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.
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13. Rotational Inertia & Energy
Conservation of Energy in Rolling Motion
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