14. Torque & Rotational Dynamics
Intro to Torque
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Hey guys. So in this video, we're gonna introduce the idea of torque, which is the rotational equivalent of force. Let's check it out. Alright. So you can think of torque as a twist that a force gives an object around an axis of rotation. So So here's the most classic example. If you have a door that is fixed around an axis here, this is the hinge of the door which is also its axis of rotation, meaning the door is free to rotate around the axis. And if you push this way with a force f, it causes the door to spin this way. I'm gonna say that the door accelerates in that direction. It gains an alpha. Okay. So when you push on a door, it rotates around its hinges. So more generally speaking, when a force acts an object as it does here, away from its axis, it produces a torque on it. So let's talk about these two parts, away from the axis. If you push on a door here, it doesn't actually cause spinning. I'm gonna say alpha equals 0. Right? If you try to push a door by pushing it, it's if you try to open the door by pushing the hinge, it doesn't spin. You have to be away from the axis of rotation. And then it produces a torque. So this is the idea that a force causes a torque which causes an acceleration. Okay? So what you're doing, you're not doing, you're not producing a torque, you're producing a force which then results in a torque on that object. Okay? So now the other point is that a force may produce a torque. We already talked about how if you push here, it doesn't cause it to rotate so you don't produce a torque. So force may or may not produce a torque and a torque may or may not produce a rotation, an angular acceleration. And that's, let's say if you're pushing this way, but then someone else, F2 is pushing this way and these 2 cancel out. In that case, it wouldn't really produce them. Okay? But the most important point I wanna make is that you have f produces a t, which produces an alpha. That's the sequence. Okay? And we just talked about this. Let's fill it in here. Similar to forces cause linear acceleration, remember, sum of all forces equals ma. Sum of all forces, as long as you have a net force, you're going to have an acceleration. It's the same thing with torques. Torques cause angular acceleration. We're not going to talk about that just yet, alpha. We're going to talk about this a little bit later. Okay? Now another difference between torque and force is that force is a straightforward number. If I tell you we push with 10, that's the end of it. But for torques, torque depends on how hard you push, it depends on how far you push, and some other stuff. So, we actually have an equation for torque. You don't have an equation for force. You're given a force, but for torque, you have to calculate it. And it is fr sin of theta, fr sin of theta, where f is the force you push with, it's a vector. R is a vector. It says here r is a vector from the axis of rotation to the point where the force is applied. Remember, little r in most rotation problems has to do with distance from the center. It's the same thing here. Okay. Now theta theta is the angle between these two vectors. See these two vectors right here, f and r? Theta is simply the angle between those two guys. Theta, oops, f and r. I meant to put an f and an r. And one thing I like to do is I like to think of like an arrow pointing towards these two guys here. Right? And that's to remind me that in that equation, the theta is the angle between the f and the r. The 2 guys are hanging out next to the theta. Okay? Theta is the angle between these two guys. Now the unit for force is newtons. The newton for the units for distance r is meters. So torque is measured in newton times meter. Okay? That's the units for torque. One last point that I wanna make here is that when we need to maximize the torque, to get the most possible torque, another way to think about this is the way to apply the least amount of force and get the most amount of results to be the most efficient with causing something to rotate is to apply the force as far as possible and perpendicular and apply the force perpendicular, perpendicular to the r vector. Perpendicular means 90 degrees. It's got the little perpendicular symbol, to the r vector. So what does that mean? So let's draw the r vector real quick on this picture here. The r vector is from the axis of the rotation, which is here, to the point where the force is applied, which is over here. This is the r vector. You want your force to make this is your force. You want your force to make 90 degrees with the vector, which in this case, it does. That's how you get the maximum torque, the easiest way to rotate. Imagine if you're instead pushing the door this way, right? This would be a little bit weaker. You have to push harder to get the same rotation. So you want to push at an angle of 90 degrees. The other thing is that you wanna push as far as possible away from the axis of rotation. So you've all opened doors before. If you try to open the door by pushing on it, let's say right here, right? I'm gonna make a little bit of a mess, but I'm gonna erase this. If you push right here, it's much harder to open the door here than than to open the door at the end. In fact, that's why the doorknob is on the opposite side of the hinge because that's where you're supposed to push. It's easiest. Okay? So, and you can relate that directly to the equation, right? So if you want the torque to be as as high as possible, you obviously wanna push as much as hard as you can, as far away or as distance from the axis, as far away as you can. And you want to maximize sine of theta. Now Now let me remind you that sine and cosine fluctuates between negative one and one. Right? So it looks like that. So the greatest possible value of of sine you can have is 1. Now where does this happen? This happens when theta is 0 I'm sorry, when theta is 90. Sine of 90 is 1. That's why if you look at the equation, that's why you get the greatest possible value for torque. Okay? So again, your torque is maxed when you push as far away from the edge as possible and when you push perpendicular, make it a 90 degree angle with the r vector. Alright? Now let's do an example. We're going to use 3 steps to font to solve all of these torque questions. We're gonna draw the r vector. You're gonna figure out what your theta is, and then you're gonna plug numbers into an equation. Okay? And here, I have 5 different forces just to show you all the different variations. So you push on you push or pull on a 3 meter wide door. So the length of this door here, let me actually just write length equals 3 meters, with 10 newtons in a bunch of different ways. So all of these forces are 10. We wanna calculate the torque that each force produces on this door. And then the rest is just explaining that f 1, f 4, and f 5 all act on the edge of the door right here. In other words, they act at a distance of 3 from the axis. This guy is halfway through the