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# Rolling without slipping problems

## Video transcript

so we saw last time that there's two types of kinetic energy translational and rotational but these kinetic energies aren't necessarily proportional to each other in other words the amount of translational kinetic energy isn't necessarily related to the amount of rotational kinetic energy however there's a whole class of problems a really common type of problem where these are proportional so that's where we're going to talk about today and that comes up in this case so imagine this imagine we instead of pitching this baseball we roll the baseball across the concrete so say we take this baseball and we just roll it across the concrete what's it going to do it's going to rotate as it moves forward and so it's going to do something that we call rolling without slipping at least that's what this baseball is most likely going to do I mean unless you really chuck to this baseball hard or the ground was really icy it's probably not going to skid across the ground or even if it did that would stop really quick because it would start rolling and that rolling motion would just keep up with the motion forward so when you have a surface like leather against concrete it's going to be grippy enough grippy enough that as this ball moves forward it rolls and that rolling motion just keeps up so that the surface is never skid across each other in other words this ball is going to be moving forward but it's not going to be slipping across the ground there's gonna be no sliding motion at this bottom surface here which means at any given moment this is a little weird to think about at any given moment this baseball rolling across the ground has zero velocity at the very bottom this bottom surface right here isn't actually moving with respect to the ground because otherwise it'd be slipping or sliding across the ground but this point right here that's in contact with the ground isn't actually skidding across the ground and that means this point right here on the baseball has zero velocity so this is weird zero velocity and what's weirder that means when you're driving down the freeway at a high speed no matter how fast you're driving the bottom of your tire has a velocity of zero it's not actually moving with respect to the ground it's just the rest of the tire that rotates around that point so that point kind sticks there for just a brief split-second that makes it so that the tire can push itself around that point and then a new point becomes the point that doesn't move and then it gets rotated around that point and then a new point is the point that doesn't move so that will take turns is very nice of them other points are moving this point up here is going crazy fast on your tire relative to the ground but the point that's touching the ground unless you're driving a little unsafely you shouldn't be skidding here if all is working as it should under normal operating conditions the bottom part of your tire should not be skidding across the ground and that means that bottom point on your tire isn't actually moving with respect to the ground which means it's stuck for just a split second it has no velocity so that's what we mean by rolling without slipping why is this a big deal I'll show you why it's a big deal this implies that these two kinetic energies right here are proportional and moreover it implies that these two velocities the center of mass velocity and this angular velocity are also proportional so that's what I want to show you here so how do we prove that how do we prove that the center mass velocity is proportional to the angular velocity well imagine this imagine we coat the outside of our baseball with paint so I'm about to roll it on the ground right roll it without slipping let's say I just coat this outside with paint so there's a bunch of paint here now let's say I give that baseball a roll forward well what are we going to see on the ground we're going to see that it just traces out a distance that's equal to however far it rolled so if it rolled to this point in other words if this baseball rotates that far it's going to have moved forward exactly that much arc length forward right because if this baseball is rolling without slipping then as this baseball rotates forward it will have moved forward exactly this much arc length forward so in other words if you unwind this purple shape or if you look at the path that traces out on the ground it would trace out exactly that arc length forward and why do we care why do we care that it it travels an arc length forward because that means the center of mass of the space ball has traveled the arc length forward so the center of mass of this baseball has moved that far forward that the distance the center-of-mass has moved and we know that's equal to the arc length what's the arc length remember we had a formula for that if something rotates through a certain angle so if we consider the angle from there to there and we imagine the radius of the baseball the arc length is going to equal R times the change in theta how much theta this thing has rotated through but note that this is not true for every point on the baseball consider this point at the top it was both rotating around the center of mass while the center of mass was moving forward so this took some complicated curved path through space it might look like that this distance here is not necessarily equal to the arc length but the center of mass was not rotating around the center of mass because it's the center of mass the center of mass here this baseball was just going in a straight line and that's why we can say the center mass of the baseball's distance traveled was just equal to the amount of arc length this baseball rotated through in other words it's equal to the length painted on the ground so to speak and so why do we care why do we care the distance the center of mass moves is equal to the arc length here's why we care check this out we can just divide both sides by the time that that took and look what we get we get the distance center mass moved over the time that took that's just the speed of the center of mass and we get that equals the radius times Delta Theta over delta T but that's just the angular speed so this shows that the speed of the center of mass for something that's rotating without slipping is equal to the radius of that object times the angular speed about the center of mass so the speed of the center of mass is equal to R times the angular speed about that center of mass and this is important this you want to commit to memory because when a problem says something's rotating or rolling without slipping that's basically code for V equals R Omega where V is the center of mass speed and Omega is the angular speed about that center of mass now you might not be impressed you might be like wait a minute didn't we already know this didn't we already know that V equals R Omega we did but this is different this V we showed down here is the V of the center of mass the speed of the center of mass this V up here was talking about the speed at some point on the object a distance are away from the center and it was relative to the center of mass so in other words so you've got some baseball that's rotating if we wanted to know okay at some distance R away from the center how fast is this point moving V compared to the angular speed well if this things are rotating like this that's going to have some speed V but that's the speed V relative to the center of mass what we found in this equation is different this is the speed of the center of mass this tells us how fast is that center of mass going not just how fast is a point on the base ball moving relative to the center of mass this gives us a way to determine what was the speed of the center of mass and it turns out that is really useful and a whole bunch of problems that I'm going to show you right now let's do some examples let's get rid of all this