https://youtu.be/FdYdectVOTE
PHYS 1101: Lecture Twenty-Three, Part One
Welcome to lecture 23. Now we need to talk about a new kind of force. It’s really quite different than what we’ve seen before. This is a force that’s caused by things that are springy. What’s different about this is that the size of the force as we watch this motion where a spring is involved is not going to be constant.
We’ve always dealt with constant forces for the simple motion we’ve look at until now. The fact the force now is going to change during the motion leads to some very different behavior. The most common behavior that we’re going to study and that you’ll see that results from this is what’s called oscillatory motion where this object because it’s attached to a spring it’s going to end up oscillating back and forth.
We’ll see as a result of this that I have a new set of kinematic equations that I have to use now. I can’t use the equations from chapter two because those only applied when a net force is constant, which of course led to constant acceleration.
So I’m going to jot down here we can’t use chapter two equations because the acceleration is not constant now from start to finish of our problem, and this of course follows because fnet is not constant.
Okay, to start the lecture, as usual I have a couple of quiz questions here that go over previous material. The first one goes back to chapter four, asking you again to go back and revisit the meaning of Newton’s Second Law and Newton’s First Law.
Question one: if a car is speeding along a rough straight road at 60 miles an hour, there is a strong wind blowing right to left, is there a net force on the car? This question’s worth six points.
Question two, the same idea as I want you to think about, but now in the context of motion along a circular path, the rotational motion. The same analog applies though, and then I have to think about if I have an angular acceleration, meaning again I’m speeding up or slowing down, that this is rotational motion now that has to result. There can only be the case if I net torque divided by something analogous to the mass. This is the moment of inertia. This that net torque is the sum of all the torques on the object.
Question two: your CD’s playing a song and it’s rotating the CD at a steady pace. Is there a net torque on the CD?
Question three: now you hit stop on your CD player. An instant later, as the CD is coming to a stop, now is there a net torque on the CD?
The next collection of questions has to do, again, with motion following a circular path. Here I show you four snapshots of a motion diagram, so the time between these points is the same. So, the distance the object gets between instant three and four is larger than between one and two.
So here’s the sequence. One happened first in time, then two is the next second, three is the next second, four is the next second. Given this kind of motion that this implies or that this tells you, at instant three, at that snapshot in time, I want you to answer the following questions about the various variables that I can talk about that describe the motion at that instant.
Tell me about the centripetal acceleration at that instant. What’s omega, the angular velocity? How about tangential acceleration, is there any? Is there a centripetal force, is there a tangential force, is there angular acceleration, is there a tangential velocity at that point? It’s all about instant three.
So we’ve acquired now a whole collection of variables that describe different features of the motion, and I want you to just go back and remind yourself of what these all physically mean. And so what would they be in this context for this scenario?
Here’s our review of the previous material, last lecture. Where were we? This is where I really summarize just the basic equations that you need to tackle these kinds of problems. We were again focused on rotational motion, but now we were looking at the dynamics aspect of that, which means how does an object speed up or slow down when it’s rotating? What causes this rotational motion to change?
And there’s this very strong analogy to Newton’s Second Law for this rotational motion and our angular variables, and that’s this equation I’m going to highlight in yellow here for you, biggest equation from the previous lecture.
It’s all about the angular acceleration being a consequence of a net torque on the object, but how big is that angular acceleration? It’s not just the magnitude of the torque, but I then have to divide by what’s called the moment of inertia. This is a measure of how difficult it is, the inertia, to get this object to rotate.
Angular acceleration, we know, is units of angle per second squared, so this angle may have units of radians, degrees, or revolutions, and then we went in and looked into the details of how do you calculate a torque and the moment of inertia. That’s next in my summary.
So here’s my equation here again, the analog of Newton’s Second Law, and we went in and looked at each of these terms. What do we plug in for these torques? Every torque is caused by a single force.
When I identify the object to consider the rotation of, or the equilibrium condition of, I need to consider a rotation axis, and then I need to imagine every single force on that object. For each force, like this one I show here in blue, I can ask, what’s the torque that this force causes to this object of leads to, what torque does this force lead to for this object given this rotation axis and the fact that the force is applied at this point at this angle.
The value of the torque caused by this force is the product of the distance r from the rotation axis straight line to where the force is applied, it’s the product of that r and what we call ftan. Think of this also as fperpendicular if it helps, perpendicular to this r line that you drew.
In other words, I need to find the component of this resultant force that’s in this perpendicular or tangential direction because that’s the only part of the force that helps or contributes to rotating or trying to rotate this object.
The other component, this would be my other component that makes up my right triangle, so these two components add together to equal the hypotenuse. This component, notice, is parallel to r. This part of the force is like pushing straight into the door or straight into this object, and that doesn’t help to rotate or contribute to the rotation, so I don’t care about this part. I only care about this component of the force.
So you have to look carefully at the geometry of the situation, where the force is applied, what the angle is for that force, and carefully draw out my triangle. Picture angles to try to do your trigonometry because you need to find the expression for this component of the force, and that’s what you have to plug into this basic equation to calculate the torque caused by this force.
You have to do that for every force, calculate its torque and then add them up. Note, very important, I have to keep track of the sign of these torques because these are direction-information quantities. Torques, each force is going to put a torque that tends to rotate the object in one direction or the other, and I’m going to keep track of those competing effects by using a plus and minus sign.
And remember, the convention is if a torque would cause the object to rotate in the counterclockwise direction, I’m going to call that a positive torque. This would be a positive torque. This force would cause this thing to rotate about this axis counterclockwise. That’s positive.
