 Motion along a straight line. Motion about an axis of rotation. These two things can happen simultaneously. Molecules can both rotate and translate move along a straight line. The wheels of a car or a bicycle can do the same thing. How do we understand these motions? And how do we describe the energy that's present when both of these things can happen? Not only this, but what directional quantity can we associate with rotational motion? After all, there was a directional quantity of motion for linear motion, momentum. Is there an equivalent for rotational motion? We'll begin to explore these questions in this lecture. We're going to take a look at the final concept that we have to revisit now in the context of rotational motion, and that is angular momentum. But before we do that, we're going to take a look at a motion that combines both linear motion and rotational motion in one situation. The key ideas that we will explore in this section of the course are as follows. We will explore rolling as an application of angular ideas, combining old ideas about linear motion with the new ideas as they connect to rotational motion. We will then revisit the concept of momentum, but now in the context of rotational motion. And we will also revisit the conservation of momentum in rotational situations, and from this we will draw some conclusions about rules that we can use to understand the universe. To begin, let's talk about a motion that combines both translation, that is motion along a line, and rotation, that is motion about an axis. This motion is known as rolling. So what is rolling? Well, let's consider something like a wheel, for instance, a bicycle wheel. It is a ring-shaped object, much of its mass is concentrated on a circumference away from the axis of rotation, which is the hub or center of the wheel. It has a radius r, for instance. Now, a wheel usually makes contact with the ground. And so the ground touches a point on the wheel that is a distance r, the radius of the wheel, away from the center of mass. The wheel is designed in such a way that it has a uniform distribution of mass about its center, and so one would expect its center of mass to be located dead center, and this is usually where the axis, that is, the place where the wheel is affixed to the rest of the bicycle is attached. Now, at the point of contact, the wheel might have a material like rubber, for instance, that makes up its outer edge, so we expect that there's going to be friction between the wheel and the ground. And in fact, the only way to make a wheel work the way we understand wheels is to make sure that there is contact friction between the wheel's outer edge and the surface on which it is resting. So we have a uniform distribution of mass in a ring shape. That's the wheel. Its center of mass is located at its physical center, but it makes contact with the ground on its edge, and that would be a distance r, the radius of the wheel, away from the axis of rotation. So for instance, as you now push on the axis of rotation of the wheel, the center of the wheel, if it's free to rotate about that line, then friction will act to oppose your push. But you're pushing on the center of mass of the wheel, and friction is acting on the edge of the wheel. So while friction is opposing your attempt to move the object, this creates a torque, because the force of friction is applied not at the axis of rotation, but away from the axis of rotation on the edge of the wheel. So static friction generates a torque. Torque is a force that causes a rotation, and this sets the wheel into rotating. So this is rolling, and to roll without slipping means that the wheel is never pushed so hard that that point of contact overcomes static friction and begins to slide. In English, there's a colloquial term for when your wheel spins so fast that the point of contact exceeds the static friction force, and you get kinetic friction, and this is known as peeling out. And it makes for a very dramatic video, because when you suddenly start to slide with the contact point between the wheel and the ground, this can generate a tremendous amount of thermal energy which can actually melt the tire rubber. This is not a good situation to be in. You never want to be in a situation where you have a wheel that's normally functioning, but then suddenly is put in a situation where static friction becomes kinetic friction. This usually leads to accidents and other problems. So we're going to talk about rolling in the context of rolling without slipping, which means that the object that can rotate and move linearly is never pushed so hard that it slides at the point of contact with the ground. Now this interplay of a force that pushes on the wheel in a specific direction, so for instance a force applied to the center of mass of the wheel, its axis of rotation, combined with static friction generating a torque that then sets the wheel into a rotational motion is what is meant more precisely by a rolling motion. Now in rolling without slipping, which is what I'm going to refer to henceforth, the wheel translates. The center of mass moves along a straight line, but the wheel itself also rotates. The edge of the wheel cycles around the center of mass, which is where the axis of rotation is located. And if you study this motion extremely closely, rolling without slipping, what you find is that the wheel, the center of mass of the wheel will translate, it will move along a straight line, a distance that is equal to the amount of circumference that the wheel rotates through on the outer edge. So in other words, if we translate the center of the wheel by a distance delta x, then we will find that the edge of the wheel moves through an arc length of delta s that's exactly equal to the translation distance of the center of the wheel. Again, this only is true when you roll without slipping. Now what's nice about this is with this condition and with the relationship between arc length and angle in radians, the angle that the bit of arc length delta s subtends, we can write that this delta x is also equal to the radius of the wheel times the angular displacement delta theta that the wheel experiences when its center of mass translates a distance delta x. So this allows us to relate the linear motion, for instance, of a bicycle wheel, so moving forward 10 meters, with its angular motion and what relates them is the radius of the wheel. So if you know the radius of a wheel and you know how far the wheel's center of mass moves forward, then you can also tell me what its angular displacement is and this is great because it lets us understand this kind of motion very carefully. Now, since we can relate rotation and translation in this special kind of rolling motion, rolling without slipping, it gives us the ability to do this analysis I mentioned in a great amount of detail and then what's particularly interesting about this is that we have an object with a certain mass, m, and it can roll, but it can also translate. It can rotate, it can translate. So that means it possesses two kinds of energy of motion. It has linear kinetic energy associated with the motion of its center of mass at a certain speed, which I've denoted as V with a subscript COM. So it's going to have linear kinetic energy and it's also going to have rotational kinetic