[MUSIC] Hello, in this video, we are going to study how we can apply the general equations of motion of rigid body, which is on the rotation about a fixed axis. Remember that the rotational equations of motion is expressed by mg movement with respect to G across it cause I Alpha, right? Which is the calculated with respect to center of mass. However, we can also derive the moment with respect to some other than the center of mass which is any like a fixed point O. In this case, your equations of motion is being switch it as M respect to the O is going to be same as IO which is obtained from the parallel axis theorem. Like I G with respect to the center of mass plus M the distance from the fixed point to the center of mass R Bar Square, okay? This equality holds. How about any arbitrary point p? So if I have a P point and the distance from the P point to the center of mass is rho bar, then is this general equation holds? Yes, this one holds, so all the moment we respect the p is equal to the Ip Alpha. P here is obtained from the parallel axis theorem. Now, if p is not fixed anymore and has some motion and then acceleration Ip, why don't happen? Okay, if I try to view the whole system, we risk the p means I'm setting the reference point at p. So I'm going to describe it as a relative reference frame. So whenever I set the reference frame at p, which is moving with the acceleration ap, you consider inertia force is applied to the center of mass with the magnitude of map, right? So when you calculate the moment part with respect to p here on the left hand side, you should also consider the moment by this fictitious force, so distance rho bar cross product map. So whenever you're arbitrary point p has an acceleration, then you have to consider the moment by those inertia force minus sign means minus ap. The opposite direction to the left hand side, which equals the ip Alpha that will actually give you the equations of motion for the rotational motion. So for the derivation for these relationship, I would recommend you to refer your either textbook or the other notes and let's work on the example. Okay, this is the case where the block of mass m is hanging through the pulley. The disc is red, which is wrapped around the center and the disc is also people read at the center. And the free to rotate, then there is an external force p is applied, okay? And the radius of gyration of the disc is given as GE radius of gyration, which is defined as this one. So whenever you have a moment of inertia, right acceleration is the approximate value that the moment of inertia can be expressed by mc square, okay? So I'm supposed to find the acceleration of a block M and the reaction force at the center joint. Since this one has 2 masses, I should define the to coordinate system. So I'm going to put one on the center of the hinge point and the other maybe at somewhere here. And then I'm going to draw the free body diagram separate for the disc in the block respectively. So how many contact does this one have? This one is a cable contact here or cable contact there and the center of mass, center rotation hinge joint. So I have 3 joints, two forces that the hinge joint, and one for each like a tension, T1 and T2 and the gravity. That will actually generate the linear acceleration here. It's a fixed one so if this turns out to be 0 later and the rotational acceleration. So this fictitious relationship between the force and the motion can be transferred as a mathematical form, which is the equations of motion. So X direction or forces with being equivalent under equilibrium. Y direction of forces are under equilibrium, and then I should add whatever the moment we respected the g at the center. So maybe due to the T1 and T2 will generate the I Alpha, the angular rotational motion. And I can also do the free body diagram for the block which only has a tangential force tension force upward regenerate the acceleration upwards. So I will have a relationship like this. Okay, so I'm supposed to find what's going to be the acceleration here A Y and the reaction force at this at the pivot joint, right? And it seems like, okay, so T is given, so and beta is given so I might be able to find this one, okay? To find this one, I have to know what's going to be the T 2. And here, Alpha is not unknown, T2 is not known and to find the AY, I have towards the figure that out the T2, so there are many unknowns. How can I solve it? Luckily, since there are Alpha T2 equations, two equations about T2 and Alpha so I can eliminate T2, I'm sorry. I can eliminate the T2 to obtain the Alpha and the A but so it has three unknowns T, Alpha and AY and two equations. So there should be some kinematic constraints that to solve this problem, since the blocks motion is not independent from the disk rotation. So disk rotation it set the positive as a counterclockwise has -R multiplied by the alpha is the same as ay which is directed upward. So if I plug that in, I'm going to have 2 equations of two unknowns of T2 and Alpha. So that I can obtain what's going to be the alpha here or the a is, okay? This is how you can solve the problem. Now, let's solve the another