Ch 11: Rotational Inertia & EnergyWorksheetSee all chapters
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Ch 01: Units & Vectors
Ch 02: 1D Motion (Kinematics)
Ch 03: 2D Motion (Projectile Motion)
Ch 04: Intro to Forces (Dynamics)
Ch 05: Friction, Inclines, Systems
Ch 06: Centripetal Forces & Gravitation
Ch 07: Work & Energy
Ch 08: Conservation of Energy
Ch 09: Momentum & Impulse
Ch 10: Rotational Kinematics
Ch 11: Rotational Inertia & Energy
Ch 12: Torque & Rotational Dynamics
Ch 13: Rotational Equilibrium
Ch 14: Angular Momentum
Ch 15: Periodic Motion (NEW)
Ch 15: Periodic Motion (Oscillations)
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Ch 20: The First Law of Thermodynamics
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Ch 25: Resistors & DC Circuits
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Ch 27: Sources of Magnetic Field
Ch 28: Induction and Inductance
Ch 29: Alternating Current
Ch 30: Electromagnetic Waves
Ch 31: Geometric Optics
Ch 32: Wave Optics
Ch 34: Special Relativity
Ch 35: Particle-Wave Duality
Ch 36: Atomic Structure
Ch 37: Nuclear Physics
Ch 38: Quantum Mechanics

