[MUSIC] Okay, in this video we are going to study angular impulse and momentum relationship. So from Newton's Second Law, F equals ma instead of just directly integrated over time. We are going to first cross product the momentum r, r crossed left and right inside and then do the time integral. In that case the left-hand side is angular impulse in the right hand side is going to be angular momentum. Okay, let's drive the relationship more in detail. So let's start with the take an analogy with the linear momentum approach, linear momentum by definition is MV, and angular momentum is the one that have r cross product, okay to the MV. So here usually the angular momentum is defined with respect to the reference like angular momentum H with respect to O with respect to P. So you have all those reference point as a subscript of the H, in which is the definition of the r, so from reference point to the body, that's the definition of this momentum r, okay. And note that those point could be point, so could be O or could be P or any point, so I return and mark it as a star. So H with respect to the star is defined by momentum r defined from the start to the body and MV. And note that V here is usually r defined from the absolute coordinate frame. Okay, so let's take a derivative of the angular momentum, so time derivative angular momentum is time derivative of R cross MV. And if you have a cross product of r dot that's going to be a v right, so you have a self v cross product v. So this turns out to be 0, and what you have left is r cross product MV dot, what's v dot? V dot is the acceleration, so what you have is r cross product MA which is a force. So r cross product of summation of the force means you have all the moment acting on the particle. So moment some will be the change for the angular momentum, that's the angular impulse momentum relationship, the previous four integral, [COUGH] excuse me. And if you take the time integral on both sides what you have is left, this one is angular impulse and this one is going to be angular momentum change. So it's integrated from status 1 to 2, what you have is r and mv1 angular momentum at status 1, and angular impulse sum is going to be angular momentum at status 2. So similar to the previous approach, you should start with defining the coordinate to solve the problem. And either you should take the integral of the displacement to obtain the work energy or displace integral of the time to obtain work in person momentum. And you should do the r cross product of the equations of motion and do the time integral to obtain the angular impulse and momentum relationship. So, your angular momentum is the first state will be equal to the second state by adding angular impulse time integral of the moment. So let's solve the problem. There's a particle of mass m is released at rest in the circular circular path. Okay, and it subsequently slide along the smooth frictionless circular path. So find the magnitude of its angular momentum about point O, as I said before you have a reference point O. So those O should be noted as a subscript here Ho, which is defined by definition r cross MV, ro is here from O to any distance from those particle, since this is this a circular this one is a constant. So you're supposed to find the angular momentum, which means you are supposed to find the v or omega, okay. So let's define the coordinate, define the coordinate the polar coordinate because is a circular motion, and I'm going to do the steps one by one. And as the time here either I would have an option just do the time integral of the equations of motion, or I can just simply use the formula as a mechanical engineering student. I would strongly recommend you to to obtain the equations of motion into the time integral, and also double check your answer by just plugging the values to the formula. Okay, let's work on the equations of motion. So let's free the body, so how many contact does this one have? There's just one surface or if you're considering the assuming the ball is contacting top and bottom just I would set at all the net forces applying at the bottom part. So I have a normal force and the friction force is the gravity will generate the acceleration in r and theta direction. So I could obtain the equations of motion in R and theta direction. And since this is a smooth surface I can simplify it by deleting our friction force, or and since this is a circular one, so there is no radial change. So our dot and our double that turns out to be zero, again simplify all the unknowns at the state once you obtain the equations of motion. That's my recommendation now, to do the r cross product and the time integral which one are you going to integrate? I have two options either this one and the the second one, when you define the Ho this is going to be r cross product MV, so r is it r direction, right? So to have a cross product, I'd rather more focus on the theta direction of motion. So I'm going to take the second equation to take a derivative to the cross product of the r, okay. So if you did the cross product of r, r is a constant mg is a constant what have left is sine theta dt, okay. And at the right hand side all those constant values comes out of the integral, what I can have the theta double dot dt, is the theta dot and dt and dt canceled. So what I have is just a simple integral of the theta dot from status 1 to 2. Okay, so wait, here I have to integral the sine theta over time t without knowing theta as a function of t, I can't proceed this integral further, right. And also for the right hand side to I know it's going to be a difference between the angular velocity at the data, but I can specify them right. Without knowing the relationship between the omega in the theta or omega over t, I can get more information from here. So, integrated equation itself is okay, but this procedure doesn't really help to get the answer so, don't disappoint it yet, I can try another integral. So let's try impulse momentum relationship just to the time integral. So if I just do the time mirror for the equations of motion in that direction, what I can hear is mg sine theta dt, and time integral of the angular velocity which turns out to be exactly same as before the angular momentum case. Well, so can proceed this integral further either, so don't disappoint. Don't be disappointed because I still have one more integral to try which is working energy relationship. So instead of overtime, I'm going to take the integral over the displacement. And since this is a polar coordinate this dx is going to be what? rd theta right? So instead of having the time integral, I'm going to do the theta integral, so it may work. So d theta and the sine theta seems like working, so I would have minus mg sine theta d theta. Wait, where is this minus sign comes from? I think this is typo, this is theta double dot d theta and d theta, theta double dot is going to be split it out to theta dot dt. And if you combine this together, you have theta dot theta dot theta dot, yes. So what I could have is gravitation work done is going to be same as a kinetic energy change. So by doing the work energy relationship, applying work energy relationship. I could obtain the omega and then if I plug into the omega here, I can find what's going to be the angular momentum at each position. So, from this example, we know that after obtaining the equations of motion. You can either do many integrals like work and energy or immerse momentum or angular impulse momentum, in this specific example I tried six, it didn't work. I tried number 5 it didn't work and it worked when I apply number 4. So how do you know which one to integrate to solve the problem? Well, usually you should do the integral because you should have the you know finite-state integral over state one to status two. If you have a status information about over time, you should do the time integral. If you have a status information over the displacement or the position like the case where we just solved, you should integral over the displacement based on its information availability. You should choose which integral to do, however, however, that's very tough for the students who just learning the dynamics for the first time. So, I would recommend try one by one like either four and five and six or the other way, whatever the order you would use until you obtain the answer. I mean, by doing so by doing the integral you could understand, how your equations of motion has a relationship over displacement integral or the time integral. So, since we want to understand the physics, we are not just one to just be a good problem solver, we want to understand the physics. So, strongly recommend you to do the integral one by one until you get used to it. So by the time you take the exam you will be really, really good at which one would use, or even at the exam just to try all the integral one by one very quick to until you get the answer. Okay, so in this video, we worked on the definition of angular impulse and momentum. Basically, this is the cross product of the momentum to the equations of motion, and to the time integral. So left hand side is angular impulse and the right hand side will give you the angular momentum change.