Your Comments The checkpoint problems about the spinning disk are slightly confusing to me still. What material does the third exam cover?? Both this unit and applications? i liked the prelecture and why is exam 3 soooo close :( I was expecting this topic to be a bit challenging but it seemed quite simple. These concepts are difficult... Please help us with examples during lectures! This probably wont be on screen anyways but is there any other sites or places we can go to read the material and maybe have some more practice problems? because the homework isn't helping neither is the checkpoints. The homework due for today was waaaay to long. Its bad enough I get 4 credit hours for this class when I have to meet 6 hours a week throw over 2 hours of homework on top of that and its just unreasonable. We do have other classes too. I am very confused about the difference in the second and third checkpoint. What would happen if the Earth SUDDENLY stopped spinning/ rotating? Every week I give a shoutout commending the explanations of the one and only Gary Gladding. Every week I never make the board. Poor Gary how early do I have to enter this comment to get on the board?  not sure if people actually read this Buckaroo Banzai was AMAZING! By the way if anybody wants another good laugh check out "Flyin' Ryan" I believe the whole movie is on Youtube

Your grade so far Prelectures + Checkpoints + Lectures 100 Labs 150 Hour exams (3 x 100 each) 300 Final Exam 200 Homework (14) + Quizzes (9) 250 Prelectures: 50 Preflight's: 25 Lecture participation: 25 You can miss up to 3 of each For each Homework assignment, a score is assigned out of 100% For each Quiz, a score is assigned out of 100% The lowest of the 14 Homework and 9 Quiz scores is dropped. The remaining scores are added together to give a number out of 2200. This is scaled to be worth 250 points. Bonus Points: You can earn up to 1 extra bonus point in every lecture (for a maximum of 25 bonus points for the semester) by getting the right answers to at least 5 of the clicker questions. At the end of the semester your bonus points are added to your HW/Quiz score (max 250) Your total score out of 1000 points determines your final grade. Its just a simple formula – the computer calculates it. Your grade is determined entirely by the your performance on the components of the course as described above. There is no other “extra credit” possible. A+ (950), A (920), A- (900), B+ (880), B (860), B- (835), C+ (810), C (780), C- (750), D+ (720), D (690), D- (610), and F (<610).

Today’s Concepts: Angular Momentum Physics 211 Lecture 19 Today’s Concepts: Angular Momentum

Linear and Rotation

Angular Momentum We have shown that for a system of particles Momentum is conserved if What is the rotational version of this? Well – to start with we know this:

Torque & Angular Momentum Total angular momentum is conserved In the absence of external torques

Example – Disk dropped on Disk Suppose we have two uniform disks having the same radius and mass, one initially suspended above the other. Both are free to rotate about the same vertical axis. The top disk has an initial angular velocity omega-0 and the bottom disk in initially at rest. Suppose the top disk is now dropped so that it lands on the bottom one, and friction between the disks causes the top disk to slow down as the bottom one speeds up until they are eventually turning with the same final angular velocity omega-f. What is omega-f? In this case the system under consideration is made up of the two disks. When the disks come into contact they will exert torques on each other, but since there are no other external torques acting on this system the total angular momentum of both disks will be conserved. This is exactly analogous to an inelastic collision in one dimension between two blocks, where the blocks exert forces on each other but the total momentum of the two blocks is conserved since there are no other external forces. Since the angular momentum of the system is the same before and after the disks collide, we can work out the final angular velocity of the system by equating the total angular momentum before the collision [show] to the total angular momentum after [show] and solving for omega final. In this case we see that the final angular velocity is exactly half of the initial, in exact analogy to the case of a one dimensional collision where the blocks have the same mass and one of them is initially at rest and the blocks stick together after colliding. We can take this analogy further by comparing the kinetic energies of the system before and after the collision. We can write the kinetic energy of the system in terms of the angular momentum, which is the same before and after, and the moment of inertia, which doubles. We see that the final kinetic energy in this example is half of the initial kinetic energy, just as it was in the analogous one dimensional inelastic collision. “is friction between objetcs within a system an external force?” No !! 7

