Hey, guys. So in this video, we're going to start solving rotation problems with conservation of energy. Let's check it out. So you may remember that when you have a motion problem between two points, meaning the object starts here and ends up over here somewhere, where either the speed v, the height h, or the spring compression x changes, any combination of those three changes, we can use most of the time the conservation of energy equation to solve these problems. So we're going to do that now to rotation questions. The only difference is that in rotation, your kinetic energy can be not only linear but also rotational. So that's the new thing that you could be spinning. And it could actually also be both. Right? It could be that our total kinetic energy is linear plus rotational. So we're going to use the conservation of energy equation, which is Kinitial+Uinitial+work nonconservative=Kfinal+Ufinal. I want to remind you that work nonconservative is the work done by you, by some external force, plus the work done by friction if you have some. Now when you do this, remember, you write the energy equation, and then you start expanding the equation. What I mean by expanding is you replace k with what it is. And k used to be simply 12mv2. But now, it could be that you have both of them. Right? It could be, let's say that or let's say instead of 12mv2, the object is just spinning. So you're going to write Iω2. Okay? The key thing to remember, and you would do this for the rest of them. The key thing to remember, the most important thing in these questions to remember is that you will rewrite v and ω in terms of each other. What do I mean by that? What I mean is that when you expand the entire equation, you might end up with 1 v and one ω, or 2 v's and one ω, whatever. If you have a v and a ω, that's 2 variables, you're going to change 1 into the other so that you end up with just one variable. For example, most of the time, v and ω are linked by this, or sometimes they are linked by this. Right? Sometimes they're not linked at all. But most of the time, they're connected by either one of these two equations, which means what I'm going to do is rewrite ω as v over r. And wherever I see a ω, I'm going to replace it with a v over r. So instead of having v and ω, I have v and v. And that means that instead of having 2 variables, I have just 1, and it's easier to solve the problems. That's the key thing to remember is rewrite 1 into the other. Let's do an example. Alright. So here we have a solid disc. Solid disc, let's stop there, means that the moment of inertia we're going to use is that of a solid disk, which is the same as a solid cylinder, and it's going to be 12mr2. And it says it's free to rotate about a fixed perpendicular axis through its center. Lots of words. Let's analyze what it's saying here. Free to rotate just means that you could rotate, right? Like you can actually spin. Some things can be spun around, others can't. But even though it says that it's free to rotate, it's around a fixed axis. Okay? Remember, it's the difference between a roll of toilet paper that is fixed on the wall and it's free to rotate around the fixed axis versus a free roll of toilet paper that can roll around the floor. Okay. Here, we're fixed in place. So we're going to say that it spins like this. So it has no v, right? Like the actual disc has no velocity v because it's not moving sideways. The center of mass doesn't change position, but it does spin. Okay. Actually, we don't know which way it spins. Let's leave it alone for now. But I'm just going to write that ω is not 0 because it's going to spin. Now what else? It says that the axis is through its center, so it's spinning around the center like this. Okay. And it's saying that it's perpendicular. Perpendicular means that it makes a 90-degree angle. Okay. Perpendicular means it makes a 90-degree angle with the object. So I got a little disc here. This sort of looks like a disc, and I wan

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# Conservation of Energy with Rotation - Online Tutor, Practice Problems & Exam Prep

In solving rotation problems using conservation of energy, remember that kinetic energy includes both linear and rotational components. The conservation of energy equation is expressed as Ki + Ui + W__{nc} = Kf + Uf. When variables like velocity (v) and angular velocity (ω) appear, convert one into the other using v = r ω. This simplification aids in solving for unknowns, such as final angular speed (ω_{f}).

### Conservation of Energy with Rotation

#### Video transcript

### Work to accelerate cylinder

#### Video transcript

Hey, guys. Let's check out this example here. We have a solid cylinder, and we want to know how much work is needed to accelerate that cylinder. So, a solid cylinder means that \( I \) is going to be half \( m r^2 \) because that's the equation for the moment of inertia of a cylinder. Mass is 10 and radius is 2. I'm going to put these here. If you want to, you could already calculate the moment of inertia, right? So \( I \) is \( \frac{1}{2} \times 10 \times 2^2 \), and the moment of inertia is 20. Okay? So, we can already get that 20 kilograms meters squared.

