All right: well, today, we’re going to continue talking about a uh, torque free motion of a rigid body and we’re going to just dive in so last time. We ended talking about torque free motion of a rigid body where all of the principal moments of inertia were different and we also made an approximation that we were mostly spinning about one direction. We’re going to drop that assumption that we’re mostly spinning around one direction, but we’ll make this other assumption that two of the moments of inertia are equal and that’s, what we mean by symmetric so – and i think i said This in a previous lecture here symmetric put in quotes. It just means that two principal axes, two principal moments of inertia, are equal, so um. It could be something that geometrically looks symmetric like uh like a tube. I’ve got some kind of tube thing here, or it could be a box or something that just happens to have a shape where, if you calculate the moments of inertia, two of them are equal, so it could be a box. I don’t know something like this square box or rectangular. Looking thing as long as the two, you got two moments of inertia equal, so we would say the moment of inertia uh this one’s i3 about the third direction and it’s not equal to i1. Here,’s i1, but here about the b2 direction. That moment of inertia is also i1 same thing. Here: i3, not equal to i1. We haven’t yet decided whether or figured out if i3 is greater than i1, because you could figure out from how these things were drawn, but uh. So you could have things that are kind of tall and are symmetric or – and i just got some kind of plate – it’s kind of. Oh, you can kind of see it.’s got green in it, so it picks up the green screen. This is uh a plate, so it’s flat, ish kind of a spanish design. This is valencia, spain, behind me, nice place in the mediterranean coast. So we go back to what are we there?’s equations for a free, rigid body written in a principal axis frame, and it’s written it a few times now, so that’s the first. This is giving the time rates of change of the components of the angular velocity. As seen in the body fixed frame that’s, what omega 1 omega 2 and omega 3 are – and here’s this one – i three omega three dot minus i one minus. I two omega three and omega one omega 2. So if we have we’re making the assumption here right that i 2 is equal to i 1 that’s, what defines a symmetric, rigid bodies. Two of them are equal. We’re choosing i1 and i2 to be equal and i3 to be the odd one out. So that would mean that over here this term is exactly zero, so that means omega 3 dot equals zero, which means omega 3 is a constant in time. So we could do uh something similar to what we did last time where okay, this here omega 3 is constant. This thing becomes we’ll use i1, so we’ve got i1 minus i3 and i3 minus i1, so those are just the reverse of each other, using the same approach as last time. Right um. We called that omega s, so we just sort of labeled that it’s omega 3, that’s the constant it’s constant during motion, and then we defined um, omega n squared just like last time, but now with i1 equal to i2. This looks even simpler, so this is i1 minus i3 squared over i1 squared times omega s squared. So if we define that then uh from these other two rigid body equations of motion, we can get, we get something that looks very similar to last time. It’s just that it’s now exact, because we’ve made this assumption of a symmetric, rigid body. Well, what does it matter that it’s exact? It means that this holds not just for small deviations from uh the axis about which we’re spinning. This will hold for any deviation, so we have this essentially using the same procedure as last time, but now it’s exact. So now the equations are exact, not approximate the the differential equations we’re going to use some other notation that follows what the book uses. So, if you do want to follow along this is at the end of chapter at the end of each chapter. They have these things called tutorials, so this is tutorial 11, 1 and that’s on page 515 and that’s, where i’m, getting my things. I’m going to follow the notation that’s used there. So there they let i 1 and i2. They call that i not and then for the other one i3. They call that j. So i’ll use that same notation, and, although we need to know now is that j is not equal to i not there.’ll be two cases later is j greater than i not or is j uh less than i naught okay, so we’ve got a rigid body and it doesn’t necessarily have to look symmetric right. You could have some weird shape here.’s. A here’s a gopro, i i don’t know what the moments of inertia are, but suppose they’re, not all equal. Well, i could start shaving off mass in different directions and i could make two of these moments of inertia equal and it will not be a symmetric looking shape but um. It’ll still follow these equations of motion. So, okay, here’s, our general looking body, we’re looking at free, rigid body motion, which means the only kind of obvious reference point, is the center of mass g and we’ve got two directions where the moment of inertia through principal axis directions Of the moment of inertia, is