Hello. In this video, we are going to make a practice how to describe the motion of the rigid body which contains the rotational components and the translational components. Okay. This is the two-bar linkage which is linked at point A from the pivoted link OA in another link ABC where the B is connected to the linear motor that has been moved side-by-side through the piston movement. Okay. Suppose the velocity of the piston B connection part is VOB, what you're supposed to find is the velocity of the C. Now when the piston moves the B's left-hand side, though Theta gets smaller and the C gets higher, but due to the downward movement, C looks like somewhat showing, somewhat flat movement. That's what we're supposed to describe. This is kinematics problems. So you don't need to draw the free body diagram. Then we are going to solve this using absolute coordinate frame approach and relative coordinate frame approach. So this is what I expected as a trajectory for the VO of C. Now, if we use the relative coordinate frame, where is the point that we should put the reference coordinate, which is the fixed point O. So let's put the absolute coordinate frame x and y here, and then the displacement from the coordinate of the point C will be x, y. So this is going to be the position vector r of C from O, the absolute coordinate frame, and those could be expressed by the Theta, actually 0.5 Theta and the r. Since we know the structure, the position vector will be three r sine 0.5 Theta r cosine 0.5 Theta. Once you take a derivative of this actual position vector, you can obtain the velocity vector which is going to be expressed in terms of Theta dot here and there, and that's it. That's what you can get. We are going to solve this again using the relative coordinate frame in terms of translation and rotation and combination. Since you have known, you're given the speed for the B point translational speed of Bv. B, if you find where you should put the new reference, relative reference frame for relative coordinate B will be a good candidate. So if we set the coordinated B, the motion of the C will be the absolute translation of velocity of the V and the relative velocity of the point C with respect to the B. So the translation relation and motion of the B, and the rotational motion which is going to be a tangential to the link because it's a pure circular motion with respect to the point B. This is going to be expressed by Omega cross r as we have covered in the previous slide. Note that this omega is Omega bar ABC. Okay? Now, yeah. Then the Omega bar is a time derivative of the Theta of this amount which is 0.5 Theta. So that's how we can get 0.5 Theta dot. So vector sum of translation, the black arrow and the rotation with respect to the B, which is a red arrow, will give you the vector sum of those two components, will give you the absolutes speed for the V of C. So if you just add them together, you can have Omega cross r term here, and then this one is the absolute velocity of point B as a translation so that you can describe in terms of S dot defined from the absolute coordinate frame. So you have a ds, dt, you can get to r Theta dot cosine 0.5 Theta. So if you sum them up, this is what you can get as absolute velocity of the VC in terms of translation and rotational combination. Note that this answer is exactly same as the one that you have obtained using the absolute reference approaches. So no matter, what each coordinate you choose, you're supposed to have the same answer. Let's solve another problem. It's also a two-linked bar system, and one with the pivoted with respect to the O, and the other is connecting to the linear guide going up and down. This left-hand side link is actually driving the motion by rotating with this constant angular velocity Omega zero. What you're supposed to find is angular velocity of Omega AB, this is the Omega for this orange bar, and the velocity of the point B. Again, this is a kinematics problem, so you don't have to draw the free body diagram, and we are going to solve this using the relative coordinate frame using translation and rotation combination, and also check that out using the instantaneous center of rotation methods. Okay. First, where you should put the new reference frame, new reference point that you want to describe as a relative motion. O point is the fixed coordinate and try to describe the B. I would set a as a new reference point to view the point B as a pure rotation. So with the coordinate x and y, I have a counter-clockwise as a positive, and this amount is going to be then minus Theta. So this is Theta AB, and it's time derivative is going to be the omega AB. Omega of the bar AB. Okay? Then, absolute velocity of the point B is going to be this combination for a translation of the point A and the rotation of the B with respect to the A. So to translation of the point A is what? This is actually the circular motion with the finite radius. So it's easily obtained by r Omega zero, tangential to this rod, tangential