Lect11 - PHYS 172 Modern Mechanics Lecture 11 – The Energy Principle Fall 2011 1 Read 6.8 – 6.14 TODAY •  Multiparticle Systems and

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

This note was uploaded on 12/07/2011 for the course PHYS 172 taught by Professor ? during the Fall '08 term at Purdue University-West Lafayette.

Ask a homework question - tutors are online