i not and some the third direction where it’s j and we’re looking at what is the evolution of the angular momentum? This is i, the angular momentum, vector of the body fixed frame with respect to the i frame and the components they’re, omega 1, omega 2 and omega 3, And so it’s we’re. Looking at the evolution of those things, we’ve already found that omega 3 is equal to a constant now we’re solving for omega 1 and omega 2, but we know how to solve this ode. It’s just simple harmonic motion and omega n. You could tell from here omega n is always greater than omega n. Squared is always greater than zero. So we don’t have any unstable motion: oops um, let’s uh, we’ll assume some initial conditions, so we’ll, say at time t equals zero omega. One is negative, omega naught and omega. Two is zero and of course, omega. Three is just we’ll call that omega s and that’s what it will be for all time, so assuming those initial conditions and using these starred equations up here, the solution is omega. 1 is negative. Omega naught, cosine, omega n t omega 2. As a function of time is omega naught sine omega n t. So now we know what omega the vector is. Oops in the body fixed frame is this, so we’ve got a part that varies sinusoidally and then we’ve got that third component, which is constant all right now in the to kind of get the idea of uh well, okay. What does this mean? Let me get a rigid body here:’s, a rigid body where’s your body and then let’s say this is the angular momentum vector so as seen in the rigid body frame that third component, this is the third component will be constant and Then the tip will just be going around in a circle with a period of 2 pi over omega n, but that’s seen in the body fixed frame. We want to know how is this thing actually tumbling and moving in space. So the proceed one procedure to do that or at least to get an idea – is to look at the angular momentum. So what’s the angular momentum um here?’s a angular momentum, the angular momentum of the body about the center of mass is what well first, we have to write. What is the moment of inertia matrix? Well, we’re in the we’ve chosen a principal axis frame, so that means we know what this is. We’ve got. I naught at two places along the diagonal and then j and everything else is zeros. So if we were to write what is the angular momentum, this is in the body fixed frame. This is ig the body fixed frame times omega written in the body fixed frame. You do that matrix multiplication and you’ll get. This is negative. I naught omega naught cosine i get mt. I naught omega naught sine omega nt and j omega s and let me draw a sketch looking in the b1b3 plane, oh check this out. Look, i’ve got a blue frame. So what is this? This is b1. B2. B3, so we’re, looking at you,’ll, be we’ll, be looking in the b3. The b1 b3 frame, okay and we’re gon na draw the vector. So let me let me draw that nicely. I got some weird paint stuff on here. I used acrylic paints, okay, so looking looking at the b1 b3 plane, there’s b1, there’s b3 um, we’ve got the angular velocity and the angular momentum, and what is so. This distance from the b3 axis is omega. Naught this projection along the b3 axis is omega s and we can. We can take the um. We can look at this angle. We’ll call this angle. Let me use black this angle. We’ll call theta the angle between the angular velocity. Sorry, the angular momentum vector and the b3 axis so cosine theta – is that unit vector. So what’s the unit vector the that vector h, divided by the magnitude of that vector, h, dotted with b3 and well. We know what that is because we’ve. Actually got check it out. We’ve got all the components of h here, so we can write this out. This is going to be j omega s and then all right that’s – the third component here now divided by the um, the magnitude i not squared omega, naught squared plus j squared omega s squared. How do i get that? Well, take this thing. Squared times this thing squared and you get because it’s sine and cosine you get just – i naught squared omega naught squared all right, so that’s what we get, and i want you to notice something about this. This right hand, side here so right hand, side, it is uh, it is independent of time, so that means um. So the angle theta is a constant throughout the motion. So, here’s a symmetric, rigid body, it’s. The angular velocity angular momentum, vector that angular momentum, vector um, is going to always make the same angle with that b3 direction. It’ll be theta, no matter how the body is. Turning that angle is going to stay the same. What do we also know? We know, because this is torque free motion. The angular velocity isn’t changing in inertial space, it’s always pointing in the same direction in inertial space. So, no matter, no matter how this body is moving around, it has an angular velocity that is constant in inertial space. So even though it it’s in a different direction, as viewed in the body fixed frame, it’s changed constant in an inertial space. So this is a constant direction in inertial in the inertial frame, so we may as well. We have a choice in how to define the uh the three directions