Dinâmica de Corpo Rígido na EFOMM: como isso realmente cai

本站添加: 2026-05-12

[00:00:27]Hey everyone, good evening. All good?

[00:00:30]Are you listening to me?

[00:00:38]Just checking to make sure everything's alright here so we can get started. Uh-huh.

[00:00:55]Here it is. Ready. Well, folks, welcome to another live stream focused on EIA. Today we're going to talk about a topic that appeared in 2023, in that first strange exam, and it has been recurring ever since. So, I'm going to teach you two special things here so you can solve the exam questions if they come up. Can you see the image there already? This is the type of problem we're going to learn how to solve today. So, we're going to learn about what's called rigid body dynamics. But before we learn how to solve this type of problem, we're going to calculate everything: speed, acceleration, traction, everything imaginable to solve it. I'll first talk a little about rotational and translational motion, especially the so- called cycloid.

[00:02:08]What's the idea? Imagine you have a uniform circular motion, a little ball like this, let's say it's a disc.

[00:02:21]This disc spins as it moves forward.

[00:02:27]So, let's assume he's spinning, moving forward like this. I just need to draw a straight line here so I can guide myself and not go too far off course.

[00:02:41]So, you see? This album is moving forward.

[00:02:45]So, in order for me to observe this movement forward, I'm going to mark some reference points. For example, let's suppose it starts here at this point, the point that 's right in contact with the ground.

[00:03:02]This point rotates and moves forward.

[00:03:06]So let's assume he'll be here in a little while. Then he's here. Then he's here.

[00:03:17]Then he comes here.

[00:03:20]Then he comes here, and so on. I think you've already grasped the idea.

[00:03:30]The movement continues indefinitely, doesn't it?

[00:03:31]If we were to trace this guy's movement, we would observe that it's something like this, see?

[00:03:39]See? Come here, then go back, then climb up again, then come back, and so on.

[00:03:49]I want to create a mathematical description of the motion of this point, which I'll call A.

[00:03:59]So, what does this body do that characterizes rotation and translation without slipping?

[00:04:06]This means that, firstly, in translation, all points have the same forward velocity, because translation means that everyone is moving together at the same speed. During translation, the distance they travel is always the same.

[00:04:31]So all the points, including these ones here, which I'm not marking, have the same speed, including the center.

[00:04:41]So, everyone here has the same translational speed.

[00:04:45]But since this thing is a disc and it's spinning, let's assume it's spinning clockwise, everyone here also has a rotational speed.

[00:05:01]Again, it's all points. I'm only marking a few here for reference. So, everyone here has a rotational speed.

[00:05:13]This means that, assuming uniform circular motion, everyone at the same distance from the center has the same rotational speed.

[00:05:23]So, what is movement?

[00:05:28]Cycloidal motion is a combination of these two factors. So I'm going to copy and match these two here in the same circle. It might be a little smaller this way, right?

[00:05:44]Make a slight adjustment here. Ready.

[00:05:47]These guys here represent cycloidal motion.

[00:05:52]So, in order for the wheel not to slip, it means that at the point of contact with the ground, it must have zero velocity.

[00:06:06]In other words, this point has a total or resultant velocity equal to zero. Since I have a velocity in one direction and a velocity in the other, this means that my rotational speed is equal to my translational speed.

[00:06:24]And what is my rotational speed? Oh, xes r.

[00:06:30]So this guy is spinning this way with angular velocity omega, but it's the defined omega.

[00:06:36]If the translational speed is greater or less than the speed of the object, this object will slide across the surface. I've never seen this come up before, but it serves as a reference for a theoretical question.

[00:06:51]Now I'd like to look at the movement these guys are performing.

[00:06:57]Since it's a translation, let's assume constant speed, this characterizes a uniform rectilinear motion (MRU).

[00:07:05]How does this guy spin in a circular motion? Circular motion is the combination of two simple harmonic motions, one horizontal and one vertical. So, on the X-axis, we have, let's say, X and Y, okay? We have the movement of translation. Therefore, it performs uniform rectilinear motion (MRU) and an oscillatory motion, because it is simple harmonic motion ( SHM) in the horizontal direction.

[00:07:37]For anyone who has doubts about this, after this lesson is over, take a look at the MHS combination here; we have a lesson on it. One of the resulting equations, I think the first one I make, is a circle; that is, the circle can be decomposed into two simple harmonic motions. When you flip the equation, it will become very clear that we should call it mhsx. And it also performs simple harmonic motion, but it's simple harmonic motion along the y-axis, so the x-axis... look what I can write. I can write that his position is given by a uniform rectilinear motion (MRU), that is, VZ T plus simple harmonic motion (MHS). What is the equation for simple harmonic motion ( SHM)? The cosine of t + theta equals 0. I'm assuming zero here for convenience.

[00:08:30]And who is my 'y' in 't'? my y of t on the y-axis, that is, the sine of t. Here we need to define something in relation to the initial position.

