In this video, we are going to again, work on impulse-momentum relationship of the rigid body for the special case where basically it's going to be the time integral of the equations of motion F equals ma and moment equals I Alpha. What if there are problems combined for the impact problem and the free motion? For example, there is a mud coming towards the bar, the stationary bar with the velocity of V1 and then after collision, it stick together and then just rotated together as a combined body. So what's going to be the angular velocity of the combined body and what's going to be the linear impulse applied at the pivot point O and what's going to be the maximum height to reach? So this is actually the Part A and B is a problem for the impact and Part C, the maximum height of, to which is going to be actually the free motion problem. So we are going to apply impulse-momentum for the first part and the work-energy for the second part. The very first step is again to set the coordinate and then I'm going to set the coordinate definitely at the pivot point and what are the external forces applied? External force is applied to the system is actually the pivot point action at the x-direction and y-direction, and the gravity Ry. How about the collision force between the mud and the rod? If I define the whole system as a combined structure of the bar and the mud, then those collision forces applied by the mud, applied to mud is going to be action-reaction pair and then those internal forces are canceled. Those combined structures that the center of mass is actually located, somewhere below the bars on center of mass due to the edit mass by the mud. So this combined structure will have a translational and rotational motion. So those free-body diagram can be translated as a mathematical form and which is equations of motion. So only reaction or force in the x-direction will generate the acceleration for the center of mass and the y-direction or reaction force, and the gravity will generate the y-directional acceleration. If I set the moment equation with respect to the G, I will have the x components. We'll have a clockwise, shouldn't this be minus? Sorry about the mistake. So have a moment which will generate the I Alpha term. I combined Alpha term. However, this Rx is unknown, usually. Therefore, I'm going to write another moment equation with respect to point O. In this case, there is no external force exists, right? These F's are internal force canceled out and at O, those moment by the reaction force doesn't count. So in this case, there is no moment applied. The ritual generate the IO combined and the Alpha. So since we are supposed to find the Omega after collision, probably I have to solve the Alpha here or the acceleration. However, there are x, the reaction force turns on it involved for the equation. So it's pretty tough to actually solve from those equations and plus those reaction force is impulsive force, like a large amount of force applied for a short amount of time during collision. Why this is impulsive force? Suppose that you are drawing the free body diagram with a bar [inaudible] and then its collision forces now turn out to be an external force. So you have a very large collision force by the mud and you have a reaction force at the pivot point, and then if you would like to have a finite acceleration, what actually happens? R of x at the reaction force should be an impulsive force as well. Otherwise, you have an infinite force applied and your acceleration cannot be a finite value. So once you obtain the equations of motion, take the integral over the time, then what you can have as time integral for the Newton's equation and Euler equation. So as I said before, those reaction forces are generally impulsive force and generally, it's integrals are known because it's non-zero. Now, for the gravity, this is at this finite value. So if you integrate over a short amount of time, it will approaches to the 0. So you're actually linear momentum is not conserved because of this unknown nonzero linear impulse. If you switch into the angular impulse-momentum relationship, since there is no angular moment, the angular impulse is going to be conserved. So your Ho later and then when you formulate the Ho, generally you have the definition of the angular momentum is r cross mV for the particle and then if you integrate the many, many particles over the rigid body, that all turns out to be the I Omega. So since your Ho plus is going to be a combined structure. So I write as I combined Omega. Initial angular impulses only by the mud particle. So it's going to be r cross mV formulation. So you have mV1 and L, and direction is to the [inaudible] y. So you have a minus sign here. So we can also solve the same problem using impulse-momentum relationship. So initial mV and the linear impulse will generate the mV change, linear I Omega and then angular momentum will generate the I Omega change. Now, where we should set the point p for the angular momentum pivot points. Okay, so let's say we'd like to set the point O, the fixed point as a reference point for the angular momentum because they have 0 acceleration is a stationary point. So you don't have to consider the moment by the inertia forces. Things turn out to be simple and then this is the system where you're including the bar and the mass, it's a two-mass system. So as we have covered in Chapter 4, depending on the situation, it will be easier for you to handle the center of mass of the total system or you can separate out each mass and calculate this linear or angular momentum one by one. So linear impulse-momentum, initial linear momentum will be only mV and if we only consider the x-direction, this is the linear impulse in the x-direction, which is unknown because it's impulsive force and integral is non-zero and for the final, it's been combined together. So it's combined mass and its center of mass velocity, which could be also expressed by the rotational motion. Since this one is unknown, those linear momentums are not conserved. So you cannot bind B bar or Omega here because due to the unknown impulse. Now then, let's switch to the angular impulse-momentum relationship. Initially, you have angular momentum by the mud, and there is no angular momentum applied, angular impulse. So final angular momentum will be I combined Omega. Also, you can express it as a radius of gyration for the combined system. So what you can have is minus mv1L is going to be I Omega and then by solving that, you can find what's going to be the Omega after collision. Note that this is equivalent to the solution obtained in the previous slides by integrating