It says it is mounted and free to rotate, on a perpendicular axis through its center. Again, you have a cylinder, which is the same thing as a disk and it has an axis. It's mounted on an axis that is perpendicular to it. So it looks like this. Right? And it's free to rotate about that axis. It just doesn't wobble like that, right? It rotates like this. Now, most of the time you actually have this where the axis is horizontal. So it's on a wall, right? So it's something that's on a wall, and you have the disc spinning like this, all right?

So it says that the cylinder is initially at rest, so the cylinder spins around itself, but it's initially at rest. \( \omega \) initial equals 0. And we want to know what the work done is to accelerate it from rest to a 120 RPM. Okay? Remember, most of the time when you have RPM, you're supposed to change that into \( \omega \) so that you can use it in the equation. So let's do that real quick just to get that out of the way. \( \omega \) final is \( 2 \pi f \) or \( \frac{2 \pi \times \text{RPM}}{60} \). If you plug 120 here, you end up with 120 divided by 60 is 2. You end up with \( 4 \pi \) radians per second.

Okay? So, I am going from 0 to \( 4 \pi \), and I want to know how much work that takes. So work is energy. Hopefully, you thought of using the conservation of energy equation, \( k_{\text{initial}} + u_{\text{initial}} + W_{\text{nonconservative}} = k_{\text{final}} + u_{\text{final}} \). In the beginning, there's no kinetic energy because it's not spinning, it's not moving sideways. The potential energies cancel because the height of the cylinder doesn't change. It stays in place. Right? \( W_{\text{nonconservative}} \) is the work done by you plus the work done by friction. There is no work done by friction, just the work done by you, which is exactly what we're looking for. And kinetic energy, which is only kinetic rotational. Right? There's no linear. It's not moving sideways. The center of mass of the disk stays in place. So \( v \) equals 0. So, the only type of kinetic energy we have is rotational, which is \( \frac{1}{2} I \omega^2 \). We're looking for this, so all we got to do is plug in this number. Work is going to be \( \frac{1}{2} \times 20 \times (4 \pi)^2 \).

Okay? So, if you multiply all of this, you get that it is 1580 joules of energy, and that's how much energy is needed to get the solid cylinder from rest all the way to a speed of \( 4 \pi \) or a 120 RPM. Cool? Very straightforward. Plug it into the energy equation because we were asked for work. Alright? Hope it makes sense. Let me know if you guys have any questions and let's keep going.

How much work is needed to stop a hollow sphere of mass 2 kg and radius 3 m that spins at 40 rad/s around an axis through its center?

