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Speed of pulleys of different radii

Patrick Ford
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All right. So here we have two police with radio I 20.3 and 0.4. So notice that this one's a little bit smaller. So are one is points three and r two is 20.4 attached the shone. A light cable runs through the edge of both police. Light means that cable has no mass runs through the edge of both police. Um, the equation for connected police. Um, when they're not, when they're fixed in place is that our one omega one are two omega, too. So we're supposed to use our which is the distance to center. Now, when it says that the cable runs through the edge of both pulleys, the word edge here tells us that the distance of the center in this case happens to be big are the radius, which that's what's gonna be most of the time. Okay, so if you're not sure, you can, um, pretty safely guessed that that's what it is, But the problem should tell you. Okay, so that means I'm gonna have big are one omega one big are to Omega, too. Um, it says you pull down on the other end, causing the pulley the police to spin. So if you're gonna pull down this way, this guy is gonna spin with Omega one. And this guy's going to spin with Omega too. And then it says when the cable has a speed of five, what is the angular speed of each? So in this cable has a V equals five. What is Omega one and what is Omega too? Okay. And what I want to remind you is that the velocity here is the same as the velocity here, which is the same as the velocity here to the same velocity at any point here. So we can write that the cable is V T one VT two. So V cable, which is five, is what equals R one omega one and equals R two Omega, too. Okay. And that's what we're going to use to solve this question. So if I want to know what is Omega one, I can look into this part of the equation right here. Okay, so to solve for Omega one, I'm gonna say five equals R one omega one. So Mega one is five. Divided by points three and five, divided by 50.3 is 16.7 radiance per second. And to find a mega chew. Same thing five equals R two omega, too. So Omega two is five, divided by 50.4, which is 12 points, five radiance per second. Okay, so that's it for W W two w one w two. Um, the key point that I wanna highlight here. The key point I wanna highlight here is that not only are these, um, these two velocity is the same at the edge, which allows us to write that are one equals R two, but also that the equal the velocity of the cable that pose them. That's what's special about this problem. It's this blue piece right here that equals the velocity of the cable as well. Okay, so please remember that just in case you see something like it. All right, so that's it. Let me know if you have any questions.