[MUSIC] Hello, in this video we are going to study how we can apply work and energy relationship to the rigid body. Especially with the shape of longer one, like a bar. Okay, basically, when we are going to apply the work-energy, it's pretty similar to the one that we apply for the particle. But the thing that there is a little bit different thing for the defining the work done. Is, suppose that you have a particle of mi. And then the work done for that specific particle is defined by the Fi applied to the particle. And the displacement of that specific particle. Suppose that the force is directed this way. And it's moved around this way like a dri with the angle phi. Then your work done for this specific particle of the i is going to be Fdr cosine phi, okay? Now, dr is actually if you're defining from some fixed point O or the center of rotation. Then it's going to be dr d theta. So if you switch dr as r d theta, and let's also switch that cosine phi here with the direction to the r. Those angle is going to be 90 minus phi. And then it's a trigonometric formula, it's sine 90 degree minus phi angle, right? So if you have Fr sine angle between the force and the partition from the O point, O to the mr, m of i, that's what? A cross product formula, right? In the direction all the work done is going to be a positive, a scalar value here. Then this underlined one will be replaced, rewrote by r cross F, r cross product F. The direction is towards the counterclockwise as a positive here at the counterclockwise direction. So that the d theta is always a counterclockwise so the dot product is going to be just a single plus one. So, this r cross F is the moment where is ought to apply to the particle mi with respect to the O, right? So the work done, which is defined for Fdr for the particle is now going to be defined by moment in d theta for the rigid body, okay? So work done could be obtained by either integrated F over dr, or integrated moment over d theta. So the work done for the whole rigid body is going to be work done for a single particle and it's summation over the whole rigid body. So if I rewrite it, the work as fi di integral, and again, it's going to be mi d theta for a single particle. And then integrated over the i, here the theta is in common. So the total work done for the rigid body is going to be simply moment and d theta integration. Which we are going to use throughout the slides to solve the problem. Now, how about the potential energy? The potential energy by the gravity is for the single particle is mi gyi. And those yi could be splitted out as a vector to the center of mass and from the center of mass. So it's going to be y bar plus rho i. If you do the sum over the whole rigid body, the front part became a potential energy with respect to the center of mass. And the latter part is going to be cancelled out due to the definition of center of mass. So finally you will have a potential energy is the same as the potential energy of the center of mass treated as a single particle. Now, let's solve the problem of the rigid body. Let's think about the bar of mass m and length l is released from the horizontal position like this when theta equals 90 degree. And if we release it, it'll just swing down and then what's going to be the angular velocity at vertical position if theta equals zero? Of course the coordinates should be placed at the hinge point. And then the force is only, this one is only contact of the bar to the environment. So there's only x and y direction of reaction force and the gravity and that's it. So that will generate the translational and rotational motion. So your equations of motion will be F = ma in x direction, and then too, y direction of forces will generate the y direction of acceleration. To obtain the moment equation with respect to the G, the center of mass, the moment is generated by the Rx and Ry. So you have this moment equation with respect to the center of mass. Or you can also write the moment equation with respect to point O. And then note that you have to match in the subscript left and right. So if that's the case, it's a little bit simple. So the only moment that'll generate the rotation with respect to O is the mg. So minus one-half l mg sine theta is going to be IO alpha here. Now, the problem asks you to find the omega. So either we should find the alpha and then make the integration to obtain the omega. So you have two options, either integrate the upper one, the moment equation with respect to the G. Or with respect to the O, which one would you choose? To integrate the moment equation with respect to the Gs, you have to solve what's going to be the Rx and Ry here is unknown. So I'm going to choose the second one which have all the known variables for the left hand side. So if you take a integral over the theta, is all constant, so it can comes out of the integral. So like what I could have ultimately is going to be the work done by the gravity will induce the change of the kinetic energy. One-half I omega square. And note that, the rigid body kinetic energy is m v square plus I omega square. And that's with respect to the center of mass, okay? Here it's I of O, and which is equivalent to the translational kinetic energy. One half MV squared, plus I omega square. Now, we can solve this using the work energy formula the simpler way. So initially, no work, no energy and then there is no work done. So the final state is going to be kinetic energy and potential energy, potential energy drop. And the kinetic energy here and v in omega are dependent because it's a rotation w.r.t to a pivot point. So, the v could be expressed by the R omega here, R is going to be a half of the length of the bar. So, everything can be expressed by the omega and then those are actually I of O. And those solution obtained from the work energy relationship will be equivalent to the work solution that you obtained by integrating the equations of motion. Now Okay, this is the final solution that you obtained by integrating the moment equation. And what if we could integrate this moment equation obtained with respect to the G? As I said before, those Rx and Ry's are known, but if we anyhow resolve those unknowns, is this equivalent to the integral, integrated value for M with respect to the