We have a sense from our own experience that rotation carries tremendous energy and potentially force with it, but also that energy and force are required to get things moving in the first place. So how do we relate our concepts about motion to concepts of energy and force? It is in this lecture that we'll begin to explore this question in a little bit more detail. Now in the context of rotational motion, let's revisit the older concepts of energy and force in light of this special kind of motion that combines a whole bunch of tiny linear motions of a whole bunch of points on an object that is rotating. This is a much more complicated situation than we've seen before, but as we're going to find out, we'll be able to adapt our rotational concepts back into the context of energy and force and rediscover some old friendly equations, but now recast in the form of angular quantities. The key ideas that we will explore in this section of the course are as follows. We will begin to think about energy and force concepts as they relate to extended, rigid, rotating bodies. These are bodies, for instance like this saw blade pictured over on the right, where if you rotate the object through some angle, like one degree, all the points on this rigid body all rotate through one degree. We were going to find out that there are rotational analogs to linear concepts like kinetic energy. In fact, we're going to apply kinetic energy first and see that we can develop a rotational version of the kinetic energy equation that more easily summarizes the complex motions of all the little points in a rigid rotating body, again like this saw blade. We will come to understand how forces result in rotations, and we will explore the concept of a torque. And finally, we will connect the force and energy concepts through work, which is how they were connected in linear motion before, and we will complete a set of rotational analogs to linear energy and force concepts. So let me begin this lecture by asking you a question. Over here on the left, I have a large ventilation fan unit that can be powered by a motor and caused to spin. We'll get to the spinning part in a moment, but let me ask you this. I'm going to go ahead and throw that whole fan at you, at your hand, and I'm going to throw it at your hand at a linear speed of 170 kilometers per hour.   Is that okay? And I would hope that you probably replied, no, that is highly unsafe, or perhaps something a little less family friendly. Why would you say that? Well, because this giant vent fan, which probably has some large mass, would also then, when accelerated up to 170 kilometers an hour, have a very large, linear kinetic energy. It's a big object. It's moving pretty fast.  So if it strikes your hand, we know from looking at collisions that it's going to transfer some of that momentum that it has into your hand, either elastically or inelastically, but either is bad, and that collision is going to hurt a lot, a lot. And so of course you wouldn't agree to this. Well, fine. Let me ask you a different question then. So the vent fan has a uniform distribution of mass. There's just as much matter at the exact same distances from the center of rotation in this blade as there are in this blade, as there are in the blade you can't see here below the level of this ceiling. Okay, so if we think about the concept of center of mass and we ask for an object with a symmetric uniform distribution of material, where's the center of mass going to be? Well, it's going to be dead center. Dead center is right here on the axis of rotation of the fan blade. Okay. No problem. I'm going to take the fan blade. I'm going to use this motor and I'm going to spin the blades so that the tips of the blades are moving at 170 kilometers per hour. But the center of mass isn't going anywhere. In fact, it's the one part of this entire thing that's standing absolutely still because it's dead center right on the axis of rotation. Technically it's not moving at all. So it has no linear kinetic energy. I argue that the linear kinetic energy of this fan is zero. I'm going to stick your hand in the path of the blades. Is that okay? And again, I would hope that you replied something like, heck no. That is highly unsafe. Well, why? I mean, it's true that the center of mass has no linear kinetic energy. But those blades are moving and they are moving fast. And so there is clearly some kind of energy of motion of the rotating fan blades. That's got to be some kind of kinetic energy. And when any of those blades strikes your hand at that speed, it's going to do tremendous damage to your hand. So we obviously have to be able to quantify the kinetic energy associated with rotational motion. It's not zero. Let's see if we can figure out what that kinetic energy is. So let's consider the picture of the bicycle wheel that I showed you in the last lecture on rotational concepts, rotational quantities, like angular speed, angular acceleration, and angular displacement. We can imagine that this wheel is moving as somebody is cycling over the ground. So we have frozen this wheel at a moment in time when it's technically in motion, but we're analyzing the motion over a very small period of time, an infinitesimal period of time. So if we imagine that indeed it's spinning and that we've frozen it at some very tiny little unit of time, we can also then think about what is really going on. What's moving here? Well, the bike wheel, the spokes, all of this stuff are made from a huge number of tiny little bits of mass. And those little bits of mass are distributed at various radial distances from the center of rotation, which is down here on the axis that goes through the center of this wheel. OK, so we have a wheel that's made from a huge number of bits of mass. Each of those bits of mass has its own radius from the axis from the center of rotation. We can label each distance of a bit of mass, r with a subscript i, and each bit of mass can be m with a subscript i. Now, i is a label. It labels all n, for instance, maybe Avogadro's number, of little bits of mass. Let's pick one of the little bits of mass here. We'll pick this little bit of mass here on the spoke, this little blue dot I've just drawn. I'm going to label that one as i equals 1. That's the first bit of mass I'm going to consider. Now, we're thinking about this wheel rotating over a very infinitesimally short period of time. So it's not in zero motion. This little point has a little bit of linear velocity in that unit of time. And so for that little blue point with its little mass, m1, we can figure out what its little speed is, v1. And from that information, m1 and v1, we could compute its linear kinetic energy. So it's going to have a linear kinetic energy, k1, which is 1 half m1 v1 squared. This is just the kinetic energy that we have looked at before. And again, v1 is its linear speed as it goes in circular motion over this very tiny little chunk of time that we're considering the motion of the wheel in. Let's choose a second point. I'm going to pick this red point over here, 0.2. I'm going to label it as such, 0.2. And it, too, will have a kinetic energy, k2, which is 1 half m2 v2 squared. Now, just a comment here. The linear velocity of this blue point is going to be different from the linear velocity, even in the same moment of time as this red point. Why? Because even though this is a rigid body, and even though these points are all making one revolution in the same amount of time, the points on the outer edge of the circularly rotating object have to go over a much longer distance in that unit of time. So the linear velocity of the red point is bigger than the linear velocity of the blue point, even though they make one revolution around the central axis in the same amount of time. That's just the nature of circular motion. Now, we can keep considering the kinetic energy of other points. We can pick another point, like this cyan point out here at a slightly different radius. We can calculate its kinetic energy. We can pick this purple point over here. We can pick this yellow point over here. We can pick this green point down here. We can pick all the points that make up this wheel, and for each one of them, we can write down its kinetic energy given its linear speed and its mass. Well, this activity that we're going through here, figuring out the kinetic energy of each point, this is nice, but of course, we want to know the total energy of this rotating body, where all these little points make up the body. Well, if you want the total energy of a situation, add up the energies of the individual things that are happening in the situation. So we have a bunch of individual points, maybe Avogadro's number worth of them. Each of them has a kinetic energy. We have to add up all their kinetic energies. So to get the total kinetic energy, we're going to sum all the kinetic energies of all the individual points on this rigid rotating body. So the total kinetic energy of the wheel will be the sum of all the little kinetic energies of all its little pieces, and there are n of those, n could be a big number. So we can write that in a more compact notation as the sum from i equals one up to big n of all the little kinetic energies k labeled with the subscript i. And we can substitute in for that kinetic energy equation. Each kinetic energy is one half mi vi squared. Now, this might at first seem pretty hopeless. That's a lot of terms, and even every one of them has a different speed, potentially, that we would have to then factor in and go figure out. So this might at first seem pretty hopeless. But remember, and I hinted at this a moment ago, there's an important fact about points on a rigid rotating body. All points on a rigid rotating body go through the same angular displacement in the same amount of time. That is, they all have the exact same angular speed, omega. Angular speed is displacement in angle divided by time, and the displacement in angle in the same unit of time is the same for all points on a wheel, as long as that wheel is rigid. That's why I'm picking rigid structures here. Structures where the relative distances between points doesn't shift as the object moves. That's a more complicated situation. So in other words, what is true about every point in this is that the ratio of its speed to its radius from the center of rotation, the axis of the bike wheel, is the same. So for instance, v1 over r1 is the same as v2 over r2, et cetera, that's the same as vn over rn, and all of those are equal to omega, the angular speed of the wheel. So if this wheel is making two rotations every second, let's say, that means it's going through four pi radians