[BLANK_AUDIO] Hi, and welcome to module 45 of two-dimensional dynamics. Today's learning outcome is to develop and give an example of the principle of conservation of momentum for bodies in two-dimensional rigid body motion. And so here is where we left off. The last topic we studied was principle of impulse momentum. We head for the an initial momentum for our body, we added an impulse to it, and we ended up with a final momentum. And I gave a graphical tool an impulse momentum diagram. This shows a body and we're restricting it to planar 2D motion where we have its initial momentum. We can add an impulse to it forces or moments acting over the body for some period of time, and we end up with a final momentum. And so what we want to look at today is if we eliminate the impulse and so we're going to zero out the impulse, and if the impulse vanishes in any direction over any period of time then you have whats called conservation of momentum. So if we eliminate the impulse in the x direction, we're going to have conservation of momentum in the x direction. So here is our initial linear momentum in the x direction and our final linear momentum in our x direction. Same thing for the y direction, and also the same thing for angular momentum if there's no moments being applied. And all these are taken about point c as we've that's where we developed the impulse momentum theory about. Okay, so here's our, our equations for conservation of momentum. Let's look at an example. Here's a skater who is initially rotating with an angular velocity of omega 1, and we'll call that the z axis up through her head and down to her tips, and we're going to say at the interface surface here, there's no friction or no losses. We're going to neglect air resistance and then the skater's going to pull their arms in and we're going to have an omega 2. And so there, we're going to say that there's no forces no moments acting on the skater so we're going to have conservation of, in this case, angular momentum for the spin. And so I omega 1 is going to be equal to I omega 2. And we recall that I is the mass moment of inertia, through the z axis, about point p. And we'll say this, in this case point p or point c is, is in the center of the, the skater's body. And so, the mass moment of inertia is the integral of x squared plus y squared dm. And so, we go out in the x and y directions to each piece of moment, mass that's, and integrate over the entire body, so it's a measure of how much mass is located how far from the axis of rotation. And so, my question to you is, in this particular case is the mass moment of inertia, the skater going to be larger in position 1 or position 2? And what you should say is position 1 will have the larger mass moment inertial because the skater has her arms extended, and so she has more mass further from the axis about which she's rotating and she draws them in, in position 2. And so, what happens then if we apply this, this inequality to our conservation of momentum equation, we see that omega 2 is equal to I1 over I2 times omega 1 and since I1 is larger than I2, omega 2 is going to be larger than omega 1 and so the skater is going to spin faster in the, in the second position. So let's go ahead and look at a demo of this again. And, I did this demo earlier in the course. And so, I'm going to have my arms out to begin with so my mass movement inertia is largest when I have it out. I've got an initial angular momentum angular velocity and then I'm going to pull my arms in, and you can see that I speed up. And I actually was able to stay on longer that time, which is good. Okay. So, we saw that omega 2 is greater than omega 1. My next question is, well what about the kinetic energy? And so, I'd like you to calculate the kinetic energy, and in this case we're only going to have rotational kinetic energy, we're not going to have any linear translational kinetic energy. So calculate for position 1 and pos-, position 2, and condition 1 and condition 2, and tell me what the results are. And, we know that rotational kinetic energy is 1 half I omega squared. Here's position 1, the initial condition. Here's at time 2. And we, if we substitute in omega 2, from what we found above, we find that T2 is equal to I, 1 half I1 omega 1 squared, but then we have to multiply it by I1 over I2, and so I1 is larger than I2, which means that the skater has an increase in rotational kinetic energy in, in, in case two. And so even though momentum is conserved, there is an increase in kinetic energy, and my question to you is how can that be? Why? That seems to violate some some law of nature. And the answer is that the kinetic energy increases because the internal energy of the skater decreases as the skater pulls the arms in. And so if you look at the overall system by first law of thermodynamics, energy can be neither created or destroyed so there has to be conservation of energy, but there's a transfer of energy from the internal energy of the skater to increasing the kinetic energy of the skater. And so that's a good discussion, and we'll come back and solve another problem next time.