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Physics 231 2013 1 Ch 9 Day 1 10 Mon. 11/4 Tues. 11/5 Wed., 11/6 Lab Fri., 11/8 9.1-.2, (.8) Momentum and Energy in Multi-particle Systems 11 Mon., 11/11 9.4-.5 (.9) The “Point Particle” approximation Tues. 11/12 9.3 Rotational Energy Quiz 8 L8 Energy Quantization Review Exam 2 (Ch 5-8) Exam 2 (Ch 5-8) RE 9.a HW8: Ch 8 Pr’s 21, 23, 27(a-c) RE 9.b bring laptop, smartphone, pad,… Practice Exam 2 (due beginning of class) RE 9.c EP8, HW9: Ch 9 Pr’s 34, 40, 43 Load Vpython Center-of-mass motion o Air Track, big & small cart bound together by spring, air blower Lab3 binary w cm.py 09_Krel.py 07_twopucks.py Wheel Baton Weeble wobble Metal man with barbell Ch. 9 Multiparticle Systems Last chapter, we zoomed way in and saw in very small systems that K+U levels were quantized. Now we’re going to zoom back out again and consider more carefully how we handle multiparticle systems – systems in which there’s obviously some net, large-scale motion, but also some inner workings. Consider a system of many particles, perhaps it is a dust cloud in interstellar space and each speck of dust is one of our “particles.” On the one hand, if we focus in, we see that each speck is naturally at a different location, has a different mass, and is moving with a different velocity. Yet, if we zoom out, we see a single cloud that behaves somewhat cohesively. If we watch the cloud for a while, as it moves through space, it makes sense to speak of the whole as having some velocity and going from some position to another – in short, we can think of it as a “particle” of its own. For that matter, if we zoom way in, each of our dust “particles” are themselves made of several, much tinier particles. How do we reasonably do this – treat a composite of several particles, each more-or-less doing their own thing, as a single object and yet properly account for the internal workings? That’s what this chapter is about. We start by considering the motion and momentum of a multi-particle system, and then we look at the work and energy. 9.1 Momentum and Center of Mass. The motion of a multparticle system The chapter opens with an every-day yet rich example: a person jumping. The flexing of the system is quite obvious, and somehow that is responsible for you flying off the ground. The momentum principle for multiparticle systems. Physics 231 2013 2 Ch 9 Day 1 Back in Chapter 3, we developed the momentum principle for multiparticle dptotal systems. It quite resembled that for the single particle: Fnet,ext dt where ptotal p1 p 2 p3 p 4 ... m1v1 m2 v 2 m3 v3 m4 v 4 ... . Looking at the right hand side, if you recall, all of the internal forces, between the particles of the system, cancel thanks to the reciprocity principle (Newton’s 3rd Law: F1 2 F2 1 ), which leaves only the external forces. This will be our starting point. So, by analogy to the momentum principle for a single particle, dp1 Fnet 1 , ptotal plays the role of the momentum of the single, composite dt system. So then, what’s a representative velocity for the whole system? A velocity such that p system m systemv system . It seems a no-brainer that m system v system M total p system msystem m1 m2 m3 m4 ... so then m1v1 m2 v2 m3 v3 m4 v4 ... m1 m2 m3 m4 ... m1v1 m2 v 2 m3 v3 M total m4 v 4 ... Weighted Average: Mathematically, this is the velocity averaged over mass. Demo: Lab_3 binary wo cm.py You may remember writing code to simulate two stars gravitationally interacting. They went looping through space weaving around each other. Yet, if we call the two stars together our “system”, since there is not net external force on the system, the average velocity of its members must be unchanging – just a constant. What representative point is moving like that? 7.1.1 Center of Mass We call the representative point in a complex system the “Center of Mass.” dr You’ll see why. So, where is it? It’s the point that satisfies v . dt m1 r1 m2 r2 m3 r3 m4 r4 ... m1 r1 m2 r2 m3 r3 m4 r4 ... rCM m1 m2 m3 m4 ... M total for if you take its derivative, you get back our relation for the system’s velocity. Demo: Lab_3 binary w cm.py o Note that the center of mass is just a mathematically defined point – it’s not a fixed part of either object. Q:In this simulation, the blue