[MUSIC] Hello today, we are going to study how we can describe the rigid bodies motion in terms of absolute coordinate frame and the relative motion relative coordinate frame. Most often when we are handling the particle, it's pretty simple. So absolute coordinate frame worked well. However, for the rigid body most of the cases the motions are combined by the translation and rotation. So it's sometimes it's often easy to use the relative coordinate. For example, like when you are handling the the disk which is rolling without slipping, sometimes it's easy to describe the motion with respect to some reference point either the center of rotation or the center of the disk. If you have a pulley or the gear or the hub disk, which has a two different radius also easy to describe in a relative coordinate frame. Sometimes you have a multiple linkage system. It's also helpful to approach using the relative coordinate frame. So let's go how we can describe the relative motion. To make the long story short the ultimate result that I want you to know is actually the V relative. At the relative coordinate is similar to what we have just gone through in the fixed coordinate like Omega cross R relative. So this is the relationship, but you can describe the relative velocity. So let's suppose that you have a particle A, which will be described from the absolute coordinate frame O. Ra, and then set the new reference frame point B, and then the position of rA can be divided by rB plus relative position with respect to the B, which is rA-B, which is equivalent to r the rest of the slides. So in Chapter 3, we learn how you can describe the relative displacement, velocity and acceleration in terms of the motion of the B, absolute motion of the B and its relative motion. Now, suppose this rigid body A is now a point of a rigid body and then it's moved by combinations of translation and acceleration from the black solid line to the blue solid line. And therefore its just different amount of movement for the point B and point A. If the movement of point A is divided by the translation, which is same as the translation of the B, and the rotation with respect to the B. Those motion displacement change relative displacement change Delta rA with respect to B could be expressed by the r. The distance between A and B and Delta Theta the rotation pure rotation with respect to B. Therefore your relative velocity will be dr/dt. And then the r is our delta theta. So if you calculate them, what you ended up is actually the omega cross r here. Here, omega is absolute angular velocity. And r is the relative displacement of the A with respect to B. So similarly to the the case where we have covered in the previous video saying that the at the fixed point your velocity v in the rigid body is expressed by the omega cross R actually applies same for the relative motion approaches. The only difference is here previously, those reference point is a fixed coordinate, pivot points. Here B, the pivot point B, is any arbitrary point you want to set it has a new reference to describe the relative motion, okay? So finally, I want you to know the B relative is could be also described by the omega cross r. Now, the rotation we can discuss further about how we can set the center rotation of the rigid body using the concept of instantaneous center of rotation. Definition of the center of rotation is it has a point where we just momentarily has zero velocity. It doesn't make sense because the center of the rotation is the point where it doesn't move, right? So it has a zero velocity. Note, that doesn't necessarily mean the zero acceleration because it's instantaneous point. So as time goes by those point would vary so even though instantaneously it has a 0 velocity 0.1 second later, it will have its own velocity as well. So the acceleration is not zero. How we can find the instantaneous center of rotation? Whenever you have a to any point of the rigid body. Draw a line perpendicular to the velocity and the intersection will be the center of rotation. So for example, if you have a rigid body, which is a different velocity profile for the point A and B, the intersection for the perpendicular to its velocity is here, C is the center of rotation in this case. You have a perpendicular to the velocity has a line a point C is a center rotation. If point A and B has an up its velocity, then sometimes the center of rotation locates within the rigid body here like the case. What is the physical interpretation of center of rotation? Center of rotation means the whole system is turns out to be just a pure rotation. We respected point, so even though this rigid body like a blue field color. The rigid body is combined by rotation and a translation. We respect to this center of rotation. You can consider this one as a whole new rigid body which is on pure rotation. Same for this one, so it's just any virtual rigid body which contains the point A and B has a pure rotation with respect to the C at that instant only, and same for this one. Since those appoint a center of rotate instantaneous center of rotation varies over time. You can actually draw a line for the convenient collection of those center rotation over the space, which is called a space centrode. And then whatever the lines which corresponds to the body we call this body centrode. So for example, if we want to describe the disk rolling without slipping instantaneously this contact point has a zero velocity. So it's an instantaneous center of rotation. And the space centrode will be whatever along the ground contact and the body centrode is actually the rim of the disk. So when you solve the kinematic problems for the rigid body, either you can use the absolute coordinate if it's straight forward. Or maybe you try to use the relative coordinate approaches. Like you can set any reference point, it will be easy to describe the motion. So the combinations of translation of that point plus the rotation with respect to that new reference point. Or if you can approach using the instantaneous