[00:08:47]Why?

[00:08:48]Let's suppose, uh, if this guy starts down here, for example, or rather, if he starts, let's draw the axis here, to give you a better reference.

[00:09:04]If I started with this theta 0 = 0, it means that this guy started in this position again, completely arbitrary.

[00:09:16]And then he'll have to start with the sine, right?

[00:09:20]The cosine function will start at the maximum position, and the sine function will start at zero. So I can just put it directly like that. I just changed the position here so you can see it better. Ready. Okay? So what does this guy show us? that they can be combined into an equation, for example, without time or by modifying the idea of ​​time.

[00:09:55]Why? If I take this guy from here and isolate him in time, look what a strange thing it becomes, but it does become something. Who is the sine of ô t y over a passes the sine to the other side, right? So, who is Omega T? Arc or sine raised -1 of y over a. A is the radius.

[00:10:24]I can take this guy and replace him here. So, I can put together an equation that doesn't depend on time. It's ugly. AND.

[00:10:36]I'm going to open the V here as an A as well, because then I'll have here, look, an x, you see? And here I have ô x t. Then it becomes arched.

[00:10:52]Then you can write it however you prefer. Y over a + a times is cosine, right?

[00:11:03]So it's the cosine of omega t, that is, the arcsine of y over a.

[00:11:16]You can also rewrite this from here.

[00:11:18]I would even prefer to do it this other way, using the fundamental relationship.

[00:11:23]Cosine² + sine² does not equal 1.

[00:11:27]So what is cosine? square root of 1 - sine squared. Who is the only one who is sine? And about a.

[00:11:40]So here, I can rewrite the cosine as the square root of 1 - y of t/a squared.

[00:12:00]Look at this final equation you 're seeing; it's quite similar to the question that appeared in EV1.

[00:12:09]It's been a while now, okay? So you can write this guy in terms of any other magnitude.

[00:12:17]Here I chose to eliminate time, just to give an example, but these two elements combined already represent the movement, which is cycloidal motion, okay? So what do I want to show you here? First, translation and rotation can be combined.

[00:12:37]And by Galileo's principle of independence, you can treat the X-axis independently of the Y-axis, and write the equations independently of each other. What characterizes it as rotating without slipping is that its rotational speed is equal to its translational speed.

[00:12:54]And with that, you can set up an equation of motion X Y, as if it were, I don't know, an oblique launch or something else. It looks bad because it has a T, it has a T, so it doesn't end up looking very good. But you can rewrite all these expressions as a function and leave t using sine² + cosine squared. Who is the cosine? Cosine is x - vt, okay?

[00:13:25]about the most important thing, who is the sine y about a.

[00:13:40]Then you can still write other relationships that also compare with time. If you isolate Y, that's the question, which is the nephonium. Okay? So you could go back to that question and rewrite the simple harmonic motion like this, the motion like this, which would give you a simpler way to work with the equations of motion.

[00:14:10]Hey everyone, how's it going so far? Are you following along? Any questions?

[00:14:31]Why did I use sine instead of cosine in the equation for Y?

[00:14:34]Because if one vibrates along that axis, the other vibrates in the opposite direction, right?

[00:14:39]Because if this guy makes a circular motion, sine + cosine squared, it has to equal r². So, if I used a sine function, the other one has to be a cosine function. The sine is out of phase with the cosine, which is pi/bre². So you could write cosine, but here it would be more like pi/bre², more or less, right?

[00:14:57]And then that becomes sine, okay? I'm going to ask that if this is your question, you take a look at a lesson we have here on the channel called "The Combination of Simple Harmonic Motion".

[00:15:13]In this lesson, I cover all the cases of circles, ellipses, parabolas, and waves, and the wave equation so you can see where sine and cosine functions come from. But here, because it's a circular motion, do the decomposition, okay?

[00:15:30]Beauty?

[00:15:32]So let's start talking about the motion of rigid bodies. So, what would characterize a movement as a move? Just write the equation for the movement. Nothing else.

[00:15:46]And the equation is move, you write the equation of the move. That doesn't change the solution to the problem. Now that you understand that it can be divided, it can be any equation of motion, okay?

[00:16:08]Beauty?

[00:16:10]I'm behind you guys here, that's why I'm checking the chat now.

[00:16:23]Beauty? So let's go.

[00:16:27]What characterizes an object as a rigid body?

[00:16:32]As the name itself suggests, right? If it is rigid, it does not undergo any deformation.

[00:16:39]So, a rigid body simply means that its size is constant.

[00:16:51]Constant size.

[00:16:55]So, if I pull it to one side, the other piece has to follow so that the size of that body remains the same.

[00:17:08]This is known by other words as a geometric bond.