star’s mass is 2e30kg and it starts out at the origin; 1.5e11,0,0 m . the yellow star’s mass is 1e30kg and it starts out at ry Initially, where’ the center of mass? CW Here are two masses located on the x axis: Physics 231 Ch 9 Day 1 2013 3 What is the x component of the location of the center of mass of the system consisting of both masses? m1r1 m2 r2 10 kg 0m xˆ 2kg 3m xˆ rcm 0.5mxˆ m1 m2 10 kg 2kg Three uniform-density spheres are positioned as follows: • A 5 kg sphere is centered at ‹ 11, 25, −6 › m. • A 10 kg sphere is centered at ‹ 7, −19, 11 › m. • A 12 kg sphere is centered at ‹ −9, 14, −15 › m. What is the location of the center of mass of this three-sphere system? cm = m o Center of mass of large objects o Now, in reality, two stars are themselves collections of ‘particles’ – atoms, electrons, etc. Each has its own center of mass, quite obviously at the geometric center. Let’s call the two stars “Yellow” and “Blue”, then noting which particles in our system are a part of which star, we have M b rb M y ry rCM Mb M y m1b r1b m2b r2b m3b r3b ... m1r r1 y m2 y r2 y m3 y r3 y ... m1b m2b m3b ... m1 y m2 y m3 y ... o So, the relation scales: you can sum over fundamental particles, or over more convenient “objects.” 8.X.2. Now for a composite system. Say we make a T of two meter sticks. Taking the origin to be at the base of the T, where’s the system’s center of mass? First, where’s the center of mass of the vertical stick? Where’s the center of mass of the horizontal stick? m1 r1 m2 r2 Now apply rcm where both objects have the same mass. m1 m 2 Center of Mass = “Center of Gravity”=“Particle” location. If you’re still feeling a bit like: “nice definition, but what does it mean / what good is it”, consider this. We have defined it such that when a net force is applied to an object, independent of where on the object it is applied, the center of mass moves Physics 231 Ch 9 Day 1 2013 4 just like a point object of the whole object’s mass. dptotal dvcm d 2 rcm Fnet M total mtotal dt dt dt 2 o Example: Tossed Baton. Toss a baton – if you trace the trajectory of any point, other than the center of mass, it follows a complicated path through space, but the center of mass follows the smooth arc you’d predict for a tossed point mass. Red = center of mass trajectory, Blue = baton end trajectory o Example: Balancing a meter stick. Gravity pulls equally on each morsel of an object. If you push up on the object’s center of mass, or in line with it, then there’s an equal amount of mass on the left and an equal amount on the right, so the object falls neither way, but balances. For this reason the center of mass is also known as the “center of gravity.” Invasion of the weeble-wobbles! Who doesn’t fall down, no matter how much they may wobble? Weebles wobble but they don’t fall down. Say you tip a weebl wobble forward, what does it do? o Wobble backward. Where must its center of mass be relative to its contact point? o Behind. Demo: metal man with bent barbell – wobbles like a weeble wobble. Motion of Center of Mass o Demo/Example: Standing Broad Jump. Here comes the calisthenics portion of the lecture. Crouch down, and then jump. What significant external forces are there on you? Gravitational force pulling you down and the force of the floor (normal and friction) pushing you up and forward (the forces associated with air pressure cancel and we’ll neglect those Ffl associated with drag.) Ffl While different parts of your body follow different, complicated trajectories (say, your arms whindmill while your legs pump in and Mtotg M g tot Mtotg out), your center of mass follows the same smooth ark that a ball would follow. Its trajectory follows from dptot Fnet.ext F fl M total g upon launch and just dt Ffl dptot F M total g during the flight. Just as would a ball of net.ext Ffl dt your mass. Mtotg Mtotg Mtotg Physics 231 Ch 9 Day 1 2013 5 A meter stick whose mass is 290 grams lies on ice. You pull at one end of the meter stick, at right angles to the stick, with a force F = 7 newtons. The ensuing motion of the meter stick is quite complicated, but what are the initial magnitude and direction of the rate of change of the momentum of the stick, dPtot/dt, when you first apply the force? Magnitude left right up down What is the magnitude of the initial acceleration of the center of the stick? 