every second, that's its angular speed, four pi radians per second. So we can use this information to try to simplify this seemingly nasty sum up here. So let's go ahead and use this information to do that. So the total kinetic energy is this sum of 1 half mi vi squared, but vi over ri is the same for all points in the sum. So we can substitute for the speed of each point with the product of its angular speed and its radial location from the center. So now I have omega times ri all squared in place of vi. Now, omega is exactly the same for every term in the sum, and 1 half is present in every term in the sum. So I can pull the 1 half and the omega squared that results from squaring this quantity out, and I'm just left with this sum sandwiched in the middle, 1 half times this sum, and I'll come back to this in a moment, times omega squared. Well, this looks dangerously like 1 half times some kind of mass quantity times angular speed squared, which rings a bell. 1 half mv squared was the kinetic energy, linearly speaking, of a body moving along a straight line. Here we have a rotating body with many points that make it up, but they're all moving at the same angular speed, so we have an omega squared that's like v squared for a rotating body, and we have a 1 half out in front. So this thing here sits kind of in the seat that's reserved for inertia or mass in the kinetic energy equation, 1 half mv squared, m is inertia. This thing seems to be taking the place of inertia in a rotating situation. And in fact, we can sort of symbolically represent this equation now in a rotational context. We have this complex rotational rigid body made of a huge number of points, but we can summarize the whole kinetic energy of this rotating body in a compact equation. The kinetic energy of rotation for a rotating body is 1 half times some inertia quantity, which I'll call i for now, times angular speed squared, 1 half i omega squared, which looks very analogous to 1 half mv squared, where m was the inertial quantity in a linear motion context. Now, i, this sum of the products of the masses and their positions from the center of rotation squared, this is known as the moment of inertia. It's an inertial quantity, and it summarizes how all the masses of each piece of this rotating body contributes to the overall kinetic energy of rotation through their inertia, their mass. Now, moments of inertia are specific to different rotating structures. You have to know the mass and how it's distributed relative to the center of rotation of an object. So for instance, a hoop going in a circle around its center is not the same as a disc, a solid disc going in a rotational motion around its center. The mass, even for the same mass, hoop and disc, the mass is distributed in a different way, and that has consequences for the kinetic energy of two bodies, even with the same mass, that are rotating with different moments of inertia, even at the same angular speed. So let's take a look at moments of inertia for a few common simple configurations involving things going in a circular motion around an axis of some kind. So for instance, and let's imagine I just have a point-like mass, mass M, that's going in a circle indicated by this red path here around some axis, which is indicated in blue. In fact, thinking about the right-hand rule that tells you about angular speed and the direction of angular velocity, if you apply the right-hand rule, curl your fingers in the direction of the circular red path, your thumb indicates the direction of the angular velocity. That's where the blue arrow points. So the blue arrow is parallel to the axis of rotation and indicates the direction of the angular velocity. This is a counterclockwise rotation. It has a positive angular velocity. Now, what's the moment of inertia of this? How could we calculate the kinetic energy of this system, a point-like mass, a distance R from the axis of rotation going around the axis? Well, we have to do this sum. We have to calculate the moment of inertia of this point-like mass going in a circle. Well, the point-like mass is one point of mass M located a distance R from the center. So the moment of inertia of this point mass is the sum from I equals one to N of all the masses times their distances squared from the center. But there's only one term in the sum and so we only have one thing to calculate. We have the one point with its mass M and its distance is R from the center, so MR squared. And so we would conclude that the moment of inertia of a point mass is just the mass times the distance squared from the axis of rotation, MR squared. And that's, in fact, correct. This is the moment of inertia for this situation. So if you were to be asked what's the kinetic energy if this thing is going at some angular speed omega, it's 1 half times M times R squared times omega squared. That's the kinetic energy of rotation of this situation, of this system. Let's look at another system. I mentioned a hoop before. A hoop is a uniform distributed circle of mass M.  So this is depicted over here on the left. We have the hoop of mass M. It's going perhaps counterclockwise around the Z axis. So it's rotating around the Z axis. This looks a lot like a bike wheel. If we ignore the spokes of the wheel for a moment, this looks very similar to a bicycle wheel, rotating around its central axis of rotation. And in fact, you can use a hoop of mass M with most of its mass concentrated at the edge of the wheel as an analog to a bike wheel. Now to calculate the moment of inertia for this, we actually have to do some calculus because we have mass M, but it's distributed over the circumference of this hoop. The hoop can be treated as made from a large infinite number of extremely infinitesimal, very tiny points of mass. Each of those points has a mass D M where D M is a differential of mass, a tiny little nonzero unit of mass, but approaching zero. Now all those points of mass D M are located at precisely the same distance from the Z axis, every D M is located a distance R from the center of rotation. That's what it means to be a point mass on the hoop's edge. So we have this moment of inertia of a hoop about the Z axis shown here in this picture. This is the sum from I equals one to N of the little masses M I times their distances R I squared. But we can convert this into an integral, a sum of infinitesimally sized mass points D M, all of them at a distance R, so we have R squared D M and we're summing them in a calculus sense from zero to M. And if we do this integral, we find that we get MR squared evaluated between the limits of the integral zero and M. And then ultimately we find out that the mass here, the moment of inertia for this is also just MR squared. It's the total mass of the hoop times the distance from the center squared. So this shows you how you can use calculus, for instance, to try to set up a sum over a large number of very tiny points that make up a continuous object, like a hoop or a disc, for instance. Now here's another one. We have a rod. This is a line of mass. You can think of it as a very thin rod so that it has only an extent in one dimension. It's only got physical extent along a line, no other dimensions. It has a length L, it has a mass M and it's rotating about its center. So it's going perhaps again counterclockwise around the blue line, which indicates its direction of its angular velocity, but also the direction of its axis of rotation. Now to do this one also requires some calculus, but I'm gonna skip it and just quote the result here. The moment of inertia of a rod depicted here on the left, rotating about its center in the sense shown, is one-twelfth times the mass of the rod times its length squared. It's a funny result, but you get it from the calculus of this particular problem. Now compare that to taking that very same rod of mass M and length L, but instead rotating it about one end. So this is, for instance, the difference between taking a baton, which is a long length of metal that you can throw into the air and get it rotating about its center. Take that same baton and instead swing it like a baseball bat from one end. It's the same object. It's got a mass M, it's got a length L, but it's rotating about a different axis. In one case, it's rotating about its center. That's twirling a baton. In another case, it's rotating about its end. It's more like swinging the baton like a bat. What changes? Well, more mass is now distributed further from the axis of rotation. Before, half the mass was on one side of the axis of rotation, and the other half was on the other side. But now all the mass is on one side of the axis of rotation. So this has a very different moment of inertia. In fact, it's not one-twelfth ML squared, it's now one-third ML squared. It's much bigger than when you spin the same rod around its center. So note that for the same object, this configuration has more moment of inertia. And that means that if you get these objects, both going at the same angular speed omega around their axis of rotation, this configuration has a lot more rotational kinetic energy. So you can get very different outcomes in energy, even for the same object, but changing the axis of rotation, which changes the moment of inertia. So generally speaking, putting more mass farther from the center of rotation, the axis of rotation, means you get a bigger moment of inertia. So again, compare the rod twirled about its center versus the rod twirled about one end. The right-hand picture has much more kinetic energy for the same angular speed omega than the left-hand picture. But that also means that it takes more kinetic energy in the first place, more energy of some kind, more work, for instance, to get this thing up to the same angular speed as the one on the left. It takes less work to spin this up to an angular speed omega than it does to get this right-hand configuration up to an angular speed omega. And that's because it has a bigger moment of inertia. It is more inherent resistance to changes in its state of motion. That's what inertia is. It's the same concept that we had from way back in Newton's second law, inertia, which is mass in linear motion. Here in rotation, it is mass, but it's mass as a function of its distance from the axis of rotation, moment of inertia. It still represents inertia. And the more mass you put farther from the axis of rotation, generally speaking, the more work is required to get that object up to the same angular speed. So it works both ways. An object with a bigger moment of inertia at the same speed has more kinetic energy of rotation, but it takes more work to get it up to that speed in the first place. So all