Hey guys. So in this video, we're going to talk about how different kinds of motion give you different kinds of energy. And I'm going to walk you through a comprehensive list of all the possibilities you might see so that for any kind of problem, you always know what kind of energy goes with the problem, what kind of energy exists in that situation? Let's check it out. So what we want to do here is make sure that we know which energies go with a particular situation. A particular situation. Now, a potential problem arises when you have point masses. And that's because point masses, if you remember, if they're going a circular path, they have rotational speed. So if you have a tiny little mass here, \( m \), and you have speed. So if you have a tiny little mass here, \( m \), and it's going around a path here, it has a distance little \( r \) to the middle, right? And let's say it's spinning that way with an angular speed \( \omega \). But it also has a tangential speed, which is linear. Right? It also has an instantaneous speed that's pointing this way. We call this \( v_{\text{tan}} \). Okay. So but does that mean, however, does that mean that it has linear kinetic energy and rotational kinetic energy? It has a \( \omega \). So does it have a kinetic rotation rotational kinetic energy? It has a \( v \). Does it have a linear kinetic energy? Does it have both? And the answer is no. Nope. It doesn't have 2 energies. We only have here one type of motion, so we can only have one type of energy. The object only has one type of motion. It only spins around a central point. Okay. \( v_{\text{tan}} \) is just the linear equivalent of \( \omega \). Think of it as like a mirror image. Right? If you look at a mirror, there aren't 2 of you. It's just the mirror is a reflection of you. So \( v_{\text{tan}} \) is just the, the linear reflection of \( \omega \) but there's only one velocity, only one motion, one type of motion, I should say. So if that doesn't make sense yet, that's cool. We're going to do 6 examples and that's going to cover every possibility. So let's let's start here. So you have a box in a straight line, does it have linear kinetic candidates? Does it have rotational kinetic candidates? Does that have both? So a box in a linear, in a straight line, so something like this, the box is moving. It has a \( v \), so he has a linear kinetic energy. The box doesn't roll around itself or around anything else. So it has no kinetic, no rotational energy, only linear. A disk spinning around itself. So a disc, here's the axis in the middle. The disc spins around itself. Does it have kinetic linear? Does it have kinetic rotational? Every time you spin around yourself, you have kinetic rotational. Now linear has to do with you moving sideways or up and down. Your axis of rotation, your I should say your center of mass, the middle of the object has to actually move. If you spin around yourself, the middle never moves. So there is no linear kinetic energy in this case. Okay. What about the earth spinning around itself? So this is a disk. The earth is a sphere roughly. Right? And if it spins around itself, it's very similar. There's no kinetic linear, and there is kinetic rotation. There's kinetic rotation because they have a shape around itself. But if you're talking about the earth around itself, just that part of the earth's motion, as the earth spins around itself, it doesn't move sideways. Now you do know the earth does move around the sun but that's a different motion. Here, we're talking about just this piece of it. Now what about the earth around the sun? The earth around the sun, the center of mass of the earth does move, right? The center of mass of the earth does move around the sun. So let's draw that real quick. The earth is doing this. Now this is where it gets complicated around the sun, but it also has an instantaneous velocity \( v \). Around the sun, but it also has an instantaneous velocity \( v \). So you could think of this as linear or rotational. In fact, if you solve for for it using \( k_{\text{l}} \), and if you solve for it using \( k_{\text{r}} \), you're going to get the same number, the same answer. Okay? So you could look at the energy either way. The problem is you have to make sure you don't count it as both. What I mean by that is if I ask you for the total energy of the earth going around the sun, so the kinetic energy of the earth around the sun, you can't do \( k_{\text{l}} \) plus \( k_{\text{r}} \). Okay. You can't double count it. Alright. So here's how I'm going to simplify this. I'm going to say whenever you have an object spinning around itself or around something else, we're going to call that rotational kinetic energy and we're going to say that there is no linear energy. Right? I just mentioned how you could look at it both ways because there's a \( v \) and a \( \omega \). You just can't count it as both. Well, we're going to forget about that. We're just going to make our lives simpler and always think of it as rotational energy and never linear energy. Okay? Even though you do have a linear velocity going around this thing. So I hope that makes sense, around itself and around the sun. Now what about the total energy of the earth? I'm going to add a little thing here. I'm going to call it I got \( C \) and \( D \). I'm going to call this \( c \) \( d \). What about the total kinetic energy of the earth? Kinetic total of the earth, meaning the kinetic energy of the earth around itself plus the kinetic energy of the earth around the sun. Well, both of these are rotational. The earth has a rotational energy around itself, and it has a rotational energy around the sun, \( k_{\text{r sun}} \). \( K \). So rotational is if you spin around yourself or you spin if you spin around something else. What about the moon spinning around the earth? What's the total, what kind of energy, does the moon have spinning around the earth? So here, you have to know that the earth, the moon doesn't spin around itself. Okay? And here, by the way, I want the total energy, the total energy. And what I mean by that is I want to know, does the earth spin around itself and does it, does the moon spin around itself and does it spin around the earth? So moon goes around the earth. Oops. What happened? The moon goes around the earth. Let's just do it like this. But you might know, you should know that the moon doesn't spin around itself. So the moon only has \( k_{\text{rote}}, k_{\text{moon}} \). As it's going around the earth, the \( k_{\text{moon self}} \) plus \( k_{\text{moon earth}} \), the moon spins around the earth, but it doesn't spin around itself. Right? And that's because the moon is it's it's not really dark. It just means that we can never see it because the moon is always looking at us. Right? It's like if you look at a mirror, you can't see your back. You can only see your front. Okay? So you should know that the moon doesn't spin around itself. So it only has a rotational energy around the earth. Now what about a roll of toilet paper rolling on the floor? Right? So you got a toilet paper. It is rolling on the floor. It's running loose. Right? It has a \( v \) and it has a \( \omega \). But here, it actually has 2 types of motion. Not only it spins around itself, but it's also moving sideways. So it's doing this. This is called rolling motion. And whenever you have rolling motion, you have 2 types of energy. So the total kinetic energy is going to be linear plus rotational. So the object actually has both types of energy. This is the only case where you have, out of the 6 that I've mentioned here, where you have linear and rotational. All the other cases you have either linear or rotational, but not both. And in the case of the Earth going around the sun and spinning around itself, it has 2 rotational kinetic energies, 1 around itself, 1 around the sun. Cool? So that's it. I hope this makes sense. This basically covers every possibility so you should be rocking from here on. Alright? So if you have any questions, let me know.

