https://youtu.be/S1HnHFECG5U
PHYS 1101: Lecture Twenty-Two, Part Five
So, we’re almost there. We’re almost ready, now, to use our basic equation now completely for general cases, equilibrium and not. If the thing is in equilibrium, angular acceleration is 0; the torques balance. If it’s not, I have to take into account, I have to appreciate, what we call the moment of inertia: the resistance or the difficulty of getting this object to rotate. When it comes to rotation, the mass certainly plays an important role in terms of that difficulty of getting it to wind up or wind down. But, it turns out that how that mass is distributed is also important. I’ll show you that in a minute.
So we’re thinking about the basic equation, we’ve thought a lot about toques, now it’s time to carefully define and consider what I is, this moment of inertia. It’s only going be important if something is not in equilibrium. It really is starting to spin faster or wind down, spin down. Unfortunately, I is the variable that is used in physics for that, and it’s, again, the moment of inertia.
Note that it’s in the denominator, just like mass was in the denominator in our A= equation. This means, in general, the bigger the value for I the smaller acceleration you’re going to get because it’s in the denominator here. So, if I’ve got a big moment of inertia, it’s really hard. It takes a big torque to get something to start spinning up or slowing down. If it’s a small moment of inertia, think of it as the analog of being very light; it doesn’t take much of a torque to get the thing to wind up or down.
So, here’s what you’ll do for I. It could be that I is given in a problem. You could be given the number for it. Let’s say the moment of the inertia of the object is, whatever, it could be a variable you have to solve for. What is the moment of inertia of the wheel? Or, you may have to calculate in it, or figure it out by other means, or calculate it directly.
And here’s what I mean: you’ve got two possibilities. Either, if you do have to calculate it directly, you can calculate it for very simple geometries with a method that I’ll show you in just a minute. Or, you’ll, for more complicated geometries, look the value up in a table.
Okay, let’s start with this one so you see what I mean. This is how you calculate I for a simple rotating geometry. By simple, it has to meet these criteria: I have some well-defined axis of rotation and I have one, two, or a collection of localized masses that are located a well-defined distance from this rotation axis. So I don’t have some big, extended object that’s rotating about it’s center, or something like that. It’s these localized, little masses that are rotating about this axis. Only for this simple scenario can you calculate it.
And, here’s what it is. The value of I is equal to the contribution from each of these masses. For each one, it’s the product of the mass and the distance from the mass straight back to the rotation axis squared. Mass times the radius squared. I have to add this up for every mass contribution. There’s in one, and I go around for this case, I only have two, so I’d have to add mr2. It’s got to be the value of the mass that’s localized out at this distance, r, the straight line, straight back, perpendicular to the rotation axis’ distance squared. And, I have to add those up for every mass that I have. Here’s just the case for this example that I show you. The units for moment of inertia would have to be the units for each of these turns, of course.
What’s the units here? It’s mass in kilograms and a distance, which, of course, would be in meters. So, it’d be kilograms, meters2. Kilogram meters2. Okay, so that’s K’s 1. If you’ve got a simple geometry, you can directly calculate the moment of inertia. Or, if it’s not simple, if the mass distribution is complicated in any way beyond this, meaning it’s actually a bar rotating about the center, or about one end, the mass is not these localized points, then I have to do this. Then I’m dealing with the moment of inertia for a complicated object.
If this were a calculus-based physics class, I would show you how to calculate that using calculus. But it’s not. We’re just doing algebra in this class, so we’re going to take advantage of the calculus that other people have done. What you’ll find is for very simple geometries, people have worked out, with calculus, what the moment of inertia is by, in essence, adding up each little mass contribution times its distance squared and adding up each of these range of different geometries.
You’ll have a table of these values. This will be given to you on the exam if you need it. And here’s what it is: The moment of inertia, for this particular object, would be one-twelfth times the total mass time the length squared. And here’s what you have to be careful of. You’ve got to match the rotation axis’ geometry, that orientation, and the geometry of your object to this picture. And then, of course, the L variable is shown here, what they mean. It’s this length; it has to go into that equation. And so, you can see different equations apply for different objects. Even the rotation axis location will change the moment of inertia.
If I try to rotate a bar about an axis at the end, this would be about like a door compared to rotating it about an axis through the center; it’s quite different. It’s ⅓ML2 for this scenario, compared to one-twelfth. It’s much harder to get the door to rotate or to get the object to rotate with this rotation axis compared to this. I just fixed a typo there; it’s supposed to be L2. Okay, so, for complicated geometry where I have this distributed object, this mass, over this general shape, I look up in a table the geometry that matches my condition and I apply this equation to calculate I.
Okay, so as I say, you have to calculate I directly, look it up in the table, or in a problem, it could be a number that’s given. Or, perhaps, you have to calculate it by using the fundamental equation where you got to get a number for this and a number for the net torque. In which case, then, you can use this equation to solve for I.
Okay, here’s quiz question eight, get you to think about that a little bit. Both of these objects I show below have the same mass and the same radius. If you wanted to achieve the same angular acceleration of 4 radians per second2, which scenario requires the larger torque?
All right, the next thing I want you to do is walk through a series of fairly straight-forward quiz questions where I’m getting you to think explicitly about the torque that each specific force causes and the direction and, therefore, the sine. You’ll have problems like this when you’re faced with, say, a board and this is a diver standing on the end of the board. It won’t, at first, be obvious, but the thing a lot of these problems that is not rotating, it’s the object of focus on, is the board, highlight that in red, there’s your object. Always identify the object.
Then, you need to define a rotation axis. For the first set of quiz questions, I want you to consider the bolt contact point here to be my rotation axis. And then you need to consider all the forces on this object, the direction and where they’re located, and I want you to answer these questions and think through what the sign of the torque would be due to each of those forces.
Okay. So the first set of questions has to do with this axis, diving board, as your object. What’s the sign of the torque due to the force the person exerts on the board? How about the force at this contact point here at the fulcrum exerts on the board? What’s the sign of the torque due to the bolt? The bolt’s putting contact on the board here at the axis point. And then, what’s the sign of the torque due to the force of gravity on the board? Where does the force of gravity go? What’s the sign of its torque, that this is the rotation axis? And then, question 13, is what’s the net torque for this board about the contact point, about the bolt, our rotation axis?
Remember, the net torque takes all torques into account. In other words, it is the sum of adding up all those individual torques. And let me further remind you that those torques have to add up to something that’s consistent with our basic equation, Alpha equals the sum of the torques divided by I. What is the net torque for this object? Is it in equilibrium or not?
Okay. Now repeat all of those questions, but do that now in the context of the fulcrum as being the rotation axis. Again, the object is the diving board. What’s the sign of the torque due to the person, the sign of the torque due to the fulcrum, contact point, what’s the sign of the torque due to the bolt contact? What the sign of the torque due to the force of gravity on the board? And, then, what’s the net torque on this object about this axis point, the fulcrum contact point?