell is small and the concentrated over and over here. And I ask, "What's the moment of inertia of this barbell?" That's technically an ill-defined. I have to say, "The moment of inertia around a point." What's the moment of inertia around the center? What if I start swinging this barbell around the middle, what's the moment of inertia? Well, it's mr^2 added up. We've got to pick some labeling system here, so let's call the center zero and let's call this distance L and this distance L. So it's 2L in length and the moment of inertia about the center would be mL^2 for this one plus mL^2 for that one, or 2mL^2.
I'm making a bit of a fuss about the statement that it matters where you're swinging the object around. Supposing that we held onto this end and started swinging this barbell around our head? What's the moment of inertia now? The new moment of inertia would be this object - remember this ideally imagined to be a little point mass m, this mass is now at r equals zero. So it contributes nothing to the moment of inertia. If you've got a tiny point and you spin it around, there's no energy associated with that. The energy's associated with this mass swinging around in a circle. That's the physical origin of this kinetic energy and therefore the moment of inertia, which is telling you about something associated with the kinetic energy. I should remind you of the connection between moment of inertia and kinetic energy, because this is what I'm bearing in mind all the time. Kinetic energy is proportional to moment of inertia and , how fast is this thing rotating. It looks like mv^2, but this is the rotational version, I^2. Moment of inertia about this axis, this one contributed nothing, this one was mass times radius, which is 2L quantity squared. So you square 2L and you get 4 mL^2. It's got a bigger moment of inertia. If you take a barbell and you swing it around its middle, it's got a certain moment of inertia. If you swing it around one end, it's got more larger moment of inertia. For a given swing rate, it will have much more kinetic energy and that kind of makes some sense, if you think about what it would feel like swinging this thing.
It's easy when you have a bunch of little objects pieced together. What happens when you have a continuous distribution of matter? What happens if you have some solid object, like a disk? I have to compute the moment of inertia about some axis, supposing I just start spinning it around its center? That's the simplest problem I can think about and I need the integral version of the formula. Instead of the sum of r^2 times m or m, now it's the integral of r^2dm, where I'm imagining my little m's have now become infinitesimal chunks and I'm integrated over all of them. It's the usual story, when you see an integral like this. You might see an integral that looks a little bit like this when you're calculating the center of mass, except then it was rdm, not r^2, so the formula is a little bit different than the center of mass. And it's the same idea; you've got to think about what this formula is telling you. You have to draw a nice picture. Here is the disk and I'm going to break it up into little chunks. And this dark band here is going to represent one of my little chunks, dm. So I've kind of got to blow that up. Here's my little band and it's got a thickness r, which is going to turn into dr when I do the integral. And I'd like to know, "What is the mass of that ring?" Well, the mass of that ring is the mass per area times the area of that little ring. So you've got a little ring and it's located at a distance r from the center and it's got a thickness r. This is a little geometry puzzle. The area of such a ring is just circumference times thickness. So it's 2rr, which is going to become 2rdr in the limit that we take r infinitesimally small. And what's the area? This is the total mass of the whole disk divided by the whole area of the whole disk and the whole are is just R^2. So this is a disk with radius R and I'm going to be integrating over little radii, working my out from zero out to R.
So what have I got? I want the moment of inertia. It's the integral of r^2dm, so it's the integral, as radius goes, from zero to R of r^2. And what I just said was the little mass of the ring is mass per area, total mass over total area of the whole thing. That's mass per area, and then you multiply that by the little d area and what I'm figuring out is the little d mass, and that's 2rdr. So what have I got here? My 's cancel. 2M over r^2, those are all constants and come out of the integral. And I've got r^3dr. The integral of r^3 is r^4 over 4, evaluated from R to zero, so it's R^4 over 4 minus zero. I've got M and two of these powers cancel. Mr^2. It's the usual story. When you have to do an integral, you should stop and look at your answer. Does it make sense? It's got the right units. It's moment of inertia, kilograms times meters squared. It's not MR^2. That's the sort of naïve formula for a little single point out at radius R. This is not a single point out at radius R. There are some chunks that have very small radius and chunks that have a big radius. And the fact that it comes out to be kind of makes some kind of physical sense to me. It's sort of the average in some integral sense. So MR^2 sit the moment of inertia of a disk about its center of mass.
Supposing that you wanted to calculate the moment of inertia of that same shape about a different axis. Instead of spinning it around its center, supposed you wanted to spin it around a point on its edge? Or, for that matter, suppose you wanted to spin it about a point over here, so you tie a little thin string and you swing it around in a big old loop. What would that moment of inertia be? You might think, "Oh, no, every time I have a new physics problem, I've got to go and do another one of those integrals and it's going to get nastier and nastier as I move over." And, yes, that's true, the integral would get nastier and nastier. There's a nice theorem, a mathematical theorem, called the parallel axis theorem and it saves us a lot of grief. It says the moment of inertia of a solid object about an axis parallel to the one - here's the axis. It's perpendicular; it's the axis about which this thing is swinging. So if I want to know the moment of inertia about another axis parallel to this one, that's why it's the parallel axis theorem, the new moment of inertia about the new axis is the old moment of inertia around the center of mass plus Mh^2. Let me draw you a picture here of sort of a general case. h is just the distance between your axes. It doesn't have to be a disk, it can be any old funky shape. So if you have a funky shape and you know what the moment of inertia is around its center of mass, the parallel axis theorem will tell you what is the moment of inertia when you swing it around some other axis far away. It doesn't have to be far away, it can be anywhere. It can be partway in the object, at the edge, outside. So this formula is just quick and convenient. For instance, if you wanted to figure out the moment of inertia of a solid disk about a point right here - I'm just going to swing it around the edge, it would be MR^2, that's the moment of inertia about the center of mass, plus M. And what's h? h is the distance between the axes in this particular problem. That's just the radius again, so I've got plus MR^2 and I get 3/2MR^2. This has a bigger moment of inertia than just spinning it around its center of mass. It's always bigger.
Calculating moment of inertia can be a little bit painful. It's the same old pain as calculating the center of mass of an object, except that you can't really even take advantage of symmetry. What you do have is you've got the parallel axis theorem and typically you can find in reference books the moment of inertia of kind of standard shapes. What's the moment of inertia of a sphere about its central axis? What's the moment of inertia of a hoop? If the hoop has no thickness, that's an especially easy one, because every chunk of mass is at radius R, so it really is just total mass times radius squared. But you can look up all sorts of funky shapes, a door shape or a rod shape, or you can do the integral and, once you know the moment of inertia, it's very important. It's important for calculating kinetic energy and it's important for calculating lots of rotational observables and quantities.
The Physics of Extended Objects
Rotational Inertia and Kinetic Energy
Calculating the Rotational Inertia of Solid Bodies Page [1 of 2]