[00:17:15]Guys, a geometric link is simply a relationship based on size. So, why is the acceleration of the string on the block twice that of the other one in the geometric constraint, in order to keep the string length constant? Why is there a relationship between sine and cosine on an inclined plane? in order to keep the length, the size of the objects constant.

[00:17:39]So, a geometric link is nothing more than a geometric relationship, of course, that will define some characteristic, usually size, that is constant throughout the movement. So, if I take this guy here, for example, and rotate him, notice, he doesn't rotate arbitrarily, he rotates in such a way that, just considering the rotation so you understand first, when this guy rotates this way, this one will rotate in a way that the length remains constant, okay? So, if I rotate around the center of this object, focusing only on the center, just so you can fix the idea, these objects have to rotate in such a way that the omega, right, remains the same so that they can maintain a constant length.

[00:18:40]Ah, but can it rotate and translate at the same time? So, yes, he can walk forward and turn at the same time. Look how cool!

[00:18:55]So, if he walks forward and turns, he'll be in that position now.

[00:19:02]He's going to have a combination of movements in such a way that the distance between this movement, right, up here and this movement down here, will keep our body the same size.

[00:19:21]This guy can't walk any further. Why can't he walk any further forward? Otherwise, he will have left his body at that point. So he has to walk in such a way that he always remains in the body's position, that he always belongs to the body. This might seem a little confusing right now, but we're going to work in a much simpler way, which is as follows.

[00:19:45]As we've seen, all motion can be broken down into rotation and translation.

[00:19:51]What characterizes a rigid body is translation.

[00:19:55]Because when this guy moves forward, all the points have to move forward in the same way.

[00:20:04]So, what do I usually do in this type of problem, folks? I usually do the following, look. I always break down the motion in this direction here, and I call this direction the parallel direction.

[00:20:25]Parallel, right? Parallel to the body, okay?

[00:20:29]There is another direction, which is this direction here, the direction perpendicular to the body, as the name itself says, right?

[00:20:41]Perpendicular means that it forms a 90º angle. It doesn't matter which way this guy is moving, let's assume he's moving this way.

[00:20:50]The other one has to move in such a way that the parallel components, pay attention to this [ __ ].

[00:21:00]So that the parallel components are always equal. Why? Because if the parallel components are not equal, it means that one piece is moving further forward than the other, and that would cause the body to stretch or compress.

[00:21:19]So, the motion of a rigid body is basically characterized by this. In the direction of the body, the velocities are always the same. So this applies not only to those two extreme points I marked, no.

[00:21:37]For all points, the parallel velocity must be the same.

[00:21:47]This is basically what characterizes the dynamics or kinematics of a rigid body.

[00:21:55]Our entire lesson will be based on the principle that the parallel motion of the body is constant, okay?

[00:22:05]Alright, professor? But if there's a parallel component, let's suppose here, a parallel component, you see it here? Let's call this point A.

[00:22:14]Here, there will be a parallel component at B.

[00:22:20]These two are equal. But if it has this parallel component, it's because it also has this perpendicular component in this design. It's possible that in some cases it won't be available, okay?

[00:22:33]So, look, I'm going to have VA perpendicular and VB perpendicular.

[00:22:42]Perpendicular velocities characterize circular motion.

[00:22:49]And we saw that for circular motion to exist, the omega has to be constant.

[00:22:57]However, they may not rotate in relation to their center, in relation to the geometric center.

[00:23:06]They can rotate relative to any other point here. Let's suppose, the center is here, and I'm rotating relative to this point G.

[00:23:15]What characterizes this point G? The pivot point, the instantaneous center of rotation, means that this guy doesn't rotate. In other words, it's this little dot here that stays still when things are spinning.

[00:23:34]Look how cool!

[00:23:37]So, what's going on here?

[00:23:39]I only drew the rotation for this guy, okay?

[00:23:43]This guy is spinning this way at the same time that this one is spinning this way.

[00:23:49]Since these angles are equal and they rotate at the same time, obviously, otherwise the body, you know, because it's in one time frame, it's in another, they have to have the same omega.

[00:24:00]So, what characterizes rotation is having a point G, which is the instantaneous center of rotation, where the perpendicular velocity of that point G is equal to zero.

[00:24:25]In other words, it is the point that will only perform the translational movement.

[00:24:33]So, this body, this point G, it only has this parallel velocity.

[00:24:39]It will only make a downward movement in the direction of the object here.

[00:24:51]And if everyone has velocity, that is, if everyone has translational and rotational velocity, in other words, a vector sum for all points, the resulting velocity is the translational velocity plus the rotational velocity, and you use Pythagoras, right, to be able to... Yeah, and you use Pythagoras to find the resultant.

[00:25:32]This point is the only one that doesn't have, oops, rotation.

[00:25:38]Therefore, it is the point with the lowest possible speed.

[00:25:51]Because it only has translational speed and not rotational speed, okay?

[00:25:58]The instantaneous center of rotation, they are distinct points, they can be distinct points along the trajectory.

