https://youtu.be/zQRKD1PYECA
PHYS 1101: Lecture Thirteen, Part One
Into lecture 13. This is our last lecture on Chapter 4. We mostly have left, covering Newton’s Third Law, and then it’s still important that I walk you through some more examples. Then, I’m going to finish the lecture with an example where it’s a collection of quiz questions, walking you through the problem solving steps applied in a different context.
So, let’s begin, as usual, with our quiz questions on previous material. The collection of questions below are all worth four points if you get it correct, versus one point. It has to do with this motion of a ball rolling over this terrain that you see. So, the ball is going to roll down the hill, around this corner, roll up a hill, and then roll along that flat area.
For each of these locations, one through five, picture the motion diagram in that region is the velocity changing. If it is, what’s the direction of that Delta v vector? Therefore, what’s the direction of the acceleration? Answer it for each of those instances, those snapshots, one through five.
Question one, two, and three are locations one, two, and three. Questions four and five are: What’s the acceleration direction, for instances four and five?
The next warm-up quiz question ties a lot of these concepts together for you. We now have this notion of a net force on an object causing it to accelerate. But by now, we should all be familiar with the idea that acceleration doesn’t mean the velocity. It means how the velocity is changing. What’s the small vector Delta v that we have to add to our velocity every second to predict the proper trajectory?
In question six, I spell out an example here for you that describes a real life scenario. With that description, you have to picture what v0 is, what the net force is, and then therefore, what the acceleration will be. It’s the combination of the initial velocity direction and the acceleration direction that then tells you what trajectory or path this object is going to take.
When this rocket fires, it’s going to lead to a thrust force that’s to the right. That force is a constant force, and it continues on during this motion that you’re watching or describing. So, when this object reaches this dash line, that’s when this thrust force turns on.
What path does that rocket on this hockey puck or the hockey puck follow beyond this dash line here, location one? This is a nontrivial problem. It’ll take some thought, careful consideration, and because of that, I’ll give you ten points if you get it correct.
Warm-up quiz question seven is an important example of Newton’s Second Law and the implications, and one that people usually get wrong with their first guess. So again, I’m increasing the points here to get you to think about it very carefully. Be sure that the picture that you choose, the free body diagram that you choose, is consistent with the acceleration is equal to the net force. [inaudible 00:04:11] divided by the mass.
So, you have a scenario where you’re stopped at a red light. The light turns green. You step on the gas, and you start to accelerate forward. Which choice, A or B, represents your free body diagram? Think about that with respect to Newton’s Second Law, which does explain what happens to you.
Also, think of it in terms of your feeling that you have as you accelerate forward, and try to reconcile those two constraints. Newton’s Second Law is a constraint, and then how is that consistent with your sensation that you feel as you accelerate forward?
For the “where were we” portion of the lecture, there’s not much to say. We didn’t learn any new material. We just put these ideas together as we walked through more examples of applying Newton’s Second Law. As part of “where were we”, though, I do want to walk you by one concept to be sure you’ve put these pieces together.
Let’s derive for ourselves and understand why we started out in the class assuming that for an object in free fall, the acceleration vector points straight down and has a magnitude of 9.8 meters per second squared. Why did we assume that? Why do we routinely use this idea that objects accelerate at minus 9.8 meters per second squared?
I’m going to have us look at that carefully by applying Newton’s Second Law to this problem. Here’s a ball flying through the air. The green dash line represents the trajectory it’s following. Let’s ignore air resistance. Make the problem fairly simple.
My first question to you about this problem is, at this instant shown, how many forces are on the ball? Let’s apply Newton’s Second Law in our f equals ma problem solving steps to this ball flying through the air.
So, I’ve written down here for us our starting equations that we would apply to this problem. Just a reminder of our first couple of steps, we want to concentrate on a visual of the acceleration and what the forces are. For this problem, though, we’re trying to demonstrate what the acceleration is. So, we’re going to consider this to be the quantity that we don’t know, that we want.
So, what are the forces on this ball? Well, if it’s flying through the air, I don’t have any air resistance. The only force I have is the force due to gravity. It’s straight down.
If I go with a standard coordinate definition with positive y up, then I only have one force in the negative y direction. So, when I go to apply f equals ma, and determine what Newton’s Second Law tells me about that motion, the x equation doesn’t do any good. I have no acceleration right/left because I have no forces right/left. There’s no x components to any forces. So, let me note here. We can conclude that a(x) is zero.
What does this equation tell me? a(y) has to be equal to all of the y forces. Well, what do I have? I have the force due to gravity, which is minus m times g. The magnitude of this force, how strongly the earth pulls down on us, depends on this value of 9.8, and it depends on our mass. People that have more molecules, or made up of more molecules, have more mass. Their gravitational force is larger, the tug of the Earth down on them.
But then, Newton’s Second Law tells me that whatever force I have here, the acceleration that I get has to be divided by the inertia of that object. For the same force, a bigger mass will end up accelerating a smaller amount. So, I have to divide by the mass of the object.
So, when you do that, you see that the size of the force due to gravity depends on the mass. But, Newton tells us that the result of that force has to be determined by the magnitude of that force, but then divided by the mass. So mathematically, the mass cancels out, leaving us with an acceleration that’s equal to only minus g.
This is why we use that the acceleration due to the force of gravity is minus 9.8, and it’s directed straight down, because that’s the direction of the force of gravity. It’s equal to minus 9.8 because the size of this force depends on the mass. If you double the mass, you double the size of this force.
But, the acceleration that you get has to be divided by that mass. If you double the mass, not only does this force get two times as big, but the mass of the object is two times as big. Those factors of two cancel, and I end up with a(y) is equal to minus g.
I just wanted to point that out to you and tie up some loose ends. Remember, when we write the value, when we write g alone, it’s always a positive quantity that simply represents the value of this surface gravity, it’s often called, or the acceleration due to gravity; 9.8 meters per second squared.