the equations of motion. Okay. Now, the problem also asked what's going to be the reaction force, reaction impulse applied at the hinged point. To find this, let's take out the relationships about linear impulse-momentum, initial impulse, initial mv, and the linear impulse. We're going to generate the final mv, and the v here is going to be expressed by the rotational motion with respect to the pivot point and the Omega is what we have just found from the previous slide. So if we can find d, the only unknown is going to be the linear impulse here. How we can find the d? From the geometry, the bar has its mass and then we know that it's a uniform bar. The center of mass is located at the center and the edit mass is going to be located l down to the O so we could find the d. So if you plug in the d, we could find the velocity of the combined system's center of mass, then we could find what's going to be the linear impulse by calculating the change from before and after linear momentum. Linear momentum change is going to be equal to the linear impulse applied. Now, see this equation. Within this parenthesis, will there be a chance that this parenthesis turns out to be zero? Yeah, probably, depending on where actually the location for the MOD heading so that it will affect the change for the I combined and this may be turned out to be zero. What does that mean? Think about it. Okay. So we found the linear impulse applied the pivot point and finally what's going to be the maximum height that is been reached by the combined system. So to do so, we do the energy conservation. There is no external force work done, so initial kinetic energy is going be equal to the final gravitational potential energy. So you can find the h of max. Let's solve another problem, different from the previous case. Previous case has a collision and the free motion. This one is like a free motion and the collision. So this bar of length 2L and the mass of m is falling down from the height h, free fall. Then it's going to hit the edge B, and then it's going to then make a rotational motion. So what is the angular velocity right after collision? That's what we are supposed to find. So first step is that the coordinate at the pivot point if there is free fall, the free fall part is easy, and then after it contacts with the pivot point B, there exists only one contact at the pivot, so there are two forces x and y direction of force by the pivot and the gravity. This is the full free-body diagram and that will generate the rotational motion and the translational motion. So you can obtain the equations of motion in x-direction to y-direction and your moment equation with respect to G is only by the vertical reaction force in clockwise, so you will have minus R_yD. Again, this is unknown, so we better have another moment equation with known variables. So I will set the moment equation with respect to the B. Since the point B doesn't have any acceleration, so you don't have to consider the moment by the initial force. So it turns out to be simple, just switching the subscript from the G to B. So the moment at this point is only by the gravity and the distance, and that will generate the I_B Alpha. Okay. So since we are supposed to find the Omega, maybe finding the acceleration or the angular acceleration will give us the answer for the Omega. But to find the Alpha and acceleration, it seems like there are unknowns involved and those unknowns are impulsive force during collision. So what should we do? So if you do a time integral during the collision, a very short amount of time for the Newton's equations and Euler equation, this equation shows that the linear impulse by the reaction force are unknown, so you're linear momentum is not conserved and cannot be solved. However, the impulse by the mg is zero because a short amount of time in a group for the final value turns out to be zero, that actually could apply it to the angular momentum as well. So even though mgd is not zero, if you are doing the time integral over a short amount of time, those angular impulse turns out to be zero. So your angular momentum for after and before the collision will be conserved. So the problem could be solved by this equation like angular momentum after collision and before collision is going to be conserved, so you will be able to find what's going to be the Omega. Note that this v_1 collision speed is going to be obtained from a free fall square root mgh. Wait. Here, to calculate the angular momentum, instead of having I Omega, I used the dmv, which is r cross mv term. Since the rigid body, if the bar is a rigid body, am I supposed to use the I_B Omega term instead of r cross mv term? What do you think? Think about that. Okay. We can also, instead of integrating the equations of motion, we can use the impulse-momentum formula here. Previous mv, linear impulse, final mv. Previous I Omega, angular impulse, final I Omega. Okay. Where should we set the pivot point for the angular momentum? So we are going to set the pivot point B as a reference point, and then the acceleration at point B is zero, so you don't have to consider the moment by the initial force. So only taking into account all the moment with respect to P is going to be developing I_p Omega change. Okay. For the linear, you have mv_1 and then you have a y directional linear impulse which is not zero generally, so your final linear momentum cannot be solved due to the unknown. However, if you switch the angular impulse-momentum, this is initial angular impulse angular momentum and you have no angular impulse because angular moment by the gravity integral over a shorter amount of time is approaching to zero. So your final angular momentum will be I omega, it could be obtained from the parallel axis theorem, and then if you equate the angular inverse momentum relationship, what you can have is a relationship about the known v_1 and the unknown Omega so that you can find the Omega after collision in terms of all the given variables. So again, this is equivalent to the solution what we have just obtained from the previous slides by integrating the equations of motion. Okay. So that ends the part 6.6, the impulse-momentum equation. Again, chapter 6 is the most important chapter of the dynamics, so I strongly recommend you to do a multiple reviews about how you can obtain the moment equation; F equals MA and moment equals I Alpha for the general rigid body motion, and then take the integral to find the work and energy relationship and take the time integral to find the inverse momentum relationship. Just keep practicing for the problem set so that you can master dynamics. Good luck
