Hey guys. So another common example of an angular collision problem has you jumping onto a disc with initial velocity. Let's check out a few possibilities. Here, we have a disc of mass 200 kilograms and 4 meters in radius. So I'm going to say big M equals 200 and big R equals 4 meters. And the moment of inertia of a disc is half ㎷x squared, same as a solid cylinder. I can actually go ahead and already plug this in here because we're going to be using this a lot. And if I do this, I get 400 or actually, I'm sorry, I get 1600. This thing is squared. Right? So 1600 is the moment of inertia of the disc. It spins around a perpendicular axis through its center, so a perpendicular axis through the center of the disc just means that the disc spins like this, like you would expect a normal usual, disc rotation to look like, with the velocity of 2. So the omega initial of the disc is 2 counterclockwise. Counterclockwise is in the direction of the unit circle. It's a positive 2. Okay. So the disc has I of 1600 and an omega initial of positive 2.
Find the new speed that the disc would have if an 80-kilogram person, which we're going to treat as a point mass, jumps onto the disk in a bunch of different situations. So the person is going to come into the picture, little m equals 80. We're going to treat the person as a point mass, which means we're going to use moments of inertia equation as ㎷x squared. Okay. And I want to know what is the new, angular, the new angular speed that the disk will have. What is omega final?
So in the first situation, the person steps onto the disc with negligible speed. Negligible speed is 0. It's the same thing as if the disc is spinning, and you like land on top of the disc. You don't have a horizontal velocity. You're not doing anything that would cause the disc to move faster or slower. So before we start, I want to talk about that a little bit. Let's say the disc is here, and if you simply enter the disc, you don't affect it, you're not adding velocity to the disc in any particular direction. The disc will become slower because there's more mass now, but you're not contributing to velocity as opposed to, let's say, if a disc is at rest and then you come running this way, right, and you jump on the disc, what's going to happen is when you jump on the disc, the disc is going to spin. So you're contributing some initial momentum. Here, you're not contributing any initial momentum.
Alright. In all of these questions, we're going to solve this using the conservation of angular momentum. So ㎢initial equals ㎢final. And remember, if you have an object which in this case, the person is moving in a linear motion, we're going to use ㎢ as ㎷xvr instead. But here, the person doesn't have anything, doesn't have a velocity. So in the beginning, all we have is, the angular momentum of the disc. Okay? So this is going to be ㎷1Ω1 initial plus now the person is linear even though the person doesn't really move, with any velocity yet. So this would be ㎷2㎷2㏃2, but this is 0 because the person enters with negligible speed. ㎵ equals 0 equals, ㎷1Ω1final plus ㎷2Ω2final. Okay? The initial ㎷ of the disc is 1600. The initial speed of the disc is positive 2. Now once a person enters the disc, they will have the same angular momentum. Right? If you step onto a disc, you will spin with the disc. So I can do this, omega final, I can combine. And then all I have is ㎷ of the disc and ㎷ of the person. ㎷ of the disc is 1600, but ㎷ of the person, I have to calculate that real quick. So let's go over here and say the ㎷ equation is going to be the moment of inertia of a point mass because that's how we're treating the person, ㎷x squared. Little m is the mass of the person, which is 80. R is distance from the center. So the person is entering the disc, you're entering the disc. So if you're entering the disc without any speed, it's because you're entering at the very edge of the disc. So you can assume that the person just steps in here. Because the person has negligible speed, it doesn't matter the direction with which the person comes into the disc. The only thing that matters is that you enter at the edge, which means your r is the entire radius of the disc. Okay. So here, r will be the radius of the disc because you enter the disc at its very edge. The disc has radius 4, so it's going to be like this. And this, there's a square here, and this is 1280. Okay. So 1280 is the number that goes here, 1280. And when I move stuff around, omega final is 32100 divided by, these two guys. You end up with 1.11 radians per second. This should make sense because it's slower. So we started with an omega of 2, positive, and now the person entered, it has more mass. Whenever you add mass to a rotating disc, it will move at a slower rate. So that's it for part A. Pretty straightforward.
