Okay, knowing that the Newton's law and work energy impulse momentums are, or similarly applied for the systems of particle. We are going to now solve the examples. Okay, let's briefly summarize solution step. The first, we have to start with the drawing the free body diagram and obtaining the equations of motion. So when you're obtaining the equations of motion, either you can just do the summation of each particle, of F = ma. Or you can handle with at the center of mass, if it's easy to find. Similarly, when you have a momentum, it's going to be time rate of angular momentum of each particle. And note that there are many different ways to use that. You could set the reference point, like a fixed point O, or general mass, in that case. You can easily calculate the angular momentum change at the relative coordinate frame. And then, you may also need to do a further step, like an integral. Time integral to obtain the impulse momentum relationship. In that case, either take the sum of each masses linear momentum, or linear momentum of the center of mass, whichever easier you choose. You will choose, and also work energy relationship. You could either sum each masses kinetic energy, or you can handle the energy at the center of mass. But note that at center of mass, the kinetic energy is not only one-half MV square. You should also consider mi, rho i dot square. For angular impulse momentum, you derive, calculate the angular momentum of each mass. And note that there are many different options for setting the reference point, okay? So the first problem is I have a three balls of mass equal mass, are distributed in an equal distance from the center, with the angle of 120 degrees. So it's kind of symmetric, mm-hm. Now, the first, and then also, it's initially stationary on a smooth surface. And when the constant force is suddenly applied, off-centered, water happened. What is the acceleration of the center point O? And what's the angular acceleration of the frame? So when the sudden force is applied, maybe the total system will move to the right while it's rotating. The problem asks you, what's going to be the acceleration? So maybe, pretty much it will be resolved by the stage, where you can obtain the equations of motion. So let's set the reference coordinate here, since the center points keep moving, and where this is rotating, with respect to that point. I'm going to set absolute coordinates toward the center of the structure. And from center to each mass, I'm going to use relative coordinate. So I just take the approach to take each masses, F = ma, and make a summation. So for a single particle here, there is a displacement to the center, and its motion with respect to the center. Only forces are from the rod, it's all welded. But if you are going to handle each mass one by one, I would consider this our internal force. Those are generating the acceleration relative to the center point O, and absolute acceleration of center point O. So the total acceleration of point A, point 1 is going to be AO plus A1, with respect to the O. And that's going to be same for all mass number 2 and 3. And for the center of rod, there are contact with each masses. So there will be an internal force equal to negative to the opposite direction, and there is external force capital F. And then since it's a massless system, right inside the mass multiplied by the acceleration turns out to be 0. There is a motion, but mass is 0. So I have a total four equations of motion. And the acceleration, the relative acceleration of each ball with respect to the center's are symmetric, right? So the vector sum of each A1 refers to the O, A2 with respect to the O, A3 with respect to the O will be cancelled out, right? So if you sum them up, what you can have is, what you would have is the total external force capital F is going to be 3M acceleration at the center. So problem solved. I might solve this problem by using the center of mass approach. So since these methods are equally distributed, the center of mass is located at the center. Right, and the only external forces are applied on the top, right? That'll generate the capital mass. The total mass, multiplied by the center of mass acceleration A bar. So simple equations of motion will be F = ma. A is the acceleration of the center, which is A of O. There are three masses, so three, M, A, O. So total equivalent, either you could calculate one particle by particle, or you can handle at the center of mass. Now, to obtain the angular acceleration, let's do a moment equals angular momentum change. Again, there are many different ways. You can set the reference point, either fixed point, or the center of mass, or any arbitrary point P here. I would choose the center of mass, because it's easy to calculate, because you could handle at the relative coordinate frame. So angular momentum of the system, with respect to the center is going to be rho i multiplied by the double rho i dot, so rho i and rho i dot. Here, with respect to the center, its tangential velocity only exists, which is r theta dot. So it's going to be our distance r, and r theta dot. So you can easily obtain the angular momentum. So from the equations like moment sum will be equal to the angular momentum change. You can take the derivative of this one. Then your moment is going to be equal to the angular momentum change, which is our theta double dot. So by equating them, all the other components are given, so you could find the angular acceleration. Okay, let's solve another one. So there are similarly three balls are welded to the massless rigid frame. And this time, there are three different masses. Okay, M1, M2, M3. And then there are different lengths from the center, L1, L2, L3, as well. And which is initially rotating with Omega naught. And there are external moment is applied at the center. Okay, external moment is applied at the center. And as time passed by T0 second, what would be the new angular velocity? Maybe this system will be accelerated. So Omega will be greater than Omega naught. So it seems like you have you have to calculate maybe the angular acceleration, and consider the angular velocity change over time, T naught. I'm going to set the reference frame at the center, because it's fixed it the center. And I'm going to try to describe the motion of the center of mass. So here, still, the center of mass is located the center, similar to the previous example. Well, no, because it has three different methods and three different lengths. So it will be towards the left, because it's heavier, it is shorter, and