12. Rotational Kinematics

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# RPM of pedals of static bicyle

Patrick Ford

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Alright, So here you lift your bike slightly and you begin to spend the back wheel. Since the bikes lifted, the spinning the back wheel will not move, will not cause the front wheel to move or to spin the middle and back Sprockets have diameters d and two D. So let's draw that real quick. I got the back sprocket, which is a little one, and I got the metal sprocket here. I'm gonna also draw the wheel and I'm going to draw the pedal just in case pedals one which causes metal sprocket to to spin which caused the back spark sprocket three to spin which caused the back wheel four to spend. I'm giving the diameters here and the diameters of the middle sprocket d two is two d and of the back sprocket d three is deep now I don't know the value of tea, but I do know that the 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 gonna change this into radius radios. Tombs gonna write this to our in the radius three. Our number radius is just half the diameter. You'd basically be dividing both of these guys by two. Um, I could do our and to our 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. Um, with an rpm at x r p m. In other words, if rpm of the back wheel, which is four is X what will be the rpm in terms of X for the pedals, which is one. Okay, so we're going all the way from 4 to 1. Um, typically, you spin one which causes to to spin, which causes three to spin, which causes for to spin. But this whole thing is connected, so there's not necessarily a sequence you could spend four, and then it goes all the way and causing one to spin. Okay, so we have to be able to trace a connection between these. Well, remember these two guys were connected. These two guys were connected and these two guys were connected. Let's write those connections. So between one and two, the connections that they have the same omega Omega one equals omega too. But in this problem. We don't have Omega's. We have RPMs. So let's change that. And I want to remind you that we can write the relationship between them like this. Omega is two pi f. But F is our PM over 60 So let's do that here. F is our PM over 60. Now. If I plug this on both sides, look what I get. I get W two pine. Our PM 1/60 equals two pi our PM to over 60. What that means is that I can just cancel everything and I'm left with our PM equals R P M. That's the first relationship. Okay, that rpm one equals R P m two because they spin together the relationship between two and three. Let's put this over here. The relationship between two and three is that you have. They have the same V's. They're connected. So the two equals V three. The tensions velocity, which means you can write this as our to Omega two equals R three Omega three. Here you can do a similar thing where you replace Omega with two pi r p m. Over 60. The two Pi MD 60 will cancel on both sides. So this becomes just are too. Our PM two equals r three r p m three. Okay, so that's the second relationship and the third relationship here, it's kind of squeezes here. Sorry about that. Is the relation between three and 43 and four spin on the same axis of rotation. So Omega three equals omega four. And as I've done here, we can just rewrite this as our PM three equals R p m four. Okay, What we're looking for is our PM one which is right here. And what I have is our PM four, which is right here. OK, Our PM four is X. So we're gonna try to connect them using, um, these three equations in green rpm forests X therefore, rpm three is X as well. So this guy here is X. What I'm going to do is solve for r p m two because our PM two is the same as our PM one. So it comes down to this equation here. I'm gonna rewrite this as our two instead of rpm to I'm gonna write rpm one because they're the same. And this is what I'm looking for. Equals R. Three and our PM three is what I know, which is X, Okay? And I want the answer to be in terms of X. So rpm one equals are three X over our to let me disappear here and or three is our are two is to our times x the arse cancel and you end up with X over to X over to. And what that means is that basically the the pedals will spend on behalf the rpm off the back wheel. Okay, now we solve this sort of well, not sort of would solve 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 two wheels, if a wheel has double the radius or double the diameter, it's goingto have half the speed. The bigger you are, the slower you are. OK, the smaller you are, the faster you go. But that that that relationship only applies between the two cylinders. So you could have thought If this guy's ex, then this guy is X. This guy here's bigger double the size that could be X over to and therefore the pedals must be X over to as well. Okay, so the back wheels 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 gonna be half the speed, half the rpm andan. These two guys have the same. That might have been a little bit easier to dio, um, 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 makes sense. Let me know if you guys have any questions.

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