 We're now going to take a look at the application where we have control volumes that are undergoing linear acceleration, rectilinear acceleration, so that's basically acceleration that is only due to translation. Okay, so we're looking at conservation of momentum due to control volumes undergoing translational acceleration, so that would be in one direction that we're looking at. And why we'd be interested in this, you can have applications whereby the control volume is not stationary, nor is it moving at constant velocity. So an example of this is a rocket, when you consider a rocket, a rocket accelerates as it's moving, it might be moving in one direction, but if you wanted to do control volume analysis on that type of system, what would you need to do and would the equations apply? Well, it turns out the equations as we currently have derived them do not apply, so we have to do a little bit of a correction, that's what we're going to be doing in this segment is coming up with a modified version of conservation of linear momentum for control volume undergoing rectilinear or translational acceleration. So if you look back, what we had was the equation for extensive property that relates a systems-based approach to the governing equations to a control volume. And so we'll be referring back to that in this lecture, in this segment. And so if you recall from the last segment or section that we looked at, cases of linear acceleration, we said that as long as you measure the velocities relative to the control volume, you're okay. So that's where we had constant velocity of the control volume. So as long as we measure the time derivatives and the velocities with respect to the control volume, we should be okay. The place where we run into trouble here is the fact that Newton's second law, so the basic law that we're dealing with actually only applies to the inertial frame of reference, the frame of reference that is not moving or not accelerating, I should say. So when we look at Newton's second law, what we see is that Newton's second law is valid only for velocities measured with respect to an inertial reference frame. And so with that, we have to do a little bit of a correction to our momentum equation. And so that's what we're going to take a look at now. And we're going to begin by writing out Newton's second law written in an inertial reference frame. And if you recall before, when we looked at control volume as moving at constant velocity, we gave that capital X, Y, Z. So Newton's second law, then we have the force and momentum I will express as p. And what I'll do, I'll bring the derivative inside of the integral and we can re-express that then as being force integrated over the mass within the system is just going to be the acceleration within the inertial reference frame and then integrate it over the mass that we're looking at. And so what we're going to do, we're going to play with this right now. And we're going to find a way to be able to re-express the acceleration in the inertial reference frame in terms of the reference frame of our control volume. And so let's go ahead and do that. Okay, so what we have, we've basically broken the acceleration vector with respect to the inertial reference frame. So that would be our fixed reference frame that is not using, we use capital X, Y, Z to express that. That would be equal to, in two parts, one part would be the acceleration of our reference frame itself, ARF plus the acceleration of the system with respect to that reference frame. So we can break it into two parts. And what we're going to do, we're going to take this and we're going to plug it into Newton's second law which is here and we're going to work with that. So let's plug that in. So that is Newton's second law re-expressed and we have an acceleration of the system with respect to the non-inertial reference frame. So the one that can be moving and accelerating and then the acceleration of the frame with respect to the inertial reference frame. So we have this equation here. What I'm going to do, I'm going to make a substitution again and I'm going to re-express this here in terms of a time derivative of the velocity and I'll pull the d by dt out of the integral. Okay, so we have that expression there. When we put this all together, what we end up with is something that looks like this. And so this is all expressed in terms of a system, so a fixed mass system. We have done nothing in terms of a control volume. So we get that. Now that looks very similar to what we had earlier when we were deriving the, when we had the basic law, f equals ma with the exception that we have this correction term here. So let's continue working with this and see what we get. And what we're going to do, we're going to look back when we did Reynolds transport analogy and we noted, if you look way back in the notes, we had the system and a control volume and it turned out that at time equals t0, the system and the control volume coincided. They were one and the same and consequently we can make a comment about the forces acting on those two. So what we can say, given that they both coincided, we can write out that the force modified with this acceleration term on the system is equal to the force on the control volume modified by the acceleration of the reference frame over the control volume. And so with that, that will enable us to rewrite the linear momentum equation, but now it is being expressed for the case of an accelerating rectilinear. So it's just going in translation control volume. Okay, so that is the equation that we end up with. And this here is conservation of linear momentum or momentum. And this is for rectilinear. So translation acceleration of the control volume. And when you look at this, when we looked at control volume, it was moving at constant velocity, we made this modification. So we said that velocities had to be with respect to the control volume that was moving. The modification that we made here for acceleration is this term and so that's a new term that we need to deal with when we do the analysis. So what we're going to be doing in the next segment, we're going to take a look at a problem involving a rocket and we're going to apply this new form of conservation of linear momentum to an accelerating control volume.