Torque on Discs & Pulleys - Video Tutorials & Practice Problems

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1

concept

Torque on Discs & Pulleys

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Hey, guys. So in this video, we're going to start talking about torques acting on disks on disk like objects such as pulleys or cylinders. And this is important because it's gonna be a key part of problems of more elaborate problems we're gonna have to solve later on. So let's check it out. Alright, so let's say you have a disk or a pulley and you're pulling on it like this with the force of F, which then causes it to spin like this. Okay, now what matters? As it says here, what matters is our, which is the distance from the access to the force from the Axis, which is usually the middle to the force. This distance here are that's what matters. Not the radius, not the radius. Now, in this particular example, here, little are is the same as the Radius because the rope pulls on the disk on the edge of the disc. But let's say if you had let's call this F one r one S o. R. One is the radius. But let's say you were pulling with another force right here F to In this case, the force doesn't pull from the edge. So what, you what matters is not the radius of the disk. But in fact, what matters is the distance in this case are two is not the radius. Okay, so it's always gonna matter is the distance. Most of the time, the distance will be the radius of the disk, but sometimes it won't be okay. Let's do an example here. So two masses m one and M two m one is four. Let's put it here. M two is five are connected by a light string which passes through the edge of a solid cylinder. So there's a string here that goes like this wraps around the cylinder. Boom. Um, the cylinder has mass. M three equals 10 and radius. Remember, radius is big. Our little are is distance. The system is free to rotate about an axis so the system can spin around. An axis that is perpendicular to the cylinder. Perpendicular to the cylinder means that again it makes a 90 degree angle with the face of the cylinder, okay to the cylinder and through its center. So basically the cylinder spins around its central axis. We want to know what is the network produced on the cylinder when you release the blocks. Okay, So torque net is the sum of all torques. And you wanna know the network on the cylinder? So we have to figure out how Maney torques act on the cylinder. Add them all up. Remember, a force may produce a torque. So what we do is we look at all the forces on the cylinder, and then we figure out which ones produce a torque. So there are three forces or four forces that I come to cylinder. I have this one here is M one g pulling down M two g. Pulling down there is the M g of the cylinder itself, m three g, and there is a tension that holds the cylinder up. So there are four forces, which means there could be as many as four torques. Let's talk about this real quick. First, I want to show you how there is no torque due to m three g and do the tension, and that's because they acts on the axis of rotation. Okay, So torque of tea, right? The talk of any forces f r f r sine of data. In this case, F is t so torque of tea is tea. Our sign of data. But the tension is pulling. The tension is pulling from the middle. Here it's holding the cylinder from the middle. So this R is zero Now, this would have been zero, even if the tension was somehow holding it up here. Okay. The other problem with this, quite with this part, is that tension pulls it up. Let me drive over here. Tension. Let's say tension was going this way. Tension pulls it up. You have to draw the our vector from the middle to the point of the force happens, our vector. And these two arrows were both going the same direction. Which means that the angle between them is zero theta is zero degrees. Which means that here you would plug in sign of zero, which is zero. So whether the tension pulls in the middle or if it pushes in the edge, it doesn't matter. They make the same. They have the same angle with each other here, Um, so this whole thing would be zero. The torque due to M G is for sure in the middle. Eso it's m g zero, and it also makes an angle of zero because for this is the are for teeth and there's an r for M G and the M G is that way. So this angle is zero as well. Okay, sign of zero. So both of these guys don't actually produce any torque. So the Onley forces that will produce a torque R M one and M two. So right away. You should know this for future problems. If you have a, uh, a pulley with a disc in the middle, the weight of the pulley is not going to cause a torque on the polling. And neither is some sort of force that holds it up, whether it's attention or it's held onto an access or something. So there's, like a normal force, right? These forces holding the disc up won't produce a torque on disk. One way to think about this is that if the disk had been held in place by by this axis axis pulling up in mg, pulling it out, it wouldn't spin on its own. And that's because there's no forces causing a torque on it. Okay, three only forces that cause historical forces that could cause it to potentially spin. And that's what you get with someone that's trying to do this to the disk on em to the trying to do this. So torque one and talk to Okay, so if you imagine a disk if you pulled from the edge of the disc, it would roll from either side. Cool. So let's now calculate the torque due to these two forces. So I'm gonna call this torch one, which is force, which is M one g are sign of feta. All right, so what's the are vector for M one g? Well, it's acting M one is acting all the way at the edge of the disc because I am one. The cable for M one is passing through the outside the edge of the disk. So the our vector is exactly the radius. So are one is the radius. And by the way, that's the same thing that happens with M two g are two is also the radius because both of these guys are all the way at the edges. The angle between these guys, both of them, is 90. So look how they are vector and the force makes an angle of 90 and the are vector over here, and it's respective force makes an angle of 90. Okay, so both of these guys, we're gonna have that. The distance is the entire radius Boom, boom. And the angle is 90 degrees. So this obviously becomes a one. And now I just have to multiply the numbers. The last thing you gotta do is also figure out the direction. Is it positive or negative? The direction of the rotation? Um, M one g is trying to do this. This is if you do a complete sort of spin with your hand, you see that this is counterclockwise. This is in the direction of the unit circle. So it's positive this one is in the direction of the clock. So it's clockwise, so it's negative. All right, so Torque one is going to be positive. M one is four g. We're going around the 10 and the radius of this thing is three. So this is gonna be 1 20 Newton meter and then for torque to negative the masses five gravity rounding to 10 and the radius is 3 m, so there's gonna be negative 1 50 Newton meter. And when you add the whole thing. The sum of all torques will be 1 20 positive plus 50. Negative. So the network is gonna be negative. 30 Newton meter, and that's it. That's it for this one. So hopefully this makes sense. I mean, if you have any questions, let's keep going.

