Catching up from Yesterday In the two cases shown below identical ladders are leaning against frictionless walls. In which case is the force of friction between the ladder and the ground bigger? A) Case 1 B) Case 2 C) Same Case 1 Case 2 1

CheckPoint In the two cases shown below identical ladders are leaning against frictionless walls. In which case is the force of friction between the ladder and the ground bigger? A) Case 1 B) Case 2 C) Same Case 1 Case 2 A) Because the bottom of the ladder is further away from the wall. B) The angle is steeper, which means there is more normal force and thus more friction. C) Both have same mass. 2

CheckPoint Suppose you hang one end of a beam from the ceiling by a rope and the bottom of the beam rests on a frictionless sheet of ice. The center of mass of the beam is marked with an black spot. Which of the following configurations best represents the equilibrium condition of this setup? A) B) C)

CheckPoint Which of the following configurations best represents the equilibrium condition of this setup? A) B) C) Demo If the tension has any horizontal component, the beam will accelerate in the horizontal direction. Objects tend to attempt to minimize their potential energy as much as possible. In Case C, the center of mass is lowest.

Stability & Potential Energy CM Demos footprint

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

Linear and Rotation

New Cast of Characters with Old Theme Remember

Angular Momentum We have shown that for a system of particles Momentum is conserved if What is the rotational version of this? The rotational analogue of force F is torque Define the rotational analogue of momentum p to be Angular Momentum: For a symmetric solid object

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

Example – Disk dropped on Disk Demo 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. Friction is an “internal force” because it is “internal” to the “system”. Friction is between two parts of the “system”. Friction may change the Kinetic Energy. But friction, being an internal force, does not change the Angular Momentum (nor the linear momentum in linear situations). 11

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. = 12

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 13

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 bigger? 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 bigger? M 2M A) Case 1 B) Case 2 C) Same Case 1 Case 2 Demo: wheel rim drop A) L is equal but bigger mass in case 2 B) Case 2 has greater mass C) Conservation of angular momentum 15

Your Comments Sort of . . . Work is done by friction so KE is not conserved. Okay; now what does that mean? Find the final KE in both cases. True but what about the final KE? That’s for omega but what about the final KE? True but what about the final KE? True but what about the final KE? ? ? ? ? ? ? ? ? This doesn’t mean anything.

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. 17

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. Demo: train on bike wheel Direction given by right hand rule (out of the page in this case) 18

Playground Example w v m M R Before After Top View Disk Kid 19 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 19

Algebra Details Disk Kid 20 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 20

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: Instead of a block, imagine it’s a kid hopping onto a merry go round A) > L B) = L C) < L Top View The block provides no angular momentum since its perpendicular distance from the axis of rotation is 0. No external forces act on this systems, so angular momentum is conserved.

CheckPoint What is the magnitude of the final angular momentum of the disk-block system: A) > L B) = L C) < L Top View The block provides no angular momentum since its perpendicular distance from the axis of rotation is 0. No external forces act on this systems, so angular momentum is conserved.

Your Comments Will it be conserved in the next CheckPoint? Good! Altho’ I wouldn’t call it an external force. Good! The block’s perpendicular distance from the axis of rotation is zero. . But the block is NOT at rest. What is the final angular momentum? What does that mean for the final angular momentum? What about the next CheckPoint? Zero!>!>! How can that be? okay Tell me more. What about the block? Tell me more. ! ! ! ! !

ACT 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 Friction decreases the angular momentum of the disk But the TOTAL angular momentum of the block+disk is CONSTANT!

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) Since the block has angular momentum to begin with, that is added into L. B) L is conserved so Linitial=Lfinal C) The block adds momentum going in the opposite direction, so the total momentum is smaller than the momentum of the disk.

Your Comments Bravo! Bravo! What about the block? Why bigger? Angular momentum has a direction connected to it. What about the angular momentum of the block? Any better by now? No, not this time. Good. Okay but tell me more. Smaller than what? In the previous CheckPoint, the lever arm was zero. Tell me more.

ACT 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

ACT 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

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

Conservation Ideas Internal forces do not change the momentum. Internal torques do not change the angular momentum. However, internal forces or internal torques can still do work and, thus, change the Kinetic Energy.

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