Hello. In this chapter, we're going to study to obtain the equations of motion of rigid body which is under general plane motion, means translation and rotation combination. Before we start, let me briefly review that Coulomb friction. Suppose this is the block on the floor and there is no force applied, there is no friction. If I apply the force, there is a friction occurs. Then the mechanics of friction, by experiment, is observed to be varying. So if I applied a force and the force is not large enough to make the movement, the block doesn't move, that's what we call the static friction. When the piece actually reached to the certain amount, a threshold, which is maximum static friction, then the movement occurs and then whatever corresponding to the friction under that movement, we call it as a kinetic friction. So the whole friction profile looks like this and this region is what we call the static friction, which is less than mu_s maximum multiplied by the normal force, and this region what we call as kinetic friction which is exerted when the movement occurs, which is approximated as a mu_k multiplied by the normal force. Then, what would happen if you have a force P, I would say the horizontal component of the P greater than kinetic friction and less than maximum static friction? It really depends where those P occur in what process. Suppose that your P is like this amount, definitely smaller than the kinetic and the maximum static friction, then no motion occur. If you are just keep increasing your force and then reaches the point where you have a greater than kinetic friction but less than maximum static friction, then still there is no motion occurs because the force is not enough to initiate a movement. When it occurs to the maximum static friction, then the movement started. Then once the movements started, if your force is less than the maximum static friction and greater than kinetic friction, the box is still in motion. Therefore, if you are applying the force to reach to the maximum friction even though it's greater than Mu_k N, no movement occurs. If you are actually pass over those threshold, then the friction force even though it's a smaller than the maximum static friction, the movement could occur. So when you are handling the disk problem depending on the friction, mechanics of the forces and the friction, either slip occurs in the case where you have a greater than a maximum static friction so it actually motion occurs, then you can just think of this disk as just a box at the surface level. So the surface level, there is a slip occurs due to the movement here and looks like a movement to the right inside, there is a friction occurs to the opposite side. However when the problem says that disk doesn't slip or roll without slipping, then this point here is instantaneously zero speed, so that is a case where static friction occurs. So no-slip condition, you should specify the magnitude of the friction and the direction of the friction based on the resultants to dynamics. So you can now really specify them as a mu multiplied by the N at this point. You should actually solve those coupled Newton-Euler equation to figure that out, the direction and the magnitude of the friction. Let's just solve the problem. This is a hoop with a radius r released from rest on the incline. The coefficient of static and kinetic friction are mu_s and mu_k respectively and you are supposed to find the angular acceleration, Alpha, and the time to travel, the distance d. The very first step you should set the coordinate. I suggest you should set the coordinate and the absolute coordinate frame which is aligned with the incline and usually at the contact point of the disk and the surface like here. How many contact does this hoop have? Only one contact with that surface. Then before we start, when you are handling this disk problem, you should figure that out. There are two cases like if there is no slip occurs, means like your contact point has a zero velocity, you have a specific kinematic relationship between the angular acceleration and angular linear acceleration, such as a equals minus R Alpha. A is a direction for positive x usually the right-hand side and Alpha is due to design conventions like counterclockwise, so those minus sign came from. So if you define the direction of Alpha differently, those minus sign may be taken off. But anyway, the magnitude is going to be a, it's going to be R multiplied by the Alpha. As I said before, in this case a friction is usually unknown, so you should actually solve dynamics problem equations of motion to specify the friction. If the slip occurs is just a simple sliding, then your angular acceleration and your linear acceleration is R independent, so you can solve it independently from the equations of motion. Instead by losing the kinematic relationship information, you can get kinetic information which is you could specify the friction force as a kinetic friction mu_k multiplied by dN. So depending on the case if slip occurs or no slip occurs, you can use either of the information to solve the problem. Coming back to the problem, draw the free-body diagram so you have one context normal force and friction force and the gravity, and those are results and make a motion for the linear acceleration and the angular acceleration. If you obtain the equations of motion like a mathematical form of this pictorial relationship between force in the motion, you can get the equations of motion in x and y direction here. Then moment equals equations of motion respect to the center of mass like a torque by the friction and will generate the angular acceleration. So the Alpha is what you are supposed to find and actually the A's they are supposed to find the t over the distance d. Then friction and normal forces are unknown. So how you can solve it? As I said in the disk problem, you