energy and that rotational kinetic energy will be given by the product of the moment of inertia of the wheel about its center of mass, times its angular speed squared, times a half. So we have a total energy of motion, a total kinetic energy that is the sum of two kinds of kinetic energies. The translation kinetic energy of the center of mass of the rolling object and the rotational kinetic energy. So what's interesting about this is that means that to get a wheel started rolling, you have to take whatever energy you have available and split it into two buckets. One bucket is the translational kinetic energy and the other bucket is the rotational kinetic energy and they're related to each other by this fact that the translation distance of the center of mass is related to the angular displacement of the wheel. So there's all kinds of interesting relationships that enter into this specific kind of motion that allows you to connect the linear velocity of the center of mass to the angular speed of the wheel as it rotates. Now, let's consider a special case and this will help us to understand how this motion can be quite complex but nonetheless, describable using these concepts. So for instance, let's consider as an example of rolling a round object that can roll down an inclined surface. We've looked at inclined surfaces before. This is just a flat surface that has been lifted up at one end through an angle theta so that gravity when it acts on anything on the surface is not generating a weight that points exactly perpendicular to the surface. It points at an angle theta, in fact, relative to the surface of incline. So gravity will exert a force on the center of mass. This is the weight that we have been studying as a force for quite a while in this course now. And that means that gravity provides both a normal force. The weight is perpendicular to the inclined surface and if you look back at problems where we've attacked this question before we see that the component of weight perpendicular to the inclined surface is given by the mass of the object times the acceleration due to gravity times the cosine of the inclined angle. Now, the weight also generates another force. It generates a component of force that's parallel to the incline moving and translating, moving in linear motion along the surface of the incline. That parallel component, which I've denoted here W with two parallel lines is the mass of the object times the acceleration due to gravity times the sine of the incline angle theta. But now that we have an object that can rotate and translate, that is roll, gravity has to do work and that work is split into these two pots, rotation and translation. And so we have two activities now to divide the same energy between. So, for instance, we can answer questions like what will actually be the linear acceleration of the center of mass as a result. It won't just be G sin theta. G sin theta, if the object was sliding without friction, that would be its acceleration parallel to the incline. But now that gravity has to put work both into rotating and translating the object, we would expect that the acceleration will be reduced by some amount because some of the energy has to go into rotational kinetic energy, not just linear kinetic energy anymore. So we already have some intuition here. There's a fixed amount of energy from gravity doing work, but now we have to split that energy into two places and we're just not going to get as much linear acceleration as we used to when an object could, for instance, slide without friction. Well, let's attack this problem using Newton's second law. Newton's second law tells us that the sum of the forces will be equal to the mass times the acceleration of this rolling object, but what acceleration are we talking about here? Well, the sum of the forces, what are the forces acting on this? Well, we've got friction at the point of contact of the object and the incline surface. And we have the force of gravity parallel to the incline surface that's acting at the center of mass of the object. So as I've said before with wheels this situation is no different. We have two forces. They act in opposite directions. One is pushing on the center of mass of the object. The other one is pushing on the edge of the object. So they're attempting to cancel each other out, but the result of this is that because one of the forces is acting far from the center of mass, it generates a torque and this induces rolling. So we have the static friction force at the point of contact that's acting up the incline so I'll call that the positive direction and we have the component of the weight parallel to the surface acting at the center of mass of the object that's pointing down the incline which I'll denote as the negative direction and so these two get deducted from each other but gravity may win out over the static friction force and in fact in the case of an object that can roll it will. And the result of this is that acceleration of the center of mass times the mass of the object. Well, let's go further. This doesn't look very useful so far. So let's take information we know about forces applied on the edges of objects that can rotate about a center and see if we can start whittling this down to a useful formula. Well, I've rewritten the Newton's second law equation up here. Again, we have the friction force acting up the incline trying to oppose the motion of the rolling object. We have the component of the object's weight acting at its center of mass pointing down the incline. These are the two forces that are trying to oppose each other and the net effect of this will be to tell us what the mass times the acceleration of the center of mass will be. Now the friction force acts at a point in such a way that it's exactly tangent to the wheel. So if we draw a line from the center of mass of the wheel out to where the friction force acts, the friction force acts perpendicular to that incline. So it's tangent to the wheel. This will generate a torque and the magnitude of that torque, due to friction tau sub f, is given by the friction force times the distance from the center of rotation. That's the radius of the wheel. So friction force times r. Now, this in turn because it's a torque will generate an angular acceleration and that angular acceleration can be determined from Newton's second law for a rotating system. So writing that down we have the torque due to friction is the friction force times the radius of the wheel or the rolling object and this must be equal to the moment of inertia about the center of mass times the angular acceleration. Now, angular acceleration is not independent of linear acceleration. We know that the angular acceleration is related to the acceleration of a point on the edge of the rotating object tangent to, again, this line that goes from the center of mass out to the edge. So we can write alpha as the tangential acceleration divided by the radius of the object. Now, keep in mind that because this is an object whose center is accelerating one way, that the friction force is generating a torque that pushes the bottom of the wheel in the opposite direction that the tangential acceleration at the point of contact will wind up being opposite to the acceleration of the actual center of mass. So we have a relationship