Hey, guys. So in this video, we're going to talk about conservation of energy in rolling motion problems. Now rolling motion, if you remember, is a special kind of rotation problem where we have an object that not only spins around itself, but it also moves sideways. So similar to if you have a toilet paper roll on a wall, it's rolling on a fixed axis, right? That's not rolling motion. Rolling motion is if you take the toilet paper and throw it on the floor, and it's going to roll on the floor. So it rolls on the floor. So let's check out how that stuff works. Alright. So remember, if an object moves while rotating, this is called rolling motion, and it does this on a surface without slipping. We can say, let me draw that real quick, usually show it like vcm (velocity of center of mass)
, and there is an ω
(angular velocity) at the same time. We say that the velocity in the middle here equals rω
, where r
is the radius of that wheel-shaped object. Okay. So this is an equation that we get to use. Alright. Now remember, that in order for an object to start rotating, for it to start rotating, in other words, to go from ω0
to an ω
of not 0
or to rotate even faster. In both of these cases, we have α
, we have an acceleration. There needs to be static friction. Okay. You have to have static friction in order to roll. Friction, static. Okay. Now the role of static friction in rolling motion is that it's essentially converting some of your velocity into ω. So think about it this way. If you have this object and there's no friction, it's going to move on a surface. It's going to move like this. Notice that I'm not rolling it. What friction does is take some of this v
here and starts to turn it into rotation. Right? If this was completely frictionless ice and you do a disc, it wouldn't roll. It would just go like this. But friction is what causes it to roll at the same time. So it's taking some v
and changing into ω. Now technically, what it's doing, it's getting some linear kinetic energy and converting it into rotational kinetic energy. Okay. Now that being said, friction static does that without dissipating any energy because you're converting from kinetic to kinetic. So it stays within mechanical energy. So we're going to say that even though there is static friction, the work done by static friction is 0. Okay. The work done by static friction is 0. Alright. So very briefly here, to summarize, if you have acceleration, if acceleration is not 0
, there has to be static friction, but the work done by static friction is 0. Okay. The phrase that I want you to remember here is that you need fs
in order to have α
. Okay. Need fs
in order to have α
in these kinds of problems. Now, the term "without slipping" means that there's going to be no kinetic friction. And a vast majority of rotation problems are going to have some acceleration, but it is going to roll without slipping. So what that means is that you have static friction, but you have no kinetic friction. And but even though you have static friction, it doesn't do any work. So when you write the work equation or when you write the conservation of energy equation, the work done by static friction is 0. Cool? Let's get started. Let's do an example here. So we have a solid cylinder. Solid cylinder is the shape. It tells me that I'm supposed to use I = 1/2 mr2
. The mass is m
. The radius is r
. So this is a literal solution. We're going to solve this with letters. We're going to derive an equation here, instead of actually getting numbers. It is released from rest. Okay. So the mass is m
, the radius is r
. It's released from rest. So vinitial = 0
from the top of an inclined plane of length l
, blah, blah, blah. Let's draw this here. So you've got a solid cylinder all the way at the top here. This plane has a length of l
and it makes an angle of θ with the horizontal. The cylinder rolls without slipping. "Rolls without slipping" means that there is no kinetic friction. No kinetic friction. There's no rubbing of the cylinder on the surface. It just rolls. And I want to derive an expression for the linear and angular speed at the bottom of the plane. So when it's here, I want to know what is vfinal
and what is ωfinal. And we're going to use conservation of energy. This is very similar to when we solved, when we use conservation of energy to find the final velocity of a block at the bottom. The big difference now, it's obviously not a block, it's rotating. So we have rotational energy as well. Okay. So it's similar to this. Alright. So kinetic initial plus potential initial plus work non-conservative equals kinetic final plus potential final. Alright. There's no kinetic energy in the beginning because it's not moving initially. It starts from rest. I do have potential energy because I have a height. So I'm going to write mghinitial + work non-conservative = 0
