 Okay, this is an angular momentum problem that I said that I would post. Maybe it's a little, I don't really know if it turns out to be too complicated or not and it's going to do it. So let's go ahead and start. So suppose I have a merry-go-round, which I can assume is a disc. And so it has a mass M and a radius R. And this is from a movie and the mitbusters did this too. It's kind of cool. So suppose I shoot a bullet and I hit a bullet that comes in like this and it hits the edge of the merry-go-round, what would be the new angular velocity after that happens? So let's say this is mass Mb and it's shot with the velocity Vb that way. And it collides around the edge of that and let's just say it sticks, okay, whether it does or doesn't. Let's just say that it does, okay. So first let me just show you something very important. Can we use momentum principle in this case and the answer is no. The answer is no because if I take my system to be the bullet plus the merry-go-round, then could I say this? Can I say then the change momentum is zero? No, no, I can't. The change momentum is not zero. We can't really use momentum principle here because what forces are acting on the system? Why have the ground pushing up this way? This is the top view and gravity pulls down and those essentially cancel even though not for the bullet but even with that it doesn't matter because this little axle is fixed into the ground. So when this bullet hits the merry-go-round then if the axle wasn't fixed, the merry-go-round slide and spin but it's fixed. So the axle exerts a force this way during some time and we have no clue what that is. We can actually estimate it but let's just try to find out the angular speed of the merry-go-round but that's why this doesn't really work. I mean it works, it does not very, we don't know this Fnet. It's not conserved. That's the key for that system. But what about angular momentum? So in this case we take our same system to be the bullet plus the merry-go-round then there's no torque on the system. The bullet exerts a torque on the merry-go-round and the merry-go-round exerts a torque on the bullet but that's inside the system. The axle if I assume there's no friction it does let's it spin, it doesn't exert any torque. So that means that the change in angular momentum is constant. So I can say this is equal to zero so L1 equals L2. So let's say this starts with an angular velocity of zero and we want to find the final. So what's the initial angular momentum? It's just due to the bullet. So L1 equals R bullet cross P bullet. And now you may be saying well that's just crazy and this is about the point right here O because here there's R and P but as I get closer P stays the same but R gets smaller so shouldn't the angular momentum change? And in fact it doesn't. The angular momentum is constant because although R gets smaller the angle between these gets smaller too. So in fact I can just calculate the angular momentum it's going to be into the board right and it's going to have a value L1 equals Rmbvb because at this point right here R is the radius of the disc and they're perpendicular so remember the magnitude of L is the magnitude of R, magnitude of P times the sine of the angle and that's a 90 degree angle sine of 90 is one. So I just get mb times the velocity of the bullet and that's it. That's my total initial angular momentum. Okay what about afterwards? Afterwards I have the disc spinning and it's a rigid object so I'm going to have to use L2 equals i omega 2 where omega 2 is the angular velocity of the disc and that's technically wrong but I'm, no I'm going to do that. I'm going to do it this way and I'm going to do this for any side of the bullets because maybe it's a basketball, maybe it's a bullet, I don't care. So what's i? We already know the magnitude of L2, it's the same as L1 so what's i? i is i disc plus i bullet. We can do it that way, okay. So this is going to be 1 half big m R squared is the moment of inertia of a disc, I just derived that in the previous video unless you're watching this video before that one then I derived that in the future. What about the bullet? It's just a point mass. So the moment of inertia of a bullet is just going to be equal to mass of the bullet R squared. So that's my moment of inertia. So what's my magnitude of the angular velocity? The magnitude is going to be equal to L1 divided by i so it's going to be Rmbvb over i, 1 half m plus mb R squared. So let's just say omega equals one of the R's cancels. I could do some little math magic here. I could say velocity of the bullet over 1 half m over mb plus 1 R. Now we should check some things because we don't want to just not check things. Does this have the right units? When this is in meters per second, no units, no units divided by R so I get 1 over seconds which is the unit for angular speed and radians per second, that's the same thing. What else should I check? What happens is I increase the mass of the bullet. As I increase the mass of the bullet, this term's going to get smaller and this term's going to get closer to, I'm going to get a higher angular speed and also if I increase the velocity of the bullet, I'm going to have a higher angular speed. One last thing, how much time have I used? Seven minutes. One last thing, what if I wanted to find the loss in thermal, the increase in thermal energy of the bullet plus the merry-go-round. If I take the system to be the bullet plus the merry-go-round, then work equals delta K rotational plus delta E thermal plus delta E K, delta K translational. So I have three real forms of energy that I can look at, rotational energy, translational energy and thermal energy. If that's my system, there's no work done because there's no really external forces, so this is equal to zero. So what's the change in rotational kinetic energy? Delta K rotational would be the final kinetic energy, one-half i omega squared minus initial which is zero. What's the change in translational kinetic energy? Okay, here's the trick. What happens to the bullet afterwards? It still has translational kinetic energy. But if I use this i right here, I'm already taking into account that motion of the bullet as rotational kinetic energy. You can't count it twice. So this would be, the final would just be zero minus one-half mb vb squared. And then so if this plus this, plus this is zero, I could find delta E thermal would be negative the rotational kinetic, change in rotational kinetic energy. So it would be this, one-half mass of the bullet, velocity of the bullet squared minus one-half i omega squared, where omega is right there and i is that i right there. Cool?