of our here:’s, a different color, the three directions of our inertial frame. We’ll choose the third direction to be along that angular angular momentum vector so we’ll call that e three right, so we’ll define e3 and then e1 and e2 are just um two other vectors that fill out the right handed coordinate system. So this is uh. The third unit vector of the iframe may as well basil, pick it to be along some convenient direction, all right, um. Now what we want to, since we want to describe how the body is moving with respect to the inertial frame. That means how is the body frame this blue frame, moving with respect to the inertial frame, the red frame uh, we’ll use euler angles and the one that’s convenient for this is the three two three boiler angles, so we’ll use the Three two three euler angles right: these are also sometimes they’re just called orientation angles. We’re using the three two three convention to find the motion in inertial space or describe the motion in the inertial frame with the third direction. This is e3 hat already defined as along the angular momentum. Okay, um, if you remember so the three two three, those angles we’ll call them psi theta and phi, and it’s. No, it’s, no uh coincidence that i chose theta. So this theta really is this theta here it’s the second euler angle, for this three, two: three sequence: if you want a reminder of what the three two three euler angle sequence is, i think i’ve got it here. The video. This is a three two three euler angle sequence. So three, we do a first rotation about the common z direction and then an angle theta about the number two direction and then the final rotation fee about the new number three direction so theta. What did we have from up here? This right hand. Side is independent of time in this expression. So because that right hand, side is independent. That means that theta is a constant of motion. It’s a constant throughout the motion, and that also means that theta dot is zero, so that’s a pretty that’s, that’s pretty nice. Now, what if we go to the go to the kinematic equations of rotation for these angles and that’s in that’s equation 1048. I think we work these out, at least for the three to one or euler angles. The book does it for the three two three as well, remember: that was the relationship between the angular velocity components and the time rates of change of the euler angles. Well, we already have this one, but that euler angle, doesn’- t change in time, but we want to know how the other two change. So this is omega one and what was omega 1. We got to look back up there. Omega 1 is negative, omega naught. Cosine omega n t, and at least according to the kinematic equations of rotation, omega 1 equals negative psi dot sine theta cosine v that’s from the kinematic equations of rotation. Omega 2 is omega, naught sine omega nt and, according to the kinematic equations of rotation, omega 2 equals uh, positive psi dot sine theta sine v. What about omega 3? Well, omega 3 is just equal to a constant omega s, but according to the kinematic equations of rotation, it is equal to b dot plus psi dot, cosine theta okay. So this will be important to remember these two. In fact, i’ll give them a name like equation square. I don’t have to use numbers. I can use symbols, okay, um. What next? Well, if we let’s, take the uh these two boxed equations, i’m, going to uh square them and add them, and if i do that, i get that omega. I want to use a different color. I get that omega. Naught squared equals psi squared sine squared theta because of some trig identities. I’ve got that, but theta is equal to a constant. So if theta is equal to a constant omega naught was an initial condition that’s equal to a constant. That means that psi dot squared is a constant and what is it so psi dot equals it’s plus or minus. We don’t yet know omega naught over sine theta, okay um, we’ve got this. We’ll see later that it’s the positive root. So i’m just going to go ahead and say: okay, it’s the positive root. So this means psi is a function of time right. We already know what fade as a function of time. Is it’s just whatever it was initially, but psi as a function of time is equal to this constant times time, plus whatever psi was initially, we’ll call that psi naught and now what now we’re gon na use the rotation matrix. That relates the two frames to write b, three in terms of the the iframe right. This is the iframe, so in terms of e1, e2 and e3 so b3, and we’ll write it this way, so the b3 direction, the body fixed third direction, third direction here as a function of time. How can that be written in terms of e3, e2 and e1, so the inertial directions, and just based on this rotation mate? We could look up what the rotation matrix is for the three two three euler angles and then uh. Remember that’s. What related the b frame and then the b unit vectors and the e unit vectors, so we could write b. 3 is equal to cosine psi as a function of time, sine theta e, 1, plus sine psi as a function of time, sine theta, e2 plus cosine theta e3, which describes uh what well b3 dotted with e3 would be cosine theta, which