2

example

Net Torque on a disc

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8m

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Hey, guys. So in this way I want to show you how to calculate torque on a disk with forces in a bunch of different directions. Now, this picture looks sort of scary at first. What's what's going on with all these arrows on? You may not see something as complicated as this. I'm putting it all together so we can talk about it all at once. But you should know how to handle all these different situations, individually or together. Let's check it out. Eso Here we have a composite disc, which means there's two different discs. You got the inner disc, which is the dark one here and then the outer disc. Your this is so you can have two different radio. Um, they are free to rotate about a fixed axis perpendicular through its center. So basically, the disk skin spin this way, right? All the forces listed here are 100 Newtons, and there's four of them, So let's just write F one actually make this one half one half one f two F three and this one's gonna have four. They're all 100 on the angles are 37. The only angle here is 37 is this one. All the other ones are sort of either flattened the X or flattened. Why the dotted lines are either exactly parallel or exactly perpendicular to each other. What does that mean? That just means that this line is the same is this line They're parallel to each other and that these lines here make a 90 degree angle. So all these lines were making 90 degree angles with each other Cope Thea inner disk has a radius of 3 m. What? That means that this distance here is 3 m and an outer radius of on the outer disk has a radius of 5 m. Which means that this line over here, this entire distance over here is 5 m. Okay, We wanna know that net torque produced on disk about its central axes. Um, we're gonna use plus reminders to indicate direction clockwise clockwise. So the net torque torque net is the same thing as the sum of all torques. There are four forces, so there's potentially four torques. Remember, a force may cause a torque. There are four forces, so you could have as many as four torques. But some forces may not cause of torque. So let's let's do one by one. Here, Torque one. This guy is F one. Let's first, actually, the space for the sign. Positive. Negative. So we don't forget. Um, f one r one sign of data one, and we'll do the sign a little later. The first thing we do I know if second plug that in there, it's pretty straightforward. First thing we do is we draw on our vector. Then we figure out data and then we figure out the sign. Okay, So the our vector, um, is the distance is the vector from the axis of rotation to the point where the force happens In this case, the our vector for F one is a narrow this way. This is our one and are one has a length of the outer disc, which is five. Okay. And then sign of feta. The angle that are one makes that F one is 90 degrees. So I'm gonna put a 90 here now in terms of sign, imagine you have a disk and you're pushing this way on the disk. So you have a disk and you're pulling like this. So the disk is going like this because you're pulling this direction, right? You can also imagine, as you're, like, sort of stroking the disk this way, right? This is gonna spend like that. Um, so this is going to be a positive torque, because it's causing this to spend in a counter clockwise in the direction of the unit circle. So it's positive this is one and you just end up with positive of torque. 500 Newton meter. All right, so talk to box F two R to sign of data to I know f choose 100 but I gotta figure out our and feta so I'll leave those blank. Um, f two is right here. F two acts in the middle of the acts on top of the axis of rotation. Therefore, the R two will be zero r two equals zero, which means there is no torque at all. When you have something that pulls on the axis of rotation, it produces no Tory because there is no are. And you can see from the equation that the whole thing becomes hero. So it doesn't matter what the angle is, and it doesn't matter. What sign is because you just have zero. Okay, for torque. Three again. Box space for positive. Negative F three R three sign of data three and the forces 100. We gotta figure out our and Fada if you look at F three F three acts on the edge of the