Hello everyone. We are going to work on work and energy relationship for the rigid body. Basically, what we have covered last time is how we can generate the equations of motion for a rigid body which mostly contains a translation and rotation, so not only F equals ma, we have to derive moment equals I Alpha and for work and energy, we can integrate it over the displacement. Before we start, let's say what's been changed for the work and energy relationship when we are changing from particle to the rigid body. For a single particle of m_i has a displacement r of i in its time derivative, v of i, the absolute velocity, then the kinetic energy for this single particle is one-half mv square. Then if you want to integrate all the particles of the rigid body, that's the total kinetic energy for the rigid body. Note that the Sigma is after you did a mv_I square. Now, v_I is actually the r_I dot and then this can be splitted out as the distance from here to the center of mass, R bar in the Rho I. You would take the derivative of r_I, you can also take a derivative of this one plus this one. Now, this one take a derivative of this one is actually the velocity of the center of mass and how about this one? When you take the derivative, the distance from the center of mass and the particle, that doesn't change because that's the definition of rigid body. So would that be turns out to be zero? No. If the rigid body is rotating with Omega, then Rho dot the displacement change is zero. However, you have Omega cross r term, so your velocity now can be expressed by center of mass linear translational velocity and Omega Rho i. If you expand it, you will have the term that mv square is the total mass and one-half Omega square Rho I square m_ and you have m_I Rho I. So the first term is a linear kinetic energy of the center of mass, the second term by definition is the moment of inertia, m Rho square, with respect where the G center of mass. The last term based on the definition of center of mass, this turns out to be zero, so what you can have ultimately is one-half mv square plus I Omega square. So for the rigid body, note that not only just handling the MV square, you have to also consider I Omega square. Here everything is defined with respect center-of-mass, here v bar and the I bar. Now, let's work on the example. We have a wheels of mass M and it has the outer rim and the center hub. Center hub is actually contact with incline, then its pulled by the external force, by P which is through the cable wrapped around the disc. Since this is not a simple symmetric shape, it has hubs with it, the problem gave you the radius of gyration as Z. Now, you're supposed to find the angular velocity, Omega, after it has been traveled over the distance d and what is the magnitude of the friction? The very first step is we have to set this coordinate and as I said before, let's set the coordinate at the absolute frame which coincide with the contact points of the disc and the surface, right here. At the center of mass located here, how many contexts of this one have. So contact with the surface and contact with external force. So there are a free-body diagram, there's no M and friction force, and the P and the mg that'll generate the translation motion and rotational motion. Now, so you have F is ma x and y direction, three forces horizontal and two forces in vertical, and if you obtain the moment equation with respect to center of mass, the force that will generate the torque is the friction F and the excellent force P so you have a torque equation which will generate the IR file. Now since you have to find the Omega, maybe you have to find the Alpha and to take the integral of it. So to find this value or you don't know the friction here, you don't know the friction there and then those normal forces are also known, so you need more information to solve it. So the problems that it's no slip condition, so in that case there is a kinematic constraints between Alpha and a. A is going to be minus or Alpha, so you plug that in back to either of the equations, so you can have the two equations about Alpha and the unknown friction. So you occupy eliminating those friction that we adopt doesn't eliminate the unknowns. We will get the equations for the Alpha. Suppose that we found Alpha and then since we were asked to find the Omega, let's say take the integral of the Alpha over time. That's because the definition of the Omega. However, to do so, you have to know the some time information that you are in agree with, but the problems that find the Omega after t_0, no, it's asking to find the Omega after it travels over distance d. So this doesn't really help so you have to find another way. So the displacement means maybe you have to integrate over the displacement. Since you have a dx integral, a is the dvdt, you have this one as a definition of the v so you will ended up getting the equations, quadratic equations, velocity when you take the integral of the acceleration over the displacement. Then once you know the velocity, maybe you can get the information about the Omega. So from the questions up to this in the previous slides, instead of solving the Alpha, let's solve it with the a linear translation or velocity of the center of mass. If we rearrange the equations with respect to the a and take the integral over the displacement here. So left hand side, this is all a constant value, that structure of the will and the incline value so this is independent of the x, so you can simply just multiply it by the distance to integrate this. So all the constant value multiply by d is going to be one-half v squared, and the v squared at the rolling conditions can be expressed by r Omega. So this how you can obtain the Omega information by integrating the equations of motion. Now we can also find the friction. So since we obtain the Alpha, we can either plug back into here to obtain the friction here, or you can use the equations of motion to find the friction force. So if you just plug that in and then just plug in the Alpha or the a where we have bound earlier like this equation, you can have finally ended up this value. This is the friction for the rolling cases, so pretty much different from like Mu normal force one. So you have friction, what is proportional to the force applied and the structure of the geometry of the wheels. So to check the direction, we have to see the sign of this value and everything is a positive and then we have to check those value one in the parenthesis and small r and capital R and the z is going to be somewhat related with the r and capital R, and then suppose that for assuming that you have only a small disk, z is going to be square