of that is self-consistent. Now, we've been talking about spinning up objects so that they rotate about some axis, some fixed line in space. But I've totally avoided so far talking about how we get them spinning at that angular speed in the first place, right? I just talked about doing work, okay, what do I mean by work? How do I translate concepts of force and displacement and work into changes in kinetic energy? Can we recover the work kinetic energy theorem, but in a rotational sense? Can I talk about how I get something accelerating by applying a force to it some distance from its axis of rotation? So we've got to get into that topic now. We've got to start thinking about forces and accelerations. How do forces cause angular accelerations? And then forces acting over some displacement must equate to a kind of rotational work. What does that look like? So we know from Newton's second law of motion that if we want to take an object from a state of motion, so for instance, zero angular speed, that's a state of motion, no motion at all, to a different state of motion, non-zero angular speed, we have to apply a force. Force equals mass times acceleration in a linear context trying to move along straight lines. That is the statement of the relationship between force and acceleration. But we also know that forces can be applied to a body that's, say, fixed at some axis in space. If we apply a force to that somewhere along its length, we can get it rotating. So think of a wheel. If I grab the edge of a bike wheel and I push on it, I can get it to start rotating. It had no angular speed to begin with. Now it has a non-zero angular speed. I've clearly caused an angular acceleration. How does the force relate to the angular acceleration? Well, the linking concept here is, when a force is applied to a body that can rotate about an axis, this generates what is known as a torque. We'll come to the formal definition of a torque in a moment, but they are essentially the rotational analog of a linear force. Now, believe it or not, we actually know a few things about these torques from our experience with doors. Doors are large, flat planes of material that can rotate about an axis. And the axis is marked by the hinges of the door. It's a vertical line on one end of the plane. And you know that if you grab the handle of a door and you push or pull, you can cause the door to start rotating about the hinges, about its axis of rotation. Opening doors is a great playground for thinking about force, rotation, and torque. So we already intuit a few things about forces and how they do or don't cause rotations. For instance, if we grab a door and we push on it or pull on it in a direction that is along the plane of the door, so toward the hinges or away from the hinges, that's a force that's parallel to the line that goes from the hinges to the edge of the door. And we know that pushing and pulling on a door like that, straight along the width of the door, toward the hinges or away from the hinges, does not cause the door to open. If you can push and pull all you want, and if your component of your force lies entirely along the plane of the door, the door will just sit there. It won't rotate. It neither closes nor opens. No rotation of the door results from that kind of application of force. So instead, what we have to do is apply some component of our force that's perpendicular to the plane of the door, that is perpendicular to the line that connects the hinges to the other end of the door. When we do that, a rotation can be achieved. We can either open the door or close the door. Now, one of the other things you may have noticed about opening and closing doors, let's think about opening a door for a second, is that if we apply the same force closer to the hinges, it results in less rotation per unit time. It takes much longer to open the same door. Then if we take that same force and apply it as far as possible from the hinges themselves, the hinges mark the axis of rotation. If we pull on the door very close to the axis of rotation, it takes a lot of force to open the door in the same amount of time as if you just grab the handle, which is located very far from the hinges on the other side of the plane, and pull on the handle. This is why door handles are placed where they are. They are most effective. That is, it costs you the least amount of force to open the door in the same amount of time. If you put the door handle as far from the hinges as you possibly can physically manage on the door. So doors give us a playground to think about forces, the distance that they're applied from the axis of rotation, and the effectiveness of our ability to rotate the door as a result. That is the effectiveness of the torque that we apply. So let's represent torques by thinking about this problem of opening a door. Let's begin by representing the door as a line. So we're looking down on the plane of the door from above. Here we have the hinges of the door at zero zero in this coordinate system. I've represented that with the blue dot, and the door is presently closed. It can open inward into the plane I've drawn here into this grid. So it's a door that opens one way. You have to push it to open it. It has some width, which I'm denoting R, and I'm gonna define a vector, which is from the hinges to the place where I'm going to apply my force. That is a distance R, and the vector R goes from the hinges to where I will apply the force. That is the definition of R vector. Now let's add our force. I'm gonna open this door in a funny way. I'm gonna push at a 45 degree angle on the handle of the door. So here's my force F applied at the handle, which is at the tip of the R vector, and it's applied at a 45 degree angle. You can clearly see that depicted here. So there's my force F vector. There's R vector. That's where the handle is right at the end of the vector R. Let's think about the components of this force as I have applied it on this door. There's a component that is parallel to the length of the door, either in the same direction R is pointing, as depicted here, or if I had directed my force to the left at 45 degrees, pointing opposite the direction of R vector. That would also be parallel, but anti-parallel. It would point against the direction of R, but nonetheless, it's a kind of parallel. Here is the component of F that is parallel to the R vector. I've drawn it kind of in purple here, and I've denoted it F vector, but with two parallel lines to indicate that it's a component parallel to the vector R. These forces do nothing to cause rotations. I am yanking on the hinges with this force. If I were pointing it in the other direction, I'd be pushing toward the hinges along the width of the door. That doesn't cause a door to open. That just looks foolish. What's gonna cause the door to rotate is the force component that's perpendicular to R vector. So here it is. It's a light blue or cyan. It's F vector with a little perpendicular symbol here from mathematics. So this means perpendicular to something, and that something, again, is the vector R. It is only this perpendicular component that causes a rotation, or in technical speak, results in a torque. Torques cause rotations to occur. They create angular accelerations, and thus, open doors. So mathematically, we could write down a formula for the torque that's represented by this picture. The torque, whatever it is, is some product of the distance from the hinges that I apply my perpendicular force and the magnitude of that perpendicular force. Well, if I think about my force F applied at an angle theta with respect to the R vector, then I can immediately use trigonometry, and I know that whatever the force F is, this component perpendicular to vector R is F sine theta. It's just the opposite component in this right triangle that forms the force F. So I can write my torque as R times the perpendicular component of F, which is just R F sine theta, where theta is the angle between the vector R and the force F. It's this opening angle between R and F. This might be ringing some alarm bells for you. Torque equals a magnitude of a vector times a magnitude of a different vector times the sine of the angle between them. This looks like one of the vector products. And in fact, it should feel familiar as a vector product because we kind of just looked at it in vectors part three where we looked at the vector cross product. It's the magnitude of the cross product of two vectors. So if I have a vector a and a vector b and I take the cross product of a and b and I get the magnitude of that cross product, that is given by the magnitude of a times the magnitude of b times sine theta. That is it's the length of a times the component of b that is perpendicular to a. Theta again is the opening angle between the two vectors and that's exactly what theta is here. It's the angle between the vector R which points horizontally and the vector F which points at 45 degrees above horizontal. So torque is also a vector. It results from a vector cross product and it's the cross product of the direction from the hinges that I apply the force and the force itself. So tau which is the symbol that's used to represent a torque is a vector and it's full vector form is given by the vector R and it's cross product with the force vector. It's magnitude is RF sine theta and it's direction is determined by applying the right hand rule for the vector product. So I can get that direction from the right hand rule for the vector product. Take my index finger on my right hand, point it in the direction of R vector, take my middle finger, point it in the direction of F vector and stick my thumb out perpendicular to both of those fingers. My thumb indicates the direction of the torque. The torque points up and out of the screen so I represent it as a circle with a dot in the middle of it. It's like a vector arrow coming straight at us out of the screen. So again, this is a good chance for you to practice your vector cross product right hand rule with this situation. The magnitude of this vector is given by RF sine theta and its direction is given for instance by using the right hand rule for the vector cross product or explicitly calculating the cross product using all the little i hat, k hat and j hat cross product rules. Now that we've seen how forces and rotation can go together, a force applied a distance from the axis of rotation generates a torque. We know that this causes an object to rotate like a door opening. So can we arrive at a simple statement about how forces generating torques then lead to angular accelerations? That was our goal in the first place. We're changing the state of motion of