Example – Disk dropped on Disk Suppose we have two uniform disks having the same radius and mass, one initially suspended above the other. Both are free to rotate about the same vertical axis. The top disk has an initial angular velocity omega-0 and the bottom disk in initially at rest. Suppose the top disk is now dropped so that it lands on the bottom one, and friction between the disks causes the top disk to slow down as the bottom one speeds up until they are eventually turning with the same final angular velocity omega-f. What is omega-f? In this case the system under consideration is made up of the two disks. When the disks come into contact they will exert torques on each other, but since there are no other external torques acting on this system the total angular momentum of both disks will be conserved. This is exactly analogous to an inelastic collision in one dimension between two blocks, where the blocks exert forces on each other but the total momentum of the two blocks is conserved since there are no other external forces. Since the angular momentum of the system is the same before and after the disks collide, we can work out the final angular velocity of the system by equating the total angular momentum before the collision [show] to the total angular momentum after [show] and solving for omega final. In this case we see that the final angular velocity is exactly half of the initial, in exact analogy to the case of a one dimensional collision where the blocks have the same mass and one of them is initially at rest and the blocks stick together after colliding. We can take this analogy further by comparing the kinetic energies of the system before and after the collision. We can write the kinetic energy of the system in terms of the angular momentum, which is the same before and after, and the moment of inertia, which doubles. We see that the final kinetic energy in this example is half of the initial kinetic energy, just as it was in the analogous one dimensional inelastic collision. = 8

What about Kinetic Energy? Suppose we have two uniform disks having the same radius and mass, one initially suspended above the other. Both are free to rotate about the same vertical axis. The top disk has an initial angular velocity omega-0 and the bottom disk in initially at rest. Suppose the top disk is now dropped so that it lands on the bottom one, and friction between the disks causes the top disk to slow down as the bottom one speeds up until they are eventually turning with the same final angular velocity omega-f. What is omega-f? In this case the system under consideration is made up of the two disks. When the disks come into contact they will exert torques on each other, but since there are no other external torques acting on this system the total angular momentum of both disks will be conserved. This is exactly analogous to an inelastic collision in one dimension between two blocks, where the blocks exert forces on each other but the total momentum of the two blocks is conserved since there are no other external forces. Since the angular momentum of the system is the same before and after the disks collide, we can work out the final angular velocity of the system by equating the total angular momentum before the collision [show] to the total angular momentum after [show] and solving for omega final. In this case we see that the final angular velocity is exactly half of the initial, in exact analogy to the case of a one dimensional collision where the blocks have the same mass and one of them is initially at rest and the blocks stick together after colliding. We can take this analogy further by comparing the kinetic energies of the system before and after the collision. We can write the kinetic energy of the system in terms of the angular momentum, which is the same before and after, and the moment of inertia, which doubles. We see that the final kinetic energy in this example is half of the initial kinetic energy, just as it was in the analogous one dimensional inelastic collision. same x2 9

CheckPoint Same initial L In both cases shown below a solid disk of mass M, radius R, and initial angular velocity wo is dropped onto an initially stationary second disk having the same radius. In Case 2 the mass of the bottom disk is twice as big as in Case 1. If there are no external torques acting on either system, in which case is the final kinetic energy of the system biggest? M 2M M A) Case 1 B) Case 2 C) Same Same initial L Case 1 Case 2

CheckPoint In which case is the final kinetic energy of the system biggest? M 2M A) Case 1 B) Case 2 C) Same Case 1 Case 2 A)The moment of inertia of case 2 is larger so the kinetic energy would be smaller since L is the same in both cases.. B) more mass, more energy. C) energy is conserved if no external torque. Both initial energies were the same 11

Point Particle moving in a Straight Line axis D So far we have only discussed the angular momentum of rotating objects, a fact that should not surprise anyone since we are studying rotations. However, when you look at our definition of the angular momentum of an objects about some axis as the cross product it’s momentum [show P] and the its displacement from the rotation axis [show R] then there is nothing that prevents us from applying this to an object moving along a straight line as well. Why would we do this? In the absence of external forces we know that a particles linear momentum is conserved. We also now that if there are no forces, then there are also no torques. This implies that the angular momentum about any axis we choose must also be conserved. To illustrate the implication of this, consider a particle moving with constant momentum past a perpendicular axis as shown. If we evaluate R cross P anywhere along the path of this particle we get the same answer – the magnitude of the resulting angular momentum is just the magnitude of the linear momentum times the distance between the path of the particle and the rotation axis, and the direction the angular momentum is given by the right hand rule, in this example out of the screen. As this example illustrates, as long as there are no external torques acting on the system the total angular momentum of a system is conserved even if the object is not moving in a circle. On the next slide we will solve a simple problem involving both the angular momentum due to rotation and due to motion in a straight line. 12