root 0.5 or so you could plot that in so you can find the range for the z by assuming z is greater than, you can actually bind the z for this structure and you can say that, this is going to be positive value. So total value for the friction is positive means you have a positive reaction of the friction force. This is how we can get the magnitude here, the eruption of the friction force. Now, we can actually solve this a simpler way by using work-energy relationship like initial kinetic and potential energy plus work done is going to be same as kinetic potential and the kinetic energy later. Initially or zero, the only work done is by external force PX and over the distance. What is this X means? Is this the X for the center of mass which is given as a d or somewhat different? Look at the structure for the wheel, the center is moving over the distance d then the top part will move larger and then this contact point is zero. So by using a triangular resemblance relationship, the ratio, we can get for the displacement of d of the center of the wheel, we can estimate what's going to be the X for the top. So your work done is the displacement of that force applied. So you can express this one as in terms of d and some coefficients. The kinetic energy later is going to be one-half mv squared plus I Omega square, bar Omega squared actually. Then we know that this V and Omega are not independent anymore because it's been related as a kinematic relationship because it's rolling without slipping condition, so we can summarize it as in terms of v or Omega. So if you equate them, you could have the relationship of what's the velocity over all the other factors like external forces and the displacement. These are the equivalent to the solution we obtained through the integrating the equations of motion. Now, we are going to then actually solve this exactly same problem, except one singular point, which is now in this case slipping occurs. The previous one has a rolling without slipping, but now, we start rolled up with slipping, and then what will change? We set the coordinate in a similar manner, and then if we draw the free body diagram, since this now slipping upward, the friction force is directing downward, that's the only difference of the free-body diagram compared to the one where we have just worked on with no slipping case. So your equations of motion only switch the sign of the friction here, top minus sign here, and here, again, the friction is now generating clockwise, the clockwise rotation so it's a minus sign here. So exactly same procedures, and since it's a no slip condition, your relationship like A equals minus R part cannot be hold anymore, but instead we have information about the friction, like simply the friction is going to be Mu_K multiplied by the normal force here is going to be Mu_K mg cosine Theta, cosine B. Remember that we found the friction for the rolling case. Static friction is the form of this one, and then later, maybe if you have a chance, that you should compare this should be actually greater than for a slipping case. Now, so by plugging in those friction information in terms of all the given known variables, what you can have is the acceleration in terms of external force and the other term, and angular acceleration in terms of external force and the other term. As you look at this, these are constant value, so you have a constant acceleration and constant angular acceleration motion. So we could say if this constants can be noted as A and B, Alpha it's going to be a constant B and A is going to be constant A. Then you can just take the integral to obtain the velocity and the Omega. The ratio between the Alpha and the A is now going to be A over B, and when it becomes to be minus R that's a special case for rolling without slipping, and generally, A over B is now a function of external force P, and it's not necessarily the R. So if we keep continuing the integration, you have two options, either integrate this one or integrate the other one, but the problem say integrated over the distance d. So let's take the integral for the acceleration to obtain the velocity and then convert to the Omega. So if you take the integral over the displacement of the A, you can get the relationship and then you can convert between the relationship in Omega and the v which is not R Omega, this is some certain other constant. What you can get finally is the relationship for the Omega. It looks a little bit complicated. You can also solve the same problem using the work energy relationship. So again, your friction force is now directly the opposite side. So first energy zero and work done is, work done by the P and work done by the friction, the displacement for the P, and displacement for friction. So how can I specify the XP and XF? Are these are same as the central half displacement d given? No. So due to the structure itself, there you have are displacement at the center, and displacement on the top, displacement on the bottom, and then these have some ratio. So d with respect to the central displacement, top parts moved more, here the Theta is minus sign, moved more with the capital R Theta. At bottom part, moved less by the small r Theta term, so you just use those value here. Now, the biennial kinetic energy and potential energy and mv square plus I Omega square, now are these all related as v equals r Omega? No, because it's two totally different independent values because the slipping occurs. So if you just equate them, but you can have this relationship and those are two independent values, those are very different. So the relationship for the two velocities v and Omega from this work and leisure relationships are equivalent to the solution obtained by the integrating equations of motion. Now, let's solve another problem which is involving other components that could hold the potential energy, the spring. The spring is much the same as a previous setup, and instead it's just pulling the external force P, we have the wheel connected by the spring. Initially, the wheel is initially stretched by the amount of X Naught, and after it was released, it rolls without slipping. So you're supposed to find what's going to be the velocity of the center. Okay. First set the coordinate as similar to the previous ones at the contact point with the wheel, and the ground is absolute coordinate. Then the contact forces are here and the spring. Okay. Again, so when you define the coordinate, there are many ways you've set the reference point. So here I set the reference point as natural length. But you can also set the initial x equals o position at the stretch it position. It's up to you. So no matter where you set the reference