Ch 11: Rotational Inertia & EnergyWorksheetSee all chapters
All Chapters
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
Ch 16: Waves & Sound
Ch 17: Fluid Mechanics
Ch 18: Heat and Temperature
Ch 19: Kinetic Theory of Ideal Gasses
Ch 20: The First Law of Thermodynamics
Ch 21: The Second Law of Thermodynamics
Ch 22: Electric Force & Field; Gauss' Law
Ch 23: Electric Potential
Ch 24: Capacitors & Dielectrics
Ch 25: Resistors & DC Circuits
Ch 26: Magnetic Fields and Forces
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: Energy of Rolling Motion (Surface vs Air)

Transcript

Hey guys! You may remember that if we have a disk-like object like a cylinder or a sphere and it is moving on a surface while rolling around itself much like a toilet paper if you throw it on the floor would do, that motion is called rolling motion. It's got a special name when you have an object on the floor like that. What's special about it is that it has a both a linear velocity because it's moving, the center of mass moves, and a rotational velocity because it spins around itself. It's got two motions so it has linear kinetic energy and rotational kinetic energy. Let's look into that real quick. As I just said, a wheel-like object rotating and moving around itself is called this type of motion. ItÕs called rolling motion. You have a V, which is often referred to as the V of the center of mass, and you have an w (omega) because you're rolling around yourself. There are two types of motion here. Wat's also important about this is that there's a relationship between Vcm and w, which is Vcm = Rw. The R is big R radius. What that means is that those two variables V and w are linked. So if one grows, the other one has to grow by the same amount. It's important to make a distinction that if you have an object, that this only happens if you are on a surface. If you have an object that is rolling on air, these two variables V and w are not tied to each other. They are not tied. Vcm is not tied to w. Basically it means that you cannot use this equation right here, the green equation. This only happens if you're rolling on a surface and if you're rolling without slipping. Lucky for you, all problems in physics at least for you guys is going to be rolling without slipping so you can just assume that to be the case. To summarize, if you are rolling on the surface, this equation applies. If you are not rolling on a surface, this equation does not apply. If you throw a ball and it rolls in the floor, that would apply. But let's say if you throw a baseball it's spinning through the air and moving, you cannot say that Vcm equals Rw. That equation doesn't work. That's it.Let's do an example here. I have a solid sphere. This is the type of shape I have. That's going to tell me the moment of inertia. I of a solid sphere, I have it here. I of a solid sphere is 2/5 mr^2. IÕm given that the mass is 2, the radius is 0.3 and it rolls without slipping on a horizontal surface. It rolls without slipping on a horizontal surface means that this is called rolling motion, and it means that the green equation is going to work. IÕm going to make this green to match up with the other green up there, with 10. This 10 is the velocity of the center of mass. If I tell you an object moves at 10 meters per second, that's the velocity at the middle of the object. The question here is let's calculate the linear, rotational and total kinetic energy. Let's find the linear first, we'll plug in the other ones. Linear energy is _ mv^2. I got all these numbers, _, the mass is 2, the velocity is 10^2. This will be 100 Joules. For kinetic rotational, it's going to be _ Iw^2. I have I. I is going to be 2/5 MR^2. I don't have Omega but I can get w because I have a V and these guys are related connected by this equation right here. Let's do that real quick. Vcm equals Rw. w is Vcm / R. Vcm is 10, R is 0.3. This guy here will be 33. I can put w over here. Notice that the 2 cancels with the 2. Then I'm going to have 1/5, the mass is 2, the radius is 0.3^2 and w which is 10/0.3. If you want, what you could also do is instead of writing 33 here, I'm actually going to write this. If youÕve got a calculator just put the 33, it's faster. But IÕm going to do 10/0.3^2 and that's because if you notice, this cancels with this. Then I'm left with 2*10^2 which is 200/5. Make sure I'm doing this correctly. This is 40 Joules. Then for the total kinetic energy, we're going to have kinetic linear plus kinetic rotational, 100 plus 40, 140 Joules. That's linear rotational and the total. It's got two types of energy so you add up linear with the rotational. Notice that they're not necessarily the same. Remember that we can use this equation here because it's rolling on a surface. That's it for this one. Hopefully it makes sense. Let me know if you've any questions.

