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Ch 13: Rotational Inertia & EnergyWorksheetSee all chapters
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Ch 01: Intro to Physics; Units
Ch 02: 1D Motion / Kinematics
Ch 03: Vectors
Ch 04: 2D Kinematics
Ch 05: Projectile Motion
Ch 06: Intro to Forces (Dynamics)
Ch 07: Friction, Inclines, Systems
Ch 08: Centripetal Forces & Gravitation
Ch 09: Work & Energy
Ch 10: Conservation of Energy
Ch 11: Momentum & Impulse
Ch 12: Rotational Kinematics
Ch 13: Rotational Inertia & Energy
Ch 14: Torque & Rotational Dynamics
Ch 15: Rotational Equilibrium
Ch 16: Angular Momentum
Ch 17: Periodic Motion
Ch 19: Waves & Sound
Ch 20: Fluid Mechanics
Ch 21: Heat and Temperature
Ch 22: Kinetic Theory of Ideal Gases
Ch 23: The First Law of Thermodynamics
Ch 24: The Second Law of Thermodynamics
Ch 25: Electric Force & Field; Gauss' Law
Ch 26: Electric Potential
Ch 27: Capacitors & Dielectrics
Ch 28: Resistors & DC Circuits
Ch 29: Magnetic Fields and Forces
Ch 30: Sources of Magnetic Field
Ch 31: Induction and Inductance
Ch 32: Alternating Current
Ch 33: Electromagnetic Waves
Ch 34: Geometric Optics
Ch 35: Wave Optics
Ch 37: Special Relativity
Ch 38: Particle-Wave Duality
Ch 39: Atomic Structure
Ch 40: Nuclear Physics
Ch 41: Quantum Mechanics
Energy of Rolling Motion
Moment of Inertia & Mass Distribution
Intro to Moment of Inertia
More Conservation of Energy Problems
Types of Motion & Energy
Parallel Axis Theorem
Intro to Rotational Kinetic Energy
Moment of Inertia of Systems
Conservation of Energy in Rolling Motion
Conservation of Energy with Rotation
Moment of Inertia via Integration
Torque with Kinematic Equations
Rotational Dynamics with Two Motions
Rotational Dynamics of Rolling Motion

Concept #1: Torque with Kinematic Equations


Hey guys! Now that we've seen how to solve some basic torque acceleration questions, we're going to add motion equations into the mix which will generate a bunch of extra questions, problems we can solve. Let's check it out. You may remember that when we had force problems, most force problems were solved using F = ma. You may remember that similar to those problems would also involve our three or four equations of motion or kinematics equations. Same thing is going to happen here with rotation where some torque problems will require both Torque = I_, which is the rotational version of Newton's second law and rotational motion equations. Motion equations are our UAM equations, uniformly accelerated motion, kinematics equations, motion equations or whatever you call it. We got three to four of these guys. Remember just like how it was with linear motion, the variable that will connect Torque = I_ to the rotational motion equations which are these guys here is going to be acceleration. Because we're talking about rotation, this means _. Notice how there is an _ here and there's an _ here, here and here. We got more equations and more variables but it's not harder. It's just more stuff. Let's check out this example here. I have a solid sphere and I give you the mass and the diameter. Solid sphere is the shape of the sphere so I know that because itÕs a solid sphere, I'm going to use the moment of inertia of a solid sphere which is 2/5 MR^2. I got the mass; the mass is 200 and I have the diameter. Remember in physics we never use the diameter. As soon as I see diameter, I convert that immediately into radius which is half of that so it's 3 meters. It spins about an axis through its center, this is the regular rotation of a sphere, a solid sphere, which is around itself. It says it does this with 180 RPM clockwise. The RPM is 180. What's up with clockwise? Clockwise is going to mean that it is negative. It's got an RPM of 180. Remember as we did in motion problems and rotational motion problems, whenever youÕre given RPM vast majority of the time you're going to immediately convert that into w because most of our equations have omega, little w, but not RPM in it. The next thing I want to do here is convert this into w. w = 2¹f or 2¹, remember f frequency is rpm/60. This is going to be 2¹(-180) / 60. This is going to be a 3 right there which means the whole thing will be -6¹ rad/s. I got that. That's the initial speed. I want to know how much torque is needed to stop this thing in just 10 seconds. I'm asking what is the torque to stop it. That means that this is my initial w and I want to have a final w of 0. I want to do this in just 10 seconds, so _t is 10. I hope you noticed here that you start seeing all these motion variables. Remember the way I solve motion problems is by setting up the curly braces and putting all five motion variables there. Let's do that. w initial = -6¹, w final equals we want it to be 0, _, __, and _t. _t is 10, and these two guys, we don't have them. They're also not what we're looking for. But since I saw all these variables, I decided hey let's start setting this up because I know this is coming. But really what we're looking for is torque. You might actually have started this question. Instead of going here, you might have just written that the sum of all torques equals I_ and that's perfectly fine as well. If anything that's a more directed way to the answer, more targeted, which is fine. There's only one torque here. We're assuming there's one torque. This thing is spinning and I guess you're applying a torque to it to make it stop so you can assume that there's only one torque. The sum of all torques will become just the torque that you're looking for and that is I_. If I can have I and I have _, I'm done and that will be my answer. Let's expand I = (2/5 MR^2) _. Notice that I have M, I have R but I don't have _. What you're going to do is you're going to go over here and try to find _. Let me plug in these numbers so the only thing we're missing is _. M is 200, R is 3^2, _. As soon as we have _, we can plug it in there. Back to the motion equations. So the basic idea is we got stuck and you go to the other side. Back to the motion equations, we have three variables which means we can solve. This is my target and this is my ignored variable. I'm going to use the only equation that does not have a __ and the only equation that doesn't have a __ in it is the first equation. w final = w initial + _t. We're looking for _, so _ is going to be (w final - w initial) / t. This by the way is the definition of _. It's the change in w over the change in t. You could have started there as well. That would have worked. This is 0 Ð (-6¹), and the time is 10 seconds. These cancel. We end up with 6¹ / 10 positive, which is 1.88. I got a positive which should make sense. Even though I'm slowing down, let's talk about that real quick. My velocity is negative. If I'm slowing down, I'm trying to make my velocity positive so my acceleration should be positive. IÕm trying to make my velocity more positive. Another way to think about this that might be even easier is you have a negative w and you're trying to slow down, so you have to go in the other direction. The acceleration has to go in the other direction so the acceleration would oppose it because you're trying to slow down. This is counterclockwise which would be positive. Anyway, my acceleration is 1.88 rad/s^2. Now I can plug this in here and we are done. If I multiply all of that and then I multiply that by 1.88, I should get 1354 Nm. That's it. That's the final answer. Just to recap real quick, there's basically two parts to this. We were asked for torque so you could have started here and then you start plugging stuff in and you realize you don't have _ but you have a bunch of motion variables, so you can find _ using one of the motion equations, plug it back in. It's the classic standard type of physics question where you get stuck with something, go look for other variable, plug it in, come back with the value you got and solve. That's it. Hopefully it makes sense. Let me know if you have any questions. Let's keep going.