D Direction given by right hand rule (out of the page in this case) axis D So far we have only discussed the angular momentum of rotating objects, a fact that should not surprise anyone since we are studying rotations. However, when you look at our definition of the angular momentum of an objects about some axis as the cross product it’s momentum [show P] and the its displacement from the rotation axis [show R] then there is nothing that prevents us from applying this to an object moving along a straight line as well. Why would we do this? In the absence of external forces we know that a particles linear momentum is conserved. We also now that if there are no forces, then there are also no torques. This implies that the angular momentum about any axis we choose must also be conserved. To illustrate the implication of this, consider a particle moving with constant momentum past a perpendicular axis as shown. If we evaluate R cross P anywhere along the path of this particle we get the same answer – the magnitude of the resulting angular momentum is just the magnitude of the linear momentum times the distance between the path of the particle and the rotation axis, and the direction the angular momentum is given by the right hand rule, in this example out of the screen. As this example illustrates, as long as there are no external torques acting on the system the total angular momentum of a system is conserved even if the object is not moving in a circle. On the next slide we will solve a simple problem involving both the angular momentum due to rotation and due to motion in a straight line. Direction given by right hand rule (out of the page in this case) 13

Playground Example w v m M R Before After Top View Disk Kid 14 Suppose your are at the playground. You notice that there is no one on the merry-go-round so you take a run at it and jump onto its outer edge along a path that is tangent to the edge. How fast do you and the merry-go-round end up spinning after you jump on? If we say that the system under consideration is you and the merry-go-round; if we choose the rotation axis to be the axle at the center of the merry-go-round, and we assume that the axle is well greased so that we can ignore any friction when the merry-go-round rotates, then there are no external torques acting on the system and the total angular momentum of the system must be conserved. On the previous slide we worked out how to calculate your angular momentum about the rotation axis before you jump on – it is just your momentum times the closest distance that your path comes to the axis [show mvR]. After you jump on, both you and the merry-go-round rotate with the same final angular velocity, so the final angular momentum of the system is this angular velocity times the total moment of inertia of the system. The total moment of inertia has two parts – one due to the merry go round and one due to you. If we say that the merry go round is a uniform disk of known mass and radius we can figure out its contribution [show 1/2MR^2], and if we treat you like a point particle at the edge if the merry go round we can figure out your contribution as well [show mR^2]. We can now use the fact that the initial and final angular momentum of the system is the same to solve for the final angular velocity. If your mass is 75 kg and you run with an initial speed of 2 m/s and if we say that the merry go round has a radius of 2m and a mass of 150 kg, we find that the final angular velocity is half a radian per second. Disk Kid 14

CheckPoint The magnitude of the angular momentum of a freely rotating disk around its center is L. You toss a heavy block onto the disk along the direction shown. Friction acts between the disk and the block so that eventually the block is at rest on the disk and rotates with it. What is the magnitude of the final angular momentum of the disk-block system: A) > L B) = L C) < L Top View Instead of a block, imagine it’s a kid hopping onto a merry go round

CheckPoint What is the magnitude of the final angular momentum of the disk-block system: A) > L B) = L C) < L Top View A) the angular velocity remains the same but the mass increases. Thus, angular momentum is greater than L. B) L is conserved and since the block is thrown toward the rotation axis it has no initial angular momentum. C) There is friction between the disk and the block.

Clicker Question The magnitude of the angular momentum of a freely rotating disk around its center is L. You toss a heavy block onto the disk along the direction shown. Friction acts between the disk and the block so that eventually the block is at rest on the disk and rotates with it. Is the total angular momentum of the disk-block system conserved during this? A) Yes B) No Top View There are no external torques The block skidding on the disk is an internal torque (within the system)

CheckPoint The magnitude of the angular momentum of a freely rotating disk around its center is L. You toss a heavy block onto the disk along the direction shown. Friction acts between the disk and the block so that eventually the block is at rest on the disk and rotates with it. What is the magnitude of the final angular momentum of the disk-block system: A) > L B) = L C) < L Top View Instead of a block, imagine it’s a kid hopping onto a merry go round Mechanics Lecture 19, Slide 18

CheckPoint What is the magnitude of the final angular momentum of the disk-block system: A) > L B) = L C) < L Top View A) It will be more than the original because the block is adding more angular momentum.. B) angular momentum is conserved. C) The block starts with an angular momentum in the opposite direction of the disk so the total is less than just L

Clicker Question A student holding a heavy ball sits on the outer edge a merry go round which is initially rotating counterclockwise. Which way should she throw the ball so that she stops the rotation? A) To her left B) To her right C) Radially outward C A B w top view: initial final Bender in Space

Clicker Question A student is riding on the outside edge of a merry-go-round rotating about a frictionless pivot. She holds a heavy ball at rest in her hand. If she releases the ball, the angular velocity of the merry-go-round will: A) Increase B) Decrease C) Stay the same w1 w2

Homework

m2 m1 w0

m2 m1 wo

m2 m1 w0 wf Solve for wf

M wf

Just like for linear momentum M for angular momentum wf