Conservation of Energy in Rolling Motion - Online Tutor, Practice Problems & Exam Prep

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: $\sqrt{\frac{4}{3}ghsin\left(\theta \right)}$. This highlights the relationship between linear and rotational dynamics, emphasizing the lower coefficients in rotational motion compared to linear motion.

Conservation of Energy in Rolling Motion

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 ω_final. 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.

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.)

• A. 0.175 m
• B. 0.350 m
• C. 1.75 m
• D. 3.50 m

Sphere on rough and smooth hills

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, \( v_a = 0 \). And it has some sort of initial height which we'll call \( h_1 \). I want to know how far up it gets on this side. We're going to call this \( h_2 \). I want to know what maximum height does it attain on the second hill in terms of \( h_1 \). So it's going to be a multiple of \( h_1 \), and that's what we want to know. It says it's a solid sphere. So the moment of inertia is, by table lookup, \( \frac{2}{5}mr^2 \), has a mass of \( m \) and a radius \( r \). This is a literal solution. We're not going to have actual numbers here. It's initially at rest. It's at the top of a rough hill of height \( h_1 \), which explains that the first hill has enough friction to cause the sphere to roll without slipping. Roll without slipping means no kinetic friction, but there is going to be some static friction which causes the object to accelerate, to have angular acceleration to roll faster and faster. For this piece, it is smooth, which means there is no angular acceleration, which means whatever velocity, whatever omega you have here, let's call this point \( b \), whatever omega you have at point \( b \) will be the same omega that you have at point \( c \). Because there's no change in angular velocity in that piece. For the third interval here, we have a long smooth hill. Smooth hill means that there is no friction. Therefore, there is no angular acceleration. Again, just like the horizontal surface. So there is alpha here but there is no alpha here and here. What that means is that imagine a block. Right? The block speeds up, moves at a constant speed and then slows down on the way up. The sphere will do that except it gets faster, same speed, and then slows down and stops. But the rotation is going to be different. It's going to accelerate on the way down because it has alpha, then it spins at a constant rate. And then as it goes up the second incline, because there is no static friction, it's a smooth hill, then there is going to be no alpha. What it means is that it's going to go up, but it's going to keep rolling at the same rate, get to the top, stop, \( v = 0 \), but it's still rolling and then it starts going down. So it's kind of weird. It goes up, but it doesn't really stop rolling. What you would usually expect is that it goes up and slows down and stops. And then it comes back down spinning faster. In this case, it's going to keep rolling, go up, keep coming down. And that's because there's no friction. So that's special. That's what's special about this question. What we're going to do is we're going to write an energy equation from \( a \) to \( b \), and then we're going to write an energy equation from \( c \) to \( d \). There's nothing, there's nothing special happening from \( b \) to \( c \). In fact, \( b \) and \( c \) are really the same thing. Nothing happens there. So let's write the first one which is the energy equation from \( a \) to \( b \). \( K_a \) \( U_a \) work non-conservative between \( a \) and \( b \) is \( K_b \) \( U_b \). There is no kinetic energy at the top of the first hill because it's initially at rest. There's potential energy because it has height. There's no work non-conservative. Remember that even though, work non-conservative is you and friction, even though there is friction, static friction does a work of 0 in these problems. There's no work at all. There is kinetic at the end because we have some velocity. But \( b \) is on the floor. It's the lowest point, so potential energy is 0. So what we have here is simply potential energy going into kinetic energy. Okay? Potential energy is \( mg h_a \), or in this case, I guess, we're calling this \( h_1 \). So let me do that, \( g h_1 = \) kinetic energy. Now what types of kinetic energy does this ball have? Well, it's not only falling, but it rolls and speeds up. So it has both types of energy. So I'm going to write \( \frac{1}{2}mv_b^2+\frac{1}{2}I \omega_b^2 \). One object has 2 motions, so we have 2 kinetic energies. Now what we're going to do, as usual, is expand \( I \) and we're going to rewrite omega. Why do we rewrite omega? Because I have \( v \) and omega. Whenever you have the 2 velocities, you want to get rid of 1 so that you only have 1. So omega will become \( v \). So instead of having \( v \) and omega, I have \( v \) and \( v \). That's better. This is rolling motion, which means I can write that the velocity of the center of mass, remember this velocity of center of mass, is linked with the rotation by this equation. So \( v_{cm} \) is \( r \omega \), therefore, \( \omega = v/r \). So I can use this to replace omega. I'm going to replace omega with \( v/r \). Let's rewrite this, plug in the I, and then keep going. So \( mgh_1 = \frac{1}{2}mv_b^2+\frac{1}{2} \), and then I have \( I \). \( I \) is the moment of inertia of the sphere, a solid sphere, which is \( \frac{2}{5}mr^2 \). Now look what happens. A bunch of stuff is going to cancel. This 2 cancels with this 2. The \( r \) cancels with the \( r \). There are \( m's \) here. They're one \( m \) in every term, and they all refer to the same mass, so we can cancel those as well. I have a 2 down here, and I have a 5 down here. So to get rid of those guys, I have to multiply this by 2 and by 5. In other words, multiply the whole thing by 10. Let me make a little space here so it's not messy. When I multiply the whole thing by 10, I get \( 10 g h_1 \). 10 times a half is 5 \( v_b^2 \). 