so let's do this one right here let's say you took a cylinder a solid cylinder of five kilograms that had a radius of two meters and you wind a bunch of string around it and then you tie the loose end to the ceiling and you let go and you let this cylinder unwind downward as it rolls it's going to be moving downward let's say you drop it from a height of four meters and you want to know how fast is this cylinder going to be moving how fast is the center of mass going to be moving right before it hits the ground that's what we want to know we're calling this a yo-yo but it's not really a yo-yo a yo-yo has a cavity inside and maybe the string is wound around a tiny axle that's only about that big we're winding our string around the outside edge and that's going to be important because this is basically a case of rolling without slipping you might be like this thing's not even rolling at all but it's still the same idea just imagine the string is the ground it's as if you have a wheel or a ball that's rolling on the ground and not slipping with respect to the ground except this time the ground is the string the cylinder is not slipping with respect to the string so that's something we have to assume we're going to assume this yo-yo is unwinding but the string is not sliding across the surface of the cylinder and that means we can use our previous derivation the speed of the center of mass of this cylinder is going to have to equal the radius of the cylinder time's the angular speed of the cylinder since the center of mass of this cylinder is going to be moving down a distance equal to the arc length traced out by the outside edge of the cylinder but this doesn't let us solve because look I don't know the speed of the center of mass and I don't know the angular velocity so we need another equation another ID in here and that idea is going to be conservation of energy there's problems crying out to be solved with conservation of energy so let's do it so we're going to put everything in our system we're going to say energy is conserved starts off at a height of 4 meters that means it starts off with potential energy so I'm gonna say that this starts off with MGH and what does that turn into well this cylinder when it gets down to the ground no longer has potential energy as long as we're considering the lowest most point as H equals zero but it will be moving so it's going to have kinetic energy and it won't just have translational kinetic energy so it will have translational kinetic energy because the center of mass of this cylinder is going to be moving the center of mass of the cylinder is going to have a speed but it's also going to have rotational kinetic energy because this cylinder is going to be rotating about the center of mass at the same time that the center of mass is moving downward so we have to add one-half I Omega squared and it still seems like we can't solve because look we don't know V and we don't know Omega but this is the key this is why you needed to know this formula and we spent like five or six minutes deriving it this is the link between V and Omega so we can put this whole formula here in terms of one variable by substituting in for either V or for Omega now I'm going to substitute in for Omega because we want to solve for V so I'm just going to say that Omega you could flip this equation around and just say that Omega equals the speed of the center of mass divided by the radius so I'm going to use it that way I'm going to plug in I just solved this for Omega I'm going to plug that in for Omega over here let's just see what happens we're gonna get V of the center of mass divided by the radius and you can't forget to square it so we square that so after we square this out we're going to get the same thing over again so I'm just going to copy that paste it again but this whole term is going to be squared I have a V of the center of mass squared over radius squared and so now it's looking much better we just have variable in here that we don't know V of the center-of-mass this I might be freaking you out this is the moment of inertia what do we do with that well the moment of inertia of a cylinder you often just have to look these up the moment of inertia of a cylinder turns out to be one-half M the mass of the cylinder times the radius of the cylinder squared so we can take this plug that in for I and what are we going to get I just copy this paste that again if we substitute in for our eye our moment of inertia I'm gonna scoot this over just a little bit our moment of inertia was one-half M R squared so I'm going to have one-half and this is in addition to this 1/2 so this 1/2 was already here there's another 1/2 from the moment of inertia term one-half M R squared but this R is the same as that are so look it up that up I've got R squared and a 1 over R squared these end up cancelling and this is really strange it doesn't matter what the radius of the cylinder was and here's something else that's weird not only does the radius cancel all these terms have mass in it so no matter what the mass of the cylinder was they will all get to the ground with the same center of mass speed in other words all yo-yos of the same shape are going to tie when they get to the ground as long as all else is equal and we're ignoring air resistance no matter how big the yo-yo or how massive or what the radius is they should all tie at the ground with the same speed which is kind of weird so now finally we can solve for the center of mass we've got this right-hand side the left-hand side is just G H that's going to equal so we end up with 1/2 V of the center of mass squared plus 1/4 V of the center of mass squared that's just equal to 3/4 speed of the center of mass squared if you take 1/2 plus 1/4 you get 3/4 so if I solve this for the speed of the center of mass I'm going to get if I multiply G H by 4 over 3 and we take a square root we're going to get the square root of 4 G H over 3 and so now I could just plug in numbers if I wanted to I could just say that this is going to equal the square root of 4 times 9.8 meters per second squared times 4 meters that's where we started from that was our height divided by 3 is going to give us a speed of the center of mass of 7.2 3 meters per second now here's something to keep in mind other problems might look different from this but the way you solve them might be identical for instance we could just take this whole solution here I'm going to copy that let's try a new problem it's going to be easy it's not going to take long let's say we take the same cylinder and we release it from rest at the top of an incline that's 4 meters tall and we let it roll without slipping to the bottom of the incline and again we ask the question how fast is the center of mass of this cylinder going to be going when it reaches the bottom of the incline well it's the same problem it looks different from the other problem but conceptually and mathematically it's the same calculation this thing started off with potential energy MGH and it turned into conservation of energy says that that had to turn into rotational kinetic energy and translational kinetic energy again if it's a cylinder the moment of inertia is one-half mr squared and if it's rolling without slipping again we can replace Omega with V over R since that relationship holds for something that's rotating without slipping the MS cancel as well we get the same calculation this cylinder again is going to be going 7.2 3 meters per second the center of mass is going to be traveling that fast when it rolls down a ramp that was 4 meters tall so recapping even though the speed of the center of mass of an object is not necessarily proportional to the angular velocity of that object if the object is rotating or rolling without slipping this relationship is true and it allows you to turn equations that would have had two unknowns in them in two equations that have only one unknown which then lets you solve for the speed of the center of mass of the object