# Types of Motion & Energy - Online Tutor, Practice Problems & Exam Prep

### Types of Motion & Energy

#### Video transcript

### Kinetic Energy of a Point Mass

#### Video transcript

Hey guys, in this video, I'm going to show you how there are 2 ways to calculate the kinetic energy of a point mass going around a circle. Let's check it out. Alright. So remember, if you have a point mass around a circle, under a circular path, it's kind of like this, around a distance of little r from the axis of rotation. You have rotational speed, omega. And you also have a linear equivalent, which is your tangential velocity. Okay? All right. But you only have one type of motion. All you're doing is this. Okay? Your only motion is really rotational motion. Your only motion is rotational motion, so you only have one type of kinetic energy. Okay. But you can calculate using KL or KR. You can use the equation for linear or for KR. And that's because these two equations, as I'm going to show you now, are equivalent. Okay. The most important thing to do here is to make sure you don't double count it, okay? When I ask you for the total kinetic energy of an object, you can't, point mass like this. You can't look at it and say, well, it's got a V, so it has a linear kinetic energy and it has an omega, so it has a rotational kinetic energy. It's got 2 kinds of energies. Let's add the two of them together. You can't do that because these guys are equivalent, right? The tangential velocity is basically a mirror of omega. It doesn't mean there's 2. It just means that, one basically reflects the other, all right? So what you can't do is double count. So let me show you how this works. A small 2-kilogram object, so mass equals 2 kilograms, is going around, with a rate of it's going around the vertical axis. So what is a vertical axis? Remember axis, you can think of it as an imaginary line that you spin around. So a vertical axis would look like this. So it means the object is going around like this. Okay? Like this. Cool. So they would actually I could draw it like this and the object is doing this. Okay. And it does this at a rate of 3 radians per second, maintaining a constant distance of 4 meters to the axis. This distance to the axis is what we call little r. Little r is 4 meters. And I want to know the object's kinetic energy, and I want to do this using the KL equation, the KR equation. And the purpose of this question is to show you how the answer ends up being the same, and I'm going to summarize it at the end. So we can do KL, we can do KL, which is going to be half MV^{2}. Okay. Remember that these 2, V and r, are related, right? V and r are related by V equals romega. So what I'm going to do is also write KR equals half I omega. And I'm going to rewrite one of these equations one of these equations, and you're going to notice how it's going to look exactly like the other. So let's rewrite this one here.
1
2
I
ω
2
remember, I for a point mass is MR^{2}. So I'm going to replace this with MR^{2}. And I can rewrite omega as well. V equals romega, so omega equals V over r. So instead of omega here, I'm going to put V over r. Now look what happens. This r squared cancels with this r squared, and we're left with half MV^{2}, which is exactly this equation. Okay? So you can go from one to the other for a point mass, You can do this, which means I could have calculated them either way. Alright? So if I go here, KL equals half MV^{2}. Let's get these numbers, omega equals 3, V equals romega, so omega equals V over r, V, I'm sorry. I'm trying to get V. So V equals r4omega3V is 12, so this is half mass is 2. They cancel 12 squared. So this is 144 joules. Cool? And if I wanted to do it using KR, I already showed you how the equations turn out to be the same. Now I'm just going to plug in numbers differently. So if I wanted to do it this way, I could have done
1
2
M
R
2
ω
2
half, right, which is this. Half the mass is 2, and the distance is 4 squared, and the omega is 3 squared. So these 2 cancel. I have 16 times 9, which is 144 joules. Okay. So if you calculate it using linear, it's 144. If you calculate using rotation, it's 144. And if I ask you what is K total, the answer is 144. Okay. And I want you to please write here not 288. You do not add the 2. You can get the same answer using the two different equations. Now, to make this simpler for you, I have a convention. I always think of an object going around a circle like this. It has one motion. I always think of this as linear motion. I always I'm sorry, rotational motion, not linear. So I would always do it like this, KL plus KR, and I would say there's no KL, there's only KR, and this will guarantee that you don't double count it. Cool? So this is just a potentially tricky thing, but once you understand it, get it out of the way, it's never going to bother you again. Cool? Let me know if you have any questions.

The Earth has mass 5.97 × 10^{24} kg, radius 6.37 × 10^{6} m. The Earth-Sun distance is 1.5 × 10^{11} m. Calculate the Earth's kinetic energy as it spins around itself. BONUS:Find the Earth's kinetic energy as it goes around the Sun.

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