Hi everyone. In this chapter, we are going to learn how we can derive the motion of the rigid body in rotating axis, in the reference frame which is rotating. You might think, what is it different from the topic of what we are going to cover is compared to what we had covered? Actually, what we have covered so far is based on the rigid body motion. So usually the distance r here, the momentum, or the distance from the pivot point, the rotational pivot point to the point what we are interested in are fixed. In this case, we are going to handle when r varies, r is not constant anymore. So conceptually, very similar concept what we have compared to what we have covered so far, but a little bit complicated to those are the time varying r term. Okay. So when you have a point A which you want to describe from the absolute coordinate frame rA. Let's say I want to describe it as a relative coordinate frame with respect to the B. So your A is going to be described by rB plus rA dash B, which I would symbolize as r here from now on, which is relative. If those coordinate frame B rotates with Omega, then your rA is going to be rB plus rA dash B. Then they're here, xi and yj. If you take the time derivative in the previous section, we assume those x and y relative those x-y coordinates, more x-y coordinate, coordinate B is constant. But here when you take the derivative, let us assume that x-y could vary. So your term for the xi plus yj will have derivative of scalar term, x and y, and derivative of the unit vector term. So those unit vector which is rotating with Omega, has a time derivative term in terms of Omega. So i is going to be Omega j and j minus Omega i. So if you put that into here, which you can get is b over b plus x.i plus y.j and Omega cross-product r which xi plus yj, which is r. What we have worked on so far. So note that this term is newly introduced since those x and y term varies over time. So your relative displacement time derivative will have Omega cross r plus v relative. So your VA is going to be VB plus relative. This is relative velocity which is Omega cross r plus V relative. So this one is added term here due to the r varying. In a similar manner, we can derive the acceleration as well. So once you have a time derivative of the VA to obtain the A of a, you can take the time derivative of this one, and this one, and this one. So this one is an absolute velocity. What you can get is just absolute acceleration of B, and Omega time derivative goes with Omega here, and the r dot here, and the velocity dot relative here. This one here, since we know that r relative dot is going to be Omega cross r plus V rel. You can plug that in here. Now, the V rel, by definition V rel is x dot I plus y dot j. If you take the time derivative, you can have a time derivative of this one and also the vector term. So the first part, if you take the time derivative of a scalar, which is going to be A relative, and then time derivative of this unit vector comes into Omega cross V relative. So you plug that in to the original equation. It's going to be a of B plus Omega dot cross product r plus Omega cross r plus V rel, and a rel plus Omega cross V rel. If you rewrite them and then you have a Omega cross V rel term here and there. So what you can get finally, is going to be a over B, and then angular acceleration and the cross product r, and Omega cross Omega cross r, 2 Omega cross V rel, and a rel. This is the acceleration described by the reference frame which is rotating with Omega. So you're supposed to memorize this to solve the problem later on. If you don't memorize it, at least you will be able to derive it. Then the thing that you have to know is when you want to describe the motion of something star relative to the coordinate frame which is rotating, note that time derivative of those term, the star, whatever the self, is going to be Omega cross that self-value, and then relative the derivative. Then even though it has many term, like five terms, those are pretty similar to the acceleration in polar coordinate. The polar coordinate r direction, our acceleration is r double dot minus Omega centripetal acceleration. The Theta term is going to be r Theta double dot and two Omega r dot, which is Coriolisic acceleration. So if you just match the acceleration term to those for the polar coordinate, it might be easy to memorize. Now, let's work on the example. Here is a disk which has a slot in it. Then in the slot, there is a slider A. I want to describe the motion of the A over time. Since the disk is rotating with angular velocity and angular acceleration, what would happen is maybe the slides will slide through down, this way. Since the disk rotates, A will have a radial motion and a rotational motion as well. Then relative displacement of the slide with respect to the disk, which means the center of the disk, is going to be given. Was given like h, v rel and a rel. I'm supposed to find what's going to the absolute velocity and acceleration. This is a kinematics problem, so you don't have to draw the free body diagram. Since the acceleration, what I want to describe is actually rotating. So let us start with the equations kinematic relationship that we just derived. Do you remember how your V, velocities have been expressed? Yes, it's V_o plus Omega cross r plus V rel. There are