Concept #1: Conservation of Energy with Rotation

Transcript

Hey guys! In this video, we're going to start solving rotation problems with conservation of energy. Let's check it out. You may remember that when you have a motion problem between two points, meaning the object starts here and ends up over here somewhere, where either the speed V, the height H or the spring compression x changes, any combination of those three guys changes, we can use most of the time the conservation of energy equation to solve these problems. 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. That's the new thing that you could be spinning and it could also be both. It could be that our total kinetic energy is linear plus rotational. We're going to use the conservation of energy equation which is Kinitial + Uinitial + Worknon-conservative = Kfinal +Ufinal. 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. 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. K used to be simply _ mv^2. But now it could be that you have both of them. It could be letÕs instead of _ mv^2, the object is just spinning so you're going to write Iw^2. The key thing to remember, and you would 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 V and w 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 two Vs and one w, whatever. If you have a V and a 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 and w are linked by this, or sometimes they are linked by this. Sometimes they're not linked at all, but most of the time they're connected by either one of these two equations, which means what I'm going to do is rewrite w as V/R. Whenever I see a w, I'm going to replace it with the V/R. Instead of having V and w, I have V and V. 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. Here we have a solid disc solid disc, let's stop there. It means that the moment of inertia we're going to use is the same of a solid disc which is the same as a solid cylinder and it's going to be _ MR^2. It says it's free to rotate about a fixed perpendicular access through its center. Lots of works. Let's analyze what itÕs saying here. Free to rotate just means that you could rotate, like you can actually spin. Some things can be spun around, others can't. But even though it says that it's free to rotate, itÕs around a fixed axis. 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 axis versus a free roll of toilet paper that can roll around the floor. Here, weÕre fixed in place so we're going to say that it spins like this so it has no V. The actual disc has no velocity V because it's not moving sideways. The center of mass doesn't change position but it does spin. Actually we don't know which way it spins. Let's leave it alone for now but IÕm just going to write that w is not 0 because it's going to spin. What else? It says that the axis is through its center, so it's spinning around the center like this and it's saying that it's perpendicular. Perpendicular means that it makes a 90-degree angle with the object. I got a little disc here. This sort of looks like a disc. I want to show you real quick what perpendicular means. Imagine this is the face of the disc. Perpendicular means 90 degrees to the face of the disc, which means I'm going to stick my fingers in here and it looks like this. Perpendicular looks like this. This is the axis of rotation which means the disc spins like this. Hopefully that makes sense. You're going to see this all the time. Perpendicular is just going to mean 90 degrees with the face. It just means that the disc spins like this which is how you would imagine a disc spins. A disc isn't going to do this or some weird stuff so it just spins around its center like that. The disc spins like that. This is sort of a top view of the disc. The disc has mass 5, m = 6, radius 6, and it is initially at rest. W initial is 0 because it starts at rest. Then you have a light, a long light cable that's wrapped several times around the cylinder, the disc. Light means that the cable has no mass. IÕm going to write here mass of the cable is 0 and you wrap it up a bunch of times. You've 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. It says here you pull on the cable with a constant 10 N. Let's draw a cable right here. Then it says force of 10 in such a way that the cable unwinds horizontally at the top of the disc. The cable unwinds horizontally at the top of the disc is exactly what I just drew here. The cable is unwinding horizontally. It doesn't say if it's the left or to the right. I just drew it to the right. The word unwind here is repeated, and then without slipping. This is key. Because it's unwinding without slipping, I can say that the velocity of the rope equals (little r) rw. This rope will have a velocity v. This thing will have an w of the disc. Vrope is rw of the disc where r is the point where the rope touches the disc. ItÕs the distance between the axis of rotation and the point where the rope touches the disc or where the rope pulls on the disc. Let me show you a 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 let's say I'm pulling right here at a distance 5. What I use in this equation Vrope = rw, I would use the 5. Just to be clear when you write these equations here, we're going to this a bunch of times. The r is not the radius that's why it's a little r, not a big r. It's how far from the center the rope pulls. In this particular case, the rope is pulling at the edge so your little r happens to be the radius. By the way, that's how it almost always happens but you could have a different situation like this so you should be ready just in case. r is typically the radius but it doesn't have to be the radius. In this case, radius is 6 meters. Without slipping is what tells us that we can use this equation right here. 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. Then we're going to use conservation of energy to find the angular speed of the pulley. I want to know what is wfinal after you've pulled a rope for 8 meters. You're going to pull the rope with a force of 10 for a distance delta x of 8 meters. The picture is a little tight here but basically it would look like this until 8 meters of rope unwinds from the disc. Let's use conservation of energy and we're looking for wfinal. Kinetic initial + potential initial + plus work non conservative = kinetic final + potential final. Is there initial kinetic energy here? There is no kinetic energy at the beginning because the disc isn't spinning. There's no linear energy and there's no rotational energy. The disc doesn't move sideways and in the beginning, it doesn't spin. There's no potential energy and that's because, remember potential energy depends on your change in height and the height doesn't change, delta h is zero. The disc keeps its same height so you can just cancel out these two guys. You do have a potential energy because you are above the floor, but the two potential energies are the same. What about work non conservative? Work non conservative is the work done by you plus the work done by friction. There's no friction here. It 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 Fdcos_. Your force is 10. You do this for a distance d of 8 meters and cos_ remember is the angle between your displacements and the force. Here you push this way and the rope moves this way so the angle here is 0. I'm going to do the cos0 and the cosine of 0 is 1 so I have 10 times 8 times 1, 80. This is 80 Joules. At the end we have kinetic final. Kinetic final could be linear and it could be rotational. Is there linear energy at the end? There isn't because the disc spins around itself but there is rotational kinetic energy at the end. Let's expand that. 80 equals _ Iwfinal^2. This is exactly what we're looking for right here, w final. Let's expand the I. It's going to be 1/2, put the I in here. I is _ mr^2 because we know it's a solid disc. IÕm going to put half. The mass is a 5 and the radius is a 6. Let me just put a 6 there. What we can do is we can move everything to the other side and solve for w. Here we have 80 equals this whole thing here. It gives us a 45 wfinal^2. W final is 80/45 and then you take the square root of both sides, you get this. If you solve this, get out of the way, you get 1.33 radians per second. That's the final answer. This 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 going to see. In the beginning I mentioned how if you have a V and a w, you're going to rewrite w in terms of V. 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. 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.

