Hey, guys. So in this video, we're going to talk about problems where we have multiple wheels or objects like wheels such as cylinders, discs, or gears. And when we have multiple of these things chained together connected to each other by either a chain or a belt much like a bicycle. Let's check out how these work. So these problems where we have two, it could be more than two, but it's almost always two, wheel-like problems. So when I say wheels, I mean things like discs, cylinders, etc. They're pretty common in rotational kinematics. So, let's check them out. There are two basic cases. And we have a case where the wheels are rotating around a fixed axis. In other words, you have something like these two wheels here. Imagine that there's a chain around them, but this wheel was bolted, let's say to the wall and this wheel was bolted to the wall. So if they start spinning, they're not going to move sideways. Okay? So in this case, we have \( \omega \) because it's going to spin, but the wheel itself, there's an \( \omega \), let's call this \( \omega_1 \). And then there will therefore obviously be \( \omega_2 \) here. But this wheel will have no velocity of the center of mass. This point doesn't move sideways. Okay. It does not move sideways. Same thing with this point, \( v_{cm2} \) doesn't move sideways. Okay. So there's no velocity of the center of mass. An example of this is if these were, pulleys or gears that are, like I said, attached to the wall. The example I gave you earlier is a toilet paper that's fixed to the wall, and it rotates around itself like toilet paper usually works. The other one's a static bicycle. So there are two basic ways you can have a static bicycle. One is you lift your bicycle from the floor and then if you spin the wheels, guess what? The rotation of the wheels doesn't cause the bike to move sideways because it's basically now fixed. It's not free to move. The other way is if you flip your bike upside down. So basically if the wheels are not touching the floor, it's a free axis. It's a free wheel. Okay. And then in other cases, we're going to have situations where two wheels are connected, and they're both free to move. That's a bicycle. Right? As the wheels spin, they spin together. So we're going to look into that later. Alright. So the big thing to remember the big thing to know here is that whenever a chain connects two wheels or connects two wheels, we have that the tangential velocity at the edge of one wheel equals the tangential velocity at the edge of the second wheel. So let's pick a point here and a point here, and I can say that this \( v_t \) equals this \( v_t \). They're all the same. Okay? So \( v_{t1} \), \( v_{t2} \). \( v_{t2} \) is the same as \( v_{t1} \). In fact, you could pick a point anywhere and because it's the same chain, because it's the same chain, the velocity, the tangential velocity has to be the same. Now remember also that when you have a fixed disk, so let me write this here, for fixed axis, remember that \( v_t = r \omega \). The tangential velocity at a point away from the center, \( dt \) is given by \( r \omega \). In this case, we're talking about one, so it's \( r_1 \omega_1 \). Okay? So this guy here, for example, would be \( r_2 \omega_2 \). Now the \( v_t \)s are the same, the \( v_t \)s are the same, but so that's good. But the \( r \)s are different. \( r_1 \) and \( r_2 \) are different. The wheels have different sizes. Therefore, because these guys are different, then the \( \omega \)s will be different as well. Okay. The \( r \)s are different, so the angular speeds will be different. In fact, the greater my \( r \), the smaller my \( \omega \) and vice versa. If you got a tiny wheel and a big wheel, the tiny wheel will spin much faster while the big