ï»¿ >> Let's take a look at a pendulum problem. Let's say we have a pendulum which is just a ball on a string, and that ball's going to swing from some initial angle theta I down to the bottom, moving along at V, and then swing back up to some theta F. And let's ask the following questions. What is theta F, and what is the max speed at the bottom of its swing? Well, theta F must be equal to theta I, right? The reason that theta F has to be equal to theta I, ignoring air resistance, of course, is because, if I reverse the problem, it looks exactly the same, right? It doesn't matter if I start it from this side and let it swing to the other side, or if I start it from that other side, and let it swing back to the first side. And so, by symmetry, theta F has to be equal to theta I, okay? We can prove that later on using conservation of energy. All right, to get V, we're going to use conservation of energy. And the initial position is going to be where it starts. The final position is going to be at the bottom, okay? So Ei equals Ef. Ei is all gravitational potential energy, okay? It is up here at some height, H. And we need to write down what the gravitational potential energy as a height H. It's, of course, just mgh. Down at the bottom, it's at a height 0, that's why we're going to set our height equal to 0. And we know it is moving, so it has kinetic energy, one-half mV squared. All right, we could solve this for V. The Ms cancel out. We get V equals square root 2gh. Okay? It's nice, we didn't really even need the mass in this calculation. But, we don't really have h, right? We know L and we know theta I, so we need to combine those somehow to get h. So this is slightly tricky, but we can do it. Let's redraw this triangle just a little bit bigger. This is L. When it's at the bottom, that's also L, okay? It's the same string. And so the height that we want is this right here, that little section h. Okay, that's how high it is above the ground when it starts. There's its starting point. What do we know? We know this theta initial, okay? And we know that the whole length of the string is L. That whole length is L. So, if we knew this section right here, then we could calculate what the height is h. But we know that section because we have the hypotenuse of the triangle. This is a right angle. We have theta I and so this is L cosine theta I. Okay, so what's h equal to? H is equal to the whole length, L, minus this little section, L cosine theta I. And I can simplify that a little bit, it becomes L times 1 minus cosine theta I, okay? And now we have all those numbers and we can plug it in and try it out. Okay we said V was square root of 2gh which is square root of 2g times h which we just said was L. One minus cosine theta I. And now we can plug in all those numbers. Two times 9.8 times L which we said was 85 centimeters, but we've got to put in SI units. So that's 0.85. And then we have a 1 minus cosine of 20 degrees. Okay? Plug in all those numbers into your calculator and let's see what we get. Okay, double check my calculation but, when I plugged it in, I got 1.0 meters per second, okay? That's how fast the thing is moving at the bottom of the swing. Okay, by symmetry, earlier we said that theta F must be equal to theta I. But let's see if we can really convince ourselves of that from conservation of energy. Well, energy conservation says whatever energy we have initially has to be whatever energy is there finally. And, if our initial position is there, and our final position is there, then we just have mgh initial on the left side is mgh final on the right side. But the m is the same, the g is the same, and so we have to have h initial is equal to h final. And, if those heights are equal, we know what that is in terms of the angle. It is L 1 minus cosine theta I equals L 1 minus cosine theta F. And those things, if they're equal, the Ls cancel out. We can simplify this and write cosine theta F equals cosine theta I. The only way that happens is if theta F, in fact, equals theta I. Okay? So conservation does, in fact, tell us that those angles have to be equal. Okay, cheers.