[MUSIC] Okay, for the variable mass we have equations of motion of the form like F= instead of ma, you should add m-dot u term only as well. And also you can express linear momentum change with that. Let's solve the example, the chain problem first. Chain of total length capital L with the mass density rho is piled on the floor. And the it's pulled up by the external force P such that the hanging part, this one, has a constant velocity v, okay. So what will be the force P as a function of x, or so what will be the energy loss during those lifting event, okay? So let's solve the problem by splitting out the total system, the variable mass system as a particle part, or the control volume, and the particle, okay. So particle part needs a hanging mass part and a piling part. And in between there is the chain is keep moving from down to up, right. So that's why we called it as a control volume. So after spilling it out, the system has a particle control volume into another particle. Let's draw all the forces applied on it, which is external force, the gravity, and the internal force R. And here the normal force from the floor, the gravity, and internal force between the control volume and the particle, and control volume R and Q as well. So obtain the equations of motion for the particle part, which is P- Mg. Well here the mass is going to be expressed by the density and the length, and the R is going to be Ma. And note that the hanging part has a constant velocity so the acceleration turns out to be 0. And for the control volume we have a force, that force is going to be m prime as flow delta v, which is rho v, a final v minus initial v. For the piling part there is a force like a Q, N, and gravity, and then it stays there, so there is no acceleration. So if you sum them up, you will have a equations about external force and the normal force expressed like as a function of x. Okay, so you need more information. So to obtain more information, we should actually make an assumption what really happens from between the bottom falling part and the hanging part. So there are just small chain mass is keep moving from the bottom to the top, and the velocity will be from defined as initial velocity, mid velocity, and the final velocity. So if you split out the control volume by half, you will have two equations of motion about force is going to be m prime, another steady flow control volume, final speed, and the initial speed. Now when the piling part is coming into this control volume, let's assume that its has 0 velocity entering, and it changes velocity to the v, which is hanging velocity right before it exits, okay. So the v0, let's assume v0 is also 0. And then there is a sudden change, the quantum jump at the end right before it enters to the hanging part. Okay, so by doing so, Q turns out to be 0, which means that the chain movement does not apply any dynamic force to the piling body, okay. So by having Q turns out to be 0, we could specify the N. By specify the N, we could specify the P. So P is going to be a function of displacement, rho x g rho v square. And N is going to be equal to the weight for the piled mess of the piled part. So here we can tell that, okay, we should apply the force proportional to the length of the hanging part to make this system moved up steady, constant speed, okay. So next step is we should find out the energy loss during those lifting procedures. So initially, there's no energy, and the work done is only due to the external force P. So I'm going to integrate the force P over the displacement, which I could get this value. And final states, there is going to be kinetic and potential energy of this hanging part, the height of the hanging part, the potential energy, and the kinetic energy of its own mass and velocity. So the work done and the final energy gain has a difference, and that's the energy loss. So energy loss is work done minus final energy, which is one-half rho L v squared term, okay. So this is how we can get for the chain problem. Now why this energy loss occurred? There is a pile the chain on the floor. I just lifted it up, and why there is the energy loss occurred? Would you try to interpret this value? To help your interpretation, let me propose another case like through the control volume there is a continuous speed change. For example, we have everything as same as the previous solution. Except that the velocity profile throughout this control volume, rather than having a quantum jump right entering the hanging part, maybe there is some continuous transition from the bottom to the top. Okay, so that when the piled chain is entering to the control volume, instead of just keep 0 velocity, maybe the velocity is set as some in between 0 and the v, okay, like a 0.5v, okay. So if we assume the smooth transition, that means there is a Q, it's not going to be 0 anymore, which means there is a force. Dynamic force is applied to the piled mess part. Assuming that you are on the floor and you are actually catching the rope on the air, then you should actually jump off, then you should actually hit the ground. You actually generate the force to the ground, the dynamic force. So this is the case. So if you plug in those changed Q value, that'll change the N and as well as the P. So compared to the previous trial, a P is going to be expressed by the proportional to the length of the hanging mass. Now we have a little bit different form. So P has been reduced by one-half of rho v square, and N, the normal force, is increased by the one-half rho v square. By plugging this P value, the external force, to calculate the external work done through the lifting process, then we have a one-half term here again. And then if we compare the energy lost like a work done minus a final mechanical energy, it turns out to be 0. So what it mean is if there's a smooth transition occurs, then there is no energy loss. So what this sort of two different assumptions and two different answer implies to us is that whenever there is a sudden changes through the dynamic procedures, you should consider it as an impact, like collision, okay. So in the previous slides, when we're assuming a quantum jump from the 0 velocity to the hanging velocity v, there should be a work done by the link to link collision. Okay, since you have to gain the speed from 0 to certain value, so that work done should be considered. And then whenever there's sudden changes for the dynamic parameters like a velocity, you should consider it as impact cases. And where you should consider energy loss, what we have covered at the end of chapter three. Okay, let's solve another example, it's a rocket problem. Its initial mass is m0, and it gains its vertical velocity by releasing the fuel. Okay, and then the constant nozzle exit velocity is u, and after burnout, the mass has been reduced to the mb. So what is the maximum rocket velocity is what we are supposed to find. Since this is only vertical motion, I'm going to set the coordinate as a vertical. Okay, let's split it out as a particle part and steady flow part. As opposed to the chain problem, there is no second particle. So it's only a particle and then mass has been released out of it. So the rocket has a capital M, and then fuel has a small m. And this particle part has only gravity and the reaction force with the control volume that will give you the mass times acceleration. For the control volume, you have a reaction force, which is going to be m prime delta v. And then final speed is going to be fuel speed v0, and first is a rocket speed v. So this is what we could have, and then in the problem the relative leaving speed has been given. So this is it actually given from the problem. And this force, m prime multiplied by the u, this is actually we called as a thrust force, make the rocket propel, okay. So how we can define m prime? The mass flow rate is depending on the mass lost by the rocket, right? So since this is losing the mass, I will put the minus sign here. So m prime is actually -M dot. So if you plug all those parameter back into this equation here and then plug back in those R value to the equation for the rocket, you will have all the forces will generate the acceleration Ma + M dot u. Now, once we obtain the equations of motion, let's solve the v, the maximum v. So let's solve it with respect to the v. So I have this form and multiply by dt on both sides. Then you could have a more simplified form like a dv is dM over M u and gdt. And if you integrate over time, you will have, not over time. If you integrate both sides, it's a integral over the v, it's integral over the M, it's integral over the time. What you can have is a final speed as a function of the initial one and deceleration by the gravity and the velocity gain due to the thrust, okay. By releasing the fuel out, this is the magnitude of the increased velocity of the rocket. Okay, let's solve the same problem using the different method, solution method number two. Remember that you can take into account all those parameter as a one single system and then examine the change of linear momentum. So the only force here is only the gravity. And then linear momentum and v has been changed. So you have Mv for the rocket and small mv0 for the fuel. So you could take the derivative of each term and also use the your rocket mass rate is going to be minus sign of the fuel mass rate. So if you apply to that, you will have -m dot v + M v dot + m dot v0. And then if you combine this together, you will have a form, this is a relative speed u. So you will end up having all the forces are going to generate Ma + M dot u, the typical equations of motion for the steady flow case. And these answers are exactly same as the one that we obtained from the solution method 1, okay. So with that, we have gone through solving the example for the rocket and chain problem, and this is it. This is the end of the chapter 4, and this is actually the end of chapter, the first part of the dynamics, the particle dynamics. I really appreciate your participation and hope to see you again in the rigid body dynamics part. Thank you for listening.