outer disk, this is what the R three vector looks like. Alright, Are three right here. So our three vector has a length of the outer radius, which is five. But the problem is thes two arrows, Macon angle of 180 degrees with each other. And the sign of 1 80 is zero. Okay, Sign of 1 80 is zero. So it doesn't matter what the sign is because, um whether it's positive or negative direction, because the torque will be zero. Imagine the disk. And if you push directly towards the middle of the disk, you don't cause the disks. It's spin. The only way to cause this to spin is to either push sort of tangentially on the disk or to, like, push it inning. Right. If you have a disk and you go like this, it's going to spin. But if you push like this on a disk. It doesn't spin. Okay, now let's do Torque four. This is the ugly one up here, and let's figure out what happened. So box. Um, I'm just gonna jump straight into it. The the F four is 100 radius sign of fade up. So this one, we're gonna slow down a little bit to be a little more careful. Notice that it's touching on the inner radius, Theo, inner disk. So I'm just going to redraw just the inner disc coming right here that this is a radius of three. Because it's the inner one. Let me make this a little bigger. Radius equals three. This force acts like this right there. This is the center. First thing we draw is the access. Um, the our vector. The our vector is from the access to the point where the force happens. Um, notice that here, this would look like this. So this dotted line is just an extension of your are vector. And there's an angle of 37 degrees here. Okay, so you got to figure out which angle to use. First of all, the distance will be the entire radius of the inner circle of the inner, um, disk. So it's three. And what about the vector? So what? You were not the angle. So what you could do is you get the art vector here, and you can extend it this way, right? And to make it easier to notice that this is, in fact, the angle, you should use its the angle between the two lines. So 37 is the correct angle. Okay. Remember, the angle given to you isn't always the one you're supposed to use. In fact, that's usually not the one you're supposed to use. Uh, but in this case, it turned out to be that way. What about the direction which this thing will spin? So you should imagine that if you're pushing a disk like this, it's actually gonna spin like this. One way that would make this easier is to think of this as a not as a, uh, not as a push on the disk, but as a pool. You're essentially pull into this. I'm sort of redrawing this f for over here, just kind of extending it down. You're pulling your pushing this way. You're causing it to go like that. Okay, um, so Hopefully, that makes sense. It's pretty. It's pretty visual thing there. But hopefully you can follow. That s o. This direction here is in the direction of the unit. Circle its counterclockwise. So it's positive. Okay, Positive. And if you multiply this whole thing, you get the torque. Four is positive. 1 80 Newton meter. And to find the network, we just add everything up. I got two of them that were zero. So it's just the positive 500 the positive 80 Which gives you positive. 6. 80 Newton meter. Okay, that's it for this one. Hopefully makes sense. Let me know if you have any questions and let's keep going.

3

Problem

Problem

The composite disc below is free to rotate about a fixed axis, perpendicular to it and through its center. All forces are 100 N, and all angles are 37°. The dotted lines are either exactly parallel or exactly perpendicular to each other. The inner (darker) and outer (lighter) discs have radii 3 m and 5 m, respectively. Calculate the Net Torque produced on the composite disc, about an axis perpendicular to it and through its center. Use +/– to indicate direction.

A

−100 N•m

B

+100 N•m

C

−800 N•m

D

+800 N•m

E

−900 N•m

F

+900 N•m

G

−1440 N•m

H

+1440 N•m

4

Problem

Problem

A square with sides 4 m long is free to rotate around an axis perpendicular to its face and through its center. All forces shown are 100 N and act simultaneously on the square. The angle shown is 30°. Calculate the Net Torque that the forces produce on the square, about its axis of rotation.

A

+207 N•m

B

+327 N•m

C

+348 N•m

D

+473 N•m

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