many parameters given, so you just have to plug that in to the kinematic relationship. So since o is the center of the disk, the V0, so how about the a? You, have to also memorize all the a relationship, a naught plus Omega dot cross r plus Omega cross Omega cross r plus 2 Omega cross r V rel and the a rel. So you just have to plug in all the values, setting the velocity for the center, it turns out to be 0. Those are all given, so you just have to plug that in. Same for the acceleration. So the acceleration of the center is 0, and then the rest of them are all given, so you just have to plug that in. So to solve the acceleration or the velocity in the moving reference frame, you have to either be able to derive this relationship or to memorize it. Now, let's work on the problem which have a multiple to by linkage system. So this has a CA linkage or hinged at the joint C with a fixed radius. So describing the point A with respect to this coordinate at C is pretty easy because it's just the finite length of just pendular motion. Same for this one. This one, if you want to describe the point A with respect to this fixed coordinate like just a simple bar here, it's easy because it's a fixed radius, so simple circular motion. However, those structure has been coupled with the slot link OB here. So if you look at the motion of the pin A, which is combined by the link CA and link OB, there is a rotational motion as well as radial motion here. Same for this one. Not only for the rotational motion of the point A, there is a radial motion as well. So in this case, when you want to describe the motion of A in respect to the O, it's easier to set the coordinate frame rotate with it, so that the motion of A is in the radial direction. So to describe the point A, there are two ways to choose the coordinate frame. The blue one, which is a fixed coordinate, is pretty simple because the point A is a fixed radius. If you set the coordinate at O, which is rotating, since the point A has a radial motion and the rotation of Theta direction of motion as well, you would better use the relative coordinate frame attached to O and rotate with the structure, so that the motion of the A could be described as somewhat easier form. That's what we're going to work on in the examples later on. Okay. So you have a two-bar linkage. The OB is rotating clockwise with the constant Omega. In here, what's going to be the velocity and the acceleration with respect to the OB coordinate C. It's a lot easier if you describe the velocity and the acceleration with respect to the C. But anyway, it's asked what's going to be the BA and acceleration A with respect to the OB. Also it asked, what's going to be, when it's rotated clockwise that is blue one also moves that the counterclockwise. So what's going to be the angular velocity and acceleration of the AC? This is a kinematics problem, so you don't have to draw the free-body diagram. So this is a coordinate that you want to rotate with it so that A motion is somehow expressed by the radio. Then you can also set the coordinate at C, which is a fixed coordinate, so that you can just describe the motion of the A via just a simple circular motion. Now this is time that you have to use the kinematic relationship. How you can. If you set the coordinate at O, what's going to be the absolute velocity of the V_A in terms of the V_O? Which is V_O plus Omega cross R plus V_rel. V_rel here is the relative velocity of the point A with respect to this bar, which is rotating. Now what's going to be the acceleration? It has five term. It's going to be acceleration of O plus Omega dot cross r plus Omega cross Omega cross r plus Coriolis acceleration to Omega cross-product V_rel and A_rel. Now I wanted to describe respect to the O, which I'm going to use this one. Then here our O is fixed points, so I have zero velocity here. Now the relative velocity of the point A with respect to O is going to be expressed by Omega cross r and V_rel. Omega cross r is actually the tangential velocity components. So since you know the Omega and the displacement, you can just write it as r. It's this direction, perpendicular to the Rho. So this orange Omega cross r term is this direction. Now the V_rel relatives is within this bar, so it'll be either back and forth along the rod. So it'll be this way. So either this way or the other way. So that actually, add it up to the VA, which is absolute velocity of the A. Absolute velocity of the A could be also obtained from this fixed coordinate. It is very simple. It's a fixed radius rotational motion. So since it rotates clockwise, this one rotates counterclockwise. So the velocity will be tangential upward. So your absolute velocity will be AC multiply by the Omega of this AC structure. So what's you're supposed to find is V relative and the Omega AC. It seems like one equation have two unknowns. However, this is a vector form, so you have a two x and y components so that you can equate them, so that you can have the unknowns solved. So by solving the relationship, vector relationship, we can obtain what's going to be the AC, Omega AC, and then what's going to be the V relative. Now let's find the acceleration. Again, we are going to bring up those kinematic relationship. By the time, you may have memorized already or you will be able to derive the