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Torque on Discs & Pulleys

Patrick Ford
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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.