is a constant. So if i’ve got, let me draw the e3 direction and then we’ve got b3. The projection with e3 is is constant and then we’ve got sine theta and then cosine and sine of psi, which is changing linearly in time. What this means is that the b3 direction is describing a cone around the e3 direction, so b3 traces out a cone around e3 and remember what e3 is e3 is along the angular momentum, a fixed inertial direction in space, so we’ve got it traces. This traces out a cone with a precession rate psi dot, so there is think of that there being some direction in inertial space given by the angular momentum and the b3 direction of the body here,’s our b3 direction of the body. This one, it’s, tracing out a cone with this constant rate, the procession rate, so we could sketch that out. You know here,’s like a b3 direction, tracing out a cone. We call this a the space cone, so it traces out a cone. The space cone now what if we want to find uh psi? No sorry phi dot. So we know we know how two of the euler angles evolve. Theta is constant and phi changes linearly in time, so i mean psi. Psi dot is constant. What about phi to find phi dot and hopefully p? Let’s write the angular momentum. We’ll write it this way, it’s the magnitude of the angular momentum, so the scalar hg in the e3 direction, and now we’ll write e3. In terms of b1, b2 and b3, using that using the rotation matrix, so this is what we’re going to do. We’re going to write e3 in terms of b1 b2 b3. So what is this? We get negative hg sine theta, cosine, v, b1 plus hg sine theta sine v b2 plus hg cosine theta b3, and we also know according to that equation up above that, this is equal to. I naught omega one in the b one direction plus i naught omega two in the b two direction, plus j omega 3 in the b 3 direction. So let’s just take the second component. I naught omega 2 equals h g sine theta sine v. We’re just equating components and this equals i not because we have another way to write omega 2 right. We can also write omega 2 using this second um kinematic equation of rotation. There’s a lot of moving parts here, no pun intended. So we’ll write this as i naught psi dot sine theta sine phi. But what do you know? We got sine phis sine thetas canceling out so hg equals. I naught psi dot, so psi dot. This is just justifying why psi dot was positive. This is h g over i naught, which is greater than zero, so that justified this choice up here of the positive root and that’ll be important because it determines what direction procession will be when we try to bring this all together and interpret it at The end so psi dot, the precession rate is always positive, and now what now we can uh? We can solve for phi dot from the third of the kinematic equations of rotation, which was that omega s is v dot, plus psi dot, cosine theta. You know everything over here is constant and we know how to write the expression for it so v dot. Is i not minus j over? I naught times omega s, so this is the the rate of the angular velocity coning around b3. Maybe we call it the spin rate there’s, the other one was the precession rate. This is the spin rate of the angular velocity. I don’t know if koning is a verb coning around b3. Okay, if you couldn’t follow all of that. Maybe you know, look at it again and the steps will make sense, but now we’re going to interpret things, maybe that will that will help um we’re going to just consider. I want you to consider two cases. What are the main two cases? If um j, the angular momentum about the uh, let me just draw it here: what are the two cases it’s? Is i not greater than j or two? This is a one. Is it two? Is i not less than j? Okay? So let’s consider these two cases suppose i naught um it’s greater than j, so that determines right. Look at look at this part of the equation. If i naught is greater than j, then that means that phi dot is co is positive, and what does this mean physically, if i naught, is greater than j? Remember j is the the moment of inertia about that third direction. So if the moment of inertia about that third direction is the smallest. That means you.’ve got a shape that looks like it:’s kind of tall and skinny it’s called a prolate body prolate. So we’ve got um. Well, i’ve got a tube, so i’ll just draw a tube, so this means it’s tall, skinny, um. Technically the term is prolate body. So j is the it’s the minimum axis j and there’s no intermediate axis. So the other two are the maximum axis right. So what does the situation look like? We can look at this in terms of cones. Let me insert a cone, so here’s, the sketch it’s pretty good. I like it. What’s going on? We’ve got the remember. I mentioned this body cone so about the body. There’s some cone, that body cone is rolling around a space cone, so we’ve got that b3 direction. The b3 direction is making a cone around the angular velocity. Let’s say the angular momentum, let’s say. The angular momentum is the z direction here. So i’ve got this thing. The b direction is actually going around that, but there’s more to it. It uh we’ve got the. Let me use the same colors as last time, so we’ve got the angular velocity