[00:26:06]If this guy is accelerating, if the speed is changing, the instantaneous center of rotation changes at each instant. That's why it's called " instant," because it might only be valid for that specific moment, for that particular photo, okay? That's why, in situations involving stiffness, we always ask: "What 's happening now?"

[00:26:30]Then you answer: "The velocity at that instant, the acceleration at that instant, everything at that instant, okay?

[00:26:38]So the center, that's why it's called instantaneous.

[00:26:45]Okay? Idea, right?

[00:26:49]Based on that, we can now basically solve any problem.

[00:26:56]But now you need to find this point G. How do you find this point G? It's quite simple.

[00:27:03]Everything rotates with the same angular velocity.

[00:27:08]Omega A is not equal to omega B.

[00:27:13]What is omega A? It's the velocity of A perpendicular to radius A, which is equal to the velocity of B perpendicular to radius B.

[00:27:26]And the radii are the distances of these points to the point of rotation.

[00:27:36]So, if the whole bar is here, point G is here, this is RB, this is RA.

[00:27:48]So, if I know these velocities and I know, for example, the sum of the radii, I can find the length of RA and RB to define where point G is, where the instantaneous center of rotation is.

[00:28:06]Here is contained the whole idea, the whole idea of ​​rigid body dynamics.

[00:28:13]If If you read this in a university textbook, it will use some more sophisticated tools. Why? To decompose, you need angles, you need to know an instant, you need to define theta. So, it may be that in very difficult problems I have to use a rotation matrix, I have to use a vector product, but that's not our case. In our case, we're going to do simpler and usually two-dimensional situations.

[00:28:46]Can Efon test in 3D? Yes. Mark D for God and move on to the next question. You're not going to learn what people in mechanics learn in a semester, okay?

[00:29:04]Let's use this now to define the geometric constraint, which is using the derivative.

[00:29:15]A geometric constraint, as I said, is a constraint of size. So, for example, if I have that classic situation, right, of the ladder leaning against the wall, I'll start with this one, because this is the simplest.

[00:29:31]Ladder leaning against the wall. The length of the ladder is constant.

[00:29:36]This point only moves along the X-axis and this one only moves along the Y-axis. So, you see?

[00:29:47]The ladder is sliding like this, and from this sliding I can define the motion, okay?

[00:30:06]Now, what I want you to notice here is something funny. This guy is big, but this one isn't so big. Why? Because the velocities aren't necessarily the same. There might be a point where they are equal, but they don't have to be equal. What needs to be equal is only the parallel velocity.

[00:30:28]But I want to use the derivative, I don't want to use the explicit geometric constraint.

[00:30:34]What's the idea? I need to find a geometric relationship that remains constant. Why? When we take the derivative of a constant, the result is zero. Here we have the length of the ladder as constant. So, if I call this guy X, this point here A, this point here B, and this one Y, we know that X² + Y² = L².

[00:31:01]Oh, professor, and if it wasn't 90º, if it was any angle theta? Law of cosines, okay? But that wouldn't change the fact that there's a constraint between They.

[00:31:12]If we do the first derivative, I'll use this notation here for the derivative, okay? Derivative with respect to time. What is the first derivative? The derivative of a polynomial will be 2x.

[00:31:28]Chain rule dx/dt plus the derivative of y/y², right? 2 chain rule dy/dt, the derivative of a constant is zero.

[00:31:48]I'll cancel the two because it's zero on the other side. This here is vx and this here is vy. That is, x vx + y x vy = 0.

[00:32:04]This is the velocity constraint.

[00:32:09]So, notice proportionally, right, the... sorry, first sign, they are symmetrical.

[00:32:18]So, if X and Y are positive, if VX is positive, Vy is negative. Damn, it makes perfect sense, right? If this ladder moves this way, the other one has to decrease its position.

[00:32:29]Second, the greater the x, the smaller the vx, right? The greater the position, the smaller the velocity, okay?

[00:32:42]Now let's do the second derivative.

[00:32:45]DDT again, Or if you prefer, d²t/dt². It's the second derivative for those familiar with this notation. Look how cool.

[00:33:02]Just so my image doesn't get in the way at any point.

[00:33:06]Here's the product rule. So it will be the derivative of x. I'll just put it directly: vx times the second plus the first times the derivative of the second.

[00:33:20]Derivative of velocity, acceleration plus the derivative of the first times the second plus the first times the derivative of the second.

[00:33:37]So, again, we arrive at a constraint for the accelerations. So, I have x = ax + vx² + y = ay + vy² = 0.

[00:33:56]Of course, if I had values, I could isolate Vx or Vy here and substitute them below and make a constraint with only one velocity. I'm not going to do that because the calculation will look kind of messy, okay? That's not the goal right now. But, for example, isolate Vy here.

[00:34:18]What is Vy? It's -x over y.

