A small object (red, m) is on a smooth table top and attached to a light string that runs through a hole in the table. The other end of the spring attaches to a hanging weight (green, M). When the small object is given some speed, it spins in a circular path around the hole, with the tension from the hanging weight providing the centripetal force that keeps it spinning. If the object spins with angular speed ω when it is a distance R from the central role, what new angular speed (in terms of ω) does it have when this distance is halved? What new mass does the hanging weight need, in terms of M, to support a circular path at the new speed?

What's up guys? I hope you try this practice problem. Here we have a literal problem of conservation of angular momentum. Literal meaning we're not going to have numbers. We're gonna solve this in terms of variables. Alright? So a small I have a small object red right here, little m, is on a smooth frictionless tabletop attached to a light string, light meaning the string has no mass like always, that runs through a hole in the middle of the table. So the idea is that the cable, you have a table, you have a block, the block connects to the hole so that we can regulate the distance, the length of the cable up here which regulates the distance from the center, that this object will have as it spins around the central points. Okay. The other end of the spring, I meant string. Fix that, please. Sorry about that. The other end of the string attaches to a hanging weights. We're gonna call this big m. And it says, when the small object is given some speed, it spins around a circular path, with attention from the hanging weight providing the centripetal force to keeps it spinning. So it's just describing to you the basic setup. So here I have mg. There's a tension that pulls this thing up. This block here is at equilibrium so I can say that tension here is big m g. The same tension that pulls the block up, pulls the rope here down. So this is tension and it's the same force that same tension that pulls this here which is the centripetal force. So I can write that mg is my same as the tension which is my centripetal force which you can also write from f equals m a. Remember, f equals m a in centripetal motion, in centripetal problem would become f equals m v, squared over r. You replace ac with v squared over r. So here I can say that not only is mg equal to t, that happens to be the the centripetal force. So because it is a centripetal force, I can also write that it equals to mvsquaredoverr which has to do with the motion of this block, therefore is little m. So this is, an important relationship here for this question. Okay. Alright. It says the object spins with an angular speed of omega when it is a distance r from the central central, hole. So when this distance here, this distance here is initially distance, remember, is little r. The initial little r is big r. And when we have this, we have an initial angular speed of omega. K. So omega is just omega. I want to know what new angular speed in terms of omega will this thing have when this distance is halved. So the idea is that if my r initial is big r, when my r final, little r final becomes half of r, what will happen to my so it becomes half of r. What would happen to my angular speed in terms of omega? So the first thing I hope you're thinking, right away is that if I'm reducing the length of the rope, this thing is going to spin faster. And a good guess or a reasonable guess is that if this is going down by a factor of 2, that this would go up by a factor of 2. You're gonna see that this is wrong but at least it's it's generally it's directionally correct, right? It is going to grow. It's just going to grow by 4 instead. So let me show you that. So what we have here is, a problem of rotation where the radius is changing. Therefore, this is a conservation of angular momentum question. And I'm gonna write l initial equals l final and I'm gonna expand this. Actually, I'm gonna write, I'm going to write this version here because this is a proportional reasoning question. So think of it as we're changing the one object and we're going to know what happens to it. So I'm gonna rewrite this as l initial equals a constant, and I'm gonna expand l initial to I omega, I initial omega initial being a constant. Okay. I'm gonna expand I further. This is a point mass. It has moment of inertia, m r squared. K. And then I have omega here, and this whole thing is a constant. K. So check it out. R is changing by a factor of 2. R is becoming half as long, so I can say that it's going down by a factor of 2. Now r is squared. So the net effect here is actually that this whole expression here, the whole left side mass doesn't change. The whole left side of this thing here, is going down by a factor of 4. But this the entire left of the equal sign is, is a constant. It has to stay the same. So to balance the fact that the blue part is going down by 4, the yellow part has to go up by 4. Okay. So this has to go up by 4. This means that my new omega will be 4 times my my, old Omega. Okay. That's it. So your new omega is 4 omega initial. Alright. So let's do the bonus part. Let me draw a line here. Bonus. The bonus question is asking bonus question is asking what new mass does the hanging weight need? So the mass of the hanging weight is this guy right here, big m. What is the new big m or the final big m in terms of the original, mass which is big m, support a circular path at the new speed. So the idea is that if you're going to if you're spinning, you spin because there's a, centripetal force. It says right here. You spin because you have a centripetal force which in this case is provided by the tension which in turn is provided by mg. Well, if you wanna spin faster, you need more of a centripetal force. Imagine if a rope is spinning faster, you have to hold on to it harder, which means you need more tension which means you need a, g doesn't change so that means you need this height this, weight here to be of a greater mass. And that's what we wanna know, how much grade does the mass have to be. Now what we're gonna do to solve this, we're gonna use this equation, the yellow box here as a relationship. And more specifically, I'm gonna use the left most, element here and the right most term because this is the the tool that draw a connection between, how big m is related to little v. The tension and the, centripetal force are just connecting really these two guys. So I'm gonna go down here and write that. I'm gonna say big mg equals little m, v squared over r. And I want to know is as the v changes, right, so the v changes, I'll put a little delta here, and the r changes as well, how is that going to affect m? This is a proportional reasoning question. What I'm going to do is I'm going to first solve for m. So that's since that's my target variable, and I'm going to say m is m, little mvsquaredovergr. Okay. Now the little mass doesn't change. The little g doesn't change either. So I can cancel those 2 guys out. What's changing is v and r. But I don't actually have I don't actually know exactly how much v is changing by. I know my new omega is 4 times my old omega. So what I'm going to do is I'm going to replace v with omega. V, remember, is r omega. Right? You can make a relationship between v and r in this problem. So what I'm going to do is I'm going to replace this with r omega and the whole thing squared. So look what happens here. I have 1, I have r squared omega squared divided by r. One r cancels here, and I have r omega squared. Now what's happening with these guys? Now we're good to plug stuff in. The r is changing by, a factor of 2. The r is going down by a factor of 2, and the omega is going up by a factor of 4. Now in this equation, r isn't squared. So it's just 2. But Omega is squared. So this 4 actually gets squared. Okay. So this whole thing is going up by a factor of or this guy here is going up by a factor of 16. Okay. But we have 16 up. Whoops. I have 16 up and 2 down. You can combine those 2 by basically doing going down by a factor of 2 means you'll be multiplied by half. And going up by a factor of 16 means you'll be multiplied by 16. If you combine those, you get 8 up. Okay? 2 down times 16 up is just 8 up. What that means is that my new, my new, that's the final answer right here, my new mass has to be 8 times greater, 8 times greater than the original mass for us to be able to shrink the this, distance by, by a factor of 2. Okay? So that's it for this one. Hopefully, it makes sense. Let me know if you have any questions and let's keep going.

A small object (red, m) is on a smooth table top and attached to ...

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