[MUSIC] In chapter four we are going to study about the kinetics of systems of particle. Means like there are many particles. In chapter three, we usually handle the case where you have only one particles or two masses. But here, we are going to generalize how those equations of motion and work energy could be expanded when there are many, many particles. Before we start, let's briefly review the table of contents of the textbook. As I said before, the first half is going to be talking about the particle dynamics and second half is going to talk about the rigid bodies. For the dynamics of the particle, first, we should relearn a we learned about the kinematics of the particle, the position, velocity, and acceleration relationship. And later, we learned how the forces are related about the motion, F equals ma, Newton's Second Law and it's integral. It actually applies to similar manner for the rigid body chapter. Like a kinematic chapters precedes the kinetics chapter for the rigid body and same for the 3-D dynamics. Ultimately, we want to teach you the force and motion relationship, like a chapter six and chapter three are the ultimate interest for the chapter of the dynamics. To do so, we need to be really familiar handling the acceleration. So chapter two covers the rotational acceleration formulation. And chapter five, we will discuss about how we can formulate the acceleration in multiple linkage systems, etc., or the rigid bodies. And same for the 3-D dynamics. Okay, so the first half is about the particle. The second half is about the rigid body. What is the difference between the particles and rigid bodies? Rigid bodies generally the system where there are many, many, many, like infinite number of particles, okay? So to transfer our knowledge from the particle to the rigid body, chapter four is located. So chapter four, we are going to briefly review how those kinetics knowledge what we have covered in chapter three will be formulated when there are more than one particle. So what we call as our systems of particles. So as I said, we are going to first review how those F equals ma, Newton's Second Law equations of motion will be applied for the particle when there are more than one particle. Can we just sum them up? Yes. And how about the work and energy? If you have a summed forces and summed particles and its acceleration, can we just do the integral over the displacement, the time to obtain the work and energy and impulse momentum? Yes. So we are going to review the basic concept for the Newton's law for the case where there are many particles. And later, we are going to learn about some special cases, like steady mass flow and variable masses. Like the the rocket problem or the chain problem you might have covered in university physics. So that's how chapter four consists of. Okay, first, let's learn how Newton's Second Law, F equals ma, could be formulated when there are many masses. When you have only one particle, we are going to describe the position of that particle from the origin, and we are going to take a derivative of this to obtain the acceleration, and formulate the force and motion relationship. What if there are many, many masses? Like masses are consist of some system, okay? To describe this systems of particle motion, we are going to define the center of mass concept. This is actually the position, okay? So the definition, center of mass, r-bar, is going to be defined by particle mass and particle displacement multiplication and its summation divided by the total mass, okay? Geometry call it the center of mass is approximately located at the center of the whole structure. And if we define the vector from the origin to the center of mass as r-bar, the particle position, which is ri, can be also formulated as r-bar plus rho-i. Rho-i is a displacement vector with respect to the center of mass. So if I change this right-hand side r vector as r-bar plus rho-i, and if I split it out, I have miri plus mi rho-i. And since this one is nothing to do with the particle integral or the summation, this comes out. So it's just going to be total mass and the center of mass position vector, which is same to the left-hand side, the definition of center of mass. So by definition, this term on the right should be 0. So mi multiply by the rho-i summation is 0. So conceptually, that does make sense because center of mass is located at the center, right? So the vector sum of the displacement from the center of the mass to each particle turns out to be 0. In a similar manner, your vector sum of mi rho-i dot, the velocity vector, and mi rho-i double dot, acceleration vector, sum will be 0. Okay, now, the Newton's Second Law F equals ma. If you have many particles, you can just put the sigma on left and right-hand side. So all the external forces you should take into account, and all the internal forces should take into account. But all the internal forces are action-reaction pair, so it should be all cancels out. So what you have left is the ma components for each particle. And since we define the center position ahead, this is going to belong to the total mass and it's acceleration. So the point here is for the systems of particle, you can describe the motion as a summation of single particles, or the total mass and its center of mass. So the summation holds for left and right side of the Newton's Second Law, and this is going to be equivalent to the total mass multiplied by the acceleration of the center of mass. Now, how work and energy would change is the total sum of the force and the motion can be just integrated to obtain the work and energy. Let's examine it. Before we go through, let's just make a self-check. So since we know that all the forces applied to the particle could be represented as center of mass behavior, would that be also similar to describe the other parameters like potential energy? Potential energy is mgh, and many, many particles you should sum them up. And would that be equivalent to the total mass and the gravity constant and the height of the center of mass? How about the kinetic energy? Kinetic energy is going to be one-half mv-square. So work done is going to be change of kinetic energy. And then, would that be mv-bar-square, the kinetic energy of the center of mass? And also, same for the inverse momentum. So linear momentum change will be mv-bar, could be equivalent to the mv-bar, and the same for the angular momentum. Angular momentum is defined by the r cross mv. And can the angular momentum of many, many particles could be represented by simply the rho, the center of some reference point to the center of mass and the center of mass velocity. What's the answer? Yes, except question number two, everything else is correct. So how we can formulate the work and energy for many particles? Basically, we want to answer if the work done for the single particle and its summation is going to be same as kinetic energy of a single particle and its summation. And it's going to be equal to mv-bar-squared. To start with, let's just call up the definition of work done. Work done for a single particle is defined by the Fdr. And Fdr can be re-written as ma and the dot product of the dr. And if I integrate it, that's going to be work done for a single particle. And that's going to be change for kinetic energy for a single particle, which is denoted as delta-Ti. If I sum all the work and energy relationship of a single particle over the whole systems of particle, that's the work done applied on the total system. And that's going to be sum of a single particle's kinetic energy. That's what I would define as a delta-T, the total kinetic energy change, okay? So by definition, that's sum of a single kinetic energy. And those vi stands for the displacement derivative from the origin to a single particle. So if I rewrite this ri vector in terms of r-bar vector and the rho-i vector, I can do the v-bar plus rho-i dot for vi term here, okay? And if I expand it, I will have a miv-bar-squared term, and v-bar rho-i dot term, and rho-i dot square term. The first one is going to be kinetic energy of the center of mass. In the second term, the v-bar comes out of the sigma, and what I have is mi rho-i sigma, which is 0. Yes, by the definition of center of mass. How about the last term? Last term is mi rho-i dot square, right? So it doesn't goes canceled. So this is the additional kinetic energy term for when you are handling the systems of particle. So not only considering the kinetic energy change for the center of mass, you should also consider the additional term, which is one-half mi rho-i dot square term. So in summary, you can just simply apply for the integral for the work of the force and the acceleration. Right-hand side will be a work done, and the left-hand side will be a work done and the right-hand side will be a kinetic energy change. And that's not only mv-bar-square, the kinetic energy of the center of mass. You should also consider the energy for each particle with respect to the center of mass, which is mi rho-i dot square.
