Hi everyone. Today, we are going to study Chapter 6, the Plane Kinetics of Rigid body. The first half of the textbook is dedicated to the dynamics of the particle and the rest of them, we're going to study about the dynamics of the rigid body. Ultimately, we are going to study about the force and motion relationship, which is kinetics, F equals ma. To do so, sometime we have to be familiar with how we can actually derive the acceleration in rotational motion or the rigid bodies which has links on that so that's what we have covered in the kinematics chapter, precedes the kinetics chapter. So personally, I think Chapter 6 is a very important core of the dynamics textbook, so I hope you would put lot of effort to understand it and digest it so that you can master dynamics. So there are three parts in Chapter 6. The first one is how we can derive the equations of motion in rigid by the version. Second, from that equations of motion, we can do the integral over the displacement to obtain the work and energy relationship, and then later we are going to integrate the equations of motion over time so that we can obtain the impulse-momentum relationship as we have done in similar way in Chapter 3, the particle case. More concretely, we are going to learn how those F equals ma could be extended for the rigid body cases, and do some examples and also work in energy and impulse as well. Before we start, let's briefly go over how translation and rotational motion could be in general combined. Suppose that, if you're only looking at the particle, if you just focus on one particle in the previous chapter. For the rigid body, there are many particle, many points that you can describe for the motion. So as time goes by, those two particles of the rigid body moves in the same displacement, then it's justly simple translation. Sometimes the motion of each point might differ. So in that case, we can describe the motion of the A as a combinations of translation with the amount of B and rotation with respect to the B. So this is translation and rotation or combination that compose of general motion as we have briefly covered in the Chapter 5, because that's what we're going to describe. Often do students ask, is the angle rotation or angle of the point A with respect the B is the same as a rotation angle of the another point, say P with respect to the B? Yes. So for example, if I made an angle of the motion of the A with respect to the B, that would be same as the angle for the P as well. That's because by the definition in rigid body, the displacement between the particles never change. So if we just graphically describe it and then just put the two parallel lines and make a rotation, all those angles are the same. So that's how you can define the angular velocity or angular motion of the rigid body with any of the reference point of the rigid body system that are actually all equal. Now let's say there, what's the general equations of motion for the rigid body? Let's start with the easy part, F equals ma. As we have gone through the systems of particle case, F equals ma whenever there are many many particles, is going to be same as ma bar, which means at the center of mass, all the external forces are equivalent to the mass and acceleration of its center of mass. In rigid body case, we would add another relationship, moment equals I Alpha form. So moment with respect to the G, center of mass, is the same as angular momentum rate with respect to the G. So all the moment about the center of mass is going to be same as angular momentum about the center mass, which will equal to the I bar a, which is also indicating respect to the center of mass. So these are the two equations of motion pair for the rigid bodies so that you can use it later on. To derive the moment equals I Alpha equation, let's briefly review how we define the angular momentum in the previous chapter. Angular momentum is r equals mv. So suppose that this is a particle defined from the fixed coordinate O with the displacement r and its derivative v_i. By definition, H of O is going to be moment m_i cross-product mv. Here note that the O means the reference point of the angular momentum and r is, even though it's now written here, is actually defined from O, and r dot which is v is also defined from O. So it's defined from the absolute velocity defined from fixed point O. Now you can actually define the angular momentum with any other reference point like a center of mass or at any arbitrary point P. So by definition, so if I define the distance from those reference points to the mass as rho i and rho i prime respectively, then the H of G defined from the center of mass is by definition is rho cross-product mv. Here note that the rho is defined from G. So if I subscribe G and then those r is actually defined from O. So here those reference frame, reference point for the rho and mv doesn't the same. So this is definition of the angular momentum with respect to the G. For the specific case at the center of mass, what would happen? Those r_i term can be also replaced by the rho i dot term, which is all defined from the G here. So for the special case for the angular momentum with respect to the center of mass, equality holds for mv cross product and then m rho dot cross-product. How about for the P? By definition, H of P is defined by rho i prime, which is defined from P cross-product of mv, which is defined by O. Then you can also make it more familiar way in terms of H of G, but if you want to review it, please review the Chapter 3. Now here we want to focus on the case of the angular momentum defined at the center of mass. So H of G is defined by rho i cross product of row i dot. So it's the case where you have mass with distance rho i from the center of mass. Then suppose that your whole rigid body rotates with respect to the center of mass. So if that's the case, when you take the derivative of rho i dot is not simply the rho i dot itself, you have to actually consider this rotational term, Omega cross those parameter. So that's d rho, dt is going to be Omega cross rho and rho dot. So if you plug in those relationship into here, what you can get is rho i cross product, mi Omega cross product rho and rho i dot. Then since this is a rigid body, rho i never changes. So time rate will be zero. So what you can get is rho i squared term and the Omega here. If you just do the cross product twice, it's going to be self. So it's going to be Omega term getting out of their Sigma term, but what you can have is Sigma mi rho i is square, which is moment of inertia with respect to the center of mass. So what you have is H of G is going to be i bar Omega. Then when you take a time derivative on left and right, that's what you can get as moment with respect to the center of mass is same as i bar Alpha, which is additional term you have to consider to obtain the equations of motion for the rest body case. So we have gone through how your F equals ma relationship could be extended in the rigid body case, which means you have additional term with moment equals i Alpha. So using that, we can actually solve the examples in the next session. Thank you for listening.