a door. It is accelerating in an angular sense. It gains an angular speed where it had none before. There must be a relationship between force, torque and angular acceleration. Can we figure out what it is? So let's begin by considering a simpler example than the door which has a uniform distribution of mass. Let's think about just a point mass m that can rotate around an axis and that axis is a distance r from where the point mass is located. So in this case, we apply a force to the mass and let's just apply a force perpendicular to the vector r that goes from the axis of rotation out to where the point mass is located. So we're gonna apply a component of a force there that's entirely perpendicular to that r vector. That's going to begin to cause an acceleration of the mass. That mass is gonna have a tangential acceleration to the circular path of motion. We've encountered this before when we were talking about angular accelerations. If I change the velocity of an object that's going in circular motion, I'm causing an angular acceleration. The angular acceleration is related to the tangential component of the linear acceleration of that object. So I can write a equation that relates the perpendicular force and the tangential acceleration of that point mass m. So here we go, it's Newton's second law. That perpendicular component of my force, f perpendicular, must be equal to the mass times the tangential linear acceleration of this point mass m. Well, how is tangential acceleration related to angular acceleration? Angular acceleration is tangential acceleration divided by r, the distance from the center of rotation. So I can substitute in for a tangential with angular acceleration alpha. And this is what I get. The mass m times the angular acceleration times the distance r from the center of rotation. Now, notice that if I were to multiply both sides of this equation, f perpendicular and m alpha r by an additional r, then I can get torque on the left-hand side. So let's do that. Let's introduce torque into the above equation by multiplying both sides by the distance r.  Because again, I multiplied the right-hand side by r as well. Well, this is a point mass going around a center of rotation, a distance r from that center. Notice I have m r squared here. And m r squared might seem a little bit familiar. It's in fact the moment of inertia of a point mass going around a center of rotation, a distance r from that center. So I can substitute into this equation using the moment of inertia of this point mass. I have torque tau, the magnitude of torque on the left-hand side. I have moment of inertia and angular acceleration on the right-hand side. And so I get a very simple equation for all of this. Torque is moment of inertia times angular acceleration. And this is the angular equivalent of Newton's second law of motion, but now no longer for linear motion, but for rotational motion. So again, instead of f equals m a, I have tau equals i alpha. I have a rotation equivalent of force torque equal to a rotational equivalent of inertia, i times the rotational equivalent of acceleration alpha. This expresses how a torque generates an angular acceleration. So now we see the link between pushing on a door, its state of motion changing, gaining an angular speed. That must be due to an angular acceleration. How do I relate the two? I relate the two through torque and moment of inertia. Well, I hinted at this earlier before, if we are changing the kinetic energy of the door from a state of zero rotational kinetic energy to a non-zero state of rotational kinetic energy. And if we are acting on this door with the rotational equivalent of a force, a torque, a force acting at a distance from the axis of rotation, then we have a relationship between a force-like thing and the change in rotational kinetic energy of this body. So can we relate the change in the rotational kinetic energy to the angular analog of work? Work was, in a linear sense, force times displacement. What's the angular equivalent of work in a rotational context? So we know how to write the kinetic energy for a rotating body. That's not the issue. We have to figure out the equation for the work in terms of forces and hopefully in terms of torques, ultimately. So the work kinetic energy theorem should still hold here. The work done by a bunch of forces on an object should equal the change in the kinetic energy of the object. So whatever that work is, it's gotta be equal to 1 half i omega final squared minus 1 half i omega initial squared, final minus initial kinetic energies. Okay, well, that's nice. I mean, that doesn't really tell us anything new. Work is equal to the change in kinetic energy. We know how to write kinetic energy for a body that can rotate. But how do we go from torque to work? That's what we really care about. We have torque equals i alpha. That's the Newton's second law equivalent for a rotating body. Well, the definition of work hasn't really changed. Our force is being applied perpendicular to the line connecting the axis of rotation to where we apply the work. And we're causing the object to go through a little displacement, ds. So work is still gonna be defined the way it was before. Work is the integral of our force and our displacement, the product of those things in a vector sense, a dot product. Okay, so if the force causes a change in displacement that is in the direction of motion, then it's a positive work. And if it causes a change in displacement opposite the direction of motion, then it's a negative work. It's causing something to slow down. It's gotta be depleting energy out of the system. Now what is S? I picked the symbol on purpose, but here S is just some little bit of path over which the force F acts. Now in a rotation, it's only the perpendicular component of a force that causes a rotation to begin. And what's the path over which that force component acts? Well, as a body begins to rotate and accelerate along some circular path, that force perpendicular to the radial line of the path is acting over a little bit of arc length of the circular path, a little bit of DS. So if S is arc length, DS is just a tiny differential bit of arc length. It's that little differential bit of arc length over which force is acting. And we can relate using angular quantities, and in particular the definition of the radian, we can relate changes in path over the arc length to the little bits of angle that those paths subtend, d theta. And so in fact, the little bit of arc length that we traverse with applying our force, F perp, is equal to the radial distance at which the forces applied are times the little bit of angle that we cause the object to rotate through, d theta. So again, if this is something we haven't practiced yet enough so that you're comfortable with this, go back to the definition of the radian and think about, oh, I move a little tiny bit of arc length, DS. I must sweep out a little bit of angle in radians, d theta, and the relationship between those two has to be DS equals the radius times d theta. That's where this comes from. Okay, so let's again start from our definition of work. Work is the integral of force times displacement. And if we utilize that information that we just thought about, that it's only the perpendicular component of the force, and that is acting over a little bit of arc length DS, then we arrive at the following equation. That work is the integral of F dot DS. Well, it's only the perpendicular component of the force that acts over that little bit of displacement DS, and we can substitute DS with r, d theta. So we have that the perpendicular component of the force times the distance from the center of rotation, the axis of rotation, times the little bits of displacement d theta. That's all to be integrated. And thus, we arrive at the relationship to torque because again, we have the product of the perpendicular component of the force and the radial distance from the axis of rotation at which that force is applied. That is the magnitude of the torque. And so then it must be true that work is the sum from some initial angular distance to some final angular distance of the torque times d theta. All I've done is substitute in here F perpendicular times r with the magnitude of the torque, and I've put limits of integration between an initial angle and a final angle on this integral. So work is the integral of torque times angular displacement. Just like in linear motion, work was the product of force in linear displacement. So again, we see that there are analogies here between the linear world and the rotational world. Now finally, a little bit of revisiting the work kinetic energy theorem will lead us to a revisitation of power. Power is the work done by a force per unit time. And so we can start relating all the things that we have so far. We have changes in kinetic energy. We know how to write down rotational kinetic energy. We now have the work in terms of torque and angular displacement. Torque can be related to moment of inertia and angular acceleration, I alpha. We have all these pieces and now we can sort of come to a final concept here that will relate back to a linear concept we visited earlier and that was power. So the change in kinetic energy must be equal to the integral of the torque times the displacements over which the torques are acting. That's just the work kinetic energy theorem written out with change in kinetic energy on the left and the work on the right. Now if we just consider a teeny tiny little bit of work done by the torque tau over some little teeny change in angular location during the rotational motion, then we would have this equation that dw is equal to tau d theta. This is sort of just undoing the integral sign. I take out the integral and I'm left with differential quantities, dw equals torque times d theta. Now we can think about the energy per unit time, that is the power that is being supplied to either drive or slow the rotation. I can accelerate a wheel by making it go faster or if it's already spinning at some angular speed, I can apply like a force, I can put my hand on the wheel, apply some friction and I can slow it down. That's a deceleration. So I can apply a positive torque or I can apply a negative torque. I can speed the motion up, I can slow the motion down. I can put energy into the system by speeding it up. I can take energy out of the system by slowing it down. One of those is a positive power and increase in energy per unit time. One of those is a negative power, a decrease in energy per unit time. Power is just the change in work over the change in time. And in the rotational context, dw is tau d theta. So this can