it's upper, so it will be either left, or a little bit to the upside, okay? So whenever it rotates, center of mass will now move to somewhere here, okay? So the trajectory will be some oval shape, will show oval shape. So to formulate the x bar and y bar by definition, it's going to be a form of a function of theta. So it's not that simple, compared to the previous one. So let's change the strategy. Let's try individual mass summation approach. Okay, so for each mass, only external forces by the internal go through the rod. And that will generate the acceleration, and same for the other masses for the rod. There is a internal force from the ball, and they're only outer contact with the outer environment. It's at the center at the pivot. There is a few odd reaction force. And mass that should be balanced. So I have three equations of motion for each ball, and one equation of motion for the rod. If I sum them up, what I can have is point reaction force will generate the acceleration for each particle. And since this is centered and rotational, so those acceleration will be dependent on angular acceleration in the lengths, alpha L, okay? But still, to figure that, I'm supposed to figure out the alpha to take down integral to obtain the Omega at time T. This reaction force is not known, so maybe I would try the other approach. So maybe I could try a moment equals to angular momentum change approach. Where should I set the reference point? Yes, at the fixed point O, because the center of mass is difficult to calculate, too complicated to calculate. So at the fixed point O, angular momentum is defined by R cross MV. And those are defined from this center 0. So the R is just simply the length of the bar. And V is going to be L theta dot. So same for L2 square theta dot and L3 square theta dot. So when you take the derivative of this, you can obtain the angular acceleration, in terms of all known variables, including the external momentum applied. So by knowing theta double dot, you can find out what's going to be the Omega at time T. Now, let's solve another one. So instead of three welded structure, now what we have two balls of mass M, and with the length L, which is initially at rest the position shown, along the smooth surface. So this part from here to here is a circular part from here to here. It's just a straight horizontal. So mass P is going to keep moving as a straight velocity. Mass 1 is sliding down through the arc. Okay, find the velocity when the structure at the horizontal position. So when this particle A, particle 1 is reaching to the point A, okay? And what's going to be the normal force applied on a sphere 1 before it reaches the horizontal position? It's the right entering [INAUDIBLE]. So it seems like I'm supposed to find out the V velocity in the normal force. So maybe, I should use the equations of motion and velocity. Either maybe time integral of the acceleration, or petition integral of the acceleration. We'll see. Where should I set the reference coordinate to describe the ball number 1, it's along the circular motion. So I would definitely set the polar coordinate. If I have to describe the motion of ball B, that'll be definitely the rectangular coordinate. So try a draw the free body diagram for the mass. Number 1, there are one contact with the surface, normal force and friction force and the gravity. And internal force through the rod. And for the second mass, there are no more friction gravity and internal force as well. And the acceleration for mass 1, it will be polar coordinate, and 2 will be a rectangular coordinate. So if I write down the equations of motion for mass number 1, that's the R directional forces at certain arbitrary theta position. And these are the theta direction on forces in acceleration. Well, if the ball is moving through the arc, that is going to be constant. So this R data, R double the terms goes that goes to 0. And frictionless, friction goes to 0. But still, I have a normal force is unknown, internal forces are known. Angular acceleration, angular velocities, there are too many unknowns. The problem says, what will be the velocity at the instant when this ball one reaches [INAUDIBLE] horizontal position? So at the horizontal position, I have only the normal force and the gravitational one. There is no vertical radial, or the vertical components for the internal force. So that would help to make this equation a little bit simpler, but still, there are many unknowns. So I might try other approaches, like center of mass. Well, the center of mass is hard to describe, because it's not really a combinations of circular and straight line motion, so that wouldn't be easy. So maybe a moment and angular momentum change might help. We'll see, definitely not the center of mass 1, because it's hard to describe. How about fixed point O? Where should I set the fixed point O to describe it? It's tough too, because it has changing velocity and position, so this may not be helpful. So let me just try other options, like integral. So if I do the work energy relationship, initially, or 0, there is no work done, it is frictionless. Finally, two bowls will gain the speed. And those speed will be the same at the horizontal position, because it's moving in the same direction a horizontal direction. And then only our potential energy change will be applied for the mass. Number 1, there is a drop of the height by the amount of R. By equating the energy loss and gain, we could have the velocity at the point A when it reaches the point A, is square root of GR, okay? Now, how can I make it as angular version? Okay, so I'm supposed to find the V. So the mission is completed, but to calculate the normal force at that instant, I should figure that out. What's going to be the R theta dot square? So note that this mass number 1 is moving through the arc with the tangential velocity R Omega, right? So this is going to be same as R theta dot at point A. So I could find out what's going to be the theta. So plug that into the original equations of motion, regarding the normal force. I could find out the magnitude of the normal force from the surface to the ball. And note that at that instant, when it's actually entering here, the y direction or acceleration is nonzero, okay? You may think maybe horizontal motion. There should be an acceleration in y direction. Turns out to be 0, but not at the moment when it actually entering to the horizontal state, okay? So we've worked on several examples about applying either F = ma, or work energy through the system has a couple of particles.