problem which is a pendulum rotating which is initially released at the horizontal position and find what's going to be the reaction force when Theta pass by Theta 0. Okay, so whenever it's a horizontal, the reaction force will be a function of motion, right? So I will set the reference point at the pivot joint and then also since this one is a rotating motion, I also using the Cartesian coordinate or entity and coordinate or the polar coordinate. In this case, let me just use deity and cord in the rotational coordinate. There is only one contact to the environment of the bar, which is the pivot join. So there is a reaction force only and the gravity that will generate the tangential acceleration and centripetal acceleration and the angular acceleration. So transfer this into mathematical form the equations of motion. So RT will generate the tangential acceleration, no more forces, some will generate the centripetal acceleration. And also I can have the moment I Alpha relationship with respect to the center of mass in this case. So the RT will generate a torque so that this one is the I Alpha and I'm supposed to find the reaction forces the center. So, let's see if I could find since I don't know what's going to be Alpha and Omega, there's nothing that I can do further, right? So instead of having this, let me just try the other relationship. Like if you can establish the momentum equation with respect to the O here, you don't have to worry about the torque moment by this one and the other one. Just the only moment is generated by the gravity. It's all known so I could have the torque by the gravity will generate the I Alpha. Note that this one is instead I of O instead of I of G. So by solving this problem, I could figure that what's going to be the alpha so I can figure that out what's going to be the T tangent or reaction forces. But still I have to figure that out what's going to be Omega to find out RN. So, how can I find the R Omega, since I know what's going to be the Omega what I'm looking for? I'm looking for the Omega at the state where theta equals Theta naught. I can integrate both sides of this moment equation over the Theta of to the Theta naught. This is or known, this one is actually Alpha is D Omega DT. So if I integrate over D Theta, this turns out to be Omega D Omega so I can do the integral. So I can find the Omega where the theta equals Theta naught, so find those value could be plugged in here to get the normal force. No more component of the reaction forces. Okay, now let's solve another problem. There is a bar of length 3 l in total. Sorry, it's a 3L and mass of M is mounted here at the O and then it was rotated by the external moment M. And what's kind of the reaction force at the connected part by the bearing when it starts to rotate? Okay, so where should I set the coordinate? Well, I was at the coordinate here. Center of mass is located in somewhat here at the half of the bar and then I'm going to set the coordinate at the joints O and here again, since this is rotating. I'm going to use either TN coordinate or R Theta coordinate. So there are only forces with the environment is at the center, the reaction force X and Y components to tangential and normal components, and the gravity in the vertical direction. So that will generate the acceleration in the horizontal plane tangential and normal and angular acceleration. Since you're only force attendance, our force will generate the tangential acceleration, no more directional forces will give you the normal direction of acceleration. And angular motion with respect to the G here is done by the tangential forces so and active movement at the shaft. So those two torque source will generate moment sources will generate the rotational angular acceleration, okay? Now, since I'm supposed to find this then I have to know Omega and the alpha. Let's see if I could find the alpha here, okay? So since Alpha is unknown initially, but I have two relationship that contains Rt in the Alpha. So if I just solve it as a series, I can eliminate the LT here and obtain the what's going to be the Alpha. Once I know the Alpha, I can find the RT. I can also obtain the equation of motion rotation or equations of motion with respect to the O. If I set the moment equation with respect to the O, the good thing is I don't have to worry about the moment by the Rt here. So the only moment is external moment, which is going to be generating rotational motion with respect to the io Alpha. So this is another way you can solve the Alpha. Okay, since you have to know the Omega to find the RN, but note that it's the bar starts rotating, bar starts rotating from the resting initially at rest. So that means Omega is about 1 or 0. So your normal reaction force turns out to be 0. Okay, so this is what we have covered that we apply the Newton's or Newton or wheeler equation here to the the rigid body, which is under rotation about a fixed axis. And we examine there are many different cases either we can apply the moment. We reduce to the G or o or some other points. Okay, next time, we are going to study combinations of translation and rotation. Thank you for listening.