we can use to our advantage here. We can actually wedge in the acceleration of the center of mass into this torque equation. The tangential acceleration of the edge of the wheel which points up the incline is opposite the center of mass's acceleration, which points down the incline. And so finally we obtain that the force of friction, solving these series of equations up here, is equal to the negative of the moment of inertia about the center of mass, times the acceleration of the center of mass, divided by the radius of this rolling object squared. So we have the force of friction in terms of moment of inertia, acceleration of the center of mass, the radius of the wheel. We can plug that back into the original Newton's second law equation and now we basically have the mass of the wheel, the moment of inertia of the wheel, the radius of the wheel, the acceleration of the center of mass winds up being the only unknown. So we can now finally figure out what the acceleration of the center of mass is going to be. So substituting that back into our original Newton's second law equation, we get something that doesn't look very pleasant here. I've just substituted in for the force of friction. Here I still have the component of the weight parallel to the incline surface, and again I have the total mass times acceleration of the center of mass. We have acceleration of the center of mass in two places, so we can do some algebra. I'm going to skip that here, but ultimately what this algebra allows us to do is to solve for an equation that tells us about the motion of the center of mass, and what we find is that the acceleration of the center of mass of this rolling object is given by the negative of g sin theta divided by this term. So if this object couldn't roll, that is if it didn't have a moment of inertia, moment of inertia was zero for instance, then in that case we would recover the old relationship between the acceleration the object would experience, and the angle of the incline. We would find that the acceleration of the center of mass would just be related to g sin theta. But in this case we have an object that can roll because of the torque exerted by friction on the edge of the object, and so this acceleration g sin theta is modified by this denominator. So we see that this number, this denominator can only ever be bigger than one, and so we're always dividing by a number bigger than one. We're always taking g sin theta and we're dividing it by a number that's one or greater, and that means that we get less of an acceleration of the center of mass when rolling is present than we would if an object could just slide. So we see here how this motion can get quite complicated, but nonetheless it admits itself to an analysis based on rotation and linear motion, translation. With this background having looked at an application of both linear motion and rotational motion in one scenario, let's turn our attention to the last concept in linear motion that we encountered, momentum, but now try to place it in a rotational context. So we've focused on the kinetic energy of rotation. We've taken a look at how if we want to rotate an object, and we put a certain amount of work into rotating that object, that if that object can roll some of the energy goes into linear motion and some of the energy goes into rotational motion. But what about a directional quantity of motion? After all, linear motion had a directional quantity momentum. It had a scalar quantity, kinetic energy, and it had a directional quantity, momentum. Is there an equivalent directional quantity for rotational motion? Indeed, we know that all the points on a rotating body rotate, and so there must be some equivalent for rotational motion to momentum. After all, in a rigid rotating structure there's clearly some sense of directional motion, right? If we use our right-hand rule to figure out where the angular velocity points, there is a sense that going counterclockwise is different from going clockwise, that the angular velocity flips direction. So we should be able to use these sort of intuitions to now think a little bit more about what it means to define momentum in the context of a rotating situation. So let's consider a single point of mass m, and I've denoted that here in the picture by this red dot. And it's going in circular motion around some axis of rotation, some center of motion. So that path that it's taking as it goes around the center of motion is indicated by the blue circle with the arrows. And it's a counterclockwise motion, so it has a positive angular velocity. Now it's doing this rotational activity a distance r from the center, and I can denote this distance by an actual vector in space. It starts on the center of rotation, the axis of rotation, it points out to where the mass m is at any given moment in the motion. So at the moment I have frozen this motion in this picture, it points down into the left from the center out to the mass m, the red dot. Our vector then is the vector that describes the length and direction of this green arrow that I have drawn here. Now at a given moment in time this rotating object has linear velocity, v, that's tangent to the circular path. And that means it has linear momentum, p equals m times v. And that linear momentum will also, if this was exactly circular motion, be tangent to the circular path. Now of course it might be possible that in a rotational motion a point on a rotating body has a component of momentum that points parallel to the radius that goes back to the center of motion. But that component won't contribute to the rotational motion, it will cause some translation of the object, but it won't cause more rotation of the object. It's only the momentum perpendicular to the r vector, that is tangent to the rotational motion that actually does anything involving rotation itself. But we can imagine that there's an analogy here. Torque was a force applied perpendicular to this radial line causing a rotation to begin. We can imagine defining angular momentum in a way that similarly we saw that torque was defined. Torque was vector r cross vector f. So angular momentum could be written down as vector r cross vector p, the momentum vector. And since we have a point mass we can write explicitly what the momentum is for this point. It's the mass of the point times its velocity. Mass is a scalar quantity, it's just a number so I can pull it out in front of all this vector multiplication that's happening here. I can't change the order of the vector multiplication without consequence. If I swap the order of the vector multiplication I pick up an extra minus sign, go back and look at the cross product if that doesn't feel familiar to you at this point. As long as I preserve the order of the cross product I'm okay and I really have to make sure that if something is in the order of A cross B I leave it in the order A cross B as I'm playing around with an equation. However, numbers like mass which are just scalars, they're directionless, they can be moved around anywhere at no consequence to the vector multiplication that's going on. So I pulled mass out in front and then I have this quantity r cross v that's left over. Okay, so I have a definition here of angular momentum. It's just a vector and it's cross product with the vector that goes