Practice: A light, long rope is wrapped around a solid disc, in such a way that pulling the rope causes the disc to spin about a fixed axis perpendicular to itself and through its center. The disc has mass 40 kg, radius 2 m, and is initially at rest, and the rope unwinds without slipping. You pull on the rope with a constant 200 N. Use the rotational version of Newton’s Second Law to calculate how fast (in rad/s) the disc be spinning after you pull 50 m of rope.

Practice: A system is made of two small, 3 kg masses attached to the ends of a 5 kg, 4 m long, thin rod, as shown. The system is free to rotate about an axis perpendicular to the rod and through its center. Two forces, both of magnitude F and perpendicular to the rod, are applied as shown below. What must the value of F be to the system from rest to 10 rad/s in exactly 8 complete revolutions?

Practice: Two rotating doors, each 6.0 m long, are fixed to the same central axis of rotation, as shown (top view). When you push on one door with a constant 100 N, directed perpendicular from the face of the door and 50 cm from its outer edge, the rotating door system takes 8 s to complete a full revolution from rest. The doors can be modeled as thin rectangles (moments of inertia for thin rectangles, around two different axes, are shown for reference). Calculate the mass of the system.

Example #1: Stopping flywheel with friction


Hey guys! In this example we have a flywheel that's spinning and we're going to press an object against the wheel to cause it to stop. Imagine you got spinning and if you squeeze an object against it, if you push an object against it, it would stop. We want to know how hard do you have to push against the flywheel so that it stops in a particular amount of time. It says your flywheel is a rotating disc, this means we're going to use the moment of inertia of the disc which is 1/2 MR^2, same as a solid cylinder and it's used to store energy. Suppose this one has a mass of 8 x 10^4 kg and has a diameter of 5 meters, which means we're going to immediately change the diameter into a radius of 2.5. ItÕs setup vertically so we got this wheel that's vertical and it's free to spin around a fixed axis. There's a fixed axis there perpendicular to the wheel. Basically the wheel spins like this around a central axis. To slow down the flywheel, I mentioned this, you push this block here so there's a force that you apply. I want to point out that once you push here, it's going to have a contact therefore there's going to be a normal force back. This normal force will have the same magnitude as your force, so action-reaction. If you push with 10, normal pushes back with 10. It says that the coefficients of friction between the block and the wheel, right here the coefficients of friction are 0.6 and 0.8. I'm going to write them here, _ static. Remember, when you have two coefficients of friction, the static coefficient is the greater one, so 0.8. _ kinetic is going to be 0.6. We want to know how hard you have to push, so I wanna know what is F. It says here that the wheel is going to come to a complete stop, so w final will be zero from an RPM of 300, so rpm initial is 300. It's going to do this in a _t of 30 seconds. Remember rpm almost always gets converted into w, so I'm going to do that real quick. I'm going to say w initial = equals 2¹ rpm/60. That's the equation to convert it to. If you multiply this, you get 10¹ rad/s. I'm going to actually move that over here and scratch this out just because I'm trying to list all my motion variables and I'm trying to make the point to you that as of now we already have three motion variables which is good news. It means we could solve for the others. The other variables that are missing here are _and __. All I'm doing is grabbing the information and sort of fixing it up. How hard do you have to push? Again, this is a force that's going to cause a torque. We're going to use the sum of all torques = I_ equation. But I want to explain to you what's going on here. The idea is that the wheel is spinning this way, let's say with an w initial. When you push, there is now a normal force here. I had that drawn earlier. What that means is because thereÕs a normal force and there's friction, there's going to be friction acting against motion. Motion is rotating that way, friction is going to try to stop the wheel which means there's going to be a force of friction this way. You got a normal this way and there's a force of friction this way. What that force of friction does, it causes a torque like this, torque of friction, which is opposite to my velocity so it's trying to slow it down. ItÕs trying to get it to stop. In this particular case, just because of the way I drew it and this is actually negative and then this would be positive. As long as they're opposite