Conservation of Energy with Rotation - Video Tutorials & Practice Problems

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1

concept

Conservation of Energy with Rotation

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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, ends up over here somewhere where either the speed, the the height H or the spring compression ex changes any combination of those three guys changes we can use most of the time a 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 you could actually be. It could 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 K initial plus your initial plus work. Non conservative equals K final Plus you find a I want to remind you that work non conservative 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 half MV squared. 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 have from the square, the object is just spinning. So you're gonna write I Omega Square? Okay, the key thing to remember and you'll 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 the end omega 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 one V and one W or to V s and one w whatever. If you have a Vienna W, that's two variables. You're going to change one into the other so that you end up with just one variable. For example, most of the time V nwr 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 I'm What I'm gonna do is rewrite W as V over r and whatever I see a w I'm gonna replace it with the view over are So instead of having v n w. I have V and V, and that means that instead of having two variables, I have just one, and it's easier to solve the problems. That's the key thing to remember is rewrite one into the other. Let's do an example. All right, so here we have a solid disc solid disc. Let's stop There means that the moment of inertia we're gonna uses the same of a solid desk, which is the same as a solid cylinder. And it's gonna be half M r squared. And it says it's free to rotate about our fixed perpendicular access to its center. Lots of words. Let's analyze with staying here free to rotate, it just means that you could rotate right like you can actually spin. Some things can be spun around others, Kintz Um, 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 axes versus a free roll of toilet paper that can roll around the floor. Okay, here were fixed in place. So we're gonna say that it spins like this, so it has no view, right? Like the actual disc has no velocity V because it's not moving sideways. The center of mass doesn't change position. Um, 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 gonna write that Omega is not zero because it's going to spin now. What else? It says that the access 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 objects. So I got a little disc here. Um, this sort of looks like this, and I want to show you real quick what perpendicular means. So imagine the this is the face of the disc perpendicular means 90 degrees to the face of the disc. Which means I'm gonna stick my fingers in here, and it looks like this Cool. So perpendicular. Looks like this. This is the axis of rotation, which means the disk spends like this. Okay, Hope that makes sense. So you're gonna see this all the time. Perpendicular is just going to mean 90 degrees with the face. Just means that the disk spins like this. Which is how you would imagine the disk spins. A desk isn't gonna do this right Or some weird stuff. So just spins around its center like that. Cool. So this expends like that, This is sort of a top view off the disc. Um Okay, so the disk has mass five and he goes five radius six, and it is initially at rest. So omega initial is zero because it starts arrest, and then you have a light along the light cable that's wrapped several times around the cylinder. The disk. Okay, light means that the cable has no mass, so I'm gonna right here. Mass of the cable is zero. And you wrap it up a bunch of times. You got a lot of problems like this, and basically what we're doing is we're saying we're setting it up to say there's all this rope around this thing So when I pull on it, it's going to unwind. So it says, here you pull on the cable with a constant 10 Newtons. So let's draw cable right here. And then it says Force of 10 in such a way that the cable unwinds horizontally at the top of the disc, the cable unwind horizontal at the top of the disk is exactly what I just drew here, right? So the cable is unwinding horizontally. It doesn't say if it's the left to the right. I just threw it to the right. Um, now the word unwind here is repeated. Sorry about that. And then without slipping, this is key because it's unwinding without slipping. I can say that the velocity of the rope equals little Are omega okay? And this thing, this rope will have a velocity V this thing will have on omega of the disc. So V rope is our omega of the disc, where R is the point where the rope touches the disk so it's where its distance between center. The axis of it, I should say axis of rotation distance between the axis of rotation and the point where the rope touches the disk or where the rope pulls on the disk. Okay, let me show you what quick example here just to be very clear here. Let's say this disc has a radius of 10. So the distance all the way to the end here is 10. But the But let's say I'm pulling right here at a distance. Five. So what I using this equation the rope equals are Omega. I would use the five. Okay, just to be clear, when you write these equations here, we're gonna do this a bunch of times. The are is not the radius. That's why it's a little are not a big are. It's how far from the center the rope pulls. In this particular case particular case, the rope is pulling at the edge, so your little are happens to be the radius, and by the way, that's how it almost always happens. But you could have it. You could have a different situation like this, so you should be ready just in case. Um are is typically the radius, but it doesn't have to be. The radius in this case rate is is 6 m. Okay, so without slipping is what tells us that we can use this equation right here. Okay. Without slipping is what tells us that we can use this equation right there. Alright, ignore any frictional forces. If you don't see that, you can just assume that you're supposed to ignore friction unless it tells you what the friction is. And then we're gonna use conservation of energy to find the angular speed of the pulley. So I want to know what is Omega final? What is Omega final after you've pulled the rope for 8 m? So you're gonna pull the rope with a force of 10 with for a distance delta X of 8 m pictures a little tight here, but basically would look like this until 8 m of rope unwinds from the disc. Alright, So let's use conservation of energy and we're looking for W final. So kinetic initial potential initial plus work, non conservative equals kinetic final plus potential final. Okay, the is their initial kinetic energy here. There is no kinetic energy the beginning because the disc isn't spinning, there's no linear energy and there is no rotational energy. The disk doesn't move sideways in the beginning, doesn't spend. Um, there's no potential energy. And that's because the remember potential energy is relative to a change. It depends on your change in height and the height doesn't change. Delta H zero. The disc keeps its same height, so you can just cancel out these two guys, right? You do have a potential energy because you are both the floor, but the two potential energies are the same. Okay, now, what about work? Non conservative work, non conservative? Is the work done by U plus the work done by friction. There is no friction here. Told us to ignore frictional effects. But you are pulling on this thing right? And the work done by you is the work done by a force F which is F d co sign of theta. Okay, your force is 10. You do this for a distance. D of 8 m and cosign of fate. A remember is the angle between your displacement and the force. Now here you push this way and the rope moves this way So the angle here is zero. So I'm going to do the co sign of zero in the CO sign of zeros one. So I have 10 times. Eight times one 80. Okay, so this is 80 Jules. And at the end, we have kinetic final, um, kinetic final could be linear. And it could be rotational is their linear energy at the end. There isn't because the disc spends around itself, but there is rotational kinetic energy at the end. So let's expand that 80 equals Half I Omega squared will make a final. This is exactly what we're looking for right here. Omega final. Okay, so let's expand the I so it's gonna be half gonna put the I in here. I is half m r squared because we No, it's a solid disk. So I'm gonna put half the mass is of five, and the radius is a six. Let me just put a six there. Cool. What we can do is we can move everything to the other side and solve for omega. So here we have 80 equals, um, things whole thing here. It gives us a 45 Omega final squared, so we'll make a final is 80 divided by 45 and then you take the square double sides. You get this. And if you solve this, get out of the way. You get 1.33 radiance per second. Okay? And that's the final answer. Just took a little while. But it's because I wanted to introduce some of the terminology for these kinds of questions. Some of the language you're gonna see now. In the beginning, I mentioned how if you have a V, um, if let me write this here, if you have a V and A W, you're going to rewrite W in terms of V, right? Well, in this question, when I expanded everything, I only had a w. So I didn't have to change one into the other. And I was looking for W. So I just solved for it. Okay, so you do that if you have the two variables so that you can simplify, that's it for this one. Let's keep going. Let me know if you have any questions.