[00:34:29]Then you can substitute it here now.

[00:34:33]So now you have a constraint of accelerations only as a function of one.

[00:34:37]Velocity, not as a function of both.

[00:34:40]Just an example of how this might appear on the test, okay?

[00:34:45]The next derivative doesn't have any physical meaning, because the derivative of acceleration is the acceleration of acceleration.

[00:34:51]We don't use that, okay? But using the derivative, you can quickly determine the geometric constraints.

[00:35:00]What's the problem? You have to know derivatives.

[00:35:04]If you don't know derivatives, you'll have to stick to this type of solution. Is everything clear so far, everyone? Any questions?

[00:35:28]Well, if there are no questions, let's do the little example.

[00:35:35]The little example is very simple, find the whole thing, okay? This is an example I invented.

[00:35:41]Hopefully it has a solution, okay?

[00:35:45]So, look, this guy is vertical. B is vertical, okay? So it's supported like this, like a ladder, right? As if it were a ladder.

[00:35:56]There's a block B here with a mass of 2.5 kg that 's attached to this ideal bar.

[00:36:05]There's this point A that has a mass of 3 kg, it's also attached to the bar, it moves to the right with velocity and length 13 cm.

[00:36:18]X here It's 12, so the conclusion here is C, right?

[00:36:22]Just use Pythagoras.

[00:36:24]And I want to find VB first.

[00:36:28]I'll do it first using the velocity of the parallel axis, okay? So let's go. What's the idea of parallel velocity?

[00:36:43]Here's the bar, here's the axis of the bar.

[00:36:57]Okay, here's the axis of the bar.

[00:37:00]I'm going to decompose this velocity va in the direction of the bar, here, perpendicular.

[00:37:10]So, here's the parallel velocity VA.

[00:37:16]So, what do I know? That B and all points must have the same velocity, okay? And how do I find this velocity? I'm going to use the angle, for example, this angle here, theta.

[00:37:34]This angle theta is the angle between the horizontal and the plane, that is, it's this one here.

[00:37:44]I'm disregarding the size of the cart, obviously, okay? So, what is this angle theta?

[00:37:51]Here it's 5. The sine of theta is 5/13 and the cosine of theta is 12/13.

[00:38:03]And I know that the parallel VA has to be equal to the parallel VB.

[00:38:11]What is the parallel VA? V cosine of theta is the projection of velocity A in the direction of the bar.

[00:38:21]VB is definitely downwards.

[00:38:27]So what is the parallel VB?

[00:38:30]Look, here's the perpendicular, right? Here's the horizontal.

[00:38:38]So this angle here is theta.

[00:38:43]So what is VB?

[00:38:45]The parallel VB, VB sine theta.

[00:38:51]Sine over cosine is the tangent, right?

[00:38:53]So, VA is equal to VB, tangent theta.

[00:38:58]What is tangent theta? Tangent theta is y over x, okay? The opposite, right? Because we found Vy in the VX function, okay?

[00:39:17]So I can relate these things regardless of the method you use.

[00:39:26]So, look, what is VA, right? I'll go back here.

[00:39:33]V is 2, cosine theta 12/13 is equal to VB. Sine of theta 5/13.

[00:39:45]So, VB 24/5 = 4.8 m/s.

[00:39:56]So, notice, this position is smaller, the speed is greater. This position is larger, the speed is smaller to compensate.

[00:40:07]Okay?

[00:40:11]Now let's find the instantaneous center of rotation. I'll start with this one, and then we'll do the derivatives to find the acceleration, okay?

[00:40:20]So let's go. How do I find this instantaneous center of rotation? I need to find the perpendicular velocities.

[00:40:29]So, look, I need to find here, the perpendicular VA and the perpendicular VB.

[00:40:37]Why? Because the amount these elements rotate relative to point G is such that the omegas are equal.

[00:40:49]So, what is omega A? V perpendicular over RA, VB perpendicular over RB.

[00:41:00]And RA + RB is the length of the bar, which is 13.

[00:41:06]I can use everything in centimeters, okay? There's no problem at all. Centimeter, centimeter. It's all in the same unit.

[00:41:12]So, what is the perpendicular VA?

[00:41:16]It's VA sine theta.

[00:41:21]And what is the perpendicular VB? It's VB cosine theta. Did you notice that I only switched here, right? The sine with the cosine, which is the other projection I'm looking for now.

[00:41:32]About RB.

[00:41:35]VA is 2.

[00:41:38]I'll put the sine of theta as 5, okay? I 'll cancel out the 13.

[00:41:44]VB is 4.8 x 12 divided by RB.

[00:41:52]So this will be RB equals 4.8 x 12 divided by 10, right? 5.76 RA.

[00:42:07]Then, RA + RB has to equal L, right? RA is 5.76.

[00:42:15]Sorry, RB, right? It's 5.76 RA.

