Lect11

# Lect11 - PHYS 172 Modern Mechanics Lecture 11 – The...

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Unformatted text preview: PHYS 172: Modern Mechanics Lecture 11 – The Energy Principle Fall 2011 1 Read 6.8 – 6.14 TODAY •  Multiparticle Systems and Potential Energy •  Relationship of Force and Potential Energy •  Energy Graphs 2 ANNOUNCEMENTS •  October Break – 10/10, 10/11! •  No lectures that week, no labs. •  You do have recitation! •  Exam 2 is Tuesday, Oct. 18, 8-10 PM! 3 Last Time: Single Particle System Energy principle (single particle system): !Esingle particle system = W + Q where energy is Esingle particle system = ! mc 2 = and work is Won particle = 0 for now mc 2 1" v / c 2 2 ! ! = Fnet on particle i! r 4 How do we generalize these results to multiparticle systems? Example: Energy in 2-Particle System ! F2,surr system ! f1,2 1 ! F1,surr ! f 2,1 !Esystem = W + Q 2 =0 Let s find the energy change of each particle: ! !! ! ! E1 = f1,2 i! r1 + F1,surr i! r1 = W1,internal + W1,surr ! !! ! ! E2 = f 2,1 i! r2 + F2,surr i! r2 = W2,internal + W2,surr We re counting the work done by internal forces. 5 Example: Energy in 2-Particle System ! F2,surr ! f1,2 1 ! F1,surr ! f 2,1 2 Thus ! (E1 + E 2 ) = Wint + Wsurr + Q where Wint = W1,internal + W2,internal and Wsurr = W1,surr + W2,surr Put system on left side, surroundings on right side: ! (E1 + E 2 ) " Wint = Wsurr + Q Now define the change in potential energy as ΔU ≡ – Wint : ! (E1 + E 2 ) + !U = Wsurr + Q 6 Potential Energy (in 2-Particle System) ! F2,surr ! f1,2 ! f 2,1 2 1 ! F1,surr ! !! ! !U = "Wint = " f1,2 i! r1 " f 2,1 i! r2 The potential energy U represents a sum of interaction energies between all pairs of particles inside the system. NOTE: U is defined to take into account both terms above. ! ! Question: If f1,2 + f 2,1 = 0 , why isn t ΔU = 0 here? Answer: Usually, ! ! r1 " !.r2 ΔU is related to a system changing shape. 7 Energy of a Multiparticle System sum of single sum of interaction Energy of system = + particle energies energies of all pairs Esys = (m1c 2 + m2 c 2 + …)+ (K1 + K 2 + …) + (U12 + …) We now write !Esys = Wsurr + Q Now W is about external forces only (internal forces show up in U). 8 Energy of a Multiparticle System !Esys = Wsurr + Q Wsurr is about external forces only Add a mass to the spring. How much work does the gravitational force do on the mass? What if the mass oscillates before coming to equilibrium? 9 Connection: Force and Potential Energy ! ! !! !! ! !! ! = "W!tU = "1Wi!U1 " f"22,1ii! r12" f1,,2 i! r1 " f 2,1 i! r2 !r 1 !r 2 in = " f , 2 int = = f ,Wint = " 2 1 ￿ ￿ f 1 ,2 = − f 2 ,1 ! !! !!! ! ! !Equal and opposite ! F = " f ,1 ( ) r " 2, i ! = " f 2,1 i(! r2 "2!ir1 !=2" f!1r1() r2 " ! r1 ) ! 2 !! !!!! !! = i! e = " f 2,1 i! = " f 2,1 i! r " f 2,1whr re r = r2 ! r1 r 2 ,surr ! f1,2 ! f 2,1 1 The combination is independent of coordinate system. ! F1,surr ! !U Thus f 2,1 = " ! !r dU # fr = " dr 10 Connection: Force and Potential Energy ! F2,surr ! !U hus f 2,1 = " ! 2 !r ! f1,2 ! f 2,1 1 dU # fr = " dr For Gravity: ! F1,surr !U dm1m2 U " ! # f =" 1 = Gravitation: Frr = -G !r drr 2 ! m1m2 U g = "G r To see this: 11 Gravitational Potential Energy Ug=0! m1m2 U g = !G r Yes, it really is negative! Note: U g = 0 at r = ! r=0! 12 Example: Planet and Star System: planet+star System: planet+star !Esys = Wsurr + Q = 0 !Esys = Wsurr + Q = 0 ! " E particles + U system \$ = 0 & # ! "E + U%ystem \$ = 0 & s # 2 particles %2 ! " mstar c + K star + mplanet c + K planet + U system \$ = 0 2 # ! " m c2 + K + m c + K planet + U%ystem \$ = 0 star planet s # star % Rest energiethese is constant !K star ' 0 & s constant; Each ne these & Rest eof rgies constant; !K staSo' 0 together r m K planet,f + U system, f = K planet , i + U system,ust be constant K planet,f + U system, f = K planet , i + Uisystem,i Thus, K planet + U system = const Thus, K planet + U system = const 13 Rest energies constant;Planet 0 & Example: !K star ' and Star K planet,f + U system, f = K planet , i + U system,i Energy Thus, K planet + U system = const K K+U r U 14 Application: Escape Speed What does it take to launch a rocket so it leaves the Earth's gravitational well? Minimal condition for escape: K +U = 0 Assume: kinetic energy of a planet is negligible, v<<c 2 mvesc ⎡ Mm ⎤ Ki + U i = + ⎢−G ⎥=0 2 R⎦ ⎣ 2 mvesc Mm =G 2 R vesc 2GM = R 15 Application: Escape Speed What does it take to launch a rocket so it leaves the Earth's gravitational well? Bound state: K +U < 0 Unbound state: K +U ≥ 0 16 Gravitational U Near Earth's Surface Mm U = −G r Are these the same? They are the same near the Earth's Surface. RE RE + h Mm ∆U = − G − RE + h ￿ U = mgh ￿ Mm −G RE Mm 1 = −G RE 1 + h/RE 1 ≈ 1 − x + ... 1+x Taylor Expansion ≈ −G ￿ Mm 1 − h/RE − 1 RE ￿ −1 ￿ ￿ ￿ GM =m h ≡ mgh = ∆U 2 RE g= 2 GM/RE ￿ 17 WHAT WE DID TODAY •  Multiparticle Systems and Potential Energy •  Relationship of Force and Potential Energy •  Energy Graphs 18 ...
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