Hello everybody. In this lecture, we're going to continue to study how to obtain the equations of motion for the general rigid body motion, which contains both translation and rotation. In this example, we are going to handle the rigidest body with the structure of the long part, like a rod, which could be expressed by the relative motion, rotational motion and translation. This is the problem with the bar AB with the mass of m, is moving through the linear and vertical guide with the roller, which means it's a very smooth surface, you can ignore the friction. So the external force P is applied so that a rod starts moving to the right in the top part, and then that means that bottom part will goes up. So you don't have a clockwise rotational motion, as well as the translation. So you're supposed to find the acceleration and the normal force between the roller and the vertical guide here. Excuse me. Where are we going to insert the coordinate frame? Somewhere here at the fixed-point. Well, the bar looks like this and that as time goes by. So if you actually set the coordinate frame as the fixed-point, there will be a little bit difficult to describe the motion. So I'm going to set the reference frame at point A, which is moving with the roller. In that case, what you should know, this may have an acceleration in it. So this is non-Newtonian coordinate frame, so you should consider the inertia force. It has two contexts, up and bottom. So there are two forces, normal and friction, normal and friction on the bottom and the gravity and the external force f. As I said, since this is a moving coordinate frame, you should also include those inertia force at the center of mass with the magnitude of ma. Those forces are generating the motion in x and y-direction acceleration and the rotational motion, sign convention as a counterclockwise is positive. So once you are done with the [inaudible] body diagram, transfer your pictorial relationship between the force and the motion into the equational form. So you have all the forces and gave you the center of mass, x-direction acceleration, y-directional forces, center of mass, y-direction acceleration. Now, you obtain the rotational equations of motion with the moment at G, that center of mass. So there are three force components generating the torques and two force component generating the force. In total five force moment components exist that'll generate the I bar, our bar, with respect to the center of mass. Now, this is smooth surface, you can simplify by eliminating the friction and you are supposed to find what? Those are unknowns, and you're supposed to find the normal force as the Alpha. But it still seems like there are too many unknowns here, so you should calculate N, you should know mg and MAY directional acceleration. But this is unknown and same for this one. If you want to find the Alpha, there are unknowns N and NB, which also contains those. So how you can solve it? In the previous disk solution, what we have gone through is, we were relating the Alpha and A by using no-slip condition, right? So like this, we may be able to relate this translational acceleration in terms of rotational acceleration using the kinematic constraints. So if we could relate them as an Alpha, maybe this equation would be solvable. Also we derive the moment equation with respect to the point G, but we can put any arbitrary points. So we can put M with respect to the O, to calculate the I with respect to the O. Or here since we set the reference frame at A, a moment with respect to the A is going to be iA Alpha. Note that in that case if A is moving with acceleration, all those moments of motion enclose the moments by the inertia force. So there are one, two, three, four force component that will generate the moment, normal force, friction force, the fictitious inertia force, and the gravity. So either you can use this one or the other one. Sometimes if you set a coordinate frame here, you can eliminate some of the moment equation by unknown such as N. Now again, so these are what we are supposed to find and then we'll see if we could relate these linear translational acceleration, in terms of rotational motion. In Chapter 5, what we had learned is we learned how to express the motion of some certain point with respect to the relative coordinate frame. So this is what we want to know, and can we express it with respect to the relative coordinate frame A and its relative acceleration in terms of pure rotation? So since if I look at point G with respect to the A, that's going to be a pure rotation. So their acceleration will be either polar coordinate or tn coordinates. So easily expressed by the rotational kinematic parameter like Omega or Alpha. Since we don't know the a of A and Alpha, cannot go further to that actually. But instead we are expressing acceleration B, which is constrained vertical, there might be a solution there. So you just rewrite this relative coordinate kinematic relationship of the point B as A of B is going to be a of A plus aB with respect to the A. B of A is purely rotational, so it'll be either tn components of acceleration, or the polar acceleration. So since this is the problem when bar start to moves from rest, so Omegas could be set to be zero here. Since this one has been constrained, aB is vertical only, aA is horizontal only. So even though we don't know the Alpha, by solving the vector equations in i and j components, line by line, we might be able to figure that out what's going to be the unknown a of A and Alpha. So if you equate them, you can solve this by having the i component to be equal. So we can express a