center of rotation approach. Let's work on through the example. This is the wheel of the radius r rolls without slipping on a flat surface. And try to describe the motion of the wheel in terms of motion of the center, okay? Since this is a kinematics problem, you don't have to draw the free body diagram. You just set the appropriate coordinate frame and just go through how you can take a derivative of the position to obtain the velocity and the acceleration. Again, in this case I'm going to try to describe the angular motion using the absolute coordinate frame. Or and relative coordinate frame which is combinations of translation and rotation and using the instantaneous center of rotation approaches. First, if I use this absolute coordinate frame, where should I put the coordinate? Watever that has fixed characteristics. So at the ground contact point C, I put the absolute coordinate x and y, which is attached to the ground, okay? A few seconds later the disc actually will rotate. So this is far away from the coordinate, right? So if I want to describe the velocity of the point C, which is the now turning to C Prime, maybe the velocity. No, it looks like this like a curvy trajectory tangential. And to obtain them I have to actually take the derivative of the position vector, which is r over r of C. How can I find the r of C? When the disk rolls without slipping those displacement change from the center or to the O to the O prime will be the same as the ground contact point displacement, which is C to the A, which is denoted s, okay? So those distance actually equivalent to the ark what has been rotated over the distance. So if I have a rotation angle from C to A is theta, s turns out to be r theta, equal to r theta because those rim has been equivalent to the distance change from C to A. So the displacement rC could be expressed by x, y, which is s minus r sine theta, this red Vector can be expressed by the s and the theta. And if you take the derivative, you can obtain the equations of motion. You can obtain the kinematic equations of the velocity in terms of theta dot. And if you take another time derivative, you can actually obtain the acceleration. Now, suppose that you are given the petition the movement of the center your unknown theta dot and theta double dot or sometimes even theta, could be expressed by the motion of the center, such as displacement s, velocity will be o and acceleration of a of a, a of O, okay? So if you substitute those unknowns into known variables such as BO and AO, you can obtain the velocity of the point C and acceleration. Which is what you can get using the absolute coordinate frame. Now, I can also describe the velocity of the point C using the relative coordinate frame, okay? Translation and rotational motion. The first step is we'll just set the new reference point of this system, okay? I'm going to set the center O, as a new reference point because by doing so I can just describe the motion of the C as a pure rotation. So it's center has its own velocity, own motion and then with respect to that if you're observing the point C prime, the rim, which is fixed to center, it's just pure rotation, right? So your velocity v of C is relative coordinate approaches like v of O translation plus relative motion, which is cC over O, which is Omega cross r. Therefore your translation, which is r theta dot what we have described in the previous slide and it's rotation omega cross product of the r vector, r actually the r with respect to this center, okay? Which is going to be R sine theta, cosine theta term. And that's what you can get is the velocity of the point C using the translation and rotation and combination and which is exactly same as what you can get using the absolute coordinates approaches. Or so you can describe the motion with using the instantaneous center of rotation. So here where is the center of rotation? Okay, instantaneously the point contact ground contact point C has a zero 0 velocity, so obviously this point is instantaneous center of rotation. Or you can say that velocity of the C is 0 plus C of O has this amount, so you have a perpendicular line, so you have an intersection at the C. And then now you can assume that whole O and C are some part of the rigid body rotating with pure rotation with respect to the point C. Now if you want to describe the point A, once you know the instantaneous center of rotation. Okay, by definition therefore, your vO is expressed by the r, r is actually the radius here. R omega, okay? And then when you want to describe point A, suppose the distance from the center of rotation to the point is L vA is now L multiplied by the omega. And since we know the omega in terms of central velocity, you can have a vA in terms of LBO divided by r, okay? So using the center of rotation once you know the omega of the whole the rigid body with respect to that point, you can easily find the of the velocity of the other any arbitrary point with the distance multiplied by the omega. Is this equivalent to the approaches that if you can obtain the velocity of the A using the translation and rotation? Yes, we can graphically to show that the translation has this amount and the rotation has a purely like risk to center it, it's a purely the tangential line. And if you vector some both you can actually have equivalent value of the absolute velocity of the point A. Sometimes this omega here, which is described with respect to the center rotation. Always same as the rotation of the disk with respect to the center, Is asked. What do you think? Is this omega or is same as theta dot here? Theta is described by the rotation of the disk we risk to the center. And omega is the angle we risk with the center of rotation. Think about that. Okay, in this session, we have covered how we can describe the general translation and rotation combined motion of the risk by the using absolute coordinate frame and relative coordinate frame. So thank you for listening. And then we are going to work on the examples in the next session.