should split it out as a two different cases like if no slip occurs or the slip occurs. Suppose that you just assumed that no slip occurs, then the kinematic constraint that you can use is the relationship between the linear acceleration and the angular acceleration, which is a equals minus R Alpha. If you plug that into the equations of motion, what you can get is the equations of motion which is expressed everything by the Alpha or the a. So here, even though those friction forces are unknown, those are two equations and two unknowns so you can solve the Alpha. So if you eliminate the friction, you can obtain the Alpha value and not stop here, you should actually go further to double-check if your assumption for no slip will hold. So you can either plug-in those Alpha value here in the Newton's equation or the Euler equation, and then obtain the friction force here and compare this with the maximum static friction. Here this mu_ s is given. You can compare your value of the friction obtained and this maximum static friction. As far as mu_k is less than 0.5, then your assumption doesn't hold and the slip occurs. If that's the case, just solve the problem again by assuming of friction forces, kinetic friction. So it's a mu_k multiplied by the N and plot this friction information to here to obtain the Alpha, and here to obtain the a. So there are many different form of the problem could be possible like if you have a disk instead of have a free released like the one we just have solved, you may have an external forces applied through the cable which is wrapped around the center of the disk, or maybe the force is exerted with some data value, or maybe force is applied on the top of the disk so that the rotational motion have the opposite direction, or maybe the disk is on the incline so maybe some part of your gravity force will also [inaudible] to make a linear motion. Then there are some cases where the force is applied in the middle of the disk, or the bottom part, or even at the top like this. So to understand what's happening in the disk that understand the physics, we should split it out the translational motion and rotational motion separate. So for example, if you are applying the force to the right on the bottom part, maybe the rotations are maybe the counterclockwise, and then there are translation to the right-hand side. So depending on the relative magnitudes of those, your total rotational motion will vary. Suppose that you applied a large force here and then that or actually relative caused a small rotational motion due to the small moment arm, you have this situation. Suppose that even though you have a same force maybe your surface is really rough, so maybe that will reduce the acceleration, translation, or direction and they may cause a large rotational motion. Depending on that, your center motion is inherited from the left-hand side. Then due to the rotational motion, your top and bottom parts over disk has its translational velocity components. So those blue part, which is translation and the red part, which is due to the rotational motion can be superposed, and then these are the resultant motion. So as you can see, there is a large counterclockwise rotation of motion will be generated. On the other hand, if you have larger rotation compared to the smart translation, there are larger backward motion on the top and then forward motion on the bottom. If you superpose those translation apart and the rotation, you have the forward motion on the bottom and the backward motion on the top. So even though, so maybe this is somewhat the standard instantaneous center of rotation. Overall, rotations are similar to the top part, that the counterclockwise, maybe the rotational magnitude will be smaller. Or sometimes, even though you have a positive, the right-hand inside translation at motion. But there's also a clockwise rotation, then your superposition will give you the rotational motion in the clockwise direction. When you are actually doing the superposition at some point, at some specific combination will give this the resultant velocity at ground contact point to be zero, that's the special case where the rolling without slipping occurs. There's a very, very special case. So when you actually practiced that disk problem by yourself, check out for many different cases like what if the force is applied are became doubled, greater or Theta becomes smaller or what if the surface is either incline or deflate and the friction coefficient gets larger, et cetera. To understand the physics underlying and also be able to solve the problem for various cases. Now, let's solve the problem here. This is the disk which has a radius of R, and then there is inner part has group, and then those are wrapped by the cable. Then I'm pulling the cable with the tension t, with the up incline angle Theta. Then the radius of gyration z is given. The radius of gyration by definition is, its moment of inertia is equal to mz square. So you are supposed to find the angular acceleration and the acceleration of the disk at the center of mass, and then also supposed to find the friction force. As I said, let's set the coordinate in absolute underground, which is coincide with the ground contact point. How many contact does this have? One with the ground contact and one with the external chord. So there are two contacts. So there are normal force, and the friction force, and the tension, and the gravity. As I said, since the motion hasn't been specify, you should split it out, you should split out the two cases, if slipping occurs or no. So those forces are generating linear and angular acceleration. So pictorial form of the free body diagram can be turned into transformed as a mathematical form, which is equations of motion. So x directional equation motion, y directional equations of motion and rotational equations of motion with