[00:42:21]So, RA divided by 10, add 1, divide by 13 and invert.

[00:42:33]It gives two, if I didn't make a mistake in the calculation, 1.90 centimeters.

[00:42:44]In other words, this point is very close here, actually, to point A, look. It's around here. This is the pivot point and it makes sense, right? This speed is lower and this one is higher. To compensate, the radius is smaller. See? That's more or less the idea.

[00:43:05]This is point G. And this point G only has translational velocity.

[00:43:12]So, point G, the velocity of G is only the parallel velocity.

[00:43:20]And the parallel velocity is here, look.

[00:43:22]We already calculated it.

[00:43:23]It's 2 x 12 over 13, which is approximately 1.8 m/s.

[00:43:42]So, look, we found VB and we found the instantaneous center of rotation.

[00:43:50]Guys, any questions so far? I'll zoom out so you can see the whole problem.

[00:44:11]Now let's find the acceleration.

[00:44:14]Hey guys, for acceleration, the derivative is the best solution, okay? Don't even bother inventing another method here.

[00:44:28]Using the derivative here is much better. Oh, but I don't know how to differentiate, professor. Damn, learn in 2 seconds how to differentiate polynomials and the chain rule, okay? If you've been studying for Efon, then for the naval academy, right, all those, you know. Okay, so let's differentiate, right? Is it possible to do it with vector products? Yes. And I said at the beginning, you learn this in the mechanics course, but we don't teach vector products for AFA in Efon, so there's no point in me showing it here using vector products, especially since it's a 3x3 matrix to determine, okay? So I'm not going to do it with vector products, unless it's extremely necessary to use this 3D, torque in 3D, then yes, we use vector products. Otherwise, no chance, okay?

[00:45:24][snoring] So let's go.

[00:45:29]We discovered this here, right, for acceleration, okay?

[00:45:43]So here, I'm going to write this constraint. I have X, it's worth 12.

[00:45:59]I don't have Ax. Here I have to be careful, right? Now I have to put it in meters, right? So, 0.12 ax + [snoring] vx, which is va, right? It's worth 2.

[00:46:18]Y is 0.05 times ay plus vy which is vb, Okay, so we just found it here. 4.8 squared, right?

[00:46:42]It's ugly, isn't it?

[00:46:46]Let's see if it's correct. It seems correct. We'll manipulate this equation later.

[00:46:52]Now let's go to the forces, right? Here, I have a weight force, which is the weight of B.

[00:47:04]Here I have a force A, which is the weight of A. I have a normal force, I'll disregard the friction force. I don't know what it says, right? But then you would mark a friction force. I'll disregard that one. And I have a tension here and here a tension in the bar itself, okay? Notice the following: This guy only has acceleration on the X-axis, right? So, on the Y-axis, the resultant will be zero. Only the X-axis will remain. The only force on the X-axis is the tension, so only that will be considered.

[00:47:44]I don't have a normal force, so there's no friction either, right? So it's no use looking at that force. So, I'll decompose the tension here. I 'll use that same angle theta. What we saw. So, t cosine theta, cosine theta is given, okay? It's the mass of A times ax.

[00:48:09]Cosine theta is 12/b and 13.

[00:48:15]The mass of A is 3 x ax.

[00:48:19]So, what is ax?

[00:48:22]4t over 13.

[00:48:26]Oh, ax.

[00:48:30]Now let's go to y, which is the other body.

[00:48:35]The other body here is theta, right?

[00:48:39]So we have uh F - PB uh T. Then I adjust the sign of the acceleration in X, okay? Because the tension is to the left, right?

[00:48:59]- T sin theta equals MB x AY.

[00:49:07]Here, since the axis is this way, it's negative, okay?

[00:49:16]Now here, look, f minus, because the axis is upwards, okay?

[00:49:21]So now I'm putting the correct sign so there's no problem when doing the calculation, okay?

[00:49:29]Uh, 15 - PB MG - Theta equals MB AY.

[00:49:48]Here It will be -10, right? It will be 10 + 5t / 13 = 2.5y. Then I'll divide everything by 2.5.

[00:50:04]Here it will be 2, right?

[00:50:11]And 10 / 2.5 is 4. Here it is.

[00:50:21]Now we're going to do this, take this Ay, put it here, take this Ax, put it here.

[00:50:31]And then we'll find the value of the tension.

[00:50:38]With the value of the tension, you go back and find Ay and Ax.

[00:50:49]That's the problem.

[00:50:55]In last year's test, I think it was the one where he asked to use the energy relationship, right?

[00:51:03]He asked to use MV² over 2, right? The conservation of energy. It was so that we wouldn't have to do this here, okay? So, uh, the way he asked to use the conservation of energy last year was so that this wouldn't need to be used.

[00:51:22]So, you could use the conservation of energy here, for example, doing mgh, using gravitational kinetics, using kinetics here, doing the conservation of Energy.

