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Intermediate Mechanics Physics 321 Richard Sonnenfeld New Mexico Tech :00 Lecture #1 of 25 Course goals Physics Concepts / Mathematical Methods Class background / interests / class photo Course Motivation “Why you will learn it” Course outline (hand-out) Course “mechanics” (hand-outs) Basic Vector Relationships Newton’s Laws Worked problems Inertia of brick and ketchup III-3,4 2 :02 Physics Concepts Classical Mechanics Study of how things move Newton’s laws Conservation laws Solutions in different reference frames (including rotating and accelerated reference frames) Lagrangian formulation (and Hamiltonian form.) Central force problems – orbital mechanics Rigid body-motion Oscillations Chaos 3 :04 Mathematical Methods Vector Calculus Differential equations of vector quantities Partial differential equations More tricks w/ cross product and dot product Stokes Theorem “Div, grad, curl and all that” Matrices Coordinate change / rotations Diagonalization / eigenvalues / principal axes Lagrangian formulation Calculus of variations “Functionals” Lagrange multipliers for constraints General Mathematical competence 4 :06