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Ch 14: Torque & Rotational DynamicsWorksheetSee all chapters
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Torque on Discs & Pulleys
Intro to Torque
Torque Due to Weight
Torque & Acceleration (Rotational Dynamics)
Net Torque & Sign of Torque
How to Solve: Energy vs Torque

Concept #1: Torque on Discs & Pulleys


Hey guys! In this video we're going to start talking about torques acting on discs, on disc-like objects such as pulleys or cylinders. This is important because it's going to be a key part of problems of more elaborate problems we're going to have to solve later on. Let's check it out. Let's say you have a disc or a pulley and you're pulling on it like this with a force of F which then causes it to spin like this. What matters as it says here what matters is r which is the distance from the axis to the force, from the axis which is usually in the middle to the force, this distance here r. That's what matters, not the radius. In this particular example here, little r is the same as the radius because the rope pulls on the disc and the edge of the disc. But let's say if you had let's call this F1r1, so r1 is the radius. But let's say you were pulling with another force right here, F2. In this case the force doesn't pull from the edge so what matters is not the radius of the disc. But in fact what matters is the distance. In this case r2 is not the radius. WhatÕs always going to matter is the distance. Most of the time the distance will be the radius of the disc, but sometimes it won't be. Let's do an example here. Two masses, m1 and m2, m1 is 4. Let's put it here, m2 is 5. TheyÕre connected by a light string which passes through the edge of a solid cylinder. There's a string here that goes like this. It wraps around the cylinder. The cylinder has mass m3 = 10, and radius, remember radius is big R, little R 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, to the cylinder and through its center. Basically the cylinder spins around its central axis. We want to know what is the net torque produced on the cylinder when you release the blocks. Torque net is the sum of all torques and we want to know the net torque on the cylinder. We have to figure out how many torques act on the cylinder, add them all up. Remember a force may produce a torque. What we do is we look at all the forces on the cylinder and then we figure out which ones produce a torque. There are three forces or four forces that act on the cylinder. I have this one here is m1g pulling down, m2g pulling down. There is the mg of the cylinder itself, m3g and there is a tension that holds the cylinder up. 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 how there is no torque due to m3g and due to tension. That's because they act on the axis of rotation. Torque of T, the torque of any force is Frsin_. In this case, F is T so torque of T is Trsin_. But the tension is pulling from the middle here, it's holding the cylinder from the middle so this r is zero. This would have been zero even if the tension was somehow holding it up here. The other problem with this part is that tension pulls it up. Let me draw it over here. Let's say tension was going this way. Tension pulls it up. You have to draw the r vector from the middle to the point of the force happens, r vector. These two arrows are both going in same direction which means that the angle between them is zero. _ is zero degrees, which means that here you would plug in sine of 0, which is zero. Whether the tension pulls in the middle or if it pushes in the edge, it doesn't matter. They have the same angle with each other here, so this whole thing would be zero. The torque due to mg is for sure in the middle, so it's mg(0) and it also makes an angle of zero because this is the r for T, and there's an r for mg and the mg is that way, so this angle is zero as well. Both of these guys don't actually produce any torque. The only forces that will produce a torque are m1 and m2. Right away, you should know this for future problems if you have a pulley with the disc in the middle. The weight of the pulley is not going to cause a torque on the pulley and neither is some sort of force that holds it up whether itÕs a tension or it's held onto an axis or something so there's like a normal force. These forces holding the disc up won't produce a torque on disc. One way to think about this is that if the disc had been held in place by this axis, axis pulling up and mg pulling down. It wouldn't spin on its own and that's because there's no forces causing a torque on it. The only forces that cause a torque are forces that could cause it to potentially spin. That's what you get with m1 that's trying to do this to the disc and m2 thatÕs trying to do this, so torque 1 and torque 2. If you imagine a disc, if you pull from the edge of the disc, it would roll from either side. Let's now calculate the torque due to these two forces. I'm going to call this torque 1 which is force which is m1grsin_. What's the r vector for m1g? m1 is acting all the way at the edge of the disc because the cable for m1 is passing through the outside, the edge of the disc. The r vector is exactly the radius, so r1 is the radius. By the way, that's the same thing that happens with m2gr2 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. Look how the r vector and the force makes an angle of 90 and the r vector over here and its respective force makes an angle of 90. Both of these guys, we're going to have that the distance is the entire radius and the angle is 90 degrees. This obviously becomes a 1 and now I just have to multiply the numbers. The last thing you got to do is also figure out the direction Ð is it positive or negative, the direction of the rotation. m1g 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. Torque1 is going to be positive, m is 4, g we're going to round to 10, and the radius of this thing is 3. This is going to be 120 Nm. Then for Torque 2, negative, the mass is 5, gravity rounding to 10 and the radius is 3 meters. This is going to be -150 Nm. When you add the whole thing, the sum of all torques will be 120 positive + 150 negative. The net torque is going to negative 30 Nm and that's it. That's it for this one. Hopefully this makes sense. Let me know if you have any questions. Let's keep going.