be written as tau d theta dt. What is d theta dt? d theta dt is angular speed. So power in a rotational context is torque times angular speed, tau omega. This ought to look familiar. The linear equivalent of this was force times velocity. Power in a linear sense is force times velocity. So let's take stock again of angular motion and linear motion and see the symmetries between the equations, between the linear context and the angular context. So let's think about linear motion in just one dimension. So just along the x-axis, for instance, we have Newton's second law, f equals ma. The forces acting along the x direction equal the mass times the acceleration along the x direction. We have the kinetic energy, one half m times v squared. We have work, which is caused by a force acting over a displacement, dx. And that force can vary with distance. So we might have to integrate f dx in order to get the correct work for that force. And power is the change in work over the change in time, but that in linear motion can be related to force times velocity at any given moment. The angular quantities that are equivalent to this are Newton's second law for a rotating body, that's tau equals i alpha, torque equals moment of inertia times angular acceleration. Rotational kinetic energy is one half times the moment of inertia times the angular speed squared, one half i omega squared. The work done on or by a rotating body is equal to the integral of the torque times the angular displacements, integral of tau d theta. This is analogous to the integral of f dx in a linear context. And finally, the power, the energy supplied to a body or taken out of a body per unit time is equal to the torque times the angular speed. So the above are just some examples of relating linear quantities to their angular equivalence. And in principle, just as with angular speed and angular velocity and angular displacement, you can map the linear quantities, which may be more familiar at this point to you in the course, onto the angular quantities. All of the 1D equations of energy and force have an angular analog. And to go from the linear equation to the angular equation, you can change x to theta. So linear position to angular displacement or angular position. V to omega, linear velocity to angular velocity or angular speed. A to alpha, so for instance, linear acceleration to angular acceleration. Mass to moment of inertia. So a moment of inertia is a distribution of mass, a distance, various distances from the center of rotation. It's more complicated than mass, but it's the mass equivalent in rotation. It's the inertia in rotation. And finally, changing force with torque, f goes to tau. If you do that in any linear equation, you will essentially gain the angular analog for free. The key ideas that we have explored in this section of the course are as follows. We began to think about energy and force concepts as they related to extended rigid rotating bodies, rotational motion. And we saw that there are rotational analogs to linear concepts, like kinetic energy. We started with kinetic energy. We developed the rotational analog of kinetic energy in terms of the moment of inertia, which is the inertial quantity and rotational motion and angular speed. We've seen how forces result in rotations. This has led us to the concept of a torque, and we have explored that concept in the context, for instance, of Newton's second law relating torques and angular accelerations. And we have connected the force and energy concepts through the work concept, and we've completed a set of rotational analogs to linear energy and force concepts. These ideas applied to a rotating system allow us to develop a whole vocabulary and problem solving toolkit to analyze seemingly complex situations involving objects and rotational scenarios. But with a fairly compact set of equations, one must master the terminology, one must master the application of these ideas, because again, we have to deal sometimes with cross products which are a bit funny compared to the kinds of vector quantities we've been dealing with so far. But once you've had enough practice and you've gained some basic mastery of these ideas, they become very powerful problem solving tools when thinking about the natural world.


Welcome to Spotlight. I'm Liz Wade and I'm Bruce Gulland. Spotlight uses a special English method of broadcasting. It is easier for people to understand no matter where in the world they live. Every October 31st, children all across the United States do something strange. They imagine they are someone or something else. They put on strange clothes. They may even paint their faces or wear false hair. They walk around from house to house. They knock on the door of the house. When someone opens the door, all the children say the same thing. The person in the house gives them a sweet candy treat. The children do this again and again for one house to the next. This is the holiday of Halloween. Today's spotlight is on Halloween and other holidays like it. The tradition of Halloween is very old. In fact, people who study history believe that Halloween traditions began as long as 2,500 years ago. They believe these traditions started in ancient Ireland with the Celtic people. The Celts of Ireland counted two parts of the year. The bright half of the year was summer and the dark half was winter. They believed that the bright half of the year ended around October 31. After this day, winter was coming. The Celtic people called this change in season sawin, which means the end of summer. Sawin was a special time. It was both a celebration of the end of summer or life and the beginning of winter or death. So it was also a frightening time. The Celts believed that there was a natural barrier between the world of the living and the world of the dead. But at sawin, the barrier disappeared. They believed that spirits could rise from the dead and come among the living. Historians say that during the celebration of sawin, the Celts dressed themselves in different clothes. They walked through their villages making lots of noise and they caused disorder in the village. They believed that this frightened the spirits away. Some historians also believe that the Celts sacrificed humans or animals during this celebration as a warning to the spirits. Other people groups in other parts of the world also celebrated similar holidays. Around the year 400, Christians came to Ireland. They began to celebrate Christian holidays at the same time the Celts celebrated their holidays. For many years, the Christians had a celebration in honour of saints, special people in the church. They called this celebration All Saints Day or All Hallows Day. It was a day to celebrate these special saints or holy people. They celebrated All Hallows Day on November 1st. The night before a special day is called an Eve. After a time, people began calling the sawin celebration All Hallows Eve. Since language is always changing, people soon shortened the name. Instead of All Hallows Eve, they called it Halloween. And that is what we call it today. In the 16th century, another tradition began. In the days leading up to the sawin celebration, the people of Ireland and Britain dressed in different clothes. They went house to house. At each house, they sang songs to the dead. Then the people of the house gave them small, sweet cakes. Irish settlers brought Halloween traditions to the United States in the middle 1800s. But people there did not really celebrate Halloween until almost 100 years later during the early 1900s. For most children today in the US, Canada, the UK and other places, Halloween is about candy and fun clothes. But other people have a day like this to celebrate the people who have lived and died before us. Many countries have a celebration to remember people who have died. Some celebrations are at different times of the year. They do not always call these celebrations Halloween. But the celebrations have similar meanings. So how do you celebrate Halloween? On Halloween in Belgium, there are parades and scary decorations like spiders and ghosts. But people also light candles. These candles help people remember their dead relatives. In the Czech Republic, they celebrate Dusíčké. People visit the place where a family member is buried. In China, people do not celebrate Halloween. But they have a very similar celebration called Uelán. It is the festival of the hungry ghosts. The people there believe that spirits visit the world for 24 hours. Some people burn pictures of fruit or money. They believe that the spirits can see these burned pictures. And the burned pictures bring calm and peace to the ghosts. This celebration is not a time for games or dressing up. Instead, it is a day to remember and respect the memory of dead ancestors. In Mexico, the people celebrate Dia de Muertos, the Day of the Dead. Some other Latin American countries also celebrate this holiday. It is not a scary day. It is a time to remember friends and family members who have died. A three-day celebration begins on October 31. During this time, families make special places in their homes to honour their dead friends and family. They place candy, photographs, fresh water, flowers and food and drink in this special place. They burn candles and strong smelling incense to help spirits find their way home. The last day of the celebration is called All Souls Day. On this day, living family members eat a meal at the gravesite where dead friends and family are buried. They tell stories and remember good times of when the person was alive. And the tradition of Halloween is still popular in Ireland too. Today, on Halloween, Irish boys and girls dress like ghosts, witches or other frightening creatures. They light big fires. Many children go around to different houses. They ask for fruits, nuts or sweets. Halloween can be a lot of fun for children. But it is also a good time to remember someone you love who has died. Do you celebrate Halloween? Do you have a tradition that helps you remember people who have died? Tell us about your experiences. You can leave a comment on our website or email us at radio at radioenglish.net You can also comment on Facebook at facebook.com slash spotlight radio. The writers of this program were Liz Wade and Adam Novus. The producer was Michiel Ozaki. The voices you heard were from the United States and the United Kingdom. You can listen to this program again and read it on the internet at www.radioenglish.net This program is called All About Halloween. Visit our website to download our free official app for Android and Apple devices. We hope you can join us again for the next Spotlight program. Goodbye.


