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Fall 2009 PHYS 172: Modern Mechanics

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Fall 2009 PHYS 172: Modern Mechanics PHYS 172: Modern Mechanics Fall 2009 Lecture 16 - Multiparticle Systems; Friction in Depth Read 8.6 Exam #2: Multiple Choice: average was 56/70 Handwritten: average was 15/30 Total average was 71/100 Remember, we are using an absolute grading scale where the values for each grade are listed on the syllabus Typically, exam grades (500/820 points) are lower than homework, recitation, clicker, & labs grades Clicker Question #1 You get in a parked car and start driving. Assume that you travel a distance D in 10 seconds, with a constant acceleration a. Define the system to be the car plus the driver (you), whose total mass is M and acceleration a. Choose the correct statement for this system: A. The final total kinetic energy is Ktotal= MaD. B. The final translational kinetic energy is Ktrans= MaD. C. The total energy of the system changes by +MaD. D. The total energy of the system changes by -MaD.. E. The total energy of the system does not change. Clicker Question Discussion You get in a parked car and start driving. Assume that you travel a distance D in 10 seconds, with a constant acceleration a. Define the system to be the car plus the driver (you), whose total mass is M and acceleration a. Choose the correct statement for this system: A. The final total kinetic energy is Ktotal= MaD.No, this is just the translational K. There is also relative K (wheels, etc.) B. The final translational kinetic energy is Ktrans= MaD. Correct, use point particle system; force is friction applied by the road on the tires. C. The total energy of the system changes by +MaD. No, total energy will decrease; fuel is consumed. D. The total energy of the system changes by -MaD.No, there are other energy losses to Krelative and Q. E. The total energy of the system does not change. No, heat goes out the exhaust pipe (from burning fuel)…. Further Discussion The center of mass of this system is accelerating. Hence, Macm must be equal to the sum of all external forces. Friction is pushing the car! (Does friction always oppose the motion?) r dP uv r total F net,ext =M a = The force of friction from the road on the tires is Ma . total cm dt cm Throw a Baton This result says that the center of mass of a complex system moves as if all forces r dP were applied to the entire mass at that uv r total point. F net,ext =M totalacm = dt Throw a baseball, and you see parabolic motion. What happens if you throw a baton? 1D-11 Trajectory of the Center of Mass Center of Mass Motion - Review Which puck will move faster? r dP r tot ! M ar = F dt cm net,surr Both pucks move at same speed! HOWEVER: Hand #2 has to pull the string farther: W2 > W1. Where does this energy go? Rotational energy. The bottom spool is spinning. Point Particle vs Real Systems When we treat a real system as a point particle system: •The net external force acts on the Center of Mass •The change in the point particle’s translational kinetic energy is equal to the work done by the net external force in moving the CM through a displacement Δr, ie, f r !K = F " drr trans # ext cm i •The point particle system has no internal degrees of freedom. Therefore it cannot undergo a change of its internal energy - of any sort! Point Particle vs Real Systems In contrast, a real multiparticle system has internal degrees of freedom that may undergo a change of energy when the system is subjected to exertnal forces. The net external force does Work on the system given by: f r W = F ! drr ext " ext i Note that the displacement in the above expression is not that of the CM but that through which the net force acts. If an external force does not act through a displacement, it does no work on the real system! (Friction, for example.) In general, this work appears as changes in the real system’s translational kinetic energy and its internal energy. The internal energy may appear as rotational motion, potential energy, thermal energy…(yo-yo: p. 290) Clicker Question - Review Through what distance did the force act on the Point Particle system? A) 0.03 m B) 0.04 m C) 0.07 m D) 0.08 m E) 0.10 m Clicker Question - Review Which is the energy equation for the Point Particle system? A) ΔKtrans = F*(0.07 m) B) ΔKtrans = F*(0.08 m) C) ΔKtrans + ΔKvib + ΔUspring = F*(0.07 m) D) ΔKtrans + ΔKvib + ΔUspring = F*(0.08 m) Clicker Question - Review Through what distance did the force act on the Real system? A) 0.03 m B) 0.04 m C) 0.07 m D) 0.08 m E) 0.10 m Clicker Question #2 Which is the energy equation for the Real system? A) ΔKtrans = F*(0.07 m) B) ΔKtrans = F*(0.08 m) C) ΔKtrans + ΔKvib + ΔUspring = F*(0.07 m) D) ΔKtrans + ΔKvib + ΔUspring = F*(0.08 m) Clicker Questions Summary Real system: ΔKtrans + ΔKvib + ΔUspring = F*(0.08 m) Point particle system: ΔKtrans = F*(0.07 m) We thus conclude that ΔKvib + ΔUspring = F*(0.01 m) or, using our previous notation, ΔEinternal = F*(0.01 m) The point particle technique may seem strange, but it’s very powerful. How else could we find ΔEinternal? A Paradox Exert a constant force F and move the block a distance d at constant velocity. You do an amount of work Fd, and friction does an amount of work -Fd (because f=-F). Apparently the net work done on the block is zero. However, the block got hotter. How can the block get hotter if no work was done on it? Reading Question - Sec. 8.6 Regarding the paradox: A. This paradox cannot be resolved. B. There is no paradox, because the block does not actually get hotter. C. It’s not actually possible to have the frictional force exactly equal the force pulling the block. D. Friction must somehow not be doing the same amount of work as the force pulling the block. E. None of the above. How can the block get hotter if no work was done on it? Analysis of Point Particle System Point Particle System: All external forces are applied to the center of mass; correctly describes the motion of the center of mass. !Ktranslational = (F " f )d Since v=constant, no change in K, so: r r F = f Can this tell us anything about internal motions of the system? Analysis of the Real System !Ktrans + !Ethermal = Fd + W friction For simplicity, assume Q=0. !Ethermal = Fd + W friction Since ΔK=0. Since this is the real system, not the point particle system, forces do not all act at the same point, and they do not have to act over the same distance: !Ethermal = Fd " fdeffective # 0 We must have F=-f, but not d=deff! A Model for Friction Actual work done by friction forces: KEY ISSUE - The point of contact where the friction force is applied moves a shorter distance than the distance the block moves. This means friction does less work than expected. A More Realistic Model Instead of teeth stuck on each other, let them drag on opposite surfaces. Each then applies force F/2. When the top block moves by d, one tooth moves by d, while the other one just bends, meaning no displacement at the contact. Only one tooth does work across the distance d, with a force F/2, so only half of the expected amount of work is done. 2 Identical Blocks Consider 2 identical blocks as shown. Lower block is fixed. Total work done on the system (2 blocks) is just Fd (top block displacement d). Divide the increased thermal energy between the two blocks: !Eth,1 + !Eth,2 = Fd Fd !E = = Fd " Fd th,1 2 eff Hence, deff = d / 2 In this simple model, friction only does half the expected work. The remaining work done by the external force goes into thermal energy..
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