angular momentum. I keep getting those two mixed up: the angular momentum and that defines this e3 direction. It’s a direction fixed in the inertial frame, so fixed in space, and then we’ve got the angular velocity and that angular velocity is moving around um following the right hand, rule it’s moving around seen from the top it’s. Moving around counterclockwise around that e3 direction direction fixed in space, the angular momentum direction, that is the precession rate okay, so this is going around with a with the precession and then, in addition to that, there’s, this other rotation that we might call the spin. So we’ve got a spin rate which is also greater than zero okay, so that’s what this will do in three space. If i got it spinning it’s going to do some behavior like that. So let’s see a video, not that video here it is remember it. Doesn’t have to be a cone. It just has to be a shape that has uh the two moments of inertia, um, equal and then the third one is the smallest moment of inertia. So in this this purple thing here is the e3 direction. It’s the angular momentum. We’ve got in blue the angular velocity coning around, so we’ve got these it’s like it’s, the it’s like two cones rolling around each other space cone and the body cone and together it makes this type of motion. So you have procession and spinning you see here. In this case the spinning is a lot faster. Spinning of that body about kind of its i1 axis here is a lot faster than the slower precession rate, so the other case, this was a right. All this had to do with was. Is i 1 greater than j, or is i 1 less than j? If i 1 is less than j, then that changes the sign of v dot and some other things so that’s the other case, so let’s now suppose i naught is less than j. This implies that v dot is less than zero and we’ve got something that’s flat short and squished like a plate or frisbee or a box, or a coin or pancake all right. So j is now his j j is the major axis or maximum axis, and then we have the other two which are smaller. So this is a case of something that is uh short squished. Well, technically, it’s called an oblate body uh. So what did i say a coin or you know frisbee or a box, a square box that’s just small in the situation here we could also draw the space cone and the body cone. So let’s do that, and here it is here. We’ve got one cone inside of the other, but that’s not hard for us, and still you could view the interpretation, as you have one cone that is rolling without slipping about the other cone so think of um. We’ve got the space cone and then the body cone is this larger cone that’s rolling without slipping. These look like the type of cones that dogs have on their heads like the cone of shame. Our dog has not had to wear the cone of shame so that’s that’s good um. Here,’s, the angular momentum right that defines this um it’s. The angular momentum is the only thing that distinguishes a direction in space compared to anything else, so we’ll use that as the e3 direction and the same thing where this intersection of the two cones is along the angular velocity direction and the angular velocity is Moving around in a cone tracing out this space cone, but the overall motion of the body is it’s rotating about its b3 direction and this body cone about the space cone? The um we still have the the precession is positive, but then the spinning is that way. So let’s take a look at the take a look at a video of this again. So this is the a box where these two directions in the kind of plane of the box are smaller than the angular uh or moment of inertia about that perpendicular direction. So this is the motion we’ve got h. The angular momentum is a constant direction and you see the angular velocity is going around that in a cone. But then what is the body doing it?’s got this procession and spin. I think it’s harder to interpret the oblate body.’s rotation. The other one with something as tall and skinny is easier, but that’s just me okay, so this should at least help you do a couple of the homework problems you have to do. You actually knew enough from last time to do problem one and now you’ve got enough to do most of problem, two uh or problem three one of tossing a coin into the air. There is so we’ve looked at this kind of specialized case of when the two moments of inertia when two moments are equal, so there’s, no intermediate axis. You don’t have to worry about that um and you get this nice interpretation of cones, but you really have to try to make something that has two moments of inertia equal. So the more general motion is all moments of inertia are different, so there’s a question about what are these cones? Is there a way to see the different cones independently? I i think i have a demo. I don’t have it ready today, but i could show it next time that makes it better to see the two cones. It’s a mathematica or wolfram demo. I just can’t get it to work on the ipad, so i’ll have to get it to work on the computer, but you don’t actually need the cone interpretation to do those homework problems, but it does help with the understanding. So i will, i will try to get that