relationship, those five relationships. So acceleration term here starts with A of O, which is fixed one, so it turned out to be zero. Then there is Omega which is constant. So there's Omega the terms goes to zero. Now this is centripetal acceleration of the point A viewed from point O. So it's directed towards the center, and it's Coriolis acceleration is perpendicular to the radial motion. So perpendicular to this way. Then A relative is going to be either back and forth along the rod through the slit. Now those are equal to the absolute acceleration of the A. Absolute acceleration of A could be find easily from this fixed coordinate here because at this fixed pivot point, acceleration of this point A is simply just circular motion. So it has a tangential and normal components of the acceleration, which is our r Alpha and r Omega square. Here Alpha and Omegas are or about the structure AC. So what you can get is vector equations of a_t and a_n to the left-hand side, which is going to be equal to the acceleration relative to the point O. So if you equate them, you can solve the unknowns. Let's work on another example similar to the previous one. Instead, this part, the solid structure AC is rotating with the Omega. Then now you are supposed to find what's going to be the angular acceleration and angular velocity of this pink structure. Similarly this is a kinematics problem, so you don't have to worry about the free-body diagram. You have two options, which is a coordinate frame rotating with it. They're rotating with it so that it'll be a lot easier to describe the A which is moving radially this direction and the rotational motion. Also, you can describe the point A with respect to the fixed coordinate here, which is a lot easier because it has a fixed radius. So let's work on start by starting the bring up the kinematic relationships of the V of A and A of A. What's V of A in terms of relative coordinate frame which is rotating? Which is V of O plus Omega cross r plus V_rel. What's going to be the acceleration? You memorize all the five terms, which is A of O plus Omega dot cross r, and Omega cross, Omega cross r centripetal acceleration, and two Omega cross V, the Coriolis acceleration, and A_rel, relative acceleration. Now the velocity V of A is O, which is zero, and then Omega cross r. Here it's going to be, since I set the reference frame at O, the Omega is Omega for the OB structure. Since OB structure at this instant is, since this is rotating this way, this one is going to be rotating clockwise. So it's directed tangential to this way. So this OB orange part is going to be r Omega, which is OA multiplied by the Omega OB. V relative is now, since you're moving with this structure is only up and down throughout this slot. We can't specify which direction yet. Those are supposed to be equal to the absolute velocity of the A. Absolute velocity of the A could be easily found from the fixed coordinate frame, which is counterclockwise rotation. So to this way, which is AC multiplied by the Omega. So if I equate these two equations in a vector form, you can obtain the unknowns. Same for the acceleration. I want to obtain the acceleration from the acceleration kinematic relationship. So you have all five terms. Since you're describing with respect to this rotating coordinate frame at O, all those Omega terms are about the OB structures. Now this pivot point O has no acceleration, so this turns out to be zero and this one is tangential components for the angular acceleration. So since this one is rotated clockwise, this Omega dot cross r product term is tangential to this. This one is a centripetal acceleration. So toward the center, Omega, Omega cross e. This one is Coriolis acceleration. This is perpendicular to the V_rel. So perpendicular to this slot. Now the acceleration is also along this slot. So I have all those four vectors either in this way or the other way as like this, which is equal to the absolute acceleration of the A which could be easily described when we use a fixed coordinate frame. Fixed coordinate frame, you have tangential and normal components of the A. So you have absolute acceleration, domain to the tangential, which is equal to the relative acceleration with respect to the O. So all those vector form should equate with each other. So you can solve the unknowns by equating x-component and the y-component with each other. To make it a little bit simple, let me specify this one is a Theta. This angle is Theta, and this angle is Theta. So if you plug that in, and since this Omega rotation is constant, this terms goes to zero. So what you can have an acceleration is only centripetal acceleration, which is going to be equal to vector components of tangential acceleration, centripetal acceleration, Coriolis acceleration, and relative acceleration as well. So if you equate them, you can solve the unknowns. So we've gone through how we can describe the motion of the rigid body is rotating axis using the relative coordinate frame. If you change the r, radial direction, equations get a little bit complicated compared to the rigid body by the R is constant cases. However, as you practice more, you will memorize the kinematic relationship or you can even be able to derive it. This ends the chapter 5 and next time we are going to study chapter 6. Thank you for listening.