24.1m/s2 Summing up We’ve learned o How to relate a system’s velocity and position to the velocities and positions of its constituents o That this point, the center of mass, moves in response to external forces just as a point mass does, according to the momentum principle. 7.2.4 Application: Pull on two hockey pucks Consider two pucks, one with a string attached to its center and one with a string wrapped around its circumference. Both are pulled with the same force, and for the same period of time. Q: What does that imply about the changes in momentum of their centers of mass? o A: the same since p Fnet t Q: What does that imply about the motion of the center of mass? o A: The same. o Physics 231 Ch 9 Day 1 2013 6 Demo: 07_twodiscs.py o While the center of mass motion is the same, how do the two pucks behave differently? One has internal motion; it spins. Internal motion doesn’t show-up on net momentum. While the motions of the individual particles of the two pucks are very different, they have different momenta from their counterparts, these differences cancel when you sum pBottom =p over the whole disc. ptop + pbottom = 2p Look at two points at either end of an axis on the red disc, and the two similar points on either end of an axis on the blue disc, at any moment, the two on the blue disc have pTop =p + dp equal momenta to each other, while one of the points one the red disc has a little more, the other has an equal amount less, so summed together, they give the same net. pTop =p pBottom =p - dp ptop + pbottom = 2p While, for both pucks, the same force was applied for the same time, thus they experienced the same p Fnet t ; what was different about the application of force? Over different distances. Of course, E Fnet r Since the system’s momentum doesn’t speak to its internal motion, we’ll turn to our other tool for quantifying motion: the work-energy relation. Separation of Multiparticle System Energy As you recall, the work-energy relation is Wnet.ext system Fi.ext dri Es Es i Where the sum reminds us that we need to add up all the work done by each individual force acting on the system. Before we get to applying it, let’s focus on the right-handside and consider the forms of energy there are. On the one hand, Es mi c 2 K i U ij - the total energy of the system is simply the sum of the rest, i kinetic, and potential energies of all its members. On the other hand, it sure would be convenient to be able to separately talk about the translational motion of the whole system: 2 pcm 2 K cm 12 M systemvcm 2M system Physics 231 2013 7 Ch 9 Day 1 And, then the energy associated with the ‘internal’ details of the system. Es K cm mi c 2 E int ernal Ki U ij i mi c 2 Where E int ernal Ki U ij K cm i We already know how to describe this down on the sub-atomic level (electronic states), on the atomic level (vibrational states), on the molecular level (rotational states), and when all the atoms are jiggling fairly randomly (thermal energy). But sometimes the internal energy manifests itself on the macroscopic scale, with macroscopic rotation, macroscopic vibration. We’ll look at another example and see how to take these motions / energies into consideration. Internal Energy Internal Kinetic Energy The difference between the total kinetic energy in the system and that of the whole-sale motion is that associated with internal motions. K int ernal K i K cm i Section 9.8 demonstrates that 2 1 v o K int ernal 2 mi v i .rel where i .rel vi vcm , i.e., how the particle is i moving relative to the center of mass. Considering only the change in internal kinetic energy, we can take this a step or two further. Essentially the internal kinetic energy is just the sum of each particle’s kinetic energy, relative to the center of mass. 2 2 1 K system K cm K int 12 M tot vcm 2 mi vi.rel all . particles Let’s see where that comes from before we make use of it. 2 1 o Total kinetic Energy: K total . We can express the velocity 2 mi vi all . particles 1 2 = = of an individual particle of the object in terms of the velocity of the center of mass and then the point object’s velocity relative to that. vi vcm vi.rel 2 2 1 1 K total vi.rel 2vcm vi.rel vi2.rel 2 mi v cm 2 mi v cm all . particles 2 1 2 mi v cm all . particles 1 2 2 M tot vcm 1 2 1 2 mi v cm vi .rel vcm 1 2 mi vi.rel all . particles all . particles mi vi2.rel 1 2 all . particles mi vi2.rel but the middle term is a sum over the relative momenta of each particle – this just gives the momentum of the center of mass, relative to itself – oh, it’s not moving relative to itself =0. 