all right. So the next next topic, because this is gon na, come in um. When we talk about energy based ways to get equations of motion, you need to know how to write the kinetic energy of a rigid body and we haven’t we haven’t discussed that there’s the kinetic energy of a rigid body – and this Is in gets introduced in section 9 5 of that book, but then it’s also in chapter 11, And why do we need this? We’re going to need this. It becomes useful for when we analyze free, rigid body motion in general. So for the case of all the axes or all the moments of inertia are unequal and then it becomes useful for another method for getting equations of motion. So this is the method of lagrangian the lagrangian formulation. It’s a way of writing the equations of motion just based on the kinetic energy, and we could also use it to analyze something uh analyze, some other situations, so it just it helps for you know: analyzing some motions uh just using i mean energy considerations And i might get to that today is something called the caber toss which is part of the highland games and they’re – probably not doing it this year, but at radford there’s this highland festival and one of the highlights is people carrying extremely heavy Logs, like trees, giant poles and then trying to flip them over, and you could do some kind of energetics about based on the initial flip. Will this thing fall over, like i, don’t know what you’re talking about it doesn’t matter right now, um all right, so we’re going to use the same kind of notation we used before. We were writing kinetic energies left and right before, like what happened to all that. Oh it’s coming back don’t worry so here is a arbitrary, rigid body here,’s its center of mass um. Well, let’s let’s, give it a velocity here here:’s the velocity of the center of mass with respect to the inertial origin, and this also has a um, some angular velocity, which we always draw some vector, but we we usually interpret it As oh, this is related to spinning or something okay, so we’ve got if you wanted to write the kinetic energy with respect to the inertial origin. This would be the kinetic energy of how the center of mass is moving with respect to the inertial origin, plus the kinetic energy of how the whole body is moving with respect to its center of mass and last when we were introducing rigid bodies, i think we Even said, this part is due to the translational motion, and then this part that’s due to all the motion relative to the center of mass once you’ve, restricted to a rigid body. So you can’t have things squishing or deforming. The only possible motion is rotation, so this is now rotational. This is the rotational motion, so we’ve got the translational motion, kinetic energy and rotational motion kinetic energy, and what are the expressions for these well for tgo? The translational motion? Is it’s like you, treat this like a giant particle, so this has mass mg, so it’s, one half mv squared one half m the magnitude of the velocity squared that’s a that’s, a g with respect to the o. Okay, so it’s the most. It’s the kinetic energy due to translational motion, and if this thing is moving through space, re read your body right, it’s. Still the center of mass might be moving, and if it is, then it’s got some kinetic energy due to that translation of motion, in addition to kinetic energy of the rotation, so we’ve got this um. If you’re following along in the book that’s page 510, all right now, what about the um? The rotational kinetic energy? Well, it’s. It might be useful here to think again of this being made up of a bunch of particles. So here’s particle. I here’s the location of particle, i, with respect to the center of mass it’s got some velocity and this velocity can’t be just anything. The velocity is intricately connected to the angular velocity. So this is the angular velocity vector cross this location from the center of mass rig okay. So i’m going to write here the rotational kinetic energy as if this was a bunch of particles, so it would be one half and then i would do i’ve got a sum mass m. I times that velocity, but i’ll just gon na substitute in omega cross r. So this is omega cross r, squared okay. So this is for a rigid body, oops and then um we can. We can write this using some identities and, if you want to you, can you could read all about it, but we write this as omega dotted with the moment of inertia matrix times the angular velocity, or we could also write this as one half the angular velocity Dotted with the angular momentum and notice we could we could do that too. For the uh translational kinetic energy. This looks like one half the velocity dotted with the momentum, since the linear momentum is just equal to the mass of the rigid body. I got to put mg here. The mass of the rigid body times its inertial velocity p equals mv. So the same it’s got the same kind of structure so that’s kind of interesting. But usually we write the kinetic energy of rotation in terms of the of just angular velocity and the three by three moment of inertia matrix. So i’m just giving you different expressions here, so you could