### Which shape reaches bottom first?

#### Video transcript

Hey, guys. Let's check out this conservation of energy example. Here we have 3 objects of equal mass and equal radius, but they have different shapes. Remember, your shape determines what equation your moment of inertia has. And it's usually something like a fraction, m \cdot r^2. But the number in here depends on the shape. So, if you have different shapes, you're going to have different I equations. They're all released from rest at the same time from the top of an inclined plane. So I'm going to have here's a solid cylinder, here is a hollow cylinder, and here is a solid sphere. They're all from rest. They all have the same mass. Right? So they have the same mass, the same radius. They all start from rest, and they all start from the top of the inclined plane. They're going to start from the same height as well. Everything is the same except the shapes. And I want to know who reaches the bottom first if they're released at the same time. And this question will depend on your moment of inertia. What I want to remind you is that moment of inertia is a measurement of angular resistance, of rotational resistance. So, you can think that the greater my I, the heavier I am, the more I resist rotation. Therefore, I will get to the bottom last because I'm slower. Okay. So more I, you can think of this as being heavier. Now it doesn't mean that I have more mass. Right? That's why I have "heavier". I have more resistance, therefore, I am slower. Okay. Now a solid cylinder has a moment of inertia of frac{1}{2}m \cdot r^2. A hollow cylinder has a moment of inertia of m \cdot r^2. A solid sphere has a moment of inertia of frac{2}{5}m \cdot r^2. So in this question, all we're doing is comparing these numbers because the m and the r are the same. Now this is a little bit easier if you use decimals. So this is 0.5. This is 1.0. And 2 over 5 is 0.4. Okay. And you can see from here that this one is the lightest one. Okay. Because the coefficient number in front of the m \cdot r is the lowest, It's the lightest one. Therefore, it is the fastest one. Therefore, it gets to the bottom first. Okay? It gets to the bottom first. So the sequence is that the solid sphere is first. I'm going to write it like this. Gets to the bottom first. The second one is going to be the solid cylinder is second. And the third one is going to be the hollow cylinder. Okay. Now there is a pattern here. There's a reason why the hollow cylinder is slower than the solid cylinder. The solid cylinder has very good mass distribution. The mass is very evenly distributed. And remember, the more evenly distributed the mass, the lighter you are, the less I. So, better mass distribution means lower I, which means you are lighter. Okay. A hollow cylinder has all of its mass concentrated on the edge. It has worse mass distribution, which means it has a higher I, which means it is heavier. So it has a worse mass distribution. Therefore, it is heavier. Now a solid sphere is even more well-distributed than a solid disk. A solid disk has all the mass on a thin layer like this. A sphere has basically the most perfect mass distribution you can have. That's why it has the most symmetrical one. That's why it has the most symmetrical one. That's why it has the lowest of them all. So the sphere is always fastest. Okay? So that's it for this one. Let's keep going.

### Cylinders racing down:rolling vs. sliding

#### Video transcript

Hey, guys. Let's check out this conceptual conservation of energy question. So here we have 2 cylinders of equal mass and radius. So basically the same cylinder. They're released from rest from the top of 2 hills having the same height. So, it's the 2 very similar or so far identical situations. Let's say I have cylinder A on hill A and cylinder B, let's put this over here, on hill B. And everything is the same except that A rolls without slipping, okay, rolls without slipping. So that means that it's not only going to have a v but it's also rolling so it has a ω. Remember, this is called rolling motion if you roll without slipping. And it means that v_{cm} = rω. But B slides without rolling. Those are the 2 options you can have. It's either going to roll without slipping or it's going to slip without rolling. You can have both. You could, but it's more complicated. You're not going to see that. Alright. What this means is that it's moving this way with a v, but it has no ω. Okay? Basically, it's falling. It's going down as if a box would. So you can think of this as a box going down because it doesn't roll. Now before we answer the question, I want to talk about how that's even possible. Well, basically, the difference is that here you have just enough static friction to cause this thing to roll. Remember, you need static friction. You need static friction to have an angular acceleration when you have a rolling problem. Here, you have a situation where the hill is different for whatever reason, and there is no static friction, which is why it doesn't roll. Okay. Now the question is, who gets to the bottom with the fastest speed? Who has is v_{af} or v_{bf} going to be the greater one? And I want you to take a guess or just sort of try to figure this out. Think about which one do you think would get to the bottom first. Now, most people get this question wrong. Most people say that this one will get to the bottom first. And I suspect it has to do, with the fact that you're going to think that objects that roll are faster, right, than objects that slide. And that's probably because you associate sliding with some friction slowing you down or you associate rolling with wheels and wheels are fast because we have wheels in cars. But in reality, the one that gets to the bottom with the greater speed is this one, greater speed. Okay. Now why is that? That's because of conservation of energy. So conservation of energy equation, k_{initial} + u_{initial} + work non-conservative = k_{final} + u_{final}. In both of these situations, the kinetic energy in the beginning is 0. I have some potential energy. Okay. The work done by nonconservative forces is 0 as well. You don't do anything. You're just watching. Now in the first problem, there is in the first case, there is static friction. But the work done by static friction is 0. Right? So it actually even though there is static friction, it doesn't do any work. There is some kinetic at the end and there's no potential at the end. So all we have is that my potential energy in the beginning goes into kinetic final at the end. So if they start from the same height, they have the same and they have the same mass and everything. They have the same potential energy for both of them. So let's say that number is, a 100 joules. In both problems, you start with a 100 joules and then you roll to the bottom. The difference is that in the first situation for a, the 100 joules is going to get split between kinetic linear and kinetic final. Okay. Between kinetic linear and kinetic rotational. Let's say 80 goes to linear, 20 goes to rotational. Totally making this up. For B, the entire amount, the entire 100 joules is going to go into kinetic linear because there is no kinetic rotational because it doesn't roll around itself. So notice how this guy ends up with a greater kinetic energy than A. And that's irrespective of the division. Right? So no matter how these two numbers get balanced out, the 100 will always be bigger. Even if this was 90/10, 100 is still greater. Okay. So one way to think about this, so the answer is that it's basically because this has one type of energy, one type of kinetic, and this one has 2 types of kinetic. One way to think about this is to think about energy as money. It's expensive to get some sort of energy going on. You're using up your potential energy and transforming it into kinetic. Now if you're falling and rolling, it takes a little bit of energy to get it to move and it takes a little bit of energy to get it to roll. So rolling costs you some energy. Therefore, some of the energy that would have gone here is actually going here to cause you to roll. Okay? In fact, the faster you roll, the more energy, the more kinetic energy you have, and then the less linear energy, the more rotational energy you have, so there's less energy that's left, for your linear, for your v. Okay? So your ω is stealing energy from your v so you end up with a lower v. That's why. Cool? That's it. It's a big, really popular question conceptually. Let me know if you have any questions. Let's keep going.