Merci beaucoup tout le monde. Good morning, everyone. Thank you for joining us here today at Fanshawe College. It's great to see so many people here, distinguished guests, students who are enjoying a morning without classes, apparently. And of course, I'm joined here by my friends and colleagues. First of all, President Devlin, thank you very much for your welcome and your incredible service to this community and this great institution. I want to thank Kate and Peter for welcoming me back to London. I also want to highlight the local MP, the NDP's Irene Matheson, who wasn't able to be here today but who sends her very, very best as well. I just wrapped up a really interesting tour for the Centre of Applied Transportation Technologies where I got a chance to try my hand working on an engine with some very talented Fanshawe students. And I found out that as soon as I left the engine, it started up just fine. So their new workspace is very impressive and I have no doubt these students are in great hands here. You know, I always enjoy touring schools like this and witnessing the kind of groundbreaking work they're doing. It really hammers home the importance of investing in students and supporting Canadian post-secondary institutions. And I'm really proud that just last month our government unveiled a budget that will do just that. A cornerstone element of our plan for students involves an investment of $2 billion in labs and buildings on college and university campuses. Students need updated facilities and equipment in order to maximize their educational experience. We know this. As Canada's economy shifts toward high skilled employment, it's more important than ever to make sure that new grads have the skills and training they need to meet the demands of the labour market. It's a wise investment in the infrastructure that supports our students' minds. To give more young Canadians greater access to good well-paying jobs, our government will invest in training for unemployed and underemployed workers. We will also develop a new framework to support union-based apprenticeship training and work to expand co-op opportunities. On top of that, we will invest over $300 million over three years in the Canada Summer Jobs Program to create good jobs for up to an additional 35,000 young Canadians. Because we all know that education is a significant financial investment and summer jobs are a part of the saving process. Now, more than ever, it's important that education remains affordable and accessible, and young Canadians shouldn't be burdened by growing student debt right after graduation. Our plan is to make post-secondary education more affordable for students from low and middle income families, and we'll do this by boosting available grant funding by up to $1,000 a year for low and middle income students. This move puts more money in the pockets of 360,000 students across the country each and every year. In addition, we will help students pay back their debt after graduation. We are proud of our plan, which is to ensure that students who obtain their college or university diplomas begin to pay back their pre-students only when they earn at least $25,000 a year. This initiative will help them start their career and help them to suffer a little, and it's a good thing to do. I am confident in the measures we have taken to support young Canadians who enter the labor market. Visits like those of today remind me that it is by investing in education and the training of young people that we invest directly in our future. Investing in young people, investing in the extraordinary institutions like Fanshawe College that support them, making sure that education is more affordable, more accessible, making sure that you don't have to start repaying your Canada student loans until you're making $25,000 a year. These are all things that will not just give opportunities to students and young people, but give opportunities to our communities to grow in our country's economy to do better in the coming years. This is what we made to Canadians in the last election. This is what we've put forward in this budget. This is how we're going to continue to work with Canadians to build a brighter future for all in the coming years. Merci beaucoup tout le monde.


Hi, good morning, and welcome to today's products and focus U.S. there is actually closed up above potential resistance at sixty five nine eight As the market still cheers good news from West Texas instruments and obviously still digesting the Apple figures a lot of Earnings have come out as a bit of a mixed bag But some of the major firms out there I'm talking about obviously IBM a lot of the other major firms seem to be doing Okay in the backdrop of this kind of macroeconomic uncertainty that we've seen However, there's still good big question marks around the Fed and what they're what constitutes a good result for them Because there are some commentators out there that feel that the Fed can only disappoint going forward So obviously a lot of pressure on those guys going forward But incidentally the two levels you want to be aware of on the US 30 16 5 9 8 and 16 7 3 8 and those are the next two potential support and resistance levels So moving forward I should also point out that a lot of the technicals you got a crossover in the MACD as well So there's a lot of technical factors there that are supporting the US 30 in the very short term And you've got that same crossover in the MACD here on the UK 100 and obviously we have already had the positive Signals to to buy it on the slow stochastic and RSI Obviously a lot of ground to make up so we don't necessarily expect that to all happen in one day But we closed towards the top end of the range yesterday We've drifted slightly lower this morning. I'm looking at 63 90 as a potential support eyeing up the next potential resistance at 64 63 Then looking at Japan 225. I'm looking at dollar yen. It's still grinding up slowly there US 30 to Japan to do 5 making up for yesterday's losses and then some a little bit 49 77 as potential support next potential resistance 15 488 Almost across over in the MACD and buying signal and slow stochastic So certainly the technicals running into today's US CPI data Which will be very very keenly watched for what the Fed is actually going to do next regarding interest rates And that's at 130 but we'll come back to that in a second. So looking at dollar yen 107 spot 36 will be potential resistance The slow stochastic's got a signal to buy RSI is neutral Whereas the MACD is hovering around about that zero line So it does look to be 107 spot 36 will be a strategic level And it's struggled and failed to break through it before so the question will be well US CPI figures today be strong enough to help it push through and bearing in mind They are they are expecting and you can see the data down here The forecast that they've got aren't exactly massively strong but a weak figure here would probably cause a US dollar tumble and The US markets take a bit of a knock at the end. So moving on to Crudwell West Texas and $84 once again remains a tough sell for crude traders and today is no different and today's Wednesdays normally they should go Know if you've got crude oil inventories, but that must be fast-forwarding on to Thursday for now But all the data there in regards to West Texas is kind of pointing to excess supply we need the China to not slow down quite so fast Otherwise long-term fundamentals and West Texas crude will always be quite negative and there's been a new report coming out from From Saudi Arabia. They're willing to accept lower prices in order to gain additional market share versus some of the other OPEC regions So rather than defend their Their slice with higher prices by cutting back supply. They're quite happy to have oil be cheap Make more deals and grab a market share So that's probably telling about why Crudwell remains at such a low level and the backdrop of So many pro oil aspects is better than at least so gold still continues to try and push on higher We actually touched a 55 period SMA yesterday. We're unable to break through 1267 still feels like the potential resistance But there seems to be a little bit less uncertainty in the markets today But then check out US CPI and you know US earnings continue to this point US CPI comes on weekend expected, you know, all bets are off again But if earnings continue to do well and some of the big bell with the stocks and CPI comes out much better than expected It's that gold to maybe take a bit of a tumble So finishing up with your dollar your dollar came off a little bit further yesterday on the back of the news that the ECB are looking to do a corporate buy bomb buying program as early as December to help support your zone Companies and to show up some of their balance sheets to help them ensure that they remain productive and That has caused the euro dollar to slide ever so slightly Obviously a little bit of dollar strength as well. It doesn't hurt, but we are now trading blow one spot 2746 I'm up 126 61 as the next potential support And we are bouncing around that 21 period SMA with the other technicals kind of just floating around a neutral territory finishing up with GBP 21 period SMA looks elusive again unable to break and close above it. We are in the middle of two ranges Probably more like that. I'm not sure you can get rid of that level just now One spot 5917 is potential support one spot 6265 potential resistance And we've still got this pattern of consolidation and curvature around here Which might be a base for springboard move higher shoot the sterling end up recouping some of his losses So as we mentioned CPI due to date 130 make sure you got your Recurring alert set for that and we fast-forward on the Thursday. We've got Chinese PMI and UK retail sales US jobless claims Which is fairly important especially for your dollar traders, and then we've got your zone CPI data as well once again if you're your dollar trader pretty much essential Now make sure you keep your eye on the chart forum make insights popular going forward and join me again tomorrow to find out What happened next?