write it this way. It’s, the angular velocity. You take the transpose right, and this is the angular. I mean the moment of inertia matrix and then this is the angular velocity as a column vector. So this is the one that you’ll see the most you might go well, which of these is correct. Well, they’re all correct. Okay, so just different ways to write the same thing. This is a it’s a column vector this is a three by three matrix, and this holds even if you’re, not in the principal axis frame. This is for any frame. This is a that same thing, but you take the transpose you write as a row vector. So this is the main version used. If you want, i don’t know if there’s any other name for it. It’s the matrix version and it is uh if you’re following along in the book page 511. We could also write the angular. I mean the uh, the kinetic energy, with respect to the inertial origin. We could do this. We could say that this is one half the angular velocity times, but now it’s the moment of inertia with respect to the inertial origin, o right. We could take this expression and write it just completely in terms of like it.’s all rotation, even though we know it’s. Not i go that that doesn’t seem right. No, it doesn’t seem right, but we can do it. So this is where the moment of inertia about the point o is related to the moment of inertia about the center of mass via the 3d parallel axis theorem. So you could, you could write it either way and just as a note for planar rigid bodies, the wrote the rotation about the center of mass right. This is just one half. I g it’s, not a three by three matrix: it’s just a scalar times the scalar angular velocity, and usually we write that as theta dot. So this is just one half i g theta dot squared, but that’s just for plane of rigid bodies now in three addict in 3d. Again, if you used a point q, let’s say we used a point: q, which is arbitrary. It’s not inertially fixed it’s, possibly moving around and doing things. So, if you use reference point q, you sort, we usually reserve the point: the notation q for something that’s not inertially fixed but could be moving. If you use a point q, which could be accelerating um – well, you’re free to do that. Then what do we get? You get a different expression for the kinetic energy. You would write. You’d still want to write the kinetic energy with respect to a reference point o uh. Maybe i’ll sketch the situation so follow along um and the point q doesn’t even have to be on the body. It could just be some kind of i don’t know maybe that’s a it’s a moving spaceship uh that’s v of q, the inertia velocity of q and then you’ve got our our arbitrary body with its center of Mass and whatnot you do need to know. Where is the point q? I’ll do that in orange, where’s the point q, this is our q with respect to g. So that shows up. So you get the expression here. T o equals the kinetic energy of q with respect to o, so it’s as if your mass was at the point q, even though it’s not plus the kinetic energy about the point q. But then there’s some other term. It has no name, but it’s good, to be aware of um it’s m times the inertial velocity of q dotted with the angular velocity cross r q g. So maybe just to conclude that there is a angular velocity here of the a body fixed frame with respect to the inertial frame. So you pick up this weird new term that has no name it’s part of the whole bookkeeping procedure, because you’ve used a point q that could be accelerating. So if this isn’t accelerating uh, something will go away here. Okay, let’s just call this our new weird term, so this would only come up if it just turns out. You want to use some moving point q or accelerating, if it’s moving, but moving with constant velocity. That’s also an inertial point. So that’s that’s – fine, okay, um – i guess to complete this uh for the arbitrary point where t q is, we’ve got one half angular velocity transpose. I q, where you would use well, you get iq. However, you want use the parallel axis theorem and then that so that’s the kinetic energy of rotation. Okay, again, this is a lot. You’re, not you haven!’t used it yet, but you will want to bookmark this for when we get to lagrange mechanics. I introduce it here because we needed to analyze arbitrary free, rigid body motion. There’s another application, and maybe we’ll do that in the last few minutes, which is this caber toss? If you have not seen the k i gang caber is just like an old fashioned word for giant log um, see which of these. Oh, i think this is a good one, so here’s a guy wearing a kilt scottish thing, and you see all these people gathered for the games. Listen to that. He’s running. He tosses it and has enough kinetic energy to go over. Hey can’t you just listen to that music all day, all right! So as an application, we’ll analyze the caber toss, it’s a c, a b e r toss. So what’s the situation here, we’re gon na we’re not gon na deal with the part where the person’s running it -‘s after they’ve, tossed it, and this thing plants into the ground. It has some angular velocity at that point, so imagine uh it’s been thrown from right to left. We’ve got the point where it plants initially, and this thing has a these. Are i think there’s some standard length like you, treat that you could treat the caber as a rigid rod, a slender rigid rod with the length l? So this is l, it’s, something like six meters. I think they have it at exactly. 