Clutch Prep is now a part of Pearson
Ch 16: Angular MomentumWorksheetSee all chapters

# Spinning on String of Variable Length

See all sections
Sections
Spinning on String of Variable Length
Angular Momentum of Objects in Linear Motion
Intro to Angular Collisions
Conservation of Angular Momentum
Angular Collisions with Linear Motion
Intro to Angular Momentum
Jumping Into/Out of Moving Disc
Angular Momentum of a Point Mass
Angular Momentum & Newton's Second Law
Opening/Closing Arms on Rotating Stool

Example #1: Spinning on a string of variable length

Transcript

Hey guys. So here's another classic example of conservation of angular momentum and here we have this weird setup where you have an object, a little block that spins on a horizontal table, so imagine I've got a table, you have an object spinning in 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's up with that setup, 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 spins, the block in the bottom is just being used to hold, to give this cable some tension. So this guy here is being pulled by mg and therefore there's a tension here and this tension is what provides the centripetal force for this guy here to spin, okay? So that's the standard setup and basically what's going to happen in these problems is the length of the cable will get reduced which means the r, the distance of rotation around the 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 we have a small object, the red one over here has a mass of 2 kilograms it is in a smooth table, no friction, attached to a light string, no mass on the string like always that runs through a hole in the table, the other end of the string attaches to a hanging weight this guy over here m2. So m1 is 2 and m2 doesn't get mentioned yet, it says, when the small object is getting some speed. So you give this guy some velocity here v, whatever, and it's going to start, right? It's not the V by the way is 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 there has some velocity, maybe I should draw it sort of like this, it doesn't look like it's going away. Alright, so when it's given some speed it spins around a circular path around the hole with attention from the hanging weights providing centripetal force, I explained that, that this mass here pulls down which the block pulls of the tent, on the Rope providing tension and this tension is what keeps you spinning because it provides the force that keeps you around the circle, it says, suppose that the object spins at 120 rpm when the radial distance, when it has a radial distance, when it is a radial distance of 10 centimeters from the hole. So when this distance l or r is 0.10 or 0.1 it has an RPM of 120. So this is how the problem starts, therefore I'm going to call this my r initial in my rpm initial, okay? So the first question is, how fast in rpm would spin if the radial distance was reduced to 6? In other words if our final is instead of 10, it is 6, so 0.06, because it's got to be 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 going to expand this to be I Omega equals I Omega initial, initial, final, final and this object is a small object which means we're going to treat it as a point mass. So instead of I we're going to expand I into mr squared and notice that I have rpms. So I don't want Omegas I want rpms remember that Omega is 2pi, F or 2pi, rpm over 60. So I'm going to rewrite Omega, expand Omega as 2 pi, rpm initial over 60, this is mass initial though the masses don't really change and the radius initial or the distance initial, on the other side I have the same thing but it's going to be m final, r final squared and then 2pi, 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 masses. So this mass cancels with this mass, the 2pi cancels is 2pi, 60 cancels at 60 so we're left with ri squared, rpm initial equals r final squared, rpm final and this is what we're looking for. So I can solve for rpm final by moving everything, by moving this guy to the other side. So it's going to be rpm final is ri squared rf squared, rpm initial. So notice how the RPM final is the initial rpm times a ratio of the squares of these distances here, right? So the initial one was 10, so 0.1 squared, the final one is 0.06 squared and the RPM initial, the RPM initial was 120. So if you do all of this you multiply it you get 2, I got it hidden here somewhere it's 333. So remember I mentioned how it was going to be faster and indeed it is faster the final rpm is going to be 333. So that's part a, let's get going for Part B it says, at this new rpm what linear speed with the object have. So I'm basically asking, remember when you spin around a circle like this you have an Omega 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 v 10 be? Let me just do this, okay? So all I got to do is multiply r times Omega. Now I want the new or final v 10. So I'm going to use the new or final R and the new or final Omega. So the new r is 0.06, I don't have Omega, I have the new rpm but I can plug in here. So it's going to 2pi RPM Final divided by 60. So this is 0.06 times 2pi 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 m/s, 2.1 m/s, that's Part B, it's another thing you might be asked since there's a relationship between those numbers and then for Part C, for Part C, we're being asked, what mass m2 does the hanging object need to have? So in other words what is m2 to maintain the small object spinning, to maintain a small object 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 m2 do we need in order to give it that speed? And the idea here is that this mass m2, this is m2, g matters because it is at equilibrium. So this tension here equals m2 g and this tension here is the centripetal force that keeps this spinning. So this, if you can 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 going to write that the sum of all forces centripetal equals ma centripetal and then I can rewrite this, the centripetal force here is, the centripetal force is the tension which comes from m2, g. So I can actually rewrite this as m2, g, m and then acceleration centripetal, I don't have it but I can rewrite it as v squared over r, remember in a lot of these problems, in a lot of these problems of centripetal force you rewrite F equals ma into F equals mv squared over r, 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 going to be m1. So what, we're doing here is we're trying to figure out what is mg. So that m1 can spin at this rpm which by the way this rpm produces this speed. So the idea is you're going to plug 2.1 here alright? So to solve for this, I have m2 equals m1 v final squared because we want to know how much mass we need to support it with the new rotation, and that is divided by g,r, okay? So m1 is 2, the velocity we found this 2.1 squared G is 9.8 and obviously if we're talking about vr, v final we're talking about r final as well, r final is 6. So it's 0.06, okay? And you multiply this whole thing and you get 15 kilograms. So the idea is that for you to be able to spin at 133 rpm which means you have a linear velocity of 2.1 at that distance from the middle, m2 has to be 15, if you want to spin even faster, if you want to spin even faster meaning this v here would grow, you would need your mass to be, the mass, the hanging weight at the bottom to have greater mass so you can have more tension so that this thing can spin faster alright? So that's the idea. So I like this example because it touches up on a bunch of different parts of this question that you can get, the most common ones Part A but you might see something where you start talking about the hanging weight and what happens with a 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.

Practice: 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?