wheel is slowly spinning. Okay? So there's an inverse relationship there. Alright. So \( v_1 = v_2 \) and \( v = r \omega \), so we can write this. We can write that \( v_{t1} = v_{t2} \), so \( r_1 \omega_1 = r_2 \omega_2 \). This is the big equation for this video, the most important one. Now what I want to do is I want to show how there are four variations or three rather variations of this equation. Okay. And that's because we not only have \( \omega \) or omega to talk about to describe how quickly something spins, we also have frequency, period, and RPM. All four of these are useful in describing how quickly something spins. So I'm going to I'm going to replace \( \omega \) with \( f \), \( t \), and RPM, and this is going to generate different versions of this equation. So \( \omega \) is \( 2 \pi f \). It is also \( 2\pi / t \). And remember, frequency is RPM over 60. So I can also say that \( \omega \) is \( 2 \pi \), and instead of \( f \), I'm going to say RPM over 60. Okay? Can you see that? Yes, you can. Alright. So we got these versions and what I'm going to do is I'm going to replace, \( \omega \) here with this, this, and this. Okay? So \( r_1 \), instead of \( \omega \), I'm going to write \( 2\pi f_1 \), and then \( r_2 2\pi f_2 \). What happens is the \( 2\pi \)s cancel, so you're left with \( r_1f_1 \), \( r_2f_2 \). Okay. If I were to do the same thing, with these other guys here, I'm going to do this one more time. If you were to do this with \( 2 \pi \), you would get \( r_1 2 \pi t_1 \), \( r_2 2 \pi t_2 \). The \( 2 \pi \)s cancel again, and we're left with, we're left with, let me put it over here, \( r_1 / t_1 = r_2 / t_2 \). And then the last piece is with RPM, and I would do the same thing. I'm going to just kind of skip here for the sake of time, but I'm going to say that you can have \( r_1 \), RPM 1 equals \( r_2 \), RPM 2. The most important of these four is the first. The other three are just sort of derivative equivalent similar versions of them. But I find that if you know all four of them, you're best suited to solve these problems very quickly. Okay? These problems are very straightforward. This is the basic idea. When you have two connected wheels, these four equations will work for you. So let's do a quick example here. I got two gears, of radius 1.23. So let's draw this here, \( r_1 = 2 \), the little one, and the bigger one, \( r_2 = 3 \). And they are free, they're free to rotate about a fixed axis. Okay? Now a little tricky here. Remember, I talk about fixed axis and free axis. So even though I use the word free here, they're free to rotate about fixed axis. So these are fixed, okay, which is what we've been talking about, which means the velocity of their center of mass is 0. What it also means is that, \( r_1 \omega_1 = r_2 \omega_2 \) and all the variations. Okay? Alright. So it says here, when you give the smaller pulley, pulley or gears, let's just rewrite this. These were gears, not pulleys. When you give the smaller gears 40 radians per second, doesnt tell me which way. So Im just going to spin it this way. I'm going to say that \( \omega \) here is 40 radians per second. Alright. I want to know what is the angular speed. So this follows this way and then it spins this way. So if this is \( \omega_1 \), what is \( \omega_2 \) that the larger one will have? Very straightforward. These are two connected cylinders, so all I got to do is write the equation for them. \( r_1 \omega_1 = r_2 \omega_2 \). We're looking for \( \omega_2 \). So \( \omega_2 \) is \( r_1 \omega_1 / r_2 \). Just solve the equation. This is 2 over 3 and \( \omega_1 \) is 40, so the answer is going to be 26.7 radians per second. Alright, that's it for this one, very straightforward. Let's check out the next example.