5. 94 meters and the mass is uh 80 kilograms. So what’s that over 160 170 pounds? These are not easy to lift the people. Do it are pretty pretty buff. So this we’ll say it plants into the ground at some initial angle and we’re going to track what that angle does. So the initial angle is theta dot with respect to the vertical and the initial angular velocity call that omega naught: okay, um! So it’s, you know heading this way and we’ll just write what’s the angular velocity squared theta dot squared. So let’s just call this: it’s theta dot, not um. Where’s, the center of mass, the center of mass, is halfway up the rigid body. So we got g so that’s actually l over two. What’s the height above of that above the ground? It’s h what’s the inertial velocity of this it’s uh. So this has some kind of inertial velocity g with respect to o. If we use an energy approach, um well, first, what’s our question? The question is for a given initial angle for a given initial angle: what initial angular velocity is needed for a successful toss? A successful toss means that it can just barely make it over the vertical position, and then it falls over for a successful um. I guess we could call it flipping of the caber and we will use an energy approach. So how will we do that? The potential energy we got to do we’re going to do a balance of you know: total energy, there’s the potential energy and it’s just due to gravity acting on the center of mass. So that is capital. U is m g and then it’s h the height, which is mg. What is h in terms of geometric things and l and theta, so it’s l over two times: cosine theta, okay, that’s. What h is the kinetic energy? We can write it two different ways and this kind of illustrates something from if we call this point here. This planting point is the point o we could write. We could use the pivot point o as our reference, and if we do that, so it’s t o, then we could just treat it as its pure rotation about the point o. So the kinetic energy of rotation about o, and so we could write that as one half angular moment of inertia for a rigid rod about the point o. And so let me just write up here. What is the moment of inertia of a rigid rod of mass m and length l about its end? If you haven’t memorized this? It’s, one third m: l, squared okay, so we have one half i naught omega squared. So this is one half one: third ml squared uh theta dot, squared okay, we could also use you know the center of mass as the reference, which means we’re. Still, writing kinetic energy with respect to that point. O kinetic energy is the kinetic energy due to translational motion of the center of mass plus the rotation of the body about its center of mass, so yeah. This is translation of g kinetic energy. This is rotation about g, so you would write one half m the magnitude of the inertial velocity of the center of mass squared plus one half ig, and then it’s the same angular velocity squared so omega squared um. What is this you could work out? What this is because we know where the center of mass is, we know what that what theta dot is. So this ends up being l over 2 theta dot squared. What is the moment of inertia for a rigid rod about its middle? It’s. 1. 12 ml squared and that’s, something that you just you either have calculated it once or you looked it up. So we have this one: half m, plus one half one twelfth ml, squared theta dot squared and this ends up being um. The same. The same thing so good it’s the same, and now we’ll, we’ll, assume um that the total energy is conserved, so that means e equals t zero. Plus. U is constant throughout the portion of the motion that we’re studying. So if we were to write the expression now, just adding these two, the kinetic plus the potential we’ll get 1 6 ml, squared theta dot, squared plus uh mg mg, l over 2 cosine theta, and we’ll say what is this initially initially, we’re at 1, 6 ml squared omega naught squared plus mg l over 2 cosine theta naught, and we want to know so when theta gets to go up here when theta gets to zero, we want there to be just enough energy to go over and so That’s it zero and theta dot is just barely above zero, but we’ll just substitute in zero, so energy zero zero. This is just mgl over two, so we could set this equal to uh what it is initially and then we could solve for omega naught. So we’ve got this and yeah so, for example, uh. If the initial angle, it looks like it’s, often about 30 degrees, then we need that the angular, the angular velocity, is greater than square root. G over l, three one minus cosine theta, which is about 0 8 radians per second or if you want 46 degrees per second. So if you start with some angle right 30 degrees, then this has to be moving at 46 degrees per second to be able to barely make it to the top with just enough energy that it then goes over. And then you’ve won the caber toss. .
