A solid wood door 1.00 m wide and 2.00 m high is hinged along one side and has a total mass of 40.0 kg. Initially open and at rest, the door is struck at its center by a handful of sticky mud with mass 0.500 kg, traveling perpendicular to the door at 12.0 m/s just before impact. Find the final angular speed of the door. Does the mud make a significant contribution to the moment of inertia?

Video transcript

Hey everyone welcome back in this problem. We have a uniform board. Okay? We're given its dimensions and its weight. It's kept in a vertical position by frictionless hinges on its length. It's unlatched and at rest and a kid is going to strike it with a lump of sticky molding plasticine. We're told information about the mass and the speed and the plastic scene is going to hit perpendicular to the window surface. Okay. The first thing we want to do is to determine the board's angular speed after the collision. All right, so, let's go ahead. The first thing we're gonna do is draw a little diagram of what's going on in this system. We're gonna take up into the right to be positive. All right now, before the plastic scene hits the window. Okay, We have the window here. We're told it has a massive 22.5 kg. They were told it has a length of 2.1 m. So, this distance here 2.1 m and a width of .9 m. So this distance here 0.9 m and it's kept in a vertical position and it's got hinges, frictionless hinges on its length. Okay, So the hinges are going to be on this side here, along the long side. Alright. And they're frictionless. Perfect. Then we have the plasticine. Okay, So we have this little ball of plasticine and it's 450 g. Okay, so 0.45 kg. We want everything in the same units. So we're going to convert that to kilograms. Okay, mask. 450 g And it's going to be traveling 12 m/s, 12 m/s perpendicular. So we're going perpendicular. You can imagine that the window is kind of on the side here. Alright, and this is going to be 12 m/s, the velocity of the pleistocene initially. All right now, after the collision, what's gonna happen? Okay, well we have the window and the plastic scene is sticky. Okay, So the plastic scene is actually going to stick to the window. So this is gonna be the same size of window, 0.9 m wide, 2.1 m high. It's still gonna be 22.5 kg. It's going to have hinges in the same place. Same window. And now it's gonna have plastic scene stuff great in the middle of it and it's going to be rotating and it will be rotating groups about these Hindus. Sorry, right here. Okay, with some angular speed. Omega, we'll call it Oh my God, Final. And this is what we're looking for when we're looking for the answer for part one, They were looking for that final angular speed when we think about the final angular speed. What's important is that the plasticine sticks to the window? Okay, so the window and the plasticine will be moving together so they'll be moving with the same speed. Alright, So let's recall when we have a problem like this, we can use conservation of angular momentum. Okay, we're told that the hinges are frictionless. So we know we have no net external torque acting on the system. And so we get our conservation of moment. Okay. All right. So in our conservation of momentum, we can say that the initial angular momentum l not it's going to be equal to the final angular momentum. L. F. What is made up? What What does this initial angular momentum in final angular momentum consist of? What? We have two things in our system. We have the window. We have the plastic scene. Okay, So for each of these, we have momentum from both. Okay, so we have the initial momentum from the plastic scene plus the initial momentum from the board. Okay. And that's going to equal the final momentum of the plasticine plus the final momentum of the board. Okay, Alright. Now let's recall when we're talking about momentum, angular momentum. If we have a linear motion. Okay, so before the plasticine hits the window, it's just moving linearly. It doesn't have angular speed only linear. Okay, so that angular momentum is going to be given by m Hey v are sign up there. Now when we're talking about angular momentum of something that is going to be moving with an angular speed or angular velocity that comes to I Oh my God. Okay, so I the moment of inertia times, omega the angular speed or angular velocity. And in this case we have I of the board initially omega of the board initially. Now on the right hand side after the collision, the plasticine stuck to the window again. And that they're gonna that's gonna that board is going to be rotating about the hinges. Okay, So both the plastic scene and the board will have angular speed, angular velocity after the collision, and they're gonna be moving with the same angular speed. Okay, So we're gonna have I the moment of inertia of the plastic scene times omega. Okay. The final speed of the plastic scene, and then similarly, I. B. F. The moment of inertia of the board, and omega B F. Okay, now, like I said, we have plastic seen sticking to the board, they're moving together. So the omegas are gonna be the same. Okay, so, the omega, the plastic scene equals the omega of the board. Okay, We're just gonna call those omega F. Okay. To simplify the negotiation. Alright, so let's move down and we can keep working here. Alright, well, we know that the board is initially at rest, it's not moving. Okay, so its initial angular speed or velocity will be zero. Omega