respect to the center of masses who need the torque by the tension T and the friction. So tension T and the friction. These are what we are supposed to find, accelerations and the friction force. So let's assume first no slip condition occurs. So if no slip occurs, there are kinematic constraints between the a and the Alpha, which is minus R Alpha. So if you plug that into the equations of motion, you can obtain the equations or in terms of the Alpha or in terms of the A. If you eliminate the unknown friction, then your Alpha could be solved, and this is what we can obtain. Do not stop here, you should actually check further if you're no slip condition assumption will hold. So check if your friction obtained from this equation of motion is less than the static friction here. So this is what I got from the equations of motion and this is the static friction, just multiply by the normal force with the static friction coefficient. Here, just by plugging the value, you can compare them, or friction is here definitely minus value. Usually, cosine is greater than sine, so minus value Mu_s is less than around 0.5. But still, it has a positive term added. So it seems as f_s max is greater than the f. So this assumption holds. So no-slip condition occurs. So the answer is fixed. Or maybe in case if your assumption doesn't hold, then you should solve the problems again by using your slipping condition. If that's the case, you're friction turns out to be a kinetic friction. So you just multiply by Mu_k and n, and use those value, plug their value here. Here, to obtain the Alpha and the A. Not solve another problem. This is, looks like a disk problem, but a little bit different from what we have just solved in the previous two slides. Disk is of mass capital M is pivoted to the cart of mass M. Small m is pulled by the external force P, which is through the cord wrapping around the rim of the disk. So you're suppose to find the angular acceleration and acceleration of the system. So where should I set the reference coordinate? Right in the incline here? No, because if that's the case, because the box keeps moving forward and the disk is rotating with respect to the center of the card. So in this case, I'm going to set the reference frame at the center of the system. So that means, this whole system is moving forward with the acceleration A0. So this is non-inertial frame. So to draw the free body diagram, the center of mass is located at the center. The whole system has only one contact to the environment, which is a inclined guide. So those total sum will be the friction force and the normal force, and external force P and the gravity, and that all jet. Since I set the reference frame which is accelerating at the center of the card, which is moving with a card, there is inertia force, fictitious force is applied at the center of mass with the amount of mass multiplied by the A. So there must be a reaction action force between the pivotal points of the card and the disk. Should I include this for the free body diagram? It depends on, if you separate out the disk part and the card part separate when you draw the free body diagram, of course, those are the contact point force for the disk, so you should draw it. However, if you are considering the card and the disk as one single system, those reaction forces are considered to be internal force, so those are canceled out, so you don't have to draw it. So those are generating the resultant motion, linear acceleration, and angular acceleration like this. So transfer this pictorial expression for force and motion relationship into mathematical form, you can get equations of motion in x and y direction here. As I said, since I put the reference frame on the card, so I don't have any acceleration. Only the acceleration term are considered to be an inertia force to the left-hand side. For the rotational motion, since the card doesn't rotate, I only have to consider the disk part. For the disk part, only external force that will generate the torque is that P term, so I have a torque generated by the external force P, will generate the moment of inertia of the disk. Now I'm supposed to find the [inaudible] and the accelerations. So if everything is given, I can just simply solve this equation to solve the [inaudible] and this equation to solve the A. Here, I have to figure that out, the friction. But if the problem said it's a smooth surface, then I can ignore the friction, then everything is done, it's pretty simple. It's not about slip condition or no slip condition because disk is not on the surface. Disk is on the on the air with the connection in our pivot with the card part. So you don't have to go through slip condition or no slip assumption part. You can just straight forward, calculating the Alpha and the A. One more. What if I could separate out the card part in the disk part separate? What will happen? Then the card has two contact, one to the environment, one with disk, one with the linear guide. So there are forces like normal and friction force from the linear guide, and the contact force with the disk and the mg. Therefore, the disk only contact through the environment is with the pivot at the card. So there's only action-reaction force and mg and the PO, external forces. So those are the free body diagram of separate card and disk. Since we set the coordinate at the non-inertia coordinate frame located at the center of the system, inertia force applied for the disk and the card. If you compare this one and this one, the equations of motion based on the whole system, you could see that those are equivalent. That ends the Part 1 of general plane motion of rigid body, mostly focused on the disk type of problem. Next time, we are going to study the rigid body, which is longer shape, so its center has been translated in the rest of the part could rotate. Thank you for listening.
