PHYS 172: Moden Mechanics Fall 2009 Lectue 6 Fundamental Foces, Recipocity Read 3.9 3.14
Exam 1 - Tuesday Septembe 15, 8:00-10:00 PM Elliott Hall of Music 1. The exam will be witten as a standad 1 hou exam, but you will have 2 hous to complete it. 2. Absences must be excused in advance, if at all possible, by filing an Absentee Repot fom in Rm 144 PHYS. (See the syllabus.) Fo emegencies, contact you instucto by email as soon as possible. 3. Students appoved fo sepaate test envionments must contact thei instucto by email ASAP. 4. You may NOT bing equations sheets, books, etc. It is a closed book exam. A few equations and constants may be povided. DO bing pencils and a calculato (of any standad type). Gaphing Calculato is ok. 5. You must show you Pudue Student ID cad when tuning in you completed exam.
Exam 1 Announcements, cont: 6. The exam coves all assigned mateial in this couse though the end of Chapte 3, including labs, ecitations, and homewok. 7. The fist exam fom last semeste can be downloaded fom the PHYS 172 web page; click on Pevious Exams.
CLICKER QUESTION #1 Reading Question (Sections 3.9 3.14) (This is a closed-book quiz, no consulting with neighbos, etc.) Choose the coect statement: A. Newton was able to explain whee gavity comes fom. B. Newton poved that the gavitational foce fom a unifom sphee is the exactly the same as if it wee a point at the cente. C. Today s Global Positioning System (GPS) expeimentally poves the complete coectness of Newton s theoy of gavity. D. Newton s theoy of gavity is not quite coect because it leaves out the Heisenbeg Uncetainty Pinciple.
Today s Lectue Unifom Cicula Motion The Electic Foce and Recipocity Pedicting Motion When Foces Change
Unifom Cicula Motion: Have one foce that pulls towads cente causes a change in velocity even though the magnitude of velocity is constant
Unifom Cicula Motion: By definition, the speed is constant, and the velocity always points tangent to the cicula path. In what diection does the change in velocity point? The change in momentum? v f! v i y v i v f v i v f R!! R x v i =!vcos",vsin",0 v f =!vcos",!vsin",0 v f! v i = 0,!2vsin",0
This is the change in velocity going though an abitay angle 2θ centeed on the y-axis. The distance taveled is 2θR (the ac length), so the time inteval is:!t = 2"R / v. v f! v i v i v f y v i! p!t = m! v!t "2mvsin# mv2 = 0,,0 = 0," 2#R / v R sin# #,0 v f R!! R! p lim!t"0!t # d p dt = mv2 0,$ R,0 x dp dt = F ext = mv2 R adially inwad, fo unifom cicula motion.
CLICKER QUESTION #2 At time t=0, a sta is at position 1e11,0,0 m and a planet is located at 2e11,0,0 m. M sta = 1e30 kg, M planet = 5e24 kg; G=6.7e-11 Nm 2 /kg 2. What is the magnitude of the gavitational foce the sta exets on the planet? A. 5.0e32 N B. 3.4e33 N C. 3.4e22 N D. 1.7e22 N E. 0.84e22 N
CLICKER QUESTION #3 At time t=0, the planet s velocity is 100,-300,0 m/s. Conside the planet s velocity 3 hous late. Choose the most coect answe: A. The change in the planet s velocity will point towads the sta. B. The change in the planet s velocity will point away fom the sta. C. The change in the planet s velocity will have both adial and tangential components.
CLICKER QUESTION #4 At time t=0, the planet s velocity is 100,-300,0 m/s. Use the momentum pinciple to detemine the magnitude of the change in the planet s velocity 3 hous late? (Remembe M planet = 5e24 kg, F =3.4e22N ). A. 73 m/s B. 128 m/s C. 28 m/s D. 0 m/s E. 100 m/s
The Electic Foce physical objects = many atoms. atomic bonds = electic foces ( + quantum mechanics) Thus, electic foces descibe almost all eveyday foces othe than gavity!
