Spinning on String of Variable Length - Video Tutorials & Practice Problems

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Spinning on a string of variable length

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Hey guys. So here's another classic example of conservation of angular momentum. And here we have this weird set up where you have an object, a little block that spins on a horizontal table. So imagine that a table you have an object spinning on top of it. But there's a tiny little hole through the table, and the object spins because it's being pulled by a string that is pulled through that hole. So you might be wondering what like What's up with that set up right? The reason we have these weird setups is because we need a way to change the length of the cable, which will in turn, change how quickly the block on top Spence, the block in the bottom is just being used to hold, um, to give this cable some tension. So this guy here is being pulled by M. G. And therefore there's attention here, and this tension is what provides the centripetal force four. This guy here to spin. Okay, so that's the standard set up. And basically what's gonna happen in these problems is the length of the cable will get reduced, which means the are the distance of rotation around a central axis will get reduced, which means the Omega will go up. The object will spin faster and faster. So let's check this out. It says. Here we have a small object. The red one over here has a mass of 2 kg. It isn't a smooth table, no friction attached to a light string. No mass on the string, like always that runs through the hole in the table, the other end of the string attached to a hanging. Wait, this guy over here M two. So M one is to and m two doesn't get mentioned yet. It says when the small object is given some speed. Um, eso you give this guy some velocity here, v Whatever. And it's gonna start right? It's not the V, by the way. It's tangential to the circle. I drew it this way. It doesn't mean that it's flying off the table. It's just going this way on the table, right? So they had some velocity. Maybe I should draw. It's sort of like this doesn't look like it's going away, all right? So when it's given some speed that spins around a circular path around the whole with attention from the hanging weights provide us centripetal force. I explain that that's this mass here pulls down which the block pulls on the tent on the rope, providing attention. And this tension is what keeps you spinning because it provides the force that keeps you around a circle. It's, I suppose, that the object spins at 120 rpm when the radio distance, when he has a radio distance when it is a radio distance of 10 centimeters from the whole. So when this distance L or R, is 0.10 or 0.1, it has an rpm off 1 20. So this is how the problem starts. Therefore, I'm gonna call this my our initial in my rpm initial. Okay, so the first question is, how fast in our PM would it spin if the radio distance was reduced to six? In other words, if our final is instead of 10 it is six. So 0.6 because it's got a big meters. What would be my rpm final? Okay, so that's part a. So let's do this real quick. Conservation of angular momentum l initial equals L final. I'm gonna expand this to be I Omega equals I Omega initial initial final. Finally, um, this object is a small object, which means we're gonna treat as a point mass. So instead of I, we're gonna expand I into m r squared and notice that I have rpm's so I don't want Omega's. I want RPMs. Remember that omega is two pi f or two pi r p m over 60. So I'm gonna rewrite omega expand omega as two pi rpm initial over 60. Um, this is mass initial, though the masses don't really change in the radius initial or the distance initial on the other side. I have the same thing, but it's gonna be m final, our final squared and then two pi rpm final, which is my target variable, by the way, divided by 60 When I write this notice that I can cancel some stuff first, the masters in change. So this mass cancers with this mass the two pi cancer with two pi 60 cancers with 60. So we're left with our I squared rpm initial equals our final squared rpm final. And this is what we're looking for so I can solve for r p m final by move everything by moving this guy to the other side. So it's gonna be rpm final is our I squared r f squared rpm initial. So notice how the rpm final is the initial rpm times a ratio of the squares off these distances here. Right, So the initial one was 10 so 0.1 squared. The final one is 0.6 square and the rpm initial Thea rpm initial was 1 20. So if you do all of this, you multiply it. You get to I gotta here somewhere it's 333 So remember I mentioned how it was gonna be faster and indeed it is faster. The final rpm is going to be 33. So that's part a. Let's get going for part beats has appear at this new rpm what linear speed with the objects have. So I'm basically asking. Remember, when you spin around a circle like this, you have a no mega, and you also have a corresponding and equivalent tangential speed v 10. And remember that V 10 is connected to an Omega by V 10 equals R omega. So we're saying at this new rpm what would be your linear, tangential speed. So what would the tan be? Let me just do this. Okay? So all I gotta do is multiply our times, Omega. Now, I want the new or final V 10. So I'm gonna use the new or final are and then you or final Omega. So then you are is 0.6 I don't have omega. I have the new rpm, but I can plug it in here, so it's gonna be two pi rpm final divided by 16. So this is 0.6 times two pi times rpm finals 333 And this whole thing divided by 60. And if you multiply all of this, I gotta hear you get 2.1 meters per 2nd, 2.1 m per second. That's part B. It's another thing you might be asked. Um, since there's a relationship between those numbers and then for part C for part C, we're being asked what mass m two Um does the hanging object need tohave? So, in other words, what is them to to maintain the small object spinning to maintain the small objects spinning at the rpm found in parts B now, this is actually rpm found in parts A right, the rpm found in part A. Okay, cool. So what am to do we need in order to give it that speed? And the idea here is that this mass m two this is m two g matters because it is at equilibrium. So this tension here equals M two G, and this tension here is the centripetal force that keeps this spinning. So this if you change this, you change the amount of centripetal force. Okay, so we're gonna This is really a centripetal force problem. So now I'm gonna write that the sum of all forces centripetal equals m a centripetal. And then I can rewrite this. The centripetal force here is the centripetal force is the tension which comes from Emma to G. So I can actually rewrite this as m two g m and then accelerations in Tripoli. I don't have it, but I can rewrite it as the square over our Remember, in a lot of these problems, um, in a lot of these problems of centripetal force, you rewrite ethical that may into f equals M V square over our Okay, So you should remember doing the stuff. I want to point out that this mass here is This is the acceleration of the block. So this is actually gonna be M one. So what we're doing here is we're trying to figure out what is M two so that m one can spin at, um, at this rpm, which, by the way, this rpm produces this speed. So the idea is we're gonna plug 2.1 here. Alright? So to solve for this, I have m two equals m one v, um, the final square because we want to know how much mass we need, um, to support it with the new rotation on that is divided by G R. Okay, so M one is to the velocity we found this 2.1 square G is 9.8, and obviously, if we're talking about VR, we're talking V final. We're talking about our final as well. Our final is six. So it's 0.6 okay, and you multiply this whole thing and you get 15 kg. So the idea is that for you to be able to spin at rpm, which means you have a linear velocity of 2.1 at that distance from the middle. Um, M two has to be 15 if you want to spend even faster. If you want to spend even faster meaning this V here would grow, you would need your mass to be the mass. The hanging weight at the bottom have a greater mass. So you can have Maura tension so that this thing can spin faster. All right, so that's the idea. Eso I like this example because it touches up on a bunch of different parts of this question that you can get. The most common one is parting, but you might see something where you start talking about the hanging weight and what happens with the hanging weight as well. Okay, so that's it for this one. Please let me know if you have any questions and let's keep going

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Problem

Problem

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?

A

ω_{f} = ^{1}/_{4}ω

B

ω_{f} = ^{1}/_{2} ω

C

ω_{f} = 2ω

D

ω_{f} = 4ω

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