. And then there's the work done by friction. And I want to be very explicit here that even though there is static friction, even though there is static friction, the work done by static friction is 0. Okay. So there is basically no work. There's nothing here. There is kinetic energy at the end. Now, what type of kinetic energy do we have at the end? Well, what's special about rolling motion is that the object is rolling around itself and moving at the same time. Right? So it's sort of doing this. So I have the one object has two types of motion. So it has two types of kinetic energy. So I'm going to write 1/2 mv2final + 1/2 I ω2final
. There's no potential energy at the end because you are on the ground. Okay? So what we're going to do is we're going to expand I
, and we're going to rewrite ω. The reason we're going to rewrite ω is because we have v
and ω. And remember, whenever we have v
and ω, what we always want to do is, instead of having two variables, v
and ω, we want to rewrite ω. So we have v
and v
, which is the same variable. So we're going to change ω into v
. And we're able to do this because in rolling motion, we have this extra equation right here. Okay. So v = rω
. Therefore, ω is v/r
. So here, instead of ω, I'm going to write v/r
. Okay. And then I'm going to plug in I
here. I
is 1/2 m r2
. Okay. And I'm just going to go ahead I know I'm writing backward here. Sorry about that. I'm going to go ahead and write this whole thing out. Okay. So if you get here, this is the most important part. As long as you can get here, the rest is just a lot of calculating. We're going to cancel out a bunch of stuff. So notice that the r2 cancels with this r
. This happens almost all the time, right, that the r
's cancel in the conservation of energy equation. So look out for that. Notice that I have m
, m
, m
. Every one of these three terms has an m
. They all refer to the same object because I only have one object, so I can cancel the masses as well. And then I'm left with, I'm left with some fractions here. I'm going to multiply this whole thing by 4 because I don't like fractions. So I'm multiplying, so this becomes 4ghinitial
, 4 times half becomes 2vfinal
, and 4 times a quarter is just 1. So this becomes vfinal2
. Okay. And obviously, these two combine to be 3vfinal2
, and I am almost ready to plug it in. Vfinal
will be √(4ghinitial/3
). There's one more thing I have to do, which is notice that I'm not given h
. Right? The problem doesn't give us h
. The problem gives us l
and θ. But I hope you remember that h = l sin(θ)
. Okay. Can you see that? Yes, you can. So I'm going to rewrite this as 4gl sin(θ)/3
. Okay. Now, so this is the final answer. I just want to make one quick point here. I want to notice how I want to actually show you, without this guy here, here. Notice what this looks like. This looks very similar to the final velocity here for a block would look like. You might remember the final velocity after an object drops a height of h
, irrespective of whether it's straight down or at an angle, is vfinal = √(2gh)
. And I want to point out that this is very similar but instead of 2gh
, I have 4/3gh
. Right? And that's because the formula, the formula of the solution, in rotation problems is similar or the same, the formula is similar or the same to linear, to the equivalent linear motion problem. Okay. What I mean by that is that you should what I mean by that is that you should expect that this final answer here will look like this. The difference is, it has a different coefficient. K. In fact, the coefficient in rotation, the coefficient is lower than it would be in linear. When you get out of here, it's lower than it would be in linear. What does that mean? Well, 2, this is 4/3, which is 1.33. And this number has to be lower than a 2. The reason why I'm making this point is so that you can feel a little more comfortable. If you remember that this is how it works, which you probably should, when you're solving a question like this, you can look at it and say, hey, this looks kind of like what it would look like in linear motion. I'm probably right. The other way this is helpful is you can check to make sure whatever coefficient you got is less than what you would have gotten in linear motion. So if it's less than 2 in this particular case, then you're good to go. If you've got something that's like 2.5, let's say you've got 5/2gh, 5/2 is 2.5. That's more than 2. So now you know that you've done something wrong. The reason why it's lower, the coefficient's lower, is because you are moving slower, right? It's a lower coefficient because you're moving at a slower pace. And that's because if you're rolling while coming down the hill, you have two types of energies. Therefore, you move overall slower than you would have a lower v
going down. Okay? So that's it. That's how these problems work. I hope this makes sense. Let me know if you have any questions.