2 2 1 . K total 12 M tot vcm 2 mi vi.rel all . particles o Rotation and Vibration Physics 231 Ch 9 Day 1 It is convenient to resolve this motion into two components – radial and tangential. 2 2 1 1 K int vi2.rad 2 mi v i .rel 2 mi v i . tan all . particles 2 1 2 mi v i . tan all . particles 2013 8 all . particles 1 2 all . particles mi vi2.rad Vibration: Radial motion would be moving toward or away from the center of mass. Take for example a diatomic molecule, the two atoms can jiggle back and forth, i.e. vibrate. 2 1 K int .vib 2 mi vi.rad all . particles Rotation: Tangential motion would be moving around the center of mass. Consider again our diatomic molecule, it can rotate. 2 1 K int .rot 2 mi vi.rad all . particles So, K int K vib K rot Of course many systems do both at the same time, but it’ll be simplest for us to consider one kind of motion at a time. I don’t really have much more to say about vibrational, but rotational motion can be rephrased a bit to make more manageable. o Rotation. Now, in the case of the puck, or say, that of this wheel. o Demo: Spin wheel. o We’ll get into much more detail about the rotational term next time. Total energy for Translating, Vibrating, Rotating system (such as a molecule). o A common place you find systems that do all three is in gas of molecules. Calling one molecule the “system”, each molecule can zip around (translate), their atoms can jiggle (vibrate), and the can spin (rotate). The total energy of such a molecule then could be broken down into terms for each kind of motion, as well as the potential energy associated with the atoms bonding. o Considering just s diatomic molecule: o Etotal K trans K rot K vib U bond m1c 2 m2 c 2 Where U bond 1 2 k sp s 2 U o Consider a system consisting of three particles: m1 = 3 kg, 1 = < 11, -8, 15 > m/s m2 = 3 kg, 2 = < -12, 11, -5 > m/s m3 = 5 kg, 3 = < -23, 36, 18 > m/s (a) What is the total momentum of this system? kg·m/s tot = (b) What is the velocity of the center of mass of this system? m/s cm = (c) What is the total kinetic energy of this system? Physics 231 Ch 9 Day 1 2013 9 Ktot = 6420 J (d) What is the translational kinetic energy of this system? Ktrans = 2910 J (e) What is the kinetic energy of this system relative to the center of mass? Krel = 3510J 7.3.1 Gravitational Potential Energy of a multi-particle system. (near the Earth’s surface) Ug m1 gy1 m2 gy2 m3 gy3 ... g m1 y1 m2 y 2 m3 y3 ... m2 y 2 m3 y3 ... gM total y cm M total o In general, as long as the object is itself small compared to the distance between it and the center of the object with which it is gravitationally interacting, you can approximate the strength of the gravitational field as constant over the whole object. M total m1 m1 m1 U GM sun ... GM sun r1 s r2 s r3 s rcm s o If the object is spherically symmetric, then inspite of gravity’s distance dependence, that dependence cancels out. o One of the nice things about momentum is that, what you see on the macroscopic scale equals the sum of what’s going on on the microscopic scale. Physically, this follows from the reciprocity principle for forces, and how forces impact momentum. Mathematically, this is a consequence of its being a vector quantity, allowing both positive and negative contributions to cancel each other. gM total m1 y1 Physics 231 201310 Ch 9 Day 1 o The individual particles, are spinning around and around the center of mass. Furthermore, the object isn’t significantly deforming, meaning that each particle stays a fixed distance from the center of mass, ri , and that all particles are moving with the exact same angular speed as each other, . So, how can we rewrite the sum relative kinetic energies in the case of rotation? 