frame in the next direction, equation of motion will be equivalent. So I'm going to just set it as a natural length here. So when will split the center of mass, there are free-body diagram, normal force, friction force, mg. I don't know the friction force here, but depending on which way do rolls down or rolls up. So I just put it as this one, the positive direction. Then once you solve the problem, you can check if your direction assumption is right. Then the positive displacement, your spring force is going to be closer to the left-hand side. In the beginning, the spring was initially stretched. So maybe should we put the initial kx as this way, the right-hand side? Well, depending on where you define the coordinate, forget about the initial condition, just assuming there is a positive displacement occur with respect to the coordinate you set. Then draw all the forces with respect to their positive deflection and then you can just plug in x initial value as a minus x_0 later. When you're up to the point where you're drawing the free body diagram. When you are handling the spring, assuming the positive deflection, and then just draw the forces corresponding to the positive deflection. Now let's think about the case. So initially it's been stretched it out and it's been released. So it's going to be rolls up or rolls down. Well, it's depending on the size of the wheel and coefficients of the equation of the spring. Suppose that you have a really light, comparable, a relatively light weight, and then you have a really stiff spring, and it's really tough to stretch it out initially. So what it means is initially you have force of zero applied downward. Then in releasing is just remove the force. So this like the case, then it will definitely rolls up to the right-hand side. Suppose that you have a really large mass compared to the very compliant spring. Then due to the weight of the disk, it will be stretched out initially. So if you don't have any forces applied, those initial stretching point is actually equilibrium. So nothing will happen if you release them. Or if you want to make a movement, either you should apply the force downward more or push it back. So there should be some F component here, edited effort component here to make some rolled up or rolls down movement here. Then when you have some comparable mass in kx, then initial displacement will be determining the magnitude of the initial stretch displacement compared to this equilibrium relationship will determine the disk will be rolling up or down. So that's physical situation. Now coming back to the original problem, let's draw the free-body diagram. You can obtain the equations of motion and let's just briefly recap. If no slip condition, you can use a kinematic constraints, and then while you're not really knowing the force friction information, and for the slip case, you have a specified friction relationship. Those relations for the a and R force are generally independent, so unknown until you solve the problem. Based on the free body diagram, you can obtain the equations of motion in x and y direction, and a rotational equations of motion with respect to the center of mass. So center of mass you have a friction force, and the spring force applied a torque, which is I Alpha. What he says was rolling without slipping. So you can apply the kinematic constraints to the top or the bottom, so that your unknown friction force could be eliminated. So what you can get is the relationship about the acceleration in terms of all the other variables. So you can rewrite it as a function of a. Now you can integrate the a to obtain the velocity of o. Can we just do the time integral or the displacement integral? Well, we see that right-hand side, all those are model parameters except the x. It has the x term here. Because x is a function of time, you cannot do the time integral. So let's do the integral over the x. So let's do the displacement over the x. But here this is the x of the spring, and this is the velocity of the center of mass. So this one is actually x-bar. So when we integrate it, should we integrate over the x or should we integrate in the x of G. Let's do it over the center of gravity because that's how you can get this form simple. So to do so you should replace the x in terms of x bar, x center of mass. So from those triangular resemblance relationship, we can express the x in terms of x bar by the coefficient Gamma here. Gamma is capital R plus r, over capital R. So let's do the integral. So try to rewrite everything in terms of x bar instead of the x, and integrate over the x bar. You can have the for equations about the center of mass, velocity in terms of other variables in all the given variables. So this is what you can get. Again, let's do the same thing using the work energy relationship. Initially, you will have stretched the spring. So you have a potential energy due to the spring, and then it's been downward with respect to the reference point, which is added as a natural length of the spring. You have a minus potential energy by gravity. Now here this is the x naught, and this is x naught bar. This is about a center of mass displacement, and this is from the top of the spring stretched length. So you can just make it as in terms of either x or x naught. So to what would you formulate it? More like it doesn't really matter. So just stick to one thing and then just stick to that value throughout the formulation. So let us stick to the center of mass displacement x bar. So your x is going to be replaced by the Gamma x naught, and there is no work done, external work because the only external forces are the friction and the normal force, and that doesn't do a work. So final kinetic energy and the potential energy is going to be expressed by mv square, plus I Omega square. Then those are related because it's been rolling without slipping. So kinematic relationship will make it as a one unified form in terms of V or Omega if you need, and your potential energy is going to be explained by the x bar and kx. So also switch as in terms of x bar. You will have a total equivalence relationship about your center of mass, velocity in terms of spring potential and gravitational potential. This closed form is equivalent to the solution obtained through the integration of the equations of motion. In this part, we've worked on some examples about how we can apply work energy relationship or versus just integrating the equations of motion to obtain the relationships about work, and the kinetic energy. Some part of kinetic energy, either Omega square or the V-square. Next, we're going to do work on the other examples about their rods. Thank you for listening.