from the axis of rotation out to where the mass is sitting at a given moment in its rotational motion. And again, I'm considering a point mass so I was able to write this as m times the quantity r cross v. Let's look at the magnitude and the direction of this quantity L, angular momentum, L vector. Well, the magnitude of angular momentum will be the same as the magnitude of any cross product. So if I want to know the magnitude of A cross B, it's just going to be the length of A times the length of B times the sign of the angle between them, theta. So if we imagine that there's indeed an angle theta between the vector r and the vector p, then it must be true that the magnitude of angular momentum is r times p times the sign of the angle theta that lies between the two vectors. Now, again, p sine theta, that's just the component of momentum that's perpendicular to r. So I could write this in this notation again, where we say that the angular momentum is just the magnitude of the radial line out to the mass m times the component of that momentum of mass m that's perpendicular to r in the first place. That's the only thing that contributes to the angular momentum because of this cross product definition for the magnitude. What about the direction? Well, the direction will be given by the right hand rule for a cross product. So what do we do? We get out our right hand, we take our index finger, that's the finger closest to the thumb on your right hand, you point it in the direction of r, that's going to go from the center of motion, the axis of rotation out toward the mass m. You then reorient your right hand so that your middle finger can point comfortably in the direction of the momentum vector itself. And then you stick out your thumb perpendicular to both your index and middle fingers. And where your thumb points indicates the direction of L vector, the angular momentum vector. So it's, again, very similar to figuring out the direction of a torque, given a force that's acting a distance r away from a center of rotation. It's the same activity because it's just a cross product and the rules for figuring out the directions of a cross product don't change. So we've been talking about a point mass m that's circulating around in some rotational motion about some axis, some center of rotation, but what about a rigid rotating body that's made from many such little points of mass? Well, in that case, it's just like what we did when we summed up all the kinetic energies of a rigid, extended rotating body, we have to add up all the momenta of all the points in the body. Well, momentum is a vector sum, so now we have to do a vector sum of all the individual angular momenta of all the little masses, m with a subscript i, where i is the index that labels all the masses, one up to n, and could be Avogadro's number, could be a big number, could be a small number, could be three or four masses, could be Avogadro's number worth of masses, but it doesn't matter, this is a generic principle. So if we want the total angular momentum of this system of masses, we take the angular momentum of one of them and add it to the angular momentum of the second of them in a vectorial body, and so on, all the way up to the last mass, the nth mass. So we can write this sum more compactly as the sum from i equals one to n of the individual angular momenta l sub i for all the little points of mass m, i. Now, we can then explicitly put the cross product into this, the original definition of the angular momentum, r cross p, for each of those little masses, m sub i. So let's do that. It's going to seem at first to make this more complicated, but what we're going to find out is that very quickly we're going to reduce this seemingly complex sum down to a fairly nice compact definition. So even having a really complicated rotating body, if we just know a couple of things about it, we can very quickly solve for the angular momentum. So the total angular momentum is the sum of all the individual angular momenta. Each of those can be written as the cross product of r i vector. That's the distance from the center of rotation out to the mass point m i and the momentum of that mass point. So p sub i vector. But again, the momentum of that mass point is its mass m sub i times its velocity v sub i vector. So let's put that in. Now we can write this term in the sum as this quantity here. Just like I did on the last couple of slides for a single point mass. Nothing really exciting is going on here yet. But it's the next step where things get more interesting. Okay, so here again is that sum. The total angular momentum of all these n masses is written out this way with the cross product of the mass of each point. Now what I'm going to do is I'm going to do something that harkens back to the original definition of a vector. A vector is a quantity with length and direction. So for instance the vector r has a length r. That's the radius of the circular motion for instance that these points might be executing. But it also has a direction. That direction points from the center of rotation out to where the mass point we're looking at is located. So it has a direction. That direction can be written using a unit vector that simply indicates the direction I just stated, r hat. So I can replace r vector and v vector with the product of their magnitudes and their unit vectors that indicate the direction where they point. So let's do that. So if I do that I pull the lengths of these vectors out in front. They don't affect the cross product but I've left the cross product of these unit vectors intact. The order is preserved. The unit vector for r precedes the unit vector for v in this cross product. I have preserved the order of the multiplication. Nothing has changed. All I've done is pull the magnitudes of r and v out and put them next to the mass. Now again as we did with kinetic energy recall that all points on a rigid rotating structure rotate with the same angular speed. That hasn't changed. That principle still holds true here. So that means they all move at some angular speed omega and it will generically be true that for every point on this rigid rotating body the quotient of v i and r i the ratio of those two things will give you omega. Doesn't matter which mass point you're looking at. They all rotate in the same sense. They all go through the same angular displacement in the same amount of time and so they all move at the same angular speed. Well awesome I can replace v in this equation with omega. V i is omega times r i and if I do that I get this the total angular momentum of this system of masses is the sum of these terms where each term is the mass times the radius squared the distance from the center of rotation to where that mass is located times the angular speed of that mass but that's the same for all masses and then I have this this unit vector cross product. Let's take a look at that and see if we can figure out what's going on there. I mean at first it seems like we have a whole bunch of these cross products that we have to do and maybe they're all different for all the Avogadro's number worth of masses but because this is a rigid rotating structure and all the points are rotating in the same angular displacement in the same time something very interesting also applies to this cross product. So