Two solid cylinders of same mass and radius roll on a horizontal surface just before going up an inclined plane. Cylinder A rolls without slipping, but cylinder B moves along a slippery path, so it moves without rotating at all times. At the bottom of the incline, both have the same speed at their center of mass. Which will go higher on the inclined plane? (Why?)

## Do you want more practice?

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

What is the conservation of energy equation for rotational motion?

The conservation of energy equation for rotational motion is similar to that for linear motion but includes rotational kinetic energy. It is expressed as:

${K}_{i}+{U}_{i}+{W}_{\mathrm{nc}}={K}_{f}+{U}_{f}$

Here, ${K}_{i}$ and ${K}_{f}$ are the initial and final kinetic energies, which can include both linear and rotational components. ${U}_{i}$ and ${U}_{f}$ are the initial and final potential energies, and ${W}_{\mathrm{nc}}$ is the work done by non-conservative forces.

How do you convert linear velocity to angular velocity in rotational motion problems?

To convert linear velocity ($v$) to angular velocity ($\omega $), you use the relationship:

$v=r\omega $

Here, $r$ is the radius of the circular path. This equation allows you to express one variable in terms of the other, simplifying the problem to a single variable.

What is the moment of inertia for a solid disc?

The moment of inertia ($I$) for a solid disc rotating about an axis through its center is given by:

$I=\frac{1}{2}m{r}^{2}$

Here, $m$ is the mass of the disc, and $r$ is its radius.

How do you solve for the final angular speed using conservation of energy?

To solve for the final angular speed ($\omega $_{f}) using conservation of energy, follow these steps:

1. Write the conservation of energy equation: ${K}_{i}+{U}_{i}+{W}_{\mathrm{nc}}={K}_{f}+{U}_{f}$

2. Expand the kinetic and potential energy terms. For rotational motion, kinetic energy is $\frac{1}{2}I{\omega}^{2}$.

3. Substitute known values and solve for $\omega $_{f}.

Example: If a disc with moment of inertia $I$ is acted upon by a force doing work $W$, the final angular speed is found by:

$W=\frac{1}{2}I{\omega}^{2}$

Solve for $\omega $:

$\omega =\sqrt{\frac{2W}{I}}$

What is the significance of 'without slipping' in rotational motion problems?

The term 'without slipping' in rotational motion problems indicates that the point of contact between the rotating object and the surface does not slide. This condition allows us to use the relationship:

$v=r\omega $

Here, $v$ is the linear velocity, $r$ is the radius, and $\omega $ is the angular velocity. This relationship simplifies the problem by linking linear and rotational motion, making it easier to solve for unknowns.

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