Alright. So here, you lift your bike slightly and you begin to spin the back wheel. Since the bike is lifted, spinning the back wheel will not cause the front wheel to move or to spin. The middle and back sprockets have diameters D and 2d. So let's draw that real quick. I got the back sprocket, which is the smaller one, and I got the middle sprocket here. I'm going to also draw the wheel, and I'm going to draw the pedal just in case. The pedal is 1, which causes middle sprocket 2 to spin, which causes the back sprocket 3 to spin, which causes the back wheel 4 to spin. I'm given the diameters here, and the diameters of the middle sprocket, D₂ is 2d, and of the back sprocket, d₃ is d. Now I don't know the value of d, but I do know that the middle one is twice the radius or twice the diameter, of the back one. Now we don't really use diameters in Physics, so I'm going to change this into radius. Radius 2, I'm just going to write this as 2r and radius 3 as r. Now radius is just half the diameter. You basically be dividing both of these guys by 2. I can do r and 2r instead. As long as this number is double this number, we're good. Okay? So I want to know if you spin the back wheel right here, with an RPM, at x RPM, in other words, if RPM of the back wheel, which is 4, is x, what will be the RPM in terms of x for the pedals, which is 1. Okay. So we're going all the way from 4 to 1. Typically, you spin 1, which causes 2 to spin, which causes 3 to spin, which causes 4 to spin. But this whole thing is connected, so there's not necessarily a sequence. You could spin 4, and then it goes all the way and causing 1 to spin. Okay. So we have to be able to trace a connection between these. Well, remember, these two guys are connected, these two guys are connected, and these two guys are connected. Let's write those, connections. So between 1 and 2, the connections that they have the same omega, ω₁ = ω₂. But in this problem, we don't have omegas. We have RPMs. So let's change that. And I want to remind you that we can write the relationship between them like this. So ω is 2πf, but f is RPM over 60. So let's do that here. f is RPM over 60. Now if I plug this on both sides, look what I get. I get ω₂π RPM₁ over 60 equals 2π, RPM₂ over 60. What that means is that I can just cancel everything and I'm left with RPM equals RPM. That's the first relationship. Okay? That RPM₁ equals RPM₂ because they spin together. The relationship between 2 and 3, let's put this over here, the relationship between 2 and 3, is that they have the same v's. They're connected. So v₂ = v₃, the tangential velocity, which means you can write this as r₂ω₂ = r₃ω₃. Here, you can do a similar thing where you replace ω with 2πRPM over 60. The 2π and the 60 will cancel on both sides. So this becomes just r₂RPM₂ = r₃RPM₃. Okay. So that's the second relationship. And the third relationship here, it's kind of squeezed this here. Sorry about that. It's the relationship between 3 and 4. 3 and 4 spin on the same axis of rotation, so ω₃ = ω₄. And as I've done here, we can just rewrite this as RPM₃ = RPM₄. Okay. What we're looking for is RPM₁, which is right here. And what I have is RPM₄, which is right here. Okay. RPM₄ is x. So we're going to try to connect them using these three equations in green. RPM₄ is x. Therefore, RPM₃ is x as well. So this guy here is x. What I'm going to do is solve for RPM₂ because RPM₂ is the same as RPM₁. So it comes down to this equation here. I'm going to rewrite this as r₂. Instead of RPM₂, I'm going to write RPM₁ because they're the same, and this is what I'm looking for, equals r₃. And RPM₃ is what I know, which is x. Okay. And I want the answer to be in terms of x. So RPM₁ = r₃x/r₂. Let me disappear here. And r₃ is r. r₂ is 2r times x. The r's cancel and you end up with x/2. x/2. And what that means is that basically the pedals will spin at half the RPM of the back wheel. Okay? Now we solve this sort of, well, not sort of. We solved this mathematically, but it might have been easier to actually just kind of think about this stuff. Okay? Now interesting here is this equation. This is a linear relationship. And what that means is that if a wheel has double the radius or double the diameter, it's going to have half the speed. The bigger you are, the slower you are. Okay. The smaller you are, the faster you go. But that relationship only applies between the two cylinders. So you could have thought, if this guy is x, then this guy is x. This guy here is bigger, double the size, so it's going to be x/2. And therefore, the pedals must be x/2 as well. Okay? So the back wheel is x, which means the back sprocket has to be x. When I cross it over to the other side, it's doubled the radius, so it's going to be half the speed, half the RPM, and then these two guys have the same. That might have been a little bit easier to do so that you don't run the risk of getting confused with the math and all the equations, whatever you prefer. Alright. That's it for this one. Tricky question. Hope it makes sense. Let me know if you guys have any questions.