O? That's what I want to ask you to think about. So to resolve this Rx and Ry, which is also the ax, function of ax and ay, maybe you can change the ax, unknown ax and ay in terms of theta, omega and theta double dot by switching the coordinate as a polar coordinate or TN coordinate, the rotational coordinate. Whenever it's rotating, those acceleration could be expressed by the centripetal, some part vector portion of a centripetal acceleration and tangential acceleration. So if you plug that in, into these equations, what you can have is everything expressed by the theta double dot or theta dot. So that's kind of solvable solution. So if you are doing this, you can have the formula in terms of the theta dot squared omega squared. So what you can get is actually the gravity work done is going to be resulting in kinetic energy change with respect to the O. So those are also equivalent, so, You can either integrated the moment equation with respect to the G or moment equation with respect to the R. Now let's solve another problem. Instead of having the pendulum, the bar is actually fixed in the center with the pivot point. And those are moving with the vertical guide. And at the end of the bar is having a roller on it. And then it's also moved in the horizontal guide. And it's been released from the 90 degree, upright position, and what's going to happen? What's going to be the angular velocity at the position theta? I can set the coordinate here, the fixed coordinate. If you set the coordinate either A or B or C, that's our moving frame, so you should consider the inertial force. So I set it as our fixed point here. Okay, here are two contacts of vertical guide and then the horizontal guide. So there are two contact forces, normal force and frictional force for each, and the gravity, and that's it. So those are generating translation and rotation acceleration. So your equations of motions are having two horizontal forces and three vertical forces. And moment equation with respect to G is only moment by the friction and the normal force from the bottom, okay? Now, since I'm supposed to find the omega, either finding the alpha or the acceleration would help. And it's a smooth surface, so the friction goes out. So what I have is only unknown, the normal force. So if I have a solving these two equations as a system, then I might be able to eliminate the unknowns in and obtain the equations for the alpha and the a. So if I integrate it and then subtract them, what I have is everything in terms of a acceleration and the alpha. If this acceleration is going to be a function of rotation like a theta, or theta dot, or theta double dot, I might be able to integrate the whole equation. So let's say how we can obtain the informations about the a of A, a at the center of mass. So it says in our center of mass. To solve this, we will use the relative coordinate approach, what we have covered in chapter five. So a of G, center of mass, is going to be a of A plus relative acceleration with respect to that pivot point, which is going to be expressed by the rotational motion. a of G is confined as a vertical motion, and a of A is confined as horizontal motion. And aG over A is relative accelerations are purely like a rotational motion with respect to the A. So you can express it as a TN coordinate or polar coordinate as a function of theta. So what I can have is unknown a of y is going to be unknown a of A and the tangential and normal components for the relative acceleration. It seems like there are too many unknowns because I don't know a of y and I don't know a of A, but this is a vector form. So if you equate the i term to the i, j component as a j component, you can replace a of y in terms of all theta rotational variable. So if you plug that into the original integral equation, what you can have is everything in terms of theta related term, theta, theta dot, and theta double dot, As an integral. So you may be able to integrate them, okay? This one is a constant. And then this one is actually the function of theta. So it look a little bit difficult to integrate over this theta. But there's a trick. Those relationship, if you actually rewrite them, rearrange them back, it's going to be expressed as a closed form of the d (theta dot cosine theta squared). So this is solvable. So what you can have finally is going to be mgl (1- sine theta) and one-half ml squared (theta dot cosine theta) squared is going to be same as one-half I omega squared. So by integrating the moment equation, you were able to obtain the relationship between the gravity energy and the kinetic energy. Now, let's do it again with the simpler form, like a work-energy relationship, like a force energy 0, there's no work done, smooth surface. Final energy will define our kinetic energy plus potential energy. In the kinetic energy here, v and omegas are not independent. They are coupled in a way with respect to the center of rotation. Okay, so this velocity is confined to the vertical. This velocity confined to the horizontal. So you can obtain the position for the instantaneous center of rotation. So the vB is going to be expressed by the omega here, which is the same omega as this omega, which we covered in chapter five, and the distance, which is l cosine theta. So everything could be expressed by the cosine theta term. So what you can have is a gravity energy, potential energy is going to be a kinetic energy with respect to the center of rotation. And this equation is equivalent to the solution you obtained in the previous slides by integrating the equations of motion. Now, let's develop the problem a little bit further. So it's exactly same problem, what we have just solved in there. What if I add a spring component at the end? Okay, spring is actually have some clearance distance from the end of the roller and the end of the spring. So initially, up to the point where this point A has actually reached the end of spring. Equations of motion, energy relationships are exactly same as what we have just solved, okay?. As far as actually touch the spring then there will be a spring resistive force applied, right? So, you have exactly same free body diagram and added forces which is k x. And same for the moment equation with respect to the g as moment by the normal