do the right hand rule draw a circle and pick a dot on the circle and let's pretend the circle that whole bunch of points that make up the circle are all rotating counterclockwise so use the right hand rule for angular velocity to figure out which way the angular velocity for a single point is pointing so curl your fingers in the direction of the sense of rotation of a given mass point your thumb sticks out and indicates the direction of the angular velocity. Pick another point it too is rotating counterclockwise at an angular speed of omega so it too will have your thumb pointing in the same direction giving you the direction of r cross v what we find out is that this cross product r i hat cross v i hat is the same for every single point on the rigid rotating structure and so as a result of that we can identify this cross product r i hat cross v i hat as being the unit vector that points in the direction of the angular velocity omega hat it's just the unit vector that points in the direction of the angular velocity of this rigid rotating body so let's substitute that in and see where we've gotten ourselves replacing that cross product now just with the unit vector omega hat that indicates the direction of that cross product which is the direction that the angular velocity points we can suddenly make a whole bunch of identifications in this equation I've got the sum over a whole bunch of point masses of m i r that looks suspiciously like the moment of inertia of a point mass about a center of motion and then every single one of them has a term omega times omega hat in it which is constant to every single term in the sum omega times omega hat is just the vector omega it's its length times its direction that is exactly the definition of the vector it's magnitude times its direction that's a vector so we just have omega vector in every term in the sum and then this other thing the sum is the moment of inertia of the rotating body it's the sum over all points of the quantity m i r i squared so thus in compact notation we can actually write the angular momentum of an extended rigid rotating body in a very simple way the angular momentum of a rigid rotating body is just the product of its moment of inertia and its angular velocity omega vector that's it the product of these two things tells us everything about the angular momentum the direction dependent quantity of motion associated with this complex extended rigid rotating structure this equation applies to a wheel that's rotating it applies to the fan blades of a wind turbine it applies to a whole planet going around in circular motion this equation can be universally applied to any rigid rotating body and what's interesting about this is we see a kind of again symmetry emerge between linear quantities and rotational quantities linear momentum was the product of mass and velocity angular momentum is the product of an inertial quantity the moment of inertia and a speed related quantity angular velocity again these are quantities in an angular context so p equals m v l equals i omega the symmetry is beautiful between these things now it's worth noting that we have a unit here for angular momentum the meter kilogram second or standard international unit of angular momentum is the joule second it's the product of energy and time so you can write this as j times s joule seconds or you can write it as kilogram meter squared per second that's what it would be in actual kilograms meters and seconds it's just energy times time now with this new concept of angular momentum in mind let's step back and ask a question what happens if we do the following what is required for angular momentum to vary in time not a very exciting question I've defined this quantity angular momentum what does it mean for it to be varying changing value in time well let's consider a simple picture again of a point mass moving in rotational motion about an axis a single point mass well in that case we can dig back and we can just write down the exact expression for the angular momentum of this point mass going around some center rotation some axis of rotation in that case the angular momentum of this mass is its mass times the cross product of r vector and v vector let's try to answer the question what does it mean for angular momentum to vary in time and to answer questions like that we have to apply calculus we want to know how thing one varies with thing two and that is exactly the question that the first derivative helps us to answer so let's take the first derivative of angular momentum with respect to time well mass is a constant it's not changing and I'm going to hold to that for this exercise so mass is a constant it's not affected by the derivative and when I substitute in for L with m r cross v I can pull m out to the left of the derivative doesn't affect the derivative however I should be careful and not assume that the radial distance isn't changing it could be maybe the object is getting further away from the center of motion over time that's one way you could change angular momentum or maybe it's speed is changing maybe the direction of the speed is changing or maybe the magnitude of the speed is changing so I could have changes in v and I could have changes in r and I've got the product of r and v here well that feels like I need to do the product rule for the time derivative so let's do that let's do the product rule for the time derivative of a product of two functions of time r could be a function of time v could be a function of time the cross product is just another kind of multiplication albeit a special one with an order dependence that matters so I'm going to keep the order of the cross product the same but I'm going to distribute the derivative to the two parts of the product in the way that you're supposed to do this in first semester calculus so I have my mass out in front and then I have the sum of two things r cross the time derivative of v and the time derivative of r cross v that's the application of the product rule and again the cross product is just a multiplication of two things and those things could depend both on time and so even if I have to apply the time derivative to a cross product of vectors the product rule for distributing the derivative is still in effect it holds true you just use it so that's what I did here okay so I've rewritten that expression here and it's not very pretty looking but let's see if we can make some sense of this let's dig way back way back to linear motion what is the time rate of change of velocity now the time rate of change of velocity is just our old friend acceleration what's the time rate of change of position well the time rate of change of position is our old friend velocity so dv dt is just acceleration and dr dt is just velocity it's the velocity vector so let's substitute those in if angular momentum is changing with respect to time that change is given by mass times this thing in parentheses r cross acceleration and v cross v added together ooh that second term looks interesting remember the rules of the cross product the cross product of a vector with itself or any vector that happens to point in exactly the same direction is zero the cross product answers the question what vector is perpendicular to the two vectors you've given me when you have a vector