to this rod. This rotation and motion will be also tangential to this bar AB. Okay? So as we have covered in the previous subsection, it's Omega cross r. Here, Omega should be Omega AB, not Omega zero, and the r is distance from here to here. Okay? So the total absolute velocity of point B is going to be a vector sum of these two components. So you can plugin V_A as r Omega 0, and sine and cosine components, and for the red arrow part, it's going to be Omega AB, cross product of the r, and that's why you can get for the V_B. You are supposed to find the velocity of the B, and Omega AB. So you have one equation having two unknowns. Can you solve it? Yes. It looks like a one equation, but this is a vector equation. So you have i component equations, and j component equations as well. Plus, since this point B is constrained to those linear guides, so the x-directional components of the velocity of B is going to be 0. So if you use that, extract out the ith components of the velocity V, this is what we can get, r Omega 0 sine Theta 0, minus 2r Omega AB sine minus Theta, which turns out to be 0. If you solve that, you can get Omega AB in terms of Omega naught, and sine Theta naught, and sine Theta. If you look at the triangle of this one, you have r and the other line is 2r, length is 2r, and if you have a Theta 0 here, and this is going to be a Theta. Then using the sine rule trigonometry, you can have the relationship between the sine Theta naught, and sine Theta, which is two. So if you plug that in, what you can get is your Omega AB is the same as your Omega zero in the counterclockwise. Minus sign indicating for the counterclockwise which means, in this special case, not in general, your Omega naught, which is the Omega for this link OA, is same as Omega for the other link, link AB. You can check it through the vector. So you have a relative rotational motion which is being shown as a red arrow, and if you added up the translation of vector, what you can get is this one, this doesn't satisfy the kinematic constraints because this blue one should be aligned with the bar. What's gone wrong? Yes. When you look at the velocity of the B with respect to the A, it could be either of this, counterclockwise, or the clockwise. So if you have a clockwise relative motion, add it up the vector for the point A then you will have a vector sum which is aligned with the guides x axis. Okay? We have just gone through the structure like this, two linked bar. Now, if I just make a little bit of modification like tilt this orange part up, and move this up and then tilt this vertical guide to the side, what would happen? What would be the main changes for your answer? Let's work on it using the translation and rotation combination or approach. Here again, I'm going to set the new reference point at A. Okay? If I see the point B here, I will have the absolute velocity of the VV is going to be the translation of the point A, and the summation for the rotational motion with respect to the A. Then I know that the tangential vector for the rotational system is going to be expressed by Omega cross r. Just note that, this Omega is always the Omega that is actual under rotation. So if you're calculating the V_A, which is also the power of rotational body, that's going to be Omega 0, and then cross product of the r. Here, this red arrow is Omega AB, cross product of the r. Okay? Now, since you are supposed to find the velocity of the VV, while you are not known what's going to be the Omega AB, it seems like one equations that's having two unknowns. However, don't worry you have a kinematic constraint to use which means, since the B is moving along this axis, whatever perpendicular to this axis will be zero. Okay? So if you have a vector sum of translational or motion of the A, that's going to be ended up resultant motion having aligned with the axis, and then if you just calculate those components perpendicular to the axis of the red arrow, the relative motion and the black arrow the translational motion using those Theta relationship, what you can get is perpendicular to the guides will be these values, and then this turns out to be zero. By solving them, you can have the relationship between the rotational angular velocity of the bar AB and the initial Omega naught of this link OA. As you can see here, this one, not generally one, therefore, this is the case where you have a different Omega zero and Omega AB, and the positive sign means it rotates counterclockwise. If I assuming the relative rotational motion vector direction is downward, what would happen if you have a vector sum? The summation doesn't satisfy the kinematic constraints. So my first guess was right, I was lucky. Okay. So with that, we have gone through several examples how we can describe the rigid body motion, which contains a translation and rotational components. So either use the absolute coordinate frame, or use the relative coordinate frame, which has a translation and rotation combination. Thank you for listening.