Example #1: Work to accelerate cylinder

Transcript

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. Solid cylinder means that I is going to be 1/2 MR^2 because that's the equation for the moment of inertia of a cylinder. Mass is 10 and radius is 2, I'm going to put these here. If you want to, you could already calculate the moment of inertia. I is 1/2 10R^2 and the moment of inertia is 20. We can already get that, 20 kilograms meters squared. It says it is mounted and free to rotate on a perpendicular axis through its center. Again, you have a cylinder which is the same thing as a disc and it has an axis. It's mounted on an axis that is perpendicular to it, so it looks like this and it's free to rotate about that axis. It just doesn't wobble like that, it rotates like this. Most of the time, you actually have this where the axis is horizontal, so it's on a wall. ItÕs something that's on a wall and you have the disc spinning like this. It says that cylinder is initially at rest so the cylinder spins around itself but it's initially at rest. _ initial = 0 and we want to know what is the work done to accelerate it from rest to 120 rpm. Remember, most of the time when you have rpm you're supposed to change that into w so that you can use it in the equation. Let's do that real quick just to get that out of the way. _ final is 2¹f or 2¹rpm/60. If you plug 120 here, you end up with 120/60 is 2. You end up with 4¹ radians per second. I'm going from 0 to 4¹ and I want know how much work does that take. Work is energy so hopefully you thought of using the conservation of energy equation. K initial + U initial + work non-conservative = K final plus U final. In the beginning, there's no kinetic energy because it's not spinning, it's not moving sideways. The potential energies cancel because the height of the cylinder doesn't change. It stays in place. Work non-conservative is the work done by you 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. The kinetic energy which is only kinetic rotational, there's no linear it's not moving sideways, the center of mass of the disc stays in place so V = 0. The only type of kinetic energy we have is rotational, which is _ I_^2. We're looking for this, so all we got to do is plug in this number. Work is going to be _, I, we already calculated I over here. It was 20 and _ is 4¹^2. If you multiply all of this, you get that it is 1580. You get 1580 joules of energy and that's how much energy is needed to get the solid cylinder from rest all the way to a speed of 4¹ or 120 rpm. Very straightforward. Plug it into the energy equation because we're asked for work. Hope it makes sense. Let me know if you guys have any questions and let's keep going.

Practice: 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?

Example #2: Which shape reaches bottom first?

Transcript

Hey guys! Let's check out this conservation of energy example. Here we have three objects of equal mass and equal radius but they have different shapes. Remember your shape is what determines what equation your moment of inertia has. It's usually something like a fraction MR^2, but the number in here depends in the shapes. If you have different shapes, youÕre gonna have different I equations. They're all released from rest at the same time from the top of an inclined plane. I'm going to have here a solid cylinder. Here is a hollow cylinder and here is a solid sphere. They're all from rest. They all have the same mass. They have the same mass, the same radius, they all start from rest and they all start from the top of the inclined plane. They're going to start from the same height as well. Everything is the same except the shapes and I want to know who reaches the bottom first if theyÕre released at the same time. This question will depend on your moments of inertia. What I want to remind you is that moment of inertia is a measurement of angular resistance, of rotational resistance. You can think that the greater my I, the heavier I am, the more resist rotation therefore I will get to the bottom last because I'm slower. More I, you can think of this as being heavier. It doesn't mean that I have more mass that's why I have heavier. I have more resistance therefore I am slower. A solid cylinder has a moment of inertia of _ MR^2. A hollow cylinder has a moment of inertia of MR^2 so you can think that there's a one in the front. A solid sphere has a moment of inertia of 2/5 MR^2. In this question, all we're doing is comparing these numbers because the M and R are the same. This is a little bit easier if you use decimals. This is 0.5, this is 1.0, and 2/5 is 0.4. You can see from here that this one is the lightest one because the coefficient number in front of the MR is the lowest. It's the lightest one therefore it is the fastest one, therefore it gets to the bottom first. The sequence is that the solid sphere is first. I'm going to write it like this. It gets to the bottom first. The second one is going to be the solid cylinder. The third one is going to be the hollow cylinder. There is a pattern here. There's a reason why the hollow cylinder is slower than the solid cylinder. Solid cylinder has very good mass distribution. The mass is very evenly distributed and remember, the more evenly distributed the mass, the lighter you are, the less I. Better mass distribution means lower I, which means you are lighter. A whole cylinder has all of its mass concentrated on the edge. It has worst mass distribution which means it has a higher I, which means it is heavier. It has a worse mass distribution, therefore it is heavier. A solid sphere is even more well-distributed than a solid disc. A solid disc 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. The sphere is always fastest. That's it for this one. Let's keep going.