to each other, you're fine. There is a torque, there is an acceleration. This is a force problem with angular acceleration, with torque, so we're going to start here and we have to find what this F is. Let's expand this equation real quick. The only torque acting on this is a torque due to friction because you're pushing against this thing. We're going to write torque of friction. The moment of inertia, I didn't read this part to you but it says here you may assume the wheelÕs entire mass is concentrated at its outer rim. This means that this is actually not a rotating disc, I apologize. This is not a rotating disc. This is going to be a hollow disc. It's going to be hollow disc. Instead of _ MR^2, itÕs going to be just MR^2 where R is the radius. I'm going to put this here, MR^2, and _. We don't have _ but we could find alpha if we wanted to. Let's take this one step further and expand this. Torque is force little r sine of theta. Force in this case is friction, so itÕs little f. R is the distance from the axis of rotation to the point where the force happens. This is the distance that we are talking about. This is my r vector because the force happens here and the axis is obviously in the middle here. The R vector is as long as the radius of the wheel because friction happens at the edge there. We're going to have F big R and then sin_. The angle will be 90 degrees. Notice how they make an angle of 90 degrees because friction is straight down as a result of the pushing against it. By the way friction is in situations like this, friction is always going to make an angle of 90 degrees and itÕs always gonna happen at the edge right, so sine of 90 so that's nice. MR^2, the radius we have that as well. We're going to be able to plug it in there and then _. We're looking for F. Don't get confused. We're not looking for little f. We're looking for big F. What we're going to do is we're going to keep expanding this equation. Friction can be expanded. Friction is _ normal. I could rewrite friction as _ normal. That's what we're going to do, except there's one change which is normal is the same thing as F. I'm going to plug in F here and I'm going to write friction as _F instead of _ normal. ItÕs the same thing. The reason we do that is because now finally our equation actually has our variable. Until then you hadnÕt seen your variable around. IÕm gonna rewrite this here. I'm going to cancel this R with this R. This is just a 1. It's going to be _F = MR_. F is MR_ / _. We know these numbers, or most of them. M is 8 x 10^4, radius is 2.5. I got to go find _ and then we do have _. Which _ do you think we use Ð kinetic or static? I hope you're thinking kinetic because the block is rubbing against the disc. It's going to be 0.6 which is kinetic. We have to go find alpha, so let's go do that real quick. We're going to go over here and look for _. To find _, I can use motion equations. I have three knowns. This is my target and this is my ignored variable which means I can use the first equation to find _, w final = w initial + _t. We're looking for _ so let's move everything out of the way. You might notice that this is the definition of angular acceleration which is change in omega over change in time. You could have started there as well. The final w is zero because we're looking to stop. The initial w is 10¹ rad/s. Notice that the way we drew it, it's a negative so let's plug it in as a negative. The time we're going for is 30 seconds. If you do this, you end up with an _ of 1.05 rad/s^2, and that's what I'm going to put right here, 1.05 right there. Once you multiply this whole thing, you end up with 3.5 x 10^5 N. That's how much force you have to push against the block with and push the block against the wheel so that this thing stops in 30 seconds. That's it for this one. This question is fairly popular. It's a little bit tricky but again I think the key thing is to realize you have a force causing a torque causing acceleration that this is the starting place. The trickiest part I think here is to try to make a connection from this initial equation and then how weÕre going to get to F. Sometimes you just have to trust that if you keep expanding the equation like we did here and then we expanded the F here, you have to trust that if you keep expanding the variable will show up. That's it for this one. Let me know if you have any questions and let's keep going.

Practice: A 1,000 kg disc that has a 5 m outer radius is mounted on a vertical, inner axle 80 kg in mass and 1 m in radius. A motor acts on the axle to speed up or slow down the system. Suppose the motor stops functioning when the system is spinning at 70 rad/s. To bring it to a complete stop, you apply a constant 200 N friction to the surface of the axle. How many revolutions will the system take to stop?