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Conservation of Energy in Rolling Motion: Study with Video Lessons, Practice Problems & Examples
In rolling motion, an object rotates while translating without slipping, requiring static friction to convert linear kinetic energy into rotational kinetic energy. The work done by static friction is zero, allowing conservation of energy to apply. For a solid cylinder rolling down an incline, the final linear speed can be derived using the equation:
Conservation of Energy in Rolling Motion
Video transcript
A solid sphere of mass M = 10 kg and radius R = 2 is rolling without slipping with speed V = 5 m/s on a flat surface when it reaches the bottom of an inclined plane that makes an angle of Θ = 37° with the horizontal. The plane has just enough friction to cause the sphere to roll without slipping while going up. What maximum height will the sphere attain? (Use g = 10 m/s2.)
Sphere on rough and smooth hills
Video transcript
Hey guys, let's check out this example of conservation of energy in rolling motion. Now what's special about this example is that you're going to have an object that's going to roll down a hill and go up the other, but on the first hill, there's going to be static friction. Therefore, there is angular acceleration alpha, which means you go from no speed to rolling faster and faster. On the second hill, however, there is going to be no static friction. So there's no angular acceleration alpha, which means that you don't slow down. So here, you spin faster and faster, but here, you're going to go up and your rotation is going to stay constant because there's nothing to slow it down. Now your velocity increases, your v grows, and then your v goes down, slows down. So the v acts as you would expect, but the omega gets faster, but then does not get slower as you go up the hill. Let's draw this real quick. I'm going to draw 2 hills like this, and we're going to call this initial point here "a". Then the initial velocity here will be 0,
You may remember that the lowest speed that an object may have at the top of a loop-the-loop of radius R, so that it completes the loop without falling, is √gR . Determine the lowest speed that a solid sphere must have at the bottom of a loop-the-loop, so that it reaches the top with enough speed to complete the loop. Assume the sphere rolls without slipping.
Do you want more practice?
More setsHere’s what students ask on this topic:
What is rolling motion and how does it differ from pure rotation?
Rolling motion is a type of motion where an object rotates around its own axis while also translating along a surface without slipping. This is different from pure rotation, where the object only spins around a fixed axis without any translational movement. In rolling motion, the velocity at the point of contact with the surface is zero, and the velocity of the center of mass (vcm) is related to the angular velocity (ω) by the equation vcm = rω, where r is the radius of the object.
How does static friction play a role in rolling motion?
Static friction is crucial in rolling motion as it prevents slipping and allows the object to roll. It converts some of the linear kinetic energy into rotational kinetic energy without dissipating energy. This means the work done by static friction is zero. Static friction ensures that the point of contact between the rolling object and the surface is momentarily at rest, enabling the object to roll smoothly.
How is the conservation of energy applied in rolling motion problems?
In rolling motion problems, conservation of energy is applied by considering both translational and rotational kinetic energies. The total mechanical energy is conserved, meaning the sum of initial kinetic and potential energies equals the sum of final kinetic and potential energies. For a rolling object, the kinetic energy includes both translational (1/2 mv2) and rotational (1/2 Iω2) components. Static friction does not do any work, so it does not affect the energy conservation equation.
What is the final linear speed of a solid cylinder rolling down an incline?
The final linear speed (vf) of a solid cylinder rolling down an incline can be derived using conservation of energy. The equation is given by:
where g is the acceleration due to gravity, l is the length of the incline, and θ is the angle of the incline with the horizontal. This equation shows the relationship between the linear and rotational dynamics of the rolling cylinder.
Why is the coefficient in rotational motion lower than in linear motion?
The coefficient in rotational motion is lower than in linear motion because a rolling object has both translational and rotational kinetic energies. This means that part of the energy is used for rotation, resulting in a lower translational speed compared to an object that only translates. For example, the final velocity of a rolling object is derived as
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