Example #3: Cylinders racing down: rolling vs. sliding

Transcript

Hey guys! Let's check out this conceptual conservation of energy question. Here we have two cylinders of equal mass and radius, so basically the same cylinder. TheyÕre released from rest from the top of two hills having the same height. It's the two very similar or so far identical situations. Let's say I have cylinder A on hill A and cylinder B on hill B. Everything is the same except that A rolls without slipping. That means that it's not only going to have a V, but it's also rolling so it has a _. Remember this is called rolling motion if you roll without slipping and it means that Vcm = R_, but B 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 going to see that. What this means is that it's moving this way with a V but it has no _. Basically it's falling, it's going down as a box would. You can think of this as a box going down because it doesn't roll. Before we answer the question, I want to talk about how that's even possible. Basically, the difference is that here you have just enough static friction to cause this thing to roll. Remember, you need static friction to 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. The question is who gets to the bottom with the fastest speed? Is VAf or VBf going to be the greater one? 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. Most people get this question, they 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 roll are faster than 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 are fast because we have wheels and cars. But in reality, the one that gets to the bottom with the greater speed is this one. Why is that? That's because of conservation of energy. Conservation of energy. K initial + U initial + Work non-conservative = K final + U final. In both of these situations, the kinetic energy in the beginning is 0. I have some potential energy. The work done by non-conservative forces is 0 as well. You don't do anything. You're just watching. In the first problem, there is in the first case, there is static friction but the work done by static friction is 0. It actually even though there is static friction, it doesn't do any work. There is some kinetic at the end and there is no potential at the end. All we have is that my potential energy in the beginning goes into kinetic final at the end. If they start from the same height and they have the same mass and everything, they have the same potential energy for both of them. Let's say that number is 100 Joules. In both problems, you start with 100 Joules and then you roll to the bottom. The difference is that in the first situation, for A the 100 joules is going to get split between kinetic linear and kinetic rotational because it has two types of energies. Let's make up some numbers here. Let's say that 80 goes here and 20. I'm sorry, I meant kinetic linear and kinetic rotational. Let's say 80 goes to linear, 20 goes to rotational. Totally making this up. For B, the entire amount, the entire 100 Joules is going to go into kinetic linear because there is no kinetic rotational, because it doesn't roll around itself. Notice how this guy ends up with a greater kinetic energy than A and that's irrespective of the division. No matter how these two numbers get balanced out, the 100 will always be bigger even if this was 10 and 90, 100 is still greater. One way to think about this, so the answer is that it's basically because this has one type of kinetic energy, 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 using up your potential energy and transforming into kinetic. 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 energy to get it to roll. Rolling costs you some energy therefore some of the energy that would have gone here is actually going here to cause you to roll. In fact the faster you roll, the more rotational energy you have so there's less energy that's left for your linear, for your V. Your W is stealing energy from your V, so you end up with a lower V. That's why. That's it. It's a big really popular question conceptually. Let me know if you have any questions. Let's keep going.

Practice: 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?)