2

example

Work to accelerate cylinder

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3m

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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 solid cylinder means that I is going to be half m r squared because that's the equation for the moment of inertia. Of a cylinder masses 10 in radius is two. I'm gonna put these here if you want to. You could already calculate the moment of inertia. Right. So I is half 10. Our square in the moment of inertia is 20. Okay, so we can already get that 20 kg meters square. It says it is mounted and free to rotate, um, on a perpendicular access 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 just doesn't wobble like that, right? It rotates like this. Now, most of the time, you actually have this where the access is horizontal, so it's on a wall, right? So it's something that's on a wall, and you have the disks spinning like this. All right, so it says that cylinders initially at rest. So the cylinder spins around itself, but it's initially at rest. Omega initially equals zero. And we want to know what is the work done to accelerate it from rest to 120 our PM. Okay, remember, most of the time when you have rpm, you're supposed to change that into W so that you could be using an equation. So let's do that real quick. Just to get that out of the way or make a final is two pi f or two pi our PM over 60. If you plug 1 20 here, you end up with 1 20 but by sixties to you end up with four pi radiance per second. Okay, so I'm going from 0 to 4 pi and I want to know how much work does that take. So work is energy. So hopefully you thought of using the conservation of energy equation K initial plus you initial plus work non conservative equals K final plus you final. In the beginning, there is no kinetic energy because it's not spinning, it's not moving sideways. Um, the potential energy is canceled because the height of the cylinder doesn't change, it stays in place. Right? Work, Non conservative is the work done by U 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 the kinetic energy which is Onley, kinetic, rotational, right? There's no linear. It's not moving sideways. The center of massive disc stays in place. So the equal zero. So the only type of kinetic energy we have is rotational, which is half Oh, May I Omega squared. We're looking for this. So all we gotta do is plug in this number. Work is going to be half. I we already calculated I over here it was 20. And Omega is four pi square. Okay, so if you multiply all of this, you get that it is 15 80 you get 80 jewels of energy, and that's how much energy is needed to get this solid cylinder from rest all the way to a speed of four pi or 120 rpm. Cool. Very straightforward. Plug it into the energy equation because we were asked for work. All right, hold makes sense. Let me know if you guys have any questions and let's keep going

3

Problem

Problem

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?