2 2 2 2 1 1 1 K int K rot mi ri 2 2 mi vi .rel 2 mi ri 2 all . particles all . particles all . particles In the case of a spinning disc or wheel, or any object with a continuous mass distribution, this sum becomes an integral. In the simple case of discrete masses, as you’ll encounter in pr.10, it’s a simple sum. o Work & Energy of Center of Mass Let’s consider the puck that spun. First let’s just think of the motion of the center of mass. Remember, the center of mass moves just like if we had a point particle experiencing the same net force as does our real system. W cm Ecm d Fnet.ext drcm d 0 d F pull drcm 1 2 2 M cm vcm .f 0 F pull drcm 1 2 2 M cm vcm .f 0 2 F pull d 12 M cm vcm .f o Work & Energy of whole Puck & String. 1 2 2 M cm vcm .i Physics 231 201311 Ch 9 Day 1 Now let’s find the work done on the real system. In this case, the point at which the external force was applied moved an additional distance, Lstring. So, an additional amount of work was done. W system Es Fnet.ext drsystem d L d 0 F pull drs 0 Fnet.ext drsystem F pull drs d L 1 2 2 M cmvcm .f 1 2 2 M cm vcm .i d d L 1 2 2 M cmvcm .f Eint 0 F pull d L 1 2 2 M cm vcm .f Eint In this case, the change in internal motion is macroscopically obvious – the disc spins, each particle is circling the center of mass. Comparing the two expressions, we find that F pull L E int The change in internal energy is the difference between the work done on the center of mass and that done on the whole system. RW 20a: Ice Skater pushing off wall & work. In RW 20a, you were asked to consider the work and energy associated with an ice skater pushing off from a wall. There were some subtleties. We’ll look at this system from a few different angles. And hopefully illuminate and explain the subtle points. Momentum and Center of Mass Motion o The force of the wall pushes back on her for some time, accordingly, dptotal Fnet.ext Fwall F floor mg Fwall dt o ptotal M tot vcm dvcm M tot Fwall dt o her center of mass moves just as would a ball of her mass. Work & Energy – Point mass vs. whole object o System = Center of Mass. A net external force is applied, Fwall and the center of mass is displaced rcm . As we’ve said, the center of mass’s Fwall motion is exactly that of a single particle of the full object’s mass, so, how much work would get done to this replacement particle, and what would the resulting change c.m in its kinetic energy? Eint Physics 231 201312 Ch 9 Day 1 W d cm E cm Fnet.ext drcm d 0 Fwall drcm 1 2 2 M cm vcm .f 1 2 2 M cm vcm .i 0 Fwall drcm 1 2 2 M cm vcm .f Fwall Fwall This is indeed what the center of mass of the person is doing; in many cases, you only care about the center of mass – for example, if you want to know how long it will take the skater to glide across the ice rink, you don’t really care what she’s doing with her arms or head – just the motion of the center of mass. o System =Whole Skater This isn’t the whole story though, and sometimes you do want to think about what’s going on within the object. Let’s consider the whole person. Now the point of application of the external force is her hand, and her hand isn’t moving off the wall as long as the force is being applied, so what does that say about the work done by the wall on her? 0. Then how is it that she goes zipping across the ice if no external work is done, and what role does the wall play? Where does her kinetic energy come from? For the first question, here’s a hint – which party has to eat three square meals a day – the wall or her, of course it’s her. She’s the one who takes in fuel. E skater K Eint ernal 0 K 1 2 M tot v Eint ernal 2 cm. f Eint ernal She transformed internal energy into translational kinetic energy. The internal energy probably changed in a few different ways: chemical bonds in her stored fuel were changed – parts of her deformed (muscles bulged, arms moved), and if she does this enough, she may notice that she’s warming up – changes in thermal energy. What roll does the wall play? Not to undersell the wall; certainly, if it weren’t there for her to push against, there’d be no change in motion – she’d just stand there on the ice thrusting her arms. The wall provides a “restraining” force – it prevents her from pushing her hands forward, and in turn allows her to drive her center of mass backward. o Similar situations: Other instances of “restraining” forces doing no work, but influencing motion are the friction of tire on road, tension in a pendulum’s string, and the track of a rollercoaster.