Additional Problems
A uniform rod of mass m and length l is pivoted about a horizontal, frictionless pin at the end of a thin extension (of negligible mass) a distance l from the center of mass of the rod. The rod is released from rest at an angle of θ with the horizontal, as shown in the figure. What is the magnitude of the Horizontal force Fx exerted on the pivot end of the rod extension at the instant the rod is in a horizontal position? The acceleration due to gravity is g and the moment of inertia of the rod about its center of mass is 1/12 mℓ2.  1. Fx = 1/13 mg sin(θ) 2. Fx = 24/13 mg cos(θ) 3. Fx = 24/13 mg sin(θ) 4. Fx = 13/12 mg cos(θ) 5. Fx = 12/13 mg cos(θ) 6. Fx = 12/13 mg sin(θ) 7. Fx = 13/12 mg sin(θ) 8. Fx = mg cos(θ) 9. Fx = 1/13 mg cos(θ) 10. Fx = mg sin(θ)
A thin-walled hollow sphere with mass 5.0 kg and radius 0.20 m is rolling without slipping at the base of an incline that slopes upward at 37° above the horizontal. At the base of the incline the translational speed of the center of mass of the sphere is v = 12.0 m/s. If the sphere rolls without slipping as it travels up the incline, what is the maximum vertical height that it reaches before it starts to roll back down?
A solid disk is released from rest and rolls without slipping down an inclined plane that makes an angle of 25.0° with the horizontal. What is the speed of the disk after it has rolled 3.00 m, measured along the inclined plane? A) 4.07 m/s B) 6.29 m/s C) 3.53 m/s D) 5.71 m/s E) 2.04 m/s
A thin-walled hollow cylinder (I = MR2), with mass M = 3.00 kg and radius R = 0.200 m, is rolling without slipping at the bottom of a hill. At the bottom of the hill the center of mass of the cylinder has translational velocity 16.0 m/s. The cylinder then rolls without slipping to the top of a hill. The top of the hill is a vertical height of 6.00 m above the bottom of the hill. What is the translational velocity of the center of mass of the cylinder when the cylinder reaches the top of the hill?
A 2.30-m-long pole is balanced vertically on its tip. It starts to fall and its lower end does not slip. What will be the speed of the upper end of the pole just before it hits the ground? [Hint: Use conservation of energy.]
A solid cylinder with moment of inertia I = 1/2 MR2, a hollow cylinder with moment of inertia I = MR2 and a solid sphere with moment of inertia I = 2/5 MR2 all have a uniform density, the same mass and the same radius. They are placed at the top of an inclined plane and allowed to roll down the inclined plane without slipping. Rank them in order of total kinetic energy at the bottom of the incline, from higher to lower.[a] sphere(1st), hollow cylinder(2nd), solid cylinder(3rd)[b] sphere(1st), solid cylinder(2nd), hollow cylinder(3rd)[c] hollow cylinder(1st), sphere(2nd), solid cylinder(3rd)[d] hollow cylinder(1st), solid cylinder(2nd), sphere(3rd)[e] they all have the same kinetic energy
A 5 kg mass hangs from a rope wrapped around the surface of a cylindrical pulley. The pulley has a mass of 20 kg and a radius of 0.8 m. If the 5 kg mass is suddenly released, falling down, what speed will it have after it has fallen 3.2 m?
A thin rod of length 2.8 m and mass 4.2 kg is held vertically with its lower end resting on a frictionless horizontal surface. The rod is then let go to fall freely. Determine the speed of its center of mass just before it hits horizontal surface. The acceleration due to gravity is 9.8 m/s2. 1. 4.53652 2. 3.42929 3. 4.84974 4. 4.92494 5. 4.02119 6. 4.3715 7. 5.07198 8. 4.2 9. 4.77336 10. 3.83406
A uniform hollow disk has two pieces of thin light wire wrapped around its outer rim and is supported from the ceiling (See the figure below). Suddenly one of the wires breaks, and the remaining wire does not slip as the disk rolls down. Use energy conservation to find the speed of the center of this disk after it has fallen a distance of 2.20 m.
A 2 kg mass hangs from a rope wrapped around the surface of a cylindrical pulley. The pulley has a mass of 12 kg and a radius of 0.4 m. If the 2 kg mass is suddenly released, falling down, what angular speed will the cylinder have after the mass has fallen 1.7 m?