Hey everyone, we're going to be talking about the Switch Pro and a little bit about the Switch Mini because we've had some conversations in the past, my theory, if you guys haven't checked it out, check it out up here where I talk about how I think the Switch Pro might be what the Switch was always intended to be versus what the Switch is today which has a multitude of issues and a lot of aspects of it feel rushed even still right now two years on and that maybe they needed more time to get the real Switch done, Wii U is failing, etc. Go watch that video for the entire premise on that. I know a lot of people started subscribing to this channel lately because of that video and I felt like it was important to follow it up because when new information comes to light I think it needs to be brought to attention because a lot of the Switch Pro rumors, Switch Mini rumors have been coming from the Wall Street Journal and the Wall Street Journal is a platform you need to pay to read and most of us, most people covering this story are not willing to pay to read the article. Well, I did that very thing today because of another YouTuber and that YouTuber is Super Metal Dave. Super Metal Dave has obviously been at the forefront of covering Switch Pro and the Nintendo Switch in general way back when it was called NX and you'll see, you know, all of his recent videos, you know, Nintendo Switch Pro is not as long a secret, you know, it's got 17,000 views and there's 21,000 views on new info. This is about the USB PD thing we brought up in a prime news episode, which by the way, we'll talk about that a little bit in this video as well because the Nintendo Switch as it today actually does support USB PD, but we'll get into that a bit. And then the new custom Nintendo Switch Pro SLC will bring blah, blah, blah. So I give credit to him because he's the one who drew attention to a couple things. First off, if you guys remember, I talked about the TAGRA X1 SLC being a 20 nanometer chip and that how Nvidia would probably like to move away from that process. Well, Nvidia earlier this year announced this, the Jetson Nano developer kit. And this is the big thing I want to give credit to Super Metal Dave on because I was not aware of this developer kit until he brought it to my attention. Now, for those who don't know, the Jetson Nano developer kit is one of three different developing kits that Nvidia is releasing. You can actually buy this thing right now on Amazon. It is an AI based kit. It is designed for robots and all that stuff. And essentially, it uses a TAGRA X1. You can go down to the technical specifications and see like the four gigabyte 64, you know, blah, blah, blah, the quad core ARM, you'll 57 running at 1.43 gigahertz, etc. But if you go through all of the announcement and all of the spec sheets, you'll find out that it is actually using a TAGRA X1, which means that Nvidia clearly has a processing line open to continue to make TAGRA X1s. And they're going to continue to make them, which does throw out the theory that Nintendo would need to potentially upgrade the Nintendo Switch because of them needing to move away from the 20 nanometer process. In fact, Nvidia is still producing TAGRA X1s for their own purposes. So obviously, they're going to continue to produce them for Nintendo. So yeah, that kind of throws out the theory that the Switch Pro and Switch Mini might replace the current Switch. Instead, the current Switch still might exist in the marketplace as is for quite some time, maybe several years to come. So hey, quite at the supermodel day for digging this up and realizing this was announced just this year. So hey, look, that's not really proof of anything in terms of them moving away from the TAGRA X1. Now for those who don't know the TAGRA X1 inside of the Nintendo Switch is running downclocked. That's right, we are not even getting the full power of the TAGRA X1 in a base Nintendo Switch. And the reason for that is heat dissipation. Even on this development kit, look at that heat sink like it is using a bigger heat sink than is actually being used inside a Nintendo Switch. And if you scroll down even further, you can watch this video and it'll show active cooling even on top of that heat sink at times. So yeah, it TAGRA X1 gets pretty hot is what I'm saying if you want to run it at its full capabilities. Now, all of that being said, why do we need to talk about the Wall Street Journal? Well, the Wall Street Journal, as you see right here, is the place where we learned about the launching of two consoles just a week and a half ago. And they've been originally reporting that we were going to see a new Switch Pro or whatever this year back in October of last year.  But as we get into this article, you'll see that the Wall Street Journal has sources both on the manufacturing side and actual developers who are using the new switches. So let's just read the article and comment along the way because I think there's a lot of interesting information in here. And again, this is information you typically are not going to find because the Wall Street Journal does paywall all of this content. I have prepaid for two months, hoping that they have more Switch Pro news heading up to E3 because they do note that one of their sources does think Nintendo is going to unveil the city three. So let's go down here. So says Nintendo plans to launch two new versions of the Switch gaming console as early as this summer. People familiar with the matter said as the company seeks to sustain sales momentum for the product going into the crucial third year. Now this is all this little beginning part is known publicly. That's the part of the article they have public. One version will have enhanced features targeted at avid gamers, although it won't be as powerful as Sony Carb's PlayStation 4 Pro or Microsoft Xbox One X according to parts suppliers and software developers for Nintendo have access to the prototype of the machine. The other version is a cheaper option for casual gamers that Nintendo sees as a successor to its aging handheld 3DS device. The supplier and developers say this is interesting because maybe it's a clamshell design. Interesting prospect here. Let's let's continue on now that that's that's all the information that's been available publicly. That's all of these videos on Switch Pro over the past week or so have been made on. But now we have all the additional information. So let's get into it. The new models are likely to be unveiled at the E3 video game expo in Los Angeles in June and possibly released a few months later. One person familiar with Nintendo's plans said Nintendo declined to comment. So yeah, E3. I've been saying it for a while. E3 is when they're going to unveil this stuff. So I'm going to be at E3. I might go hands on with this stuff. Exciting. So here we go. The Switch introduced in March 2017 has been central to Nintendo's recent business success. This is true. The company earns more than 80% of its total revenue from Switch hardware and software sales and investors are forced to focus on how long the console can remain a profit driver. Again, all of this is true and easily verifiable by looking through Nintendo's financials. Industry Watchers expect sales for the Switch to begin declining soon. Analysts polled by Visible Alpha, a provider of market forecasts, anticipate an average of 17.9 million units of the Switch were sold for this past fiscal year. And then it's going to have 17.4 million for the new fiscal year, which is happening right now. It runs basically April through March. So we're about to get the update for last fiscal year at the end of this month. So that's notable. But yeah, we're in a new fiscal year right now. And they think the sales are actually going to decline. It is notable 17.9 million is actually higher than the full year one sales. But you know, Nintendo systems do tend to show a little decline in year three. We kind of begin the exception where it just kept exploding. But anyways, let's continue. Video game hardware makers generally come out with new consoles every five to six years. And it's common for them to update the devices in the middle of their life cycles to keep the momentum going. Obviously, you know, this happens. This is just a general statement of facts. You know, Sony introduced the PlayStation 4 November 2013, updated it with the less expensive version in September of 2016 in a high end model two months later. And the sales of the PlayStation 4 made strong. It's been selling like 20 million a year like clockwork PlayStation 4 is just killing it. Nintendo suppliers and game developers have been talking with their investors about the new switch machines for several months. People who have used that's right. People who have gone hands on with the devices, say they aren't just similar new versions with a higher or lower performance. Now, remember, not just not instead of not just that does mean that they are higher and lower performance potentially. But that's not the only thing. There's some significant differences with these platforms compared to the original switch. Now, we actually learn what some of these differences might be or will be. In fact, we learned one particular item in for the switch mini. You would be wrong to think the version is similar to what Sony did with the PlayStation 4 Pro. And the other is just a cheap alternative that looks very similar to some past handheld machines. Say Sony's PlayStation Vita as a comparison. One person who has used the new device has said the switch is a hybrid device that can be played in either handheld or connected to a larger screen prices for the new switch model couldn't be learned to cut costs for the cheaper version. Nintendo plans to eliminate some of the functions used in the original switch console. I think this is a given we all