will be zero. And so that contribution for momentum will be zero. Mhm. So, on the left hand side we're just gonna be left with M. P. The P. Not R. P. Not sine theta. Okay. And on the right hand side, if we substitute omega F. For both the plastic scene and the board's final velocity. Because they're going to be the same than we have or make it f times the moment of inertia of the plastic scene. Plus the moment of inertia of the board. Hey, plugging in the values that we know. We have 0. Hey, m/s. Now, our how that's the distance we're going to be from the axis. Okay? Now, because we have a board that's 0.9 m and were striking in the middle of the board, we're going to be 0.45 m. Okay, in our sign theater, we're told that we are going to let me go back up to it. We are told that we are going to hit the plasticine is gonna hit perpendicular early to the surface of the board. Okay. And so that means that the data is going to be 90°. They are omega F. Well, that's what we're looking for. We want to know that final speed and then we have I of the plasticine. The moment of inertia of the plasticine and I the moment of inertia of the board. Now let's go back and work those out using our diagram. So the moment of inertia of the board while the board is like a slab. Okay, it's a rectangular slab. So using your moment of inertia table. Either from your textbook or that your professor has provided. The formula here is going to be 1/ a squared and a is going to be the length along the side that doesn't contain the hinges. Okay, so in this case we're gonna have 1/3 The mass of the board is 22.5 kg And the length of a is 0.9 m all squared. And working this out, we get 6.075 kgm squared. That quantity we need for a moment of inertia. Alright, so now we have this quantity IVF which we can use down here to find omega F. All right. And now what we need is I P. F. Okay, so this quantity here in green down here, I P F. We need that value as well in order to find omega F. The angular speed after the collision that we're looking for. Alright, so I P F. Well, we just have this small piece of plasticine so we can consider that as a point mass. Okay, when we're looking at a point mass, we get the moment of inertia given by M r squared. Okay, the mass of the plastic scene is 0.45 kg. The are the pleistocene is going to stick in the middle. Okay, so its distance to the axis of rotation is 0. m. So 0.45 kg times 0.45 m square. And that's gonna give us 0. kg meters squared. Okay, so that's our moment of inertia for each of those, both the board and the plastic scene after the collision. Okay, so getting back to our equation that we've written down here, We can substitute those values in. Okay, so we have six Oops, we've written them in the opposite order here, so 0.091125. Okay, that's the moment of inertia of the plastic scene that we just found and 6.075 the moment of inertia of the board that we also found. Alright, so on the left hand side, if we work this all out, we get 2.43. And on the right hand side we have omega F 6.16625. All right. In our units again, for ir kilogram meters squared kilogram meters squared. Okay? And when we divide We will get omega f equals 0.394. And the units we are left with a radiance her second. Okay, So we have on the left hand side we have this meter per second. So we get kilogram meters squared per second. And then on the right hand side we get kilogram meters squared. And so when we divide we're left with a per second. So we get radiance per second. And let's see if we were looking for. Okay, that is the final speed we wanted. So we go back up, The final angular speed is 0.394. So we're looking at either answer a or B. And now what we need to determine Is we need to determine part two. Will the plasticine significantly contribute to the system's moment of inertia? Okay, well let's think about that. Okay, So for part two will the plastic seem significantly contribute to the moment of inertia? Okay, so we know that the total moment of inertia I total. Okay. That we found here in blue was the moment of inertia of the plastic scene plus the moment of inertia of the board. Okay, When we're talking about final moment of inertia in this case it was 0. Plus 6.075. Well this gives us six 0.16625. And again our unit is kilogram meter squared. And if we're using three significant digits. Okay, so the general rule here is three significant digits. So if we're going to use three significant digits we can write this as 6.17 kilogram meter. Sweat. Okay. Alright, now let's consider if we ignored the plasticine. If we say that it's not significant and we ignore it. Are I total will just be equal to the moment of inertia I of the board. Well this is just gonna be 6.75 kg meter squared. And again, if we're considering three significant digits then we have 6.8 kilogram meters squared. So we're talking about significant in this sense we're talking about when we're looking at are significant digits doesn't make a difference. And in this case it does, we have 6.17 kg meters squared here in 6.8 kg meter squared. So it does in fact make a difference whether we consider that processing or whether we ignore it. Okay, so the answer to this problem Will be a okay. The speed, the angular speed after the collision will be .394 radiance per second. And yes, it is significant to include that plastic scene in the moment of inertia calculation. Thanks everyone for watching. See you in the next video.