Practice: : A 150-g baseball, 3.85 cm in radius, leaves the pitcher’s hand with 30 m/s horizontal and 20 rad/s clockwise. Calculate the ball’s linear, rotational, and total kinetic energy.

Example #1: Ratio of energies of cylinder on surface

Transcript

Hey guys! Some of these rotational questions will ask you to find the ratio of one type of energy over the other. I want to show you how to do one of these. Here we have a solid cylinder. Solid cylinder tells us that we're supposed to use I equals _ MR^2. It's not actually giving us the numbers. This is going to be literal question or just put letters, variables. It rolls without slipping on a horizontal surface. This is called rolling motion and it means I can use this equation, Vcm equals Rw because it's rolling like this. Vcm is tied to w. I want to know the ratio of its rotational kinetic to its total kinetic energy. What you do is you follow what's saying here and you set up a ratio like this. It's saying rotational kinetic energy so we're going to write K rotational that's the top, two, total kinetic energy that's the bottom, ratio of top to bottom. Ktotal which is Klinear plus Krotational. Now what we're going to do is weÕre gonna expand these equations as much as possible. What I mean by expanding is, what does KR stand for? KR stands for _ Iw^2. KL is _ mv^2 and KR is _ Iw^2. WeÕre gonna expand this as much as possible, meaning we're not going to stop there. We can replace I with this right here. I can replace I with this right and I can also replace w with something else. The problem here is I have these and w. There's too many variables. Whenever you have a V and a w, you usually want to replace one into the other. V equals Rw. What we're going to do is whenever we have V or w, whenever you have V and w, we want to get the w to become of V. IÕm going to write w is V/R and we're going to replace this here. The reason we do this so that we have fewer variables so it's easier to solve this question. Before I start plugging stuff in, I want to warn. You cannot cancel this with this. That's not a thing, so don't get tempted to do that. What you can do is you can cancel the _ over here because they exist in all three of these guys here. You can cancel the _ and this simplifies a little bit. What we're going to do now is expand I. I is going to become _, itÕs the _ from here, not this half. That half is already gone. It's the one inside of the I, MR^2. Remember, we're going to rewrite w as V/R and then this whole thing is squared. I wanna do the same thing at the bottom. But before I do that, you might notice right away that this R cancels. That's another benefit of doing this thing here. Another benefit of doing this thing here is that it's going to cause Rs to cancel. At the bottom I have simply MV^2 plus I which is _ MR^2 and then w which is (V/R)^2. Again, just like it did at the top, the Rs cancel. Let's clean this up a little bit and see what we end up with. I end up with _ MV^2 divided by MV^2 + _ MV^2. You may already see where this is going. There's an M in all three of these and there's a V in all three of these, so everything goes away and you end up with some numbers left here. You have half up here and then this, there's a one here right that stays there, one plus half. The mass doesnÕt matter, the velocity doesn't matter. All you have to do is do this thing here. There's two ways you can do this. If you like fractions, you can do with fractions. I'm going to do that first so IÕm going to rewrite this as 2/2 and then I have _ divided by. I got a 2 at the bottom here and then I can add up the top here so it's 2+1 = 3. I have _ divided by 3/2. I can cancel this 2 and then I end up with 1/3. If you don't like fractions, one thing you can do with this particular case is you can rewrite this like this or half is 0.5, this is a 1, this is a 0.5. This is better maybe if you have a calculator. 0.5 divided by 1.5. If you do this in the calculator, it's 0.33, which is the same thing as this okay. That's it. The ratio is one third. By the way, that ratio will change if you have a different I because this half here ends up showing up here and here, or actually that _ ends up showing up here and here. If you have a different shape, this would be a different fraction and then your final answer will be different. The ratios change depending on what kind of shape you have. That's it for this one. Let me know if youÕve got any questions.

Practice: A hollow sphere of mass M and radius R rolls without slipping on a horizontal surface with angular speed W. Calculate the ratio of its linear kinetic energy to its total kinetic energy.