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# Intro to Connected Wheels - Online Tutor, Practice Problems & Exam Prep

In problems involving connected wheels, such as gears or pulleys, the tangential velocity at the edge of one wheel equals that of another. This relationship is expressed as ${r}_{1}{\omega}_{1}={r}_{2}{\omega}_{2}$. The angular speeds are inversely related to their radii, highlighting the importance of understanding rotational kinematics in mechanical systems. Mastery of these concepts aids in solving real-world engineering problems efficiently.

### Intro to Connected Wheels

#### Video transcript

### Speed of pulleys of different radii

#### Video transcript

Alright. So here we have 2 pulleys, with radii 0.3 and 0.4. So notice that this one's a little bit smaller. So r1 is 0.3 and r2 is 0.4. Attached, a light cable runs through the edge of both pulleys. "Light" means the cable has no mass, runs through the edge of both pulleys. The equation for connected pulleys, when they're not, when they're fixed in place is that r1ω1 equals r2ω2. So we're supposed to use r, which is the distance to the center. Now when it says that the cable runs through the edge of both pulleys, the word "edge" here tells us that the distance to the center in this case happens to be big r, the radius, which that's what's going to be most of the time. Okay? So if you're not sure, you can, pretty safely guess that that's what it is, but the problem should tell you. Okay. So that means I'm going to have big R1ω1, big R2ω2. It says you pull down the other end causing the pulleys to spin. So if you're going to pull down this way, this guy is going to spin with ω1 and this guy is going to spin with ω2. And then it says, when the cable has a speed of 5, what is the angular speed of each? So when this cable has a v equals 5, what is ω1 and what is ω2? Okay. And what I wanna remind you is that the velocity here is the same as the velocity here, which is the same as the velocity here, which is the same as the velocity at any point here. So we can write that vcable2 is vcable1. So vcable, which is 5, is what equals r1ω1 and equals r2ω2. Okay? And that's what we're going to use to solve this question. So if I want to know what is ω1, I can look into this part of the equation right here. Okay. So to solve for ω1, I'm going to say 5 equals r1ω1. So ω1 is 5 divided by 0.3, and 5 divided by 0.3 is 16.7 radians per second. And to find ω2, same thing, 5 equals r2ω2. So ω2 is 5 divided by 0.4, which is 12.5 radians per second. Okay? So that's it for ω2, ω1. The key point that I want to highlight here is that not only are these velocities the same at the edge, which allows us to write that r1 equals r2, but also that they equal the velocity of the cable that pulls 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. Alright? So that's it. Let me know if you have any questions.

## Do you want more practice?

More sets### Here’s what students ask on this topic:

What is the relationship between the tangential velocities of two connected wheels?

The tangential velocities at the edges of two connected wheels are equal. This relationship is crucial in problems involving gears, pulleys, or any connected rotational systems. Mathematically, this is expressed as:

${v}_{\mathrm{t1}}={v}_{\mathrm{t2}}$

Since the tangential velocity $v$ is given by $v=r\omega $, where $r$ is the radius and $\omega $ is the angular velocity, the relationship can be written as:

${r}_{1}{\omega}_{1}={r}_{2}{\omega}_{2}$

How do you calculate the angular speed of a larger wheel connected to a smaller wheel?

To calculate the angular speed of a larger wheel connected to a smaller wheel, you use the relationship between their radii and angular speeds. The formula is:

${r}_{1}{\omega}_{1}={r}_{2}{\omega}_{2}$

Solving for the angular speed of the larger wheel ${\omega}_{2}$, we get:

${\omega}_{2}=\frac{{r}_{1}}{{r}_{2}}{\omega}_{1}$

For example, if the smaller wheel has a radius of 2 units and an angular speed of 40 radians per second, and the larger wheel has a radius of 3 units, the angular speed of the larger wheel is:

${\omega}_{2}=\frac{2}{3}*\; 40\; =\; 26.7$ radians per second.

What are the different ways to describe how quickly something spins?

There are four main ways to describe how quickly something spins:

**Angular Velocity ($\omega $):**Measured in radians per second (rad/s).**Frequency ($f$):**Measured in Hertz (Hz), it represents the number of rotations per second.**Period ($T$):**Measured in seconds (s), it is the time taken for one complete rotation. It is the inverse of frequency: $T=\frac{1}{f}$.**Revolutions Per Minute (RPM):**It represents the number of rotations per minute. It is related to frequency by: $f=\frac{\mathrm{RPM}}{60}$.

How do you convert angular velocity to frequency and period?

To convert angular velocity ($\omega $) to frequency ($f$) and period ($T$), you can use the following relationships:

**Frequency:**

$\omega =\; 2\pi f$

Solving for $f$, we get:

$f=\frac{\omega}{2}$

**Period:**

The period is the inverse of frequency:

$T=\frac{1}{f}$

Combining the two equations, we get:

$T=\frac{2}{\pi}\frac{1}{\omega}$

What is the significance of the equation r1ω1=r2ω2 in connected wheels problems?

The equation ${r}_{1}{\omega}_{1}={r}_{2}{\omega}_{2}$ is fundamental in problems involving connected wheels, such as gears or pulleys. It states that the product of the radius and angular velocity of one wheel is equal to that of the other wheel. This relationship ensures that the tangential velocities at the points where the wheels are connected are equal, which is crucial for the system to function correctly. Understanding this equation helps in solving real-world engineering problems efficiently, such as designing gear systems in machinery or understanding the mechanics of bicycles.