Fo the magnitude only: F gavity = G mm R 2 Cavendish (1798) F electic = 1 Qq 4!" o R 2 Coulomb (1795)
The Gavitational Foce and the Electostatic Foce have almost identical foms: uv m m F =! G v ˆ 2 1 gav on2 by1 2 2! 1 The Electic Foce 2! 1 uv F = 1 q q ˆ 2 1 elec on2 by1 2 2 1 4 v #!" 0 2 # 1 G = 6.7x10 N m 2 kg! 11 " 2 Note that fo both thee is ecipocity: Switch 1 and 2: ˆ " ˆ =! ˆ 2! 1 1! 2 2! 1 1 1 whee is a univesal constant: = 9x10 4!" 4!" 0 0 uv F # m m $ # m m $ # m m $ %! G & " %! G & = %! G & 2 1 1 2 2 1 2 2 2 % & % 2 1 & %! 1! 2 & ' v ( '! v ( ' v 2! 1 ( on uv =! F 1 by 2 on 2 by 1 same mag., opposite diection 9 N # m 2 C 2
The Electic Foce Both gavity and the electic foce ae invese squae laws. Note, howeve, the minus sign is only thee fo gavity: Gavity is always attactive, but Electic foces can be attactive o epulsive. uv F = 1 q q ˆ 2 1 elec on2 by1 2 2 1 4 v #!" 0 2 # 1
Non-Constant Foces m m F ˆ gav =! G v 2 1 on 2by1 2 2! 1 F sping = 2! 1 k s s these foces change as the object s positions change How do we deal with foces that change? We apply momentum update (etc.) epeatedly, known as iteatively pedicting the motion (by hand o by compute). NOTE: The mathematical limit of this iteative pocedue employs integal and diffeential equations fom calculus.
Iteative Pediction of Motion Fist divide the total time inteval into many smalle time intevals (so that F is oughly constant ove the inteval). Then Calculate the (vecto) foces acting on the system Update system momentum: Update position: Repeat p f = pi + Fnet! t x = x + v! t f i ave You ll do this many times in lab this semeste.
Sample code fom last week s Vpython lab, bouncingball.py: deltat = 0.01 t = 0 while t<4.0: ball.p=ball.p+deltat*fgav ball.pos = ball.pos+ball.p/ball.m*deltat t=t+deltat if ball.pos.y <-0.95: ball.p.y=-ball.p.y Note that this is fo a constant foce, due to gavity.
Iteative Pediction of Motion Constant Foce (mg) + Momentum Pinciple = Pojectile Motion (cuved path) But, can also add ai esistance = non-constant foce
Iteation: Moon Landing (Lab #3) Ship z y x S, E = S S, M Eath Moon F = F + F net Eath moon! m m "! m m " = # % G ˆ $ + # % G ˆ $ E S M S 2 S, E 2 S, M # $ # S, E $ & ' & S, M ' We assume Eath and moon don t move. Why is this okay? Thei positions and masses ae known.
Iteation: Moon Landing (Lab #3) You know momentum and position of ship at some instant t 1 : Find net foce at this instant:! m m "! m m " F t = # % G ˆ t $ + # % G ˆ t $ ( ) ( ) ( ) E S M S net 1 2 S, E 1 2 S, M 1 # S, E ( t1 ) $ # S, M ( t1 ) $ & ' & ' Update momentum: p t +! t = p t + F t! t Update position: ( ) ( ) ( ) S 1 S 1 net 1 ( ) ( ) p t t p t ( +! ) + ( ) S 1 S 1 S t1 +! t = S t1 +! t 2" ms Knowing new position, find new foce and go back to step 1.
Impotance of Choosing The Time Step Δt 470 3.8 < F < 6.4 (not vey constant!) Δt 60 5.2 < F < 5.6 (bette)
Why Not Just Use Calculus? Fist Answe: When you iteatively update motion with small time intevals, you ARE doing calculus: you e pefoming numeical integations. % = $ i! i #! xi " 0 i f x dx lim f x x ( ) ( ) f ( x)
Why Not Just Use Calculus? Second Answe: We can always do " ( ) but we can t usually evaluate i f x! x i! i by compute, f x dx ( ) by hand. The thee-body gavitational poblem: with only 3 bodies (!!), no exact solution.
Deteminism and Chaos We neve know initial conditions exactly. Two vey simila initial values give wildly diffeent behavios: chaos.
A Famous Non-linea System dx dt dy dt dz dt =! y $ x ( ) = x " $ z $ y ( ) = xy $ # z The Loenz attacto: A simple (!?) weathe simulation using deteministic, non-linea diffeential equations. It s a chaotic system with some sot of stuctue (a factal).