[00:51:34]But since we wanted acceleration, and also the tension of the body, then you need to use the complete geometric constraint.

[00:51:47]Guys, did you understand the idea? Any questions?

[00:51:57][snoring] There's not much else to do now, okay?

[00:52:07]Now it's time to let go. So it will be 0.12ax + 4 + 0.05.

[00:52:28]I'll change the sign here now.

[00:52:35]+ 4.8² = 0. Then you find T. I'll leave that calculation for you.

[00:52:48]Professor, do you think it 's certain to be on this year's exam? No, it's impossible for me to say that. Don't take the exam, guys. There's no way I can guarantee which questions will or won't be on the exam, okay? This has been on the exam in the last three years, since the exam started.

[00:53:03]This is the only reference I'm using. What hasn't been on the exam yet?

[00:53:07]Acceleration and tension haven't been on the exam, which is what I asked for here, okay? But there's no way I can say: "Oh, it's certain to be on the exam." I can't confirm any of that, okay?

[00:53:21]What I can't say, what hasn't fallen yet, hasn't fallen asking for attraction or asking for acceleration.

[00:53:28]That didn't fall.

[00:53:32]That's why I'm suggesting this as a possibility, okay?

[00:53:38]This has already fallen.

[00:53:41]Finding speed, finding minimum speed, all of that has already come up, okay? He was asking, he can ask what the omega is, he can ask various things here.

[00:53:53]See? We did several calculations, and some of them are still unfinished here. Ah, I didn't calculate the omega exactly, but it's here. So, you can calculate the omega, you can calculate the speed of point G, you can calculate the radius of rotation, right? It's possible to calculate the point around which these guys are rotating. It's possible to calculate the speeds. You can calculate the speed from any other point as well, okay? Because any other point rotates with the same omega, only the radius changes.

[00:54:17]Then I could find the speed from any other point. And what I think is new, something that hasn't yet come up, is this calculation of traction, okay? Calculate the tension in the bar. This one has n't fallen yet. This could be a good option, since it hasn't appeared yet. How are you all doing? Doubts?

[00:54:47]He can ask for the coordinates, man. He can ask for anything.

[00:54:52]You can ask for anything, ask for the coordinates and find point G, right? Like we did here, man.

[00:54:59]If I know where point G is, I know its coordinates. Just to make a comparison, okay? He can order anything here, you understand?

[00:55:16]In fact, these are the quiet things he can ask for. He can put the deal in place, I don't know, with some other connection that isn't Pythagoras, damn it. He can.

[00:55:29]I think at most he'll demand a contract, at most, okay? That's not 90 here.

[00:55:37]You need to use the law of cosines there.

[00:55:40]I think that's the most he can charge that differently. Ah, here, I don't know, if the angle theta is different from 90.

[00:55:49]Then you do l² = x² + y² - 2xy cosine theta.

[00:55:59]Less because I'm looking at the opposite angle, the opposite side, right?

[00:56:04]So, calculating the derivative isn't so trivial.

[00:56:12]Okay, but calculating the derivative isn't difficult, right? Zero because it is a constant. 2x 2y here, look, cosine theta is a constant, so it becomes derivative of x y, derivative of x vx, derivative of y v.

[00:56:37]Ready. Unless the angle is constant, okay?

[00:56:44]Or sorry, the angle isn't a close-up of my breasts, my apologies.

[00:56:47]So you really have to do one, you have to do a better derivative, and then you 'll be careful and all that, okay? But even if you know how to differentiate again, if you're going to take the iPhone test, the minimum you need to know is polynomial derivatives and the chain rule. So it 's not absurd for you to do this during a physics exam either, okay? A good average for physics is in the previous video. Take a look over there. The previous video was exactly about what grade you should get, okay? Will the live stream be recorded? It's the first line of the description. Yes, a live stream. I think I put it like this. Yes. The live stream will be recorded. It's in the first line of the description, okay everyone?

[00:57:43]That's what I'm guessing is what makes that angle appear. The angle can vary, but it's difficult to vary the angle; the wall needs to be opening and closing, which makes it difficult. What varies is the angle of the bar.

[00:57:57]So, like, this angle here varies, obviously, right? It keeps decreasing, increasing, it's pulling upwards. This angle varies, the others don't, okay? The others I'm talking about, this one here, not this one, this one is fixed, okay?

[00:58:19]But what I would suggest again for this year is calculating the traction.

[00:58:25]Actually, calculating accelerations might be a little more complicated, right? You have to calculate the tension, then go back.

[00:58:31]But I would say that this could be a good solution, a good question, because last year it came up, and I almost disagreed a little with the drawing, because the drawing of last year's question implied that it was standing upright and at an angle. If you looked at it, it looked like it was on an inclined plane, right? And then I said, "Wow, there's going to be gravitational potential energy," right? Because the drawing he was making was exactly like that, except it was supported on a horizontal plane; he didn't explicitly say so.

