A large wooden turntable in the shape of a flat uniform disk has a radius of 2.00 m and a total mass of 120 kg. The turntable is initially rotating at 3.00 rad/s about a vertical axis through its center. Suddenly, a 70.0-kg parachutist makes a soft landing on the turntable at a point near the outer edge. (b) Compute the kinetic energy of the system before and after the parachutist lands. Why are these kinetic energies not equal?

Video transcript

Hey everyone in this problem. We have a family dining table that has two circular wooden tabletops. There's a small top that spins about a vertical axis through its center. Okay, on top of a large stationary tabletop below it, meals are taken from the larger tape top, while foods and drinks are circulated from the rotating smaller top. We're told to suppose that the smaller top has a mass of 11.4 kg and a radius of two m and is rotating at 2.3 radiance per second when a pot of mass 4.5 kg is placed near its outer edge. Now, we're asked to determine the system's kinetic energy before and after the pot is placed down. Okay? And we're told to treat the small top like a disc and the pot is a point mass. All right, so let's do a quick diagram and we're going to do it as if we were looking at the table from above. Okay, So we have the large table and then we have this smaller table. Okay, This large table does not rotate. Okay, so it's angular speed is going to be zero. We're going to be concerned about this smaller table. Now, we know it rotates about its center. Okay, so we're just gonna draw this Um access through its center. Okay? We know that this rotates with a speed at 2. radiance per second. So Omega is 2.3 radiance per second. It has a mass of 11.4 kg and the radius is two m. Okay, so the distance from the edge to that center axis is two m. Alright. Now, after some time we're gonna have the same table. We're gonna have a pot placed right on the edge and we're gonna treat it as a point mass. Okay? So the pot is going to be placed on the edge And it's going to have a mass of 4.5 kg. Okay. We want to know the kinetic energy in each of these states. Alright, So, let's start with the kinetic energy before that pot is placed down. So, we're gonna call this K Not. Okay. And recall that the kinetic energy when we have angular motion is given by one half I omega squared. Ok. And in our case when we're talking about K not we're gonna have I not And omega nought. Alright. So just like in linear motion when we have the kinetic energy is one half Mv squared with angular motion. We have one half moment of inertia. Omega squared. Ok. All right. So what is I not? Let's go ahead and calculate in on And let's actually calculate. And I f because we're gonna need both of those for our kinetic energy. So, let's calculate those on the side and see what we get. Okay, So I not the moment of inertia for the table before that pot is placed down, we were told to treat that top like a disc. Okay, For treating it like a disc. Okay, We can look in the table in our textbook or that a professor provided in the moment of inertia for a disc. I is gonna be one half M R squared. Okay. And again, we're using M not and are not because we're talking about I not. So this is gonna be 1/ times the mass. 11.4 kilograms Times a radius of two m squared. All right. So if we work this out okay, our unit is gonna be kilograms times meters squared. So we get kilogram meters squared And we're gonna get an eye not of 22. kilogram meters squared. Alright, So that's why not. Let's work out I. F. As well. Just so we have it for our KF calculation. Okay. Now I F. Is going to be composed of two things because we're going to have the moment of inertia from the table and we're gonna have the moment of inertia from the pot. Now the moment of inertia from the table. This is the same tabletop, the same shape. It hasn't changed. And so this is just going to be equal to R. I. Not in the moment of inertia of the pot. We're told to treat this as a point mass. Now if we're looking at the moment of inertia for a point mass. Okay, We can look at the table in our textbook or that our professor provided when this is going to be given by M. R squared. Okay. And in our case this is the mass of the plot times the radius of the plot squared. All right. So, our I not we know is 22.8 kgm squared. Now the mass of the pot is 4.5 kg. And the radius this is a point mass. And we're putting it on the very outer edge. So the radius is going to be the same as the radius of that table. It's 2m all squared. This is gonna give us 22.8 kilogram meters squared plus 18 kilogram meter squared. Which gives a final value of 40.8 kgm squared. Alright, so we have our two moment of inertia is let's get back to our calculation for the kinetic energy. Okay, So for the initial kinetic energy, we had one half I not omega naught squared. We know I not now. So we have one half Times 22. kilogram meter square times omega not which is 2.3 radiance per second K. The initial angular speed of the tabletop, all squared. And this is going to give us AK. Not of 60.3 jewels. Alright, so, our initial kinetic energy before the pot gets it down at 60.3 jewels. And so we're looking at answers A C or E. We can eliminate the other two at this point. Now let's go ahead and find our final kinetic energy. So we can get that full answer. Okay. And the final kinetic energy is going to be equal to same as the initial one half I omega squared. In this case we're looking at I. F. In omega F. Alright, so we have one half our moment of inertia. After the pot has been put down, we found to be 40.8 kilogram meter squared. What about omega F? We don't actually know omega F yet. So, let's try to calculate omega F and then we can come back to this equation and fill it. Alright, so we need to find oh my God, how can we do that? Well, let's recall that we have the conservation of angular momentum. Okay, we have no no external torques acting on this system. So we know that we have angular momentum conserved. This tells us that we have l not the initial angular momentum is equal to L. F. The final angular momentum. Okay. After the policy in place down. Well, what is the angular momentum equal to? Well, it's equal to I omega. Okay, so we have I not omega on the left hand side and I. F. Omega F on the right hand side. Well, these moment of inertia is we found. So the eye not is 22.8 kg meters squared and omega not our initial angular speed. 2.3 radiance per second. Okay, and that's going to be equal to I F. That final angular final moment of inertia story after the protestant place down. So 40.8 kg meters squared times omega F. That final angular speed that we're looking for. All right. So on the left hand, socked we get 52. kg meters squared, radian per second. On the right hand side, we have 40.8 kg meter squared omega F. And when we divide, we get that omega F is going to be equal to the kilogram meters squared. Will cancel. We're gonna be left at the unit of radiant per second, which is what we want for the angular speed. We get 1. radiance per second. Okay. Alright. So this is our omega F. That we can now use in our calculation for the final kinetic energy And we get 1/2 times 40. kg meter squared times 1.28 radiance per second. All squared, Which gives us a final kinetic energy. K kinetic energy after that pot has been placed down Of 33.7 jewels. All right. So if we go back up to the top and we look at our answer choices. Now, we see that we have answer choice A. K. A. The initial kinetic energy before the pot has been placed down is 60.3 jewels. And the final kinetic energy after the pot has been placed down is 33. jules. Thanks everyone for watching. I hope this video helped see you in the next one.