door and this guy is at the hinge. So I'm gonna write in all this information here. I'm gonna say this is 1.5 meters. The rest here is 1.5 as well. So the whole thing is 3. What else? It says that f 5 is directed 60 degrees below the x axis. So if you do this, this guy is 60 degrees right there. So I got all the information. Let's do this. Torque 1, remember the equation, is just f 1 r 1 sine of theta 1. Okay. The force is 10, but I gotta figure out the r, and I gotta figure out the sign. So the first thing you do, here are the 3 steps. You're gonna draw the r vector. Okay. So the r vector is from the point where the axis is to the point where the force happens. The force happens right here. So the r vector for force 1 looks like this. Okay. So I'm going to say that this is my r one. And how long is r one? It's 3 meters. So 3 meters. Okay. Now sine is the angle is the angle between, r and f. Okay. F 1 is this way. Let me actually draw this down here. F 1 is this way and r 1 is this way. The angle between these two guys is simply 90 degrees. I'm gonna put it here. And remember, the sine of 90 is 1. I actually need you to remember that. That's going to make your life easier. But if you forget, just check-in the calculator real quick. So this is going to be 10 times 3 times 1, which is 30, and the unit is Newton meter, 30 Newton meter. Cool. We're going to do this 4 more times. So t 2, again, f 2r2sine of theta 2. The force is 10. I like to leave spaces for r and theta. So what do you think the r is here? Right? So what would the r vector look like? I want you to draw that real quick and tell me how long that r vector is. And I hope you're thinking that it is 2 meters, sorry, 1.5 meters, because it's just half. Right? So r 2 looks like this. This is r 2. I'm gonna draw it down here so I don't make a huge mess. Actually, we put it here as well. R 2 looks like that. Okay. And the force is f 2 right here, and the angle between them is also 90 degrees. Okay? Now we'll talk a little bit more about angles because there's some places where it might get confusing, but for now we're just going to keep going. If you multiply everything, the answer is 15 newton meter. So right away, you see how when you pushed farther, which was in the first situation, you got a bigger torque than when you pushed closer to the edge. Alright. Let's keep going. What about torque 3? F 3 which is 10 r sine of theta. What do you think the r vector looks like here? And what do you think the length of the vector is? So I hope as you're thinking about this, you're thinking, well, if you push at the hinge, it doesn't really move at all. So the torque should be 0. There is no you're not producing anything that causes acceleration. You're not causing acceleration. And that's correct. And that's because the r vector is going to have a length of 0. Okay? The r is the distance between you can't even draw it. It would look like this, right? This is like your r 3. It's a dot. Because it's from the axis of rotation which is here to where the force acts which is also here. So there's no distance because it's the same two points. Okay? In that case, it doesn't even matter what the sign is. You can't even do technically draw a sign because there's no arrows to, there's no arrows for you to figure out what the angle between them is. So it doesn't matter. This is just 0 and that's because you act on the axis. Whenever you act on the axis, torque is 0. End of story. Okay? Now what about torque 4? It's 10 r sine of theta. What do you think the r vector hit here is? So and how long is it? Right? So the r vector for 4, 4 acts over here. So the r vector is the same one as this guy here. This is r one and this is the same as r four. The length of this vector is the entire length of the door which is 3. So far so good. What about the angle? Right? What do you think you put for the angle? And do you think there's a torque here? So the second question might be easier to answer. If you pull in a door in the direction of f 4, the door doesn't move side the door doesn't open or close. Right? The door doesn't spin. So you should expect that the torque is 0, and it is actually 0. But the reason why that happens in terms of the equation, you can see this on the equation, it's because your r vector looks like this. This is r 4, and this is your force 4. Okay? The angle between these two guys is 0 degrees. They're both going the same direction. Right? If you got this wrong and you thought that the angle is 180, that's okay. You got the same answer. But the angle is 0 degrees and that's because you can put them sort of side by side, and you can see how the angle is 0. Okay. So the angle here is 0 degrees and if you do sine of 0, the answer is 0. Alright. And the last one, t 5, it's at an angle. The force is 10. How long is the r vector and what is the angle that we use? So again, t 5 is at a distance of 3. So this vector here, the long arrow is the same for all 3 of these guys. What about the angle? The angle has to be be careful here. Right? A lot of torque problems will give you an angle and people would just sort of blindly put a 60 here and sometimes that's wrong. They do that on purpose. So you gotta watch out for that. You should actually assume that they're trying to trick you and make sure that you're using the correct angle. K. So how do you know this? And I'm gonna draw this. I'm gonna draw our 5 right here. This is where you've slowed down. Make sure you do this correctly. And I'm gonna draw f 5 here and here is the 60. Is that the angle between those 2? And the technique that I like to use is to make it where the two arrows point from the same dots. I wanna have something like this, f and r, because then it's easy to confirm that this is in fact the angle. Now to do this, what I do is I shift the arrow around. So for example, see this r five right here, I'm going to move it. I'm going to move it this way. I haven't changed the direction of the r five. I'm just shifting it from pointing into the dot to coming out of the dot. It's the same thing but the nice thing about this is that it becomes very easy to see that this angle is in fact the one between f5 andr5, because I have the 2 vectors like this, and it's just the angle between them. Cool? So sometimes we're gonna do a lot of the shifting, to make sure that the angle that they gave you was the correct angle. Okay? So for example, if they had set 30 here, it wouldn't be the correct angle. The correct angle is instead the angle between the r and the f, which is 60. So that's good. Sine of 60 goes here. The and then if you multiply this whole thing, you get 20 6, I'm rounding it, 26 newton meters as the answer. Cool? It's really important that you know how to calculate torques. You should make sure you understood this and you can do all all 5 of these things on your own. Alright? Well, this is just the beginning. We're gonna do tons more ton more stuff, but you gotta make sure you know how to calculate basic torques. Let's get going.
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