10 times \( \frac{1}{5} \) is 2 \( v_b^2 \), and that's it. So this adds up to 7 \( v_b^2 \). And then I have \( v_b = \sqrt{10 g h_1 / 7} \), take the square root of both sides. \( \sqrt{10 g h_1 / 7} \), that's what we get for that. Now, so we got that. Cool. So that's \( v_b \). I found the velocity here. So now what I'm going to do is I'm going to write an equation from, I'm going to move over here, and I'm going to write a conservation of energy equation from \( B \) to \( C \) to show you how this looks like. You can think of this problem as really 2 questions that are merged together. And I want to show you separately what these parts look like. And I'm sorry, it's actually from \( C \) to \( D \). Okay. So \( C \) to \( D \) is going up. So \( K_c \) \( U_c \) work non-conservative, \( K_d \) \( U_d \). Alright. So is there kinetic energy at point \( C \)? The answer is yes. \( C \) is at the bottom of the hill. What kind of energy? Well, not only do you have velocity, but you also were rolling and you got to the bottom rolling and then you kept rolling. So you have both. You have \( \frac{1}{2}mv^2 + \frac{1}{2}I \omega^2 \). And I want to make the point that this velocity here at point \( C \) is really the same as the velocity at point \( B \). So \( \omega_c \) is the same as \( \omega_b \) and \( v_c \) is the same as \( v_b \) because you're on a smooth surface, nothing happened there. It's as if you basically just went immediately up another hill. So I'm going to put a \( v_b \) here and I'm going to put an \( \omega_b \) in here. What about potential energy? There's no potential energy at the bottom of the hill. There's no work non-conservative because there's no work done by you and there's actually no friction at all. What about kinetic energy at point \( D \)? This is the trickiest part of this whole question is to realize that even though you are at the highest points, even though you are at the highest points, you do have rotational energy. Your linear stops because the object does this and stops. But it keeps rolling as it goes up. So you still have this. This is the most important part of this whole question. This entire question basically exists for this one purpose. And there is potential when there's that point. So I'm going to write \( mg \), the height at point \( D \), we're calling that \( h_2 \). And by the way, \( h_2 \) is our target variable. So we're going to simplify a bunch of stuff and we're going to be left with \( h_2 \). What we got to do here, we got to expand this guy and we've got to expand this guy, and we've got to rewrite omega. We have \( v \) and (omitting) \( omega \). So we're going to rewrite omthega into \( v \) by writing \( omega = v/r \), which we can do because it's a rolling motion problem. \( \frac{1}{2}mv_b^2 + \frac{1}{2} \), this is the \( omega \), so it's \( \frac{2}{5}mr^2 \) \( I \), the moment of inertia, so it's \( \frac{2}{5}mr^2 \) \( omega \), we talked about how we can rewrite this into \( vb/r \). So \( \omega_b = vb/r \) squared. That's the left side. On the right side, I can expand the rotational energy here, which is going to be \( \frac{1}{2} \) \( \frac{1}{2}I \) \( I \) is \( \frac{2}{5}mr^2 \) \( omega^2 \). I'm just doing this in one shot. Rotation is half \( I \omega^2 \). I'm replacing the \( I \). It's the \( I \) right there. And I'm replacing (omitting) \( w \), with \( vb/r \) squared plus \( mgh2 \). Now I didn't really have to do this, this one last step here. And I'll tell you what I mean. But I wanted to do this to make it painfully clear. Notice that this thing here is exactly the same as this thing here. It's exactly the same. And it's because this rotational kinetic energy at point \( B \) is the same as the rotational kinetic energy at point \( D \). Let me go up to the drawing real quick. So you're rolling at \( B \). You're rolling, and this rolling never slows down. So whatever energy you have here, you have it here at \( C \), and then you go up and you have it here at \( D \). So what you could have done is you could have canceled these guys, right here. I'm going to not cut it out there because it's going to be messy. I'm going to cut it over here. But again, you could have canceled those 2 because those energies are the same. They don't change. So really, all you have is this and this. I can cancel the masses, and I'm solving for \( h_2 \). \( H_2 = v_b^2 / 2g \). Notice that the \( g \) went down there. Now I do have an expression for \( v_b \), so I can plug it in. \( V_b \) is right here. So it's going to be \( 1/2g \) and then \( v_b^2 \). Notice that \( v_b \) is inside of a square root. So \( 10g h_1 / 7 \), which means that the square root cancels with the 2. And then I'm left with, \( 1/2g \) times \( 10g h_1 / 7 \). And this will cancel nicely. This cancels with this. And then I have 10 divided by 14, \( 10/14 h_1 \). And 10 divided by 14 is 0.71 \( h_1 \). So we're done here. The final answer is that \( h_2 \) is \( 0.71h_1 \). Now one question, that was asked here as part of the bonus question is, why are these different? Right? So conservation of energy usually would dictate that you start at a height and you come back to the same height. Well, the reason they're different is because of the energies, right? All of the potential energy went into kinetic at the bottom, right? So there's a full energy conversion, but on the way up, only some of the kinetic energy went back into potential because some of it is still in spinning. Here you are very high with no spinning. Here you are a little bit lower because you're spinning. So energy spinning costs energy, right? So basically, what happened is you went from, an initial potential but no initial no initial rotational to a potential at the end and rotational at the end. So what happens is some of this went into here and some of this went into here. Because your initial height got split into height and rotation, you don't get as much height. Right? So if we did this in terms of energies, let's say \( 100 \) joules got split where, like, \( 70 \) joules went here and then \( 30 \) joules went here. Right? That's why you see a lower height because you spend some of that height energy into rotation energy, so you didn't recover as much of your height. So that's it for this one. Hopefully, it made sense. Let me know if you have any questions.

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.

• A. $\sqrt{5gR}$
• B. $\sqrt{\frac{13gR}{5}}$
• C. $\sqrt{\frac{27gR}{7}}$
• D. $\sqrt{10gR}$