Professor Anderson

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>> Hello, class. Professor Anderson. Welcome back. Let's talk about this idea of using conservation of energy to figure out a little bit about free fall in terms of gravity. So let's say we're going to do the following experiment. Here we have the surface of the earth, and now we're going to stick an object up here at a height h, and we're going to let that object fall. It's going to start at rest and now it's going to fall, and we want to know how fast is it moving at some other height, which we will call y. Okay. How can we figure out how fast it's moving at that y? Well, again, we can always go back to the kinematic equations, but there is probably an easier way, and the easier way is conservation of energy. So energy tells us that Ei equals Ef. We have gravitational initially. We have kinetic energy initially. That's going to equal gravitational final plus kinetic final. And now let's see what all these terms are. So when we start up there at height h, we are mgh in gravitational potential energy. We said we're going to start at rest, so that term is 0. The final is going to be what? Well, we're down at the surface of the earth. That would be 0, but we're not quite there. We're up here at height y. So this is mgy. And the last one is, of course, one-half mvf square. And now we can solve this for Vf, say. Vf is equal to what? We can cross out all the m's. If I multiply everything by 2, I cross out the half. I put a 2 there, I put a 2 there, and we get 2gh from this guy. I have to subtract 2gy, so I can slip that in right there, and that is it. If y goes back to 0, then we're back to our good old square root of 2gh. If y goes all the way up to h, it says that Vf is 0, and that makes sense. You haven't moved anywhere yet. Okay. So this is a very nice way to get to the final speed at any height y that you want. Let's see if we can take this now and generalize it a little bit more. Let's generalize the problem and let's do it like this. The object is moving at speed Vi and it starts at a height y initial. It ends at a height y final. And let's see if we can calculate what this V final, is what we're going to call it now, is equal to at this particular height. All right. How do I do this? Well, we go back to conservation of energy and we say, all right, we've got mg times y initial. That's our potential energy when we're up there. But we also have kinetic energy now and I know what that is. It's one-half mVi squared. All right. That looks pretty good. Down here I've still got some height. mgy final is my potential energy because of that height, and we have kinetic energy one-half mvf squared. And now let's see if we can put this together and simplify it a little bit. The first thing you notice is there is an m everywhere, so I can cross out the m everywhere. The other thing you notice is I've got a half and I've got a half. If I multiply by 2, I can do that. And let's just rewrite this line. 2gy sub i, plus Vi squared equals 2gy sub f plus Vf squared. And now let's rearrange some of these terms. Vf squared equals what? Vf squared is going to equal Vi squared plus 2g yi minus yf. And this should look really, really familiar. Remember one of those kinematic equations? It looked like this. Vx final squared equal Vx initial squared plus 2a sub x times delta x. It's the exact same equation. It's exactly the same as we had before. And you might have thought where did that equation come from. Well, there's a couple places it came from, but one of those is conservation of energy. If you conserve energy, this equation has to be true. Okay. You can also re-derive this equation just by combining the other kinematic equations, but I like the conservation of energy derivation of that kinematic equation. Okay. All right. Any questions about this stuff? Everybody more or less okay with this? All right.

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>> Hello, class. Professor Anderson. Welcome back. Let's talk about this idea of using conservation of energy to figure out a little bit about free fall in terms of gravity. So let's say we're going to do the following experiment. Here we have the surface of the earth, and now we're going to stick an object up here at a height h, and we're going to let that object fall. It's going to start at rest and now it's going to fall, and we want to know how fast is it moving at some other height, which we will call y. Okay. How can we figure out how fast it's moving at that y? Well, again, we can always go back to the kinematic equations, but there is probably an easier way, and the easier way is conservation of energy. So energy tells us that Ei equals Ef. We have gravitational initially. We have kinetic energy initially. That's going to equal gravitational final plus kinetic final. And now let's see what all these terms are. So when we start up there at height h, we are mgh in gravitational potential energy. We said we're going to start at rest, so that term is 0. The final is going to be what? Well, we're down at the surface of the earth. That would be 0, but we're not quite there. We're up here at height y. So this is mgy. And the last one is, of course, one-half mvf square. And now we can solve this for Vf, say. Vf is equal to what? We can cross out all the m's. If I multiply everything by 2, I cross out the half. I put a 2 there, I put a 2 there, and we get 2gh from this guy. I have to subtract 2gy, so I can slip that in right there, and that is it. If y goes back to 0, then we're back to our good old square root of 2gh. If y goes all the way up to h, it says that Vf is 0, and that makes sense. You haven't moved anywhere yet. Okay. So this is a very nice way to get to the final speed at any height y that you want. Let's see if we can take this now and generalize it a little bit more. Let's generalize the problem and let's do it like this. The object is moving at speed Vi and it starts at a height y initial. It ends at a height y final. And let's see if we can calculate what this V final, is what we're going to call it now, is equal to at this particular height. All right. How do I do this? Well, we go back to conservation of energy and we say, all right, we've got mg times y initial. That's our potential energy when we're up there. But we also have kinetic energy now and I know what that is. It's one-half mVi squared. All right. That looks pretty good. Down here I've still got some height. mgy final is my potential energy because of that height, and we have kinetic energy one-half mvf squared. And now let's see if we can put this together and simplify it a little bit. The first thing you notice is there is an m everywhere, so I can cross out the m everywhere. The other thing you notice is I've got a half and I've got a half. If I multiply by 2, I can do that. And let's just rewrite this line. 2gy sub i, plus Vi squared equals 2gy sub f plus Vf squared. And now let's rearrange some of these terms. Vf squared equals what? Vf squared is going to equal Vi squared plus 2g yi minus yf. And this should look really, really familiar. Remember one of those kinematic equations? It looked like this. Vx final squared equal Vx initial squared plus 2a sub x times delta x. It's the exact same equation. It's exactly the same as we had before. And you might have thought where did that equation come from. Well, there's a couple places it came from, but one of those is conservation of energy. If you conserve energy, this equation has to be true. Okay. You can also re-derive this equation just by combining the other kinematic equations, but I like the conservation of energy derivation of that kinematic equation. Okay. All right. Any questions about this stuff? Everybody more or less okay with this? All right.