Visualizing the component of the force that you care about can be tricky, so let me show you some different scenarios here to help you visualize that. I’ve got three scenarios of an object that would rotate about this axis, that’s the black point in each of these cases, and I show you where the force is applied, and you can see the orientation of the force. What is this ftan? What’s the component of f I would need for each of these scenarios?
Here are the steps. Step one, start out and draw the r line. That’s from the rotation axis out to where the force is applied. Do that for every combination. I’m going to label that r. Next go in and picture, visualize, the direction that’s perpendicular to r, so draw a perpendicular line out where the force is applied, nice and perpendicular to r.
Now where the force is applied nice and perpendicular to r. Now where the force is applied perpendicular to r.
Okay, I need to think of this now as an x-axis, if you will. I need to find the component of this force along this axis. If I shine a light toward this axis, I need to know the shadow that this force casts. I’m going to grab my blue pen. Shining a light toward this axis, this is the shadow that I cast onto this nice line that’s perpendicular or tangent here to r.
For this case, again, I need to shine my light toward the axis, and so it’s this component that I need. In this case, I’m going to shine my light toward this axis. The whole force is already aligned to that axis, so this entire force, I only have one component for this force in this tangential direction, if you will.
Okay, so this is what I need to find. This is ft, this is ft, and this is ft. That’s what you have to plug into the calculation of the torque as being the distance r for each of those problems times ft, the length of this component, the tangential force, and then you put the sign in front of this torque that captures either the clockwise or counterclockwise sense. So, I’ve emphasized that here with my sign convention.
For example, this force has a tangential component in this direction. That would cause this thing to rotate clockwise. That’s a negative torque. This causes this thing to rotate clockwise. That’s also a negative torque, et cetera.
Okay, we’ve thought about the numerator now for this fundamental equation. Now what do we plug in for I, what’s this moment of inertia? That reminder again, what’s this term, is some bullets here that tell you how you calculate I. If you have a simple geometry where I have a well-defined, this, I’ve got this rotation axis, and my object has just a small collection of masses that are localized here around my rotation axis, then I can calculate I directly. What do I mean by directly?
I is equal to a sum of every mass times it’s r squared. This means m1r12+m2 r22, et cetera. You have to go around and add up every mass that you see in your problem.
What are these rs? They have to be the distance from the rotation axis straight out to where the mass is. That is r2. This would be r4. This would be r3, et cetera. This, you’d have to go in and look at the geometry and be given say distances, some geometry information, to figure out what those distances are.
I guess I should complete the picture here. This would be the distance r1. That’s what has to get plugged in here. These are all positive quantities. They just add up to give me a total moment of inertia.
Okay, in this class it’s algebra based. We don’t have the tools for handling more complicated objects like solid objects. When you apply calculus to the problem you determine that this moment of inertia, the sum over all these mass parts of these objects you can calculate using calculus and arrive at these values.
So, there is a table in your book, this’ll be provided for you on the exam, that tells you for a given geometry what the moment of inertia is. That’s I. It would be for example one-third times the total mass times the length squared, and what those variables mean would be clearly indicated in the drawing. So, you need to match the geometry of your problem, where your rotation axis is, the shape of your object, to these pictures to decide which I you need.
So you either look up the right expression for I from the table, because you’re given the mass and you’re given the length for example, so you could plug these numbers in to calculate I. In general, in a problem it could be that I is given to you. They’ll tell you what the moment of inertia is for your wheel or your tire, et cetera.
Or it could be you’re asked what the moment of inertia is. Maybe you don’t have the mass or the length so you can’t calculate it directly, but you could get it from this very basic starting equation. In other words, maybe you know, you’re told, what alpha is, you know what all the torques are. You then could treat this as the variable in this fundamental equation to solve for because the I that makes this true for these knowns would be the correct value for I. So that’s another option.
Here’s a quick summary of the problem-solving steps, the general flow of solving problems based on this fundamental equation. This is Newton’s Second Law for rotation now in terms of the angular variables, torques and moment of inertia.
As always, the first step is to be sure you’ve identified the right object. Think of drawing that in red. What is the object that’s rotating or could be rotating in this problem? It’ll be the diving board, it’ll be the beam, it’ll be the board. Pick a rotation axis. You have to define that in order to calculate your torques properly. That rotation axis will either be obvious, it’ll be there’ll be a hinge in the problem, it’ll be given to you, or if it’s an object that’s in equilibrium, like this diving board that’s sitting there, you can use whatever rotation point you want because the equilibrium condition there would have to be true for any possible rotation point whether it the bolt or the fulcrum, for example.
The next step in being sure you’ve identified all the forces on the object, I need to be able to do that to properly calculate all the torques that add up to a net torque. For every force, I need to know the direction and where that force is applied. That’s the only way I’ll calculate the proper value for the torque that results.
You then have two possibilities. My fundamental equation tells me one of two things. Either this object is in equilibrium, alpha is zero. If it’s zero, my net torque is zero. Equilibrium is the case if something is spinning at a steady pace of it’s just stationary, it’s sitting there. Either case means no net torque.
Remember if something’s spinning at a steady pace that’s our counter-intuitive scenario where I don’t need a net torque for something to spin at a steady pace, okay? That’s equilibrium. The other option is that something is speeding up or slowing down in this circular motion. Then I have an alpha. The thing is what I call winding up or winding down. It’s starting to spin faster or it’s spinning slower, it’s spinning down, in which case the left side’s not zero; I’ll have a number for that or maybe I have to calculate what alpha is. I have to include the moment of inertia then, so my denominator doesn’t cancel out now, and I do have to worry about I.