plus friction and the spring forces. And then u, this one is a unit step function that all turns out to be zero when the theta is less than theta dot. Greater than theta dot, sorry. And the way as far as theta is less than theta dot, theta star. Which means the contact, the theta where the end of the roller here is actually touching the spring. And then those kx part will kick into the equation. And those are smooth so you can actually integrate them. This to obtain the work energy relationship. Or, more simpler way you can apply the work energy relationship. Exactly same like your final state in a kinetic energy doesn't change. And your potential energy, not only for the gravitational potential energy, you will now add a spring potential energy. That's the only difference. Now, let's develop the problem a little bit further again. Now, it's not a free motion anymore, now you're applying the constant torque M at the end of the bar, z.z. And initially, it's been located at this tilted position by a stopper here, located here. And there are friction exist of the surface like horizontal, vertical guides has a friction. Then what's going to velocity at upright position? That's the problem you want to solve. Again, set the coordinate at the fixed point here, it's in line with the vertical guides. So the free body diagram is exactly the same, except you should add now the active moment applied on it, active torque applied on it. And those will generate the translational motion and the rotational motion. So, this one is the only difference compared to the problem that we set to solve. Now, since we are supposed to find the velocity, either acceleration or alpha formulation or help. And then there are unknowns and then here, this is not a smooth surface. Those friction components are always your unknowns. So if we integrated the moment equations like we have done before, the problem is this one is unknown and this one is also unknown. The friction energy laws were done by the F, the friction, the horizontal guide and the vertical guide as well. So just by integrating this one there are too many unknowns. So you need more information. So in this case, let's also use the Newton equations as well here, which contains another friction, the vertical friction as well. So if we put the integral sign on it, you will have mg F2 normal force and the friction force over dy. And you have a moment equation integral, normal force, friction force and the moment over the d theta. And note that this term, okay, this friction and normal force change over y, the partition y. It varies, is not a constant value anymore. This N and F also varies over theta. Therefore, you can't actually solve the integral for this part so skip it as it is. And then to actually have some simplification, here this is N term, this is N term again. This one is integral over dy and this one is integrated over d theta. So let's express the dy in terms of d theta, so that we may be able to simplify the equations again. So if we rewrite them, you have integrated value for the constant mg and constant moment to torque. And still keeping those integrated part of the friction and the normal force. But here this Nl cosine theta d theta is here and there so you can eliminate them. So if you sum them up, what you can have is a friction work. They should be formed to be an integral, this is a mistake. Integrated friction force which is given as dE here, okay. And the energy work done by the gravity and the moment will change. Will result in the kinetic energy change here. And we square translational plus I omega square rotational energy. Okay, instead of integrating the equations of motion. We can also obtain the same relationship using to simply applying to work energy relationship. So the initial zero, energy zero, work done by the active torques applied in the friction. And that'll generate the later part for the kinetic energy and potential energy. Again, we restrict the center rotation those v and omegas are not independent anymore. So, for the direction, for the velocity here and there the horizontally guide and vertically guide, you can find the center of rotation. And the omega here which is the same omega with this one. And then the distance will be like l cosine theta here. So your kinetic energy at the end will be expressed by everything in terms of omega, okay? One step further is if the theta is a specifical in 90 degrees, those term goes to zero. Okay, so your work done by the moment and the friction force is gonnao be energy at the end. The potential energy and the kinetic energy of the system. And this relationship is exactly equivalent to the solution where we have just obtained by integrating the equations of motion. Okay, let's work on ME question. Okay, so if the previous problems we set the coordinate here as a fixed coordinate. But what if we set the coordinate here, okay? See, since when you actually obtain the moment equation it doesn't matter where you set the reference frame, right? You can have it as O or you can have it as G, the center of mass or you can have any arbitrary point. So, I can actually set the moment equation, with respect the point Z. But note that it reads that the coordinate here, this is a moving coordinate with the acceleration. So I have to consider inertia force. So, if I set the coordinate here instead of the bar in this the ground floor, then I should add a force with the amount of ma of c applied to the center of mass. Okay, if you obtain the moment equation with respect to the to the m of c, then if you integrate this, or those relationship will be equivalent to the moment equation what you have with respect to the G, right? So, as a mechanical engineer, I would strongly recommend you to obtain the moment equation to the other point here, point Z. And check if those your answer is going to be equivalent to the moment equation with respect to the G. Okay, so that's the part two, what we had just covered. So either integrating F equals ma over the displacement, or the moment equals I alpha over d theta. We can obtain the work energy relationship. I strongly recommend you to go over two steps, obtain the equations of motion and integrate it. Or simply apply the work energy formula and check your answer. Thank you for listening.