https://youtu.be/TxCBPP0aBeE
PHYS 1101: Lecture Fifteen, Part One
Welcome to lecture 15. We’re finished with Newton’s Second Law now as our primary tool for solving problems and we’re onto a whole new concept. This notion has to do with energy conservation. So again, we’re going to be focused on a single object, and we’re going to be asking, now, what’s the energy of that object and how does that energy change. That consideration is going to allow us to determine a lot of things about the motion.
At the heart of this new tool is, again, always a fundamental starting equation. For energy conservation, that equation is what we call the net work. This ends up being the net energy that’s added to this object. Whatever gets added or taken away from this object has to balance, or be equivalent to, the change in the kinetic energy of this object. Kinetic energy is nothing more than a measure of how much energy something has simply because it has mass and it has speed.
This is the very basic starting equation. In lecture 16, I’m going to introduce you to a slight variation of that equation, which is just a little easier to work with. It doesn’t look easier at first glance, but in the end, it is.
So, let’s get started with the lecture by going over a couple of quiz questions about previous material. The first one is a typical brain teaser, if you will, that people run into. This has to do with Newton’s Third Law. And the idea is that if you’re told that the force on the carriage, according to Newton’s law, is equal and opposite to the force on the horse, then how can the horse pull a carriage at all? Which of the choices, there, do you think best describes what has to be happening? Is this a contradiction of Newton’s Third Law? Plus 8 points if you get it correct.
Quiz question 2 goes back and has you looking a little more closely at uniform circular motion. You have four scenarios here below where the mass of the object changes between these scenarios. The radius, as shown, is different. All, in all cases, though, they’re going around the circle with the same speed. So, notice the v is the same for all 4 of these scenarios.
You need to rank, in order from largest to smallest, the tension in the line. So, this rope is attached to this ball that’s moving in this uniform circular motion. So you need to focus on the equation that relates the tension to the speed, the radius, and the mass.
We derived this in the previous lecture. Go back and have a look at that equation. Or, derive it again for yourself. Solve for what the tension has to be by starting with Newton’s Second Law. ay, for example, is equal to some of the forces in y over m. This is what you would use if you want to consider the snapshot of the ball, say, in this orientation at that instant. Set up this equation, derive what the tension has to be, and then see how it compares to these 4 scenarios.
Here’s a quick overview for you of last lecture, the key points. The new thing we learned about in lecture 14 was this idea of uniform circular motion. The uniform means my speed has to be constant for this object as it goes around in a circle. And circular motion, that’s straightforward.
There were some key relationships, or equations, that had to be true if something is undergoing this very special case of motion. One was that the acceleration is never 0. The acceleration vector always has to point to the center of the circle. And the size of it depends on the speed and the radius of that circle. In fact, it’s the speed squared divided by the radius.
We also then learned that the speed is not completely independent either. It depends on the circumference of the circle and the time that it takes to go around once. Speed is distance divided by time. So this is a new variable for us, this fancy capital t, and it has units of time. It’s going to represent the time to complete one cycle.
For uniform circular motion, our starting place is still Newton’s Second Law. This is just a special case of that. The key concepts to apply Newton’s Second Law, or the tricks to it, are that you have to pick a snapshot in time to apply the law because you have to have a clear picture of what the direction of a is at that instant and the direction of the net force. So you have to have a clear picture in your mind where it is when, when you want to solve Newton’s law. And that’s usually clearly spelled out for you in the problem. It will say at the bottom, at the right, etc.
And often, too, we notice that drawing the free body diagram to focus on the forces and see if it’s consistent with the a, that that usually works best if you imagine a side view of this motion. So not a top view like you see, but you imagine looking from the side. That’s why I always drew the kind of this oval trajectory, trying to get you to picture the side view and then what the forces would be.
In the context of uniform circular motion, we looked at satellites orbiting planets, the moon orbiting the Earth, for example. And for cases like that, these objects are undergoing uniform circular motion, a very simple example of that. The only force that’s on these satellites is going to be the gravitational pull caused by the large mass of the planet that this satellite is orbiting. The direction of that force is always to the center of the other mass, and the magnitude is given by this expression. In other words, the force due to gravity, which we’ve called w, has, of course, magnitude and direction.
On the magnitude side, we can continue to use the mass of the object times our 9.8 meters per second squared as long as we’re at the surface of the Earth. But once we get away from the surface of the Earth, say, a satellite, the moon, it’s far away or it’s a satellite, something that’s around or next to a different planet, we need to go to this more general equation for gravitational force.
This equation says that the size of the force due to gravity is actually equal to, very generally, a universal constant that has this value. I then need to multiply by 2 masses. The first mass is the mass of the agent. So if you’re focused on the satellite as the object, the agent causing this gravitational force is this big mass, the planet that the satellite is orbiting around. I need the mass of that planet. Then, I need to multiply by the mass of the object. And then, I need to divide that by r2, where r is the distance between the centers of these 2 objects.
In the case of a satellite orbiting, that distance will be the radius of the orbit, if you can see that from the drawing. This is the same distance r from the center of the two masses and then from the center of circle out to the path, the circular path.
Okay, the force due to the gravity, if you’re at the surface of the Earth, it just appears to us to be straight down and more generally said the force due to gravity always points to the center of the opposite mass. And so for a small object here on the surface of the planet, that’s always going to look like straight down to us.