and itself there's no one direction that is exactly an only perpendicular to that pair of vectors there's an infinite number of them so the cross product returns zero the cross product of a vector with itself is zero remember the rules of the unit vectors in the cross product i cross i equals j cross j equals k cross k equals zero so it must be true that v cross v is also zero and so one of these terms just goes away and we're finally left with this if angular momentum is changing in time that quantity dL dt must be equal to mass times r cross acceleration alright well let's rewrite that equation here and let's consider the magnitude of the above vector alright so if we do that what's the magnitude of this vector well the magnitude of the left side is just the time derivative of the magnitude of the angular momentum l the right hand side however we have a cross product the magnitude of the cross product is r a sine theta where theta is the angle between them and a sine theta is just the component of the acceleration that's tangent to the circular path perpendicular to r so we can write this more compactly as dL dt is m times r times the tangential acceleration again only acceleration tangent to the circular path that is perpendicular to the vector r contributes to the cross product anything that's parallel or anti-parallel to r doesn't contribute to the cross product digging back a little bit here recall that the tangential acceleration is related to the angular acceleration we looked at this in rolling earlier in this video alpha the angular acceleration is the tangential acceleration divided by the radius of the rotational motion around the center of motion the axis of rotation so we can substitute in we can eliminate the tangential acceleration and we can put in alpha and r and if we do that we found out that the time rate of change of angular momentum is equal to mr squared alpha this is a point mass moving in rotational motion about a center of rotation an axis of rotation mr squared is its moment of inertia mr squared is the moment of inertia of a point of mass m distance r from a center of rotation going in rotational motion so we can substitute this equation with one more piece recognizing that this mr squared is the moment of inertia we then get the following equation the change in angular momentum with respect to time must be equal to the moment of inertia times the angular acceleration of this mass well what is i times alpha i times alpha is the angular equivalent of mass times acceleration and the angular equivalent of forces is torque so in other words angular momentum can only change in time when a torque is present a torque generates a change in angular momentum in a certain amount of time just like a force generates a change in linear momentum dp in a certain amount of time dt so f forces can be written as changes in angular momentum with respect to time torques can be written as changes in angular momentum with respect to time see the analogy has a beautiful symmetry between these things it's really quite amazing that this is even possible but what this also tells us is if there are no net external torques on a system that is if you have a closed and isolated system external forces can't reach in and accelerate components of the system then we have to draw the following conclusion that the time rate of change of the angular momentum of that system is zero for a closed and isolated system that is to say in a closed and isolated system angular momentum is conserved well this is nice total energy is conserved in a closed and isolated system total momentum in a linear sense is conserved in a closed and isolated system and now we find out that total angular momentum is conserved in a closed and isolated system where no external forces can generate external torques that can affect the angular momentum of the system so if in a closed and isolated system it is true that the time rate of change of angular momentum is zero that is the initial momentum in an angular sense is equal to the final momentum in an angular sense then this has interesting consequences for physical systems that are closed and isolated from external entities first of all if a body is rotating and only forces internal to the closed system can act then angular momentum must remain fixed for the entire system even if it's parts change their orientations for instance by breaking apart or just moving around so that's an interesting consequence of this this is akin to the idea that because in a system where no external forces are acting momentum must be conserved that if that system explodes or breaks apart in some way it will be true that the momentum before the explosion will be equal to the momentum after the explosion and this lets you do all kinds of interesting analysis of the system now since angular momentum is equal to the moment of inertia times the angular speed if a closed and isolated rotating system alters its moment of inertia then angular speed must change an interesting demonstration of this consequence is as follows take a chair that can rotate and sit on it so you can spin yourself around and you have an axis of rotation that goes sort of from under your seat up through your head you can rotate around that axis on the stool hold two masses a certain distance away from your axis of rotation make sure they're really heavy start yourself spinning and then take your feet off the ground so now no external forces can really act on you yes there's friction but if it's a good enough stool it won't matter draw the masses closer to your chest you have now decreased your moment of inertia if you decrease your moment of inertia your angular speed must increase pull the masses further away from the axis of rotation increasing your moment of inertia and your angular speed must decrease draw them in closer to the axis of rotation your speed increases push them out further from the axis of rotation your speed decreases this is a nauseating demonstration and you have to be extremely careful when doing it but nonetheless it's a very quick and effective demonstration of the conservation of angular momentum and its consequences now conversely such a rotating system alters its angular speed in some way for instance by using chemical energy to suddenly speed itself up moment of inertia must change in response you can't alter one without altering the other so in a system where angular momentum is conserved changes in things like angular speed have consequences for the moment of inertia and vice-versa now here's an application of the conservation of angular momentum to a large cosmic phenomena and that is the laws of planetary motion I hinted at these in the lecture on gravitation the person who's credited with establishing the three laws of planetary motion is the astronomer, mathematician and physicist Johannes Kepler and as I mentioned in the earlier lecture on gravitation he used an extensive observational data set to establish these laws of planetary motion the first law is that the shapes of the orbits of planets for instance about the Sun are not circles as was believed but rather ellipses these are elongated circles with a long axis and a short axis the second law is that a planet in orbit around another body like a star like our Sun will sweep out as it moves equal areas