of A in terms of Alpha. Also if we know those A of A in terms of the Alpha, we could find the A of G, the center of mass in translation and acceleration in terms of Alpha here. So by having those in terms of the Alpha, all those [inaudible] force and [inaudible] could be expressed by the Alpha. So if we plug that in back into this moment equation, this whole equations about the Alpha in some known variables like external forces and length of the bar. So we could solve the Alpha and then later, find the normal forces. Let's solve another problem. Now, this one has a two more links to problem. So the bottom part has a roller on it and then moved on the horizontal plane, so it's this motion is constraint as a holding on all, and then this point A is connected to the other bar which is pivoted at O and then rotating with a constant of angular velocity omega note. So we are supposed to find the normal force from the floor or normal force here, and assuming someone surface means no friction. Where's you should reset the coordinate, coordinate the fixed coordinate frame? Since this is what we are supposed to analyze bar A, B, because we are suppose to find the normal force is here. If we set the coordinate here, it's really just typical to describe the motion of AB because it's been linked to the point A which is moving. So I'm going to set the new reference frame at A so that I can describe this in terms of relative coordinate frame. Note that this one is also moving with the acceleration, so you have to consider the inertia fictitious inertia force. Now, if you focused on bar A, B, how many contexts? One and two. So two forces per each, one and two here in the gravity. Then note that there is inertia, fictitious inertia force because we set the coordinate frame at A. Here, since omega naught is clockwise and then data we define as a clock wise direction, even though we have an x and y as a conventional way. Lets us do the positive Alpha and the app was away. It's just up to you define which direction as a positive. Just stick to it throughout the solution. So based on this pictorial relationship between the forces and the motion, you can get the equations of motion in x and y direction, and you can get the moment equation with respect to the G there are four force is component that will generate the rotational torque. Now, when you are supposed to find the normal force at the bottom and we'll see this is a free smooth so friction less and RFI's unknown, all the translational accelerations are known and reaction forces X and Y's are unknown. Wow. There are too many unknowns. How we can solve it? If similar to the previous problem, if the rigid body is expressed by the translation and rotation, maybe this translational acceleration could be combined with the angular motion. So let's work on with the kinematic constraints in the next slides. Before we go further, we could also instead of setting the moment equation, we [inaudible] the G, we can also set the moment equation with respect to the A, which is same as IA alpha as far as you consider all the inertia force if the point A is moving with the acceleration. If we set the coordinate at A, it will be good because you don't have to worry about the moment equation by this reaction force RX and RY, which are unknown. So it looks a little bit simpler like you have on MA term here and term Alpha term here. Again, you are supposed to find the normal force and we don't know the A of A and Alpha. However, A of A is a part of the body pivoted at O which is rotating with constant omega naught. So the acceleration at point A is only existing only in the central pure acceleration, so this is known. So what we are supposed specifying the alpha from the kinematic constraints. What we have learned in Chapter 5. So now problem is Olivia switches like a find the Alpha. So let's use the relative coordinate kinematic relationship. So to describe the velocity at p we want to express it in terms of A and P relative to the A which can be expressed as a pure rotation. So velocity here is tangential to this people did far OA and velocity B is somewhat to the horizontal direction, and velocity relative with respect to the A is pure rotation is tangential to the bar. So if you express them all you can have a vector equation, and if you could match the I component and this I component and the J component with the J component, you could find out what's going to be the unknown omega AB and the velocity at B in terms of known values. Similarly, we can do the acceleration as well. Acceleration B can be expressed by the acceleration A in its relative acceleration with respect to the A which is a pure rotation. So there are TN components of the acceleration here, and then A acceleration point A this is the circular motion with constant angular velocities. Only centripetal acceleration exis, t and also the acceleration P to somewhat horizontal direction. So we can express this one as a pure rotation relative motion with the finite directional constraints here. So if we write it as a vector form, we could find the I components, and so J components separately, and obtain the unknown Alpha in terms of omega AB here, which is again expressed by the all the given variables. So recall that we want to use this Alpha to find the normal force here. From the moment equation, we are finally able to solve the normal force acting on the point B from the surface. This ends the general plane motion examples about the body which is a longer shapes. So we had to combine translation and rotational motion. Next time we're going to study work energy relationship. Thank you for listening.