expect them to eliminate certain features to get the cost down on it. And one of those features is going to be the vibration feature in its controllers, according to suppliers of Nintendo parts. The company judge the new switch models won't need the vibration feature because there won't be many games released using the full benefit of it. So HD rumble. That's going away at least in the switch mini. Now, they are saying the switch models, the new switch models, does this mean they're going to get rid of HD rumble in the switch pro to that is pretty significant to me. Granted, I do think there should be rumble of some type. But HD rumble is massively underutilized. It's a brilliant piece of technology that even games that take advantage of it. Most people can't tell the difference between that and standard rumble. The few games that do take major advantage of it one to switch being a prime example are very few and far between and aren't typically why people buy the system. Breath of the wild and as my Odyssey don't really make good uses of it and they're the best selling games. Same with Smash Bros. So I can see them actually deciding that HD rumble is just not going to be part of the future of the new devices. So that would be interesting. I don't know if this means there won't be any rumble, but obviously the expense of HD rumble is probably being cut at least in the switch mini, if not also in the switch pro. That is what they are saying. This is according to the Wall Street Journal. I'm just reading what they say and extrapolating you guys, you know, can extrapolate for yourselves and come up with your own opinions on this. Alright, so then it says Nintendo is also adjusting its suppliers for the new models. Sharp Corp. An Osaka based company that had long supplied Nintendo components to Nintendo video game machines, but they failed to win the deal for the original switch. They are expected to provide liquid crystal displays, so the screens for the new switch machines. This is important too because Nintendo has had issues with the current supplier. This is actually something we've known for a while. They've had issues with the current supplier because their current supplier works with Apple and Apple isn't ordering LCD kind of screens anymore. So like they've been trying to fight with Nintendo to get more lines to make things for Apple because they have a bigger contract with Apple. So this is actually something we've known for quite some time. Not surprising that Nintendo might have been looking to go back to their old LCD supplier just because that company was obviously willing to continue to make LCDs at a decent price. So that's kind of a minor here nor there. Outside the fact that Wall Street Journal has proven in the past to actually have contacts at this place. So that's good news for Wall Street Journal's verifiable information here anyways because they have prior leaks that were linked to them having contacts at that company. Anyways, suppliers and third party game developers say Nintendo is preparing to release new game titles from well known franchises. We already know this Animal Crossing, Pokemon, etc. That's obviously Nintendo's goal this year. Slam the system of games and sell a ton of them. Some people say sales of Nintendo devices usually weaken in the third year. This is actually true. Usually, not always, but it is usually true. Usually year two is the peak year for most Nintendo systems. Anyways, well others say games to be released this year are stronger than ever before. One executive added a Nintendo supplier said, I honestly don't know how things will turn out, but I have my fingers tightly crossed. So and Yang Yang here also contributed to this article that was written by Takahashi Machiozuki. Alright, so what do we get from all of this information? Like what do we get? So what's the takeaway from the Wall Street Journal article? The takeaway is that the Nintendo Switch Pro and Nintendo Switch Mini are not going to just be this PlayStation 4 style upgrade. And the PlayStation 4 style upgrade obviously is the PlayStation 4 Pro and that was a decent leap forward in that technology. Well that leap forward is basically what the new Nintendo 3DS did. If you remember the Eurogamer article kind of said it would be similar to a new Nintendo 3DS style jump. Well the new Nintendo style 3DS jump was like you know two more CPU cores, increasing the speed of the CPU by like nearly 400% or 300% whatever it was. It was a pretty big increase, doubling the VRAM. Like it was a pretty big jump. It was a PlayStation 4 Pro style jump. Unfortunately because of the technology we're talking about at the time, that still isn't a huge leap forward just because things were already like really, really slow on the OG 3DS. So even with all that extra cores and the extra speed and the extra VRAM and the extra RAM they actually doubled the RAM in the Nintendo 3DS, there really wasn't much that people noticed a difference between the two platforms. Granted, I did. Faster frame rates, that was a big thing. The UI was much more usable on the new Nintendo 3DS but again it was, it felt minor on the consumer end just because the 3DS was already such a weak platform that even doing all that, it was still, the new Nintendo 3DS was still a really, really weak platform. Now, so what does this mean then? It means that whatever we're getting with the Switch Pro and the Switch Mini for that matter is something that is not going to be perfectly predictable by us in the public. So while I do think we're looking at a custom SOC here, I think the custom SOC that we're going to get is going to be because we're going to get something that is unexpected. If it was going to be the expected change, an expected change, it literally could just redesign the Switch to somehow run the Tegra X1 at clock, like at base clocks. But I think a custom SOC comes in the play because that also can lead to Nintendo potentially getting cheaper chips. Like if NVIDIA is going to repurpose the Tegra X1 now for their AI platforms, maybe that means that now Nintendo wants their own custom SOC out of NVIDIA, one that is specifically targeted for Nintendo's use case. Because now, as you see the Tegra X1 getting repurposed for other things, Nintendo might be like, hey look, you know, we've been working together, we have a 10-year contract. That's NVIDIA. NVIDIA said they have a 10-year contract. So that's like verifiable. So the big deal for here is that I think that NVIDIA and Nintendo have been combined to be working on a custom SOC this whole time. And that custom SOC is going to be a part of the major differences between whatever the Switch Pro and Switch Mini are doing and the original Switch. I also think because there is such a stark difference in what's going to be happening with these platforms that the OG Nintendo Switch is still going to be sold for quite some time. So I don't know that the Switch Pro or the Switch Mini are actually going to replace the Switch. If you remember, way back when, when they were talking about NX and then eventually Switch in the very early days of the platform, Nintendo kept calling it a family of systems. A family of systems. This means we could see many different forms of the Switch. And the form we have now, where it is handheld and able to be used, docked and undocked and all this stuff, is one form. Another form could be handheld only, a Switch Mini, which could have some changes to it because it's only handheld. It's possible that a Switch Pro might not be able to be handheld at all. It's possible a Switch Pro is only for use connected to your TV. And that you'll be able to cross share saves across all of the platforms. Just some food for thought. Now I don't know that this is where Nintendo is going, but when they talk about a platform that is not just your typical upgrade, you start to wonder, what's Nintendo thinking here? What's this family of systems thing they have going on? They're not going to do something traditional? I think most of us expect a traditional upgrade, right? You know, minor bump or major bump, whatever the case, in what the Switch is already doing. But what if, what if they're making a true family of systems to target every single type of demographic? You have the hybrid target, you have the handheld only target, and you have the home console style target. I'm, I think this is an interesting prospect, especially if they all cross communicate, all use the same cartridges. I honestly think that this could be a very interesting future. And as Wall Street Journal says, we're probably finding out about this at E3. So lots of, lots of food for thought. And I want to see what you guys think about all this information down in the comments below, because there's a lot to take in here. Now that we have the full context of that Wall Street Journal article, now that we know the Tegra X1 is still going to be in active development with or without the OG Switch around, there's a lot to think about. And I just want to put the thought out there that Nintendo could be doing something completely unexpected. And even what I'm saying now could be so far off base from what that unexpected thing is. Anyways, I'm Nathan Roberts from Nintendo Prime. Be sure to enter our Nintendo Switch Super Smash Bros. Ultimate Bundle giveaway through the Gleam.i only down in the description. Drop a like in this video, subscribe for more content. This won't be the last time we're talking about these platforms. As more information comes out, as more is uncovered, we'll be right there talking about all of it. And E3 cannot come soon enough. We'll be there in person. So if these devices are there to go hands-on with, you know, we'll be there going hands-on and explaining exactly what you can do with these platforms. Alright folks, we'll catch you in the next video.