and equal times we'll look more closely at what I mean by that in a moment this sweeping out of equal areas and equal times is independent of when in the motion you observe this sweeping effect to occur and then finally the square of the period of any planets orbital oscillatory motion as it cycles around its say parent star will be proportional to the cube of the long axis the so-called semi-major axis of its elliptical orbit so that is to say the period squared of a planet orbiting around a star will be proportional to the length of its semi-major axis cubed now these are empirical laws that Kepler established by studying extremely detailed samples of data on planetary motion but at the time he discovered these laws he didn't have the benefit of Isaac Newton's work he didn't have the benefit of the law of gravitation or the laws of mechanics and so he could not know the reasons for these planetary laws they were clear from the data but the reason they existed was not clear but now we understand their origin its conservation of momentum and angular momentum and the law of gravitation combined together this establishes where the three laws of planetary motion come from as an example of the application of the conservation of angular momentum let's look at the origin of the second law of planetary motion this picture will help you to understand a little bit better what I meant a moment ago in equal times so imagine we follow a planet for instance Mars Mars's orbit could be represented by this exaggerated elliptical shape shown here in purple so this is the planet labeled planet 1 in this drawing now if we let the planet complete one orbit this for instance for Earth would be one year we would see that the planet would start here and then after one cycle about the Sun it would return to that point and we would call that a year on that planet that is one orbit around the parent star now what do I mean by sweeping out equal areas in equal times well let's imagine that we watch the planet from this point to this point shown here so it starts off here over a little higher and a little above into the left on the ellipse and then after some time T1 ends up down here where it is currently indicated in the picture what you would find out is that the area swept out in that time T1 would be A1 okay fine well we wait a little bit and then we wait until the planet gets to this point over here on the ellipse and then again we time it the same time T1 and this time it moves over here in that time T1 so it sweeps out this grey wedge A2 with a seemingly different area so we've observed the planet moving along its elliptical orbit in two times they're equal times both T1 and we see this grey wedge that represents area 1 and we see this grey wedge that represents area 2 and what Kepler discovered is that these areas are the same if you watch the area swept out by planetary motion in equal times they're the same areas now this one is sort of shorter and wider and this one is longer and thinner but the areas wind up being exactly the same this was what having a tremendous data set made Kepler able to do these are sort of wedge shaped things with curved sections at the end they look a lot like slices of pizza with straight line edges and then a curved crust so let's consider a wedge of a planet's orbit so for instance this purple planet planet 1 which you know could be Mars and it's going around the sun and it's in different times say you know T1 here and T1 here it's traveling along the curve of its ellipse now let's consider a very short time DT over which it moves a very short length ds now this ds is so slight that it basically just forms a triangle two sides of the triangle or the radius are of planetary motion and the third side is the arc length s the area of the triangle that represents this wedge that swept out in this very short time DT will be one half times the base times the height of the triangle well the base is the arc length s and the height is the radius are okay so fine during that little time DT we're in an orbital orbital radius are we've swept out in a triangle with area one half s are now since ds is a little bit of arc length it's related to the angle that's subtended by that arc d theta and that's according to our old friend the definition of the radian ds equals r d theta so now we can write the change in the area as a function of time and then think about what it means to change area in some amount of time okay so the change in the area is a function of time sounds like we need calculus again so let's do that let's take the first derivative with respect to time of the area of this triangle well that's the first derivative with respect to time of one half times the little bit of arc length times r okay well I can substitute with r theta for s s equals r theta so I substitute that into this equation I have the first derivative with respect to time of one half r squared theta well one half is a constant and r is a constant over these short times the radius of the orbit is not changing appreciably over very short times but theta on the other hand is not zero theta has changed over time so we pull out the one half we pull out the r squared and we're left with the first derivative of the angular position with respect to time d theta dt what is the change in angular position with respect to time that's just the angular speed so the change in area as a function of time for a planet moving in this elliptical shape is just one half times r squared times omega again over a very short period of time okay well if we treat the planet as a point of mass m and on the scale of a solar system planets are really not that big so we can get away with this approximation I mean Mars is not very big compared to say the whole solar system or certainly to its orbit so we can treat it as a point with a concentration of mass m at that point and we know because it's going in rotational motion around some center that it's got angular momentum and that angular momentum is given by the product of its moment of inertia and its angular speed well if it's a point that's just m r squared omega we had an r squared omega in our equation for the change in area with respect to time so we can make a substitution we can divide both sides of this equation by the mass we can replace r squared omega with l the angular momentum divided by m the mass of the planet so let's do that so now we have that the time rate of change of the area swept out by the planet is one half l over m so we have this thing that you naively might think would be different different areas swept out in the same amount of time in different places in the orbit but in fact the change in area swept out by a planet as it moves around in its orbital path its elliptical path around a star for instance is equal to a constant because planets are closed in isolated systems there's no friction there's no drag there's just gravity and gravity is holding those planets in orbit around the parent star so we have one half times the angular momentum divided by the mass of the planet with the mass of the planet if that's not changing then da dt is not changing if the angular momentum of the planet is not changing then da dt is not changing and so the reason Kepler's second law of planetary motion is true that planets sweep out equal areas