Example #1: Net Torque on a disc


Hey guys! In this video I want to show you how to calculate torque on a disc with forces in a bunch of different directions. This picture looks sort of scary at first Ð what's going on with all these arrows. 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. 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 here. This is so you can have two different radii. They are free to rotate about a fixed axis perpendicular through its center. Basically the discs can spin this way. All the forces listed here are 100 N and there's four of them so let's just write F1, F2, F3, and this one's going to be F4. They're all 100 and the angles are 37. The only angle here is 37, itÕs this one. All the other ones are sort of either flat in the X or flat in the Y. 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 as this line. TheyÕre parallel to each other and that these lines here make a 90-degree angle. All these lines are making 90-degree angles with each other. The inner disc has a radius of 3 meters. What that means is that this distance here is 3 meters and the outer disc has a radius of 5 meters, which means that this line over here this entire distance over here is 5 meters. We want to know the net torque produced on disc about its central axis and we're going to use plus or minus to indicate direction Ð clockwise or counterclockwise. 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 can have as many as four torques. But some forces may not cause a torque. Let's do one by one here. Torque 1, this guy, is F1. Let's first actually leave a space for the sign Ð positive or negative so we don't forget. F1r1sin_ and we'll do the sign a little later. The first thing we do, I know F so I can plug that in there. It's pretty straightforward. The first thing we do is we draw an r vector then we figure out _, and then we figure out the sign. The r vector is the distance is the vector from the axis of rotation to the point where the force happens. In this case the r vector for F1 is an arrow this way. This is r1. r1 has a length of the outer disc which is 5, and then sin_. The angle that r1 makes with F1 is 90 degrees, so I'm going to put a 90 here. In terms of sign, imagine you have a disc and you're pushing this way on the disc. You have a disc and you're pulling like this, so the disc is going like this because you're pulling this direction. You can also imagine as you are like sort of stroking the disc this way, the disc is going to spin like that. This is going to be a positive torque because it's causing the disc to spin in a counterclockwise in the direction of the unit circle, so it's positive. This is 1 and you just end up with +500 of torque, 500 Nm. Torque 2, box F2r2sin_2. I know F2 is 100, but I got to figure out r and _, so I'll leave those blank. F2 is right here. F2 acts on top of the axis of rotation therefore the r2 will be zero, r2 = 0, which means there is no torque at all. When you have something that pulls on the axis of rotation, it produces no torque because there is no r and you can see from the equation that the whole thing becomes zero. It doesn't matter what the angle is and it doesn't matter what the sign is because you just have zero. For Torque 3, again box space for positive or negative F3r3sin_3 and the force is 100. We got to figure out r and _. If you look at F3, F3 acts on the edge of the outer disc. This is what the r3 vector looks like. r3 vector has a length of the outer radius which is 5, but the problem is these two arrows make an angle of 180 degrees with each other and the sine of 180 is zero, so it doesn't matter what the sign is because whether it's positive or negative direction because the torque will be zero. Imagine the disc and if you push directly towards the middle of the disc, you don't cause the disc to spin. The only way to cause the disc to spin is to either push sort of tangentially on the disc or to like push at an angle. If you have a disc and you go like this, it's going to spin. But if you push like this on a disc, it doesn't spin. Let's do Torque 4. This is the ugly one up here. Let's figure out what happens. I'm just going to jump straight into the F4 is 100, rsin_. This one we're going to slow down a little bit and be a little more careful. Notice that it's touching on the inner radius, the inner disc. I'm just going to redraw just the inner disc. I'm going to write here that this has a radius of 3 because it's the inner one. This force acts like this right there. This is the center. First thing we draw is the r vector. The r vector is from the axis to the point where the force happens. Notice that here this would look like this, so this dotted line is just an extension of your r vector. There's an angle of 37 degrees here. 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 disc, so it's 3. What about the angle? What you could do is you get the r vector here and you can extend it this way and to make it easier to notice that this is in fact the angle you should use. It's the angle between the two lines, so 37 is the correct angle. Remember, the angle given to you isn't always the one you're supposed to use. In fact, it's usually not the one you're supposed to use. But in this case, it turned out to be that way. What about the direction in which this thing will spin? You should imagine that if you're pushing a disc like this, it's actually going to spin like this. One way that would make this easier is to think of this not as a push on the disc but as a pull. YouÕre essentially pulling the disc. I'm sort of redrawing this F4 over here, just kind of extending it down. You're pushing this way, you're causing it to go like that. Hopefully that makes sense. It's pretty visual thing there but hopefully you can follow that. This direction here is in the direction of the unit circle, it's counterclockwise so it's positive. If you multiply this whole thing, you get the Torque 4 is +180 Nm. To find the net torque, we just add everything up. I got two of them that were zero, so it's just the +500 and the +180, which gives you +680 Nm. That's it for this one. Hopefully it made sense. Let me know if you have any questions and let's keep going.

Practice: 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.

Practice: 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.