A

+9600 J

B

−9600 J

C

+5760 J

D

−5760 J

4

example

Which shape reaches bottom first?

Video duration:

4m

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Hey, guys, let's check out this conservation of energy example. Um, here we have three objects of equal mass and equal ratings, but they have different shapes. Remember, your shape is what determines what equation your moment of inertia has. And it's usually something like a fraction m r squared. But the number in here depends on the shape. So if you have different shapes, you can have different I equations. They're all released from rest at the same time from the top oven inclined plane. So I'm gonna have Here's a solid cylinder. Um, 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 starts from the top of the inclined plane. They're gonna start from the same height as well. Everything is the same except the shapes. And I wanna know who reaches the bottom first if they're released at the same time. And this question will depend on your moments of inertia. What I want to remind you, that moment of inertia is a measurement of angular resistance of rotational resistance. So you can think you can think that the greater my eye, the heavier I am, the more I resist rotation. Therefore, I will get to the bottom last because I'm slower. Okay, So, um or 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 mawr resistance. Therefore, I am slower. Okay. Now, a solid cylinder has a moment of inertia of half m r square, a hollow cylinder at the moment of inertia of m r squared. So you can think that there's a one in the front in a solid sphere, has a moment of inertia of 2/5 and Mars Square. So in this question, all we're doing is comparing these numbers because the m and they are the same. Now, this is a little bit easier if you use decimals. So this is 0.5. This is 1.0 and to over 50 point four. Okay. And you can see from here that this one is the lightest one. Okay, because the coefficient number in front of the M 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 V that the solid sphere his first. Um, I'm gonna 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 hostile cylinder is slower than the solid cylinder. Solid cylinder has very good mass distribution. The masses 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. The whole 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 mawr 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 lowest of them all. So the sphere is always fastest. Okay, so that's it for this one. Let's keep going.

5

example

Cylinders racing down:rolling vs. sliding

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6m

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Hey, guys, let's check out this conceptual conservation of energy question. So here we have two cylinders of equal mass and radius. So basically the same cylinder, um, the release from rest from the top of two hills having the same height. So it's the two very similar or so far identical situations. So let's say, have cylinder A on hill A and cylinder. Be 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 gonna have a V, but it's also rolling. So it has a W. Remember, this is called rolling motion if you roll without slipping, and it means that the C M equals are omega. But be slides without rolling. Those are the two 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 gonna see that. Alright. What this means is that it's moving this way with a V, but it has no Omega. Okay, basically, it's falling. It's going down as if as as a boxwood. 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 toe 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 of the fastest speed, who has his VA f or V B f 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 to guess 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 role are faster right then 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 air fast because we have wheels and cars. But in reality, the one that gets to the bottom of 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 plus you initial plus work non conservative equals K Final Plus you. Finally, in both of these situations, the kinetic energy, the beginning is zero. I have some potential energy. Okay, The work done by non conservative forces is zero 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 zero, right? So it actually, even though there is static friction, it doesn't do any work. There is some Connecticut, 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 heights, 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 100 jewels in both problems. You start with the 100 jewels and then you rolled to the bottom. The difference is that in the first situation for a D 100 jewels is going to get split between kinetic linear and kinetic final. Okay, between kinetic linear kinetic final because it has two types of energies. Let's make up some numbers here. Let's say that 80 goes here in 20. I'm sorry, man. Kinetic, linear and kinetic rotational. Let's say 80 goes to linger. 20 goes to rotational, totally making this up for be the entire amounts. The entire 100 jewels 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 this these two numbers get balanced outs, um, the 100 will always be bigger. Even if this was 10 and 90 100 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 two 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 you're using up your your potential energy and transforming 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 manager 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 to have s so there's less energy that's left for your linear for your V. Okay, so you're w 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 get going

6

Problem

Problem

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?)

A

Cylinder A

B

Cylinder B

C

Both reach the same height

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