in equal time regardless of where you look at them in their orbital path is because the angular momentum is constant in a closed and isolated system which for the most part a planetary system is it's largely free of external forces because solar systems are so far apart from one another and so it must be true that in a planetary system like our own da dt must be constant that is planets will sweep out equal areas in equal times regardless of where they are in their orbital motion it's a little counterintuitive but it's true because angular momentum is conserved in a closed and isolated system so you can begin to see how many regular phenomena in the universe could arise from the application of simple smaller sets of rules and maybe you're beginning to see now what motivates the physicist what motivates physicists is to try to find all those rules that underlie the structure of the cosmos be it material science plasma physics gravitational wave astronomy particle physics my area of study or many others we see regularity in the universe it has underlying causes we've come to understand many but not all of those we want to understand more and that is ultimately what drives the physicist so let's step back and take a big picture look at conservation laws because we've essentially completed a set now we have a set of conservation laws that if we can apply them or at least understand why they're not holding allows us to analyze even seemingly complex systems now that analysis may be very difficult but nonetheless the underlying equations are quite elegant and simple so in any closed and isolated system where either translation linear motion and or rotation are possible then the following statements will be true based on everything we've studied so far in this course the total energy of that system the sum of its linear kinetic potential energy from conservative forces u potential rotational energy rotational kinetic energy internal energy such as chemical energy in the system that could be converted to other forms and thermal energy the energy of the jittering motion of atoms that won't change over time for a closed and isolated system so now where before we might have been hiding the rotational energy in this vague term internal energy we've now busted it out because we know how to write it down the kinetic energy of a rotating body so if you have an object that's spinning and moving along a straight line and doing so in a gravitational field but otherwise this system is completely closed and isolated I can tell you just about everything about the energy of that system and what its energy is going to be in the future it might change from one bucket to another but the total amount won't decline or increase the time rate of change of the total energy is zero so whatever total energy this system has at the beginning at some time zero will be the same total energy it has at some later time again it might get subdivided into different places some rotational energy might get slurped up and put away in thermal energy but none the less if I can account for all of that this statement will hold total energy is conserved in a closed and isolated system so that is one of our laws what's another one that we've looked at? well in a closed and isolated system free from external forces because force is equal to the change in linear momentum with respect to time if there are no external forces acting there is no change in the total linear momentum of a system and so dp total dt must be zero in which case the initial momentum of a system in total will be equal to the final momentum of a system in total now you might have to sum up the linear momentum but if you can do that this statement will hold total momentum will be conserved under these circumstances and finally we have our new law that we have just explored a little bit with the planetary example that is the conservation of angular momentum so if there is rotation possible in the system whatever the initial angular momentum is at the beginning of a system that becomes closed and isolated that angular momentum cannot change in total and so we have a situation where the time rate of change of the total angular momentum of the system will be zero the initial angular momentum in total will be equal to the final angular momentum in total total angular momentum is conserved these three principles together form an incredible foundation for studying the cosmos and perhaps even unraveling some of the seemingly most complicated puzzles that the universe can throw at us we have a basis for a physical analysis of the natural world so these principles combined with Newton's law of motion and the law of gravitation are really the five statements that lie at the heart of everything we have done this semester in this course if a system is closed and isolated all of the above apply we can use Newton's laws to analyze the details of the motion we can write down the different kinds of energy we can see how they might change from one to another if collisions are occurring orientation and a conservative force is changing so we have different potential energy we can do all that it might be complicated but we can do it if a system is not closed and isolated and we can recognize that that's the case then as long as we can account for the fact that energy might enter into the system or momentum might be taken out of the system that analysis can nonetheless continue if we can account for that in our analysis we still know something about what's going on and we can make some predictions about what might happen to that system in the future so let me conclude this lecture by reviewing the key ideas that we have explored in this section of the course we've looked at rolling as an application of angular ideas where both a linear and a rotational motion are possible and they're linked to each other rolling without sliding that is rolling where the point of contact always is acted on by a torque that's given by static friction we then revisited the concept of momentum but now in the context of rotational motion we revisited the conservation of momentum in rotational situations and we've accounted now for a number of conservation laws that can help us to understand the universe and in fact I would argue that you are now armed with five tools that would let you study in detail even complicated systems now it may require computers to do the analysis but if you can faithfully execute that analysis using these five ideas conservation of energy conservation of linear momentum conservation of angular momentum Newton's laws of motion and the law of gravitation in a situation where those are the only things that matter and there are other forces in nature but we're going to ignore them for right now and I argue that you have all the tools you need to pick apart what's actually going on in a situation and not only make predictions but maybe even make discoveries about what's actually happening in a system where there's order and regularity but no pre-existing explanation as to why that might be so taken together all of these ideas linear motion, rotational motion and all of these momentum and energy and force concepts are the basis of a foundation for analyzing all of nature or at least figuring out what we don't know already about the universe as a whole