Hello. Last time, we learned how we can describe the motion of rigid body in terms of translation and rotation. So there are angular velocity comes in. So before we actually start with the examples, I'm going to briefly go over the review of the angular velocity. Suppose that you have a rigid body point A and B and their velocity profile looks like this. Then you can tell that it looks like there is a translation and rotations combined. If you translate it in terms of pure point B, relative point B, then you can describe the point A as a rotational motion with respect to the B. If you interpret this as a translation of the A and the rotation with respect to that point, that's going to be angular velocity with respect to the A. So in this case, this two Omegas, like Omega at the pure point B is going to be same as Omega as pure point A. That's what we're going to go over in this slide. So we actually learn that a rigid body motion can be interpreted as a relative coordinate frame. We suppose that VA could be splitted by the velocity B in its relative velocity with respect to the point B, like this. So in this case, those V of A with respect to the B is described by pure rotation with respect to the pure point B. So it's going to be Omega A with respect to the B cross-product r. Now we can also interpret the motion of the point B of the rigid body with respect to the A relative to the A. So it's been translated with A and then there is another new pivot at the point A and the rotation r with respect to that. So in that case, the relative velocity V_B over A could be interpreted as a pure rotation with respect to the A. That's going to be expressed by the Omega B with respect to the A cross product of r. But in this case, the r vector is minus r for the blue one starting from B to A. So if you go further, if you just rewrite V_A in terms of relative to V_B, you can have V_B plus V_A over B plus V_B over A. Since V_B's are on the left and right hand side, it's been canceled out. So what we can have is relative velocity V_A over V is the same as V_B over A. Then if you express it as a pure rotation, Omega A over B cross-product R is going to be same as omega B over A cross process minus r. So if you equate them, you will end it up getting the relationship between the Omega defined from the point B is going to be same as Omega defined from point A. So in the rigid body, if you pick out two points, Omega from one is going to be same as the Omega from the other point. Now, once I know that there is two points in the rigid body, then Omega A is going to be same as omega B. Then how about the another, the third point on the rigid body? Is it going to be also same as Omega O as well? Starts to drive it. We are going to go through the similar step. Well, we have just gone through under the previous slides, try to express the V of A relative to the point O. So if you split it out, the velocity of V_A as a translational of V_O and the pure rotation reverse to the O. So if that's the case, V_A over O is going to be Omega O cross-product to r_A. Similarly, you can express the V_B with respect to the point O. So split it out as a translation of V_O and a pure rotation reverse to the point O. So the relative velocity of point B with respect to the O, is going to be Omega O cross-product of r_B. Now subtract this two and what you can get is V_A minus V_B is going to be, you just canceled out this term, and what you can get is Omega not cross-product R of A minus R of B, which is our vector starting from B to A. So is going to be Omega O cross r vector. Now we can derive V_A minus V in a different way by setting the reference point as V here. So by definition, V of A is going to be expressed by relative velocity with respect to the B, like V_B plus V_A over B. Now, this is pure rotation, we reverse the B, So it's going to be Omega A over B cross-product r. So if you rewrite them in terms of V_A minus V_B, what you can have is Omega A over V cross-product r. Since this is the same value, this is going to be equal. So what you can get is this going to be equal. So if you make an equation, then what you can have is Omega naught is going to be same as omega AB in Omega BA. So as a result, if you are picking any point of the rigid body, the Omega defined by that point is going to be the same. Now, I know that any two points in the rigid body is going to be same. But how about the Omega, which you could define from the center of instantaneous center of rotation? Okay, suppose that your point A and B of the rigid body has a velocity profile like this. How you can get the instantaneous center of rotation? Okay, make V_r perpendicular lines for each factor and that intersection will be your center instantaneous center of rotation. Now, let's go over the similar step that we have just gone through in the previous slides. Try to express a V_A with respect to the point O. Okay. It is a relative velocity. So V_A is going to be V_O plus relatives velocity of the point A with respect to the O. But here, this point here has a zero instantaneous velocity. So this goes to zero. So V of A is actually the pure rotation with respect to the O. So it's going to be Omega CoR cross-product R_A. Same for velocity V. Velocity V, write them down with respect to the V_O and then the relative velocity V_B over O is going to be a pure rotation with risk to that point expressed by Omega CoR cross-product r_B. If you subtract these two, what you can have is V_A minus V_B, which is Omega CoR cross-product r_A minus r_B vector which is r vector starting from B to A. So V_A minus V_B is now Omega CoR cross products r, and we can also derive same a relationship about the V_A minus V_B using the V_A at the relative coordinate at V. So since we defined pure point V as on your reference, relative velocity V_A over B is going to be Omega A over B cross-product R. So what you can have is V_A minus B is going to be Omega A over V cross product to r. Since these two terms are equal, what you can have is these two terms are equal. So what you can have is Omega CoR, whatever the Omega defined round the CoR is going to be same as Omega of the any points of the rigid body that you have used to generate the instantaneous center of rotation. Okay. So this is actually pretty make sense because the instantaneous center of rotation means instantaneously, those are the point of the virtual rigid body that contains point A, point B and point O. So since these are the three points in the same rigid body, Omega defined from any of these points should be equal. So let's do an example like suppose that you have a two bar linkage. One bar is rotating with Omega naught and the other bar is rotating with Omega AB, then in general, those are not necessarily the same because these are to two different rigid body. Suppose that you have a velocity vector A and B like this, and then your center of rotation will be the intersection of this point, and suppose this is coincide with the pure point O. So what they mean is we can think of point A,B and C are the same three points of the virtual rigid body at that instant pivoted at C point. So in this case, Omega CoR is going to be same as Omega AB which is Omega BA, and here, same as Omega naught. If I just a little bit tilt here that the bar to bar linkage like this, then is a same manner in general Omega naught and Omega AB are not supposed to be the same in general, and then suppose that the velocity here is like this and velocity there is like that, then that's the point that the instantaneous center of rotation. So what that mean is point A, B and C are instantaneously three points in the same rigid body, virtual rigid body. Therefore, any Omega defined from point C or point A or point B will be the same. However, here in this case, we missed the point O here when we established the virtue of rigid body at the instantaneous center of rotation and C therefore, Omega CoR is going to be same as Omega AB and even Omega BA, but not necessarily the Omega naught. Okay. So this is a brief overview for the different angular velocities of the rigid body, and the next time, we are going to actually work on the examples.