Let's do part B. Part B, the person will jump onto the disc at its edge, with 9 meters per second directed towards the center. So in the first one, you did this. You jumped directed in a direction. You jump at the edge direct towards the center. So in the first one, you did this. On the second one, let's say you jumped over here and you jumped with the direction that points towards the center. Okay? So let's think about what would happen here. Right? So I was talking about how if you have a disc and you come running this way and you jump, you're going to add a momentum this way because you jumped like this. Think of you as pushing a force. If you go the other direction, it may cause you to go the other direction as well. Now if you come running into the disc, if you come running into a disc in a direction that's towards the center, like a radial direction, you actually don't contribute any momentum to the disc. It's just like if you are trying to push on a disc, towards the radius. Right? You can push but it's not going to cause rotation. So the idea is that jumping with the velocity, if directed towards the center, is actually the same thing as if you just stepped in. You're not contributing any momentum even though you have, even though you have linear velocity, you're not doing anything. So you just that's all you need for part B. You just have to realize that you're not adding any momentum which means that this is also 0 for you and the answer is the same as part A. Omega final is just 1.11 radians per second because you don't contribute any momentum.
Now for parts C and D, you will contribute momentum, because you're jumping as I explained. Here, you're jumping in a direction that will cause the disc to spin. Okay? Alright. So for part C, you're jumping, directed tangentially up. So this is part C and this is part D. In both cases, you have the same velocity, just different directions. So let's do this real quick. Alright. So same setup. I'm going to have ㎢initial equals ㎢final. Whoops. Let's make some space here for part C. ㎢initial equals ㎢final which means I can write this equation here. ㎷1Ω1initial plus, ㎷2Ω2 ㏂2㏁2 equals ㎷1Ω1final plus ㎷2Ω2final. In all of these questions, you will rotate with the disc once you land on it. So your Ω1 and Ω2 at the end will be the same, which means we can do this, omega final, ㎷1+㎷2. This doesn't change. Okay? The initial omega of the disc is the initial ㎷ of the disc is, 1600. The ㎷, this doesn't really change. So the not the initial but just the ㎷ of the disc is 1600. The initial omega, that's what I meant, is plus 2. K. Now here, the person which has a mass of 80, is jumping in with a velocity of 9. I wanna leave a little bit of space here so we can put positive or negative. We'll talk about that in just a second. And you are jumping tangentially at the edge of the disk, which means your R is the entire radius here. Okay. Your R is the radius and the radius is 4. So I'm going to put a 4 here. Now we got to talk about this velocity. That's going to be the difference between part C and D. It's just the sign of that velocity. Okay. So initially, the disc is rotating like this with a positive 2 rad/s. Okay. Look at the C arrow here, right, for part C, you're jumping that way. And I hope you can see how jumping that way, would go against which cause the disc to spin. It's basically going against the direction of the disc. Right? So you're the disc spinning like this and then you're jumping in, you jump the disc spinning like this. I got my disc is broken here, and you're running that way. So if this is positive, this velocity has to be negative because that velocity is trying to produce a counter rotation, a rotation that's going to be clockwise opposite to the disc, therefore, would be negative. So you can think of this almost as if this was a force that would produce a torque in this direction. But in this case, it's just a force that's trying to produce a rotation in that direction. This direction is clockwise. Therefore, this linear velocity has to be negative even though it's going up because it's trying to cause negative rotation. Okay? D is going to be the same thing but it's going to be positive because it's trying to help the disc. Let me get my my broken disc again. This disc is rotating a particular direction and D is jumping in the same direction as the rotation so it's adding. So if this is positive, then this is positive as well because they're both spinning in the same direction. Cool. So parts C and D are exactly the same except that for C, this is negative and for D, that is positive. So it's going to be the same thing except positive. Okay. So let me just put a little plus here, so we remember to just copy and paste that stuff.