[00:59:06]So when I did it, I put some MGH in there at the beginning. There was no answer then, so I did it differently. I even requested the cancellation, right? Write an appeal for this issue because the drawing he made was tilted, the drawing he asked for was perpendicular, but in the end it was all on a plane. Well, if it's on a flat surface, the difference is that it doesn't have that weight, right? It's just going to be f and traction.

[00:59:34]That's the only change that would happen, of course, right? That would be an easier solution. There can be friction, of course there can be friction, look. Add some friction here. Add some friction here. There might be friction, there might be all sorts of things here. I didn't score, did I? We have the normal one here. Of course. Okay. I didn't mark that. That's true too. Okay, everyone.

[00:59:58]Okay, before we wrap up, for those who don't know yet, this is the last week that the Efon Mission will be offered, okay? Because the Efon Mission will have live classes to answer questions, right? A live Q&A session will begin as soon as sales close, okay? For those of you who just stumbled upon this, Missão Efon is a physics review and in-depth study project specifically for the exam. So, the lesson you just watched is on Missão Efon, along with some other things discussed. These are the 14 most frequently tested topics on the exam, both for review and in-depth study.

[01:00:52]This covers about 90% of the content over the last 4 years. So, if you only study missions and fum from now on, you have a chance of getting 90% on the test. It could be even more so in recent years, 90% of the exam. I have over 20 hours of classes.

[01:01:13]But it's not a class, not one of those random theoretical classes teaching you the basics. These are theoretical classes, like the one you saw. It's a lesson showing you how to apply the theory to specific types of questions, okay? These are the 14 most frequently tested topics, with over 20 hours of theory dedicated to teaching these specific applications. There are over 300 solved problems in video and PDF format, right? That's my solution manual. Starting next week, which is when sales end, we'll have a bi-weekly live Q&A session with me on Google Meet.

[01:01:56]You will have access to nine EPON simulations. Six are already available on the platform.

[01:02:02]One will be released next week, and two more will be released according to the schedule.

[01:02:10]This project is not, pay attention to this, okay? I had three people, including some who gave up on the project, bought it, and asked for a refund because they thought the project was going to save their lives. It won't work.

[01:02:29]This project isn't for you if you don't care about physics, if you listened to those prep course gurus saying, "Ah, since physics is so hard, forget about physics and study everything else." If you did that, this project isn't for you, okay?

[01:02:47]This project is for you if you've already covered the syllabus or are about to, are studying physics, and now need guidance for the iPhone exam. You know the material, you've already done level-level questions, OK? But now you need some guidance to arrive at the iPhone test prepared.

[01:03:08]This is the iPhone mission. So, if you 're having trouble figuring out how to review right now, you've studied for this exam for 3 years, you can't pass physics, you can't do well on the test, maybe this project is the solution for you to arrive well- prepared for the exam.

[01:03:28]He will review the 14 main topics and delve deeper into them.

[01:03:33]So, even if you feel you don't have the necessary depth of knowledge, you can still do the review section. And if you think you already have a good level of review, you're at a good basic level, you can do the in-depth part. Or if you think you need both, you'll get both, okay?

[01:03:52]Below in the description there's a link for you to access our exclusive group, where I guarantee a 20% discount for you, okay? Access is granted until the exam date, which is tentatively scheduled for July 25th and 26th. If that doesn't change, you'll have access until the following week, actually, the week of the exam. But if the exam date changes, you will have access until the new exam date is set. This investment amount is a one-time fee. You pay once and you get access to 14 theory lessons, 300 video-solved questions, nine practice tests, study tips, and, for those who use the coupon in the discount group, access to the question bank. It's a database with over 10,000 solved physics problems, right? In other words, with an answer key and explanations, so that you can, in addition to these 14 topics, study and review the subjects that you feel you need to do this work on during this final stretch of preparation. So, in addition to what 's written here, you also gain access to a question bank with over 10,000 physics questions with answer keys and explanations, and more than 13 questions with video answers so you can also practice and learn the most important solutions for this iPhone exam, okay? So this week is the last week that this project will be offered. Well, I couldn't, I didn't want to, I thought about it, I'll only want to buy it next week, that's it, okay? Monday will be the last day to purchase the project, because on Monday itself we will begin our live Q&A session. So no one else comes in after this shift is over, okay everyone? Well, I hope you enjoyed this lesson. Even if you're not going to buy the project, take a look at the previous lessons we did on iPhone ideas, right? One on Simple Harmonic Motion ( SHM), and we're going to do another one talking about hydrostatics. I think we did one on hydrostatics, actually, right? Yes, we covered hydrostatics, and we're going to have another lesson on one of these frequently tested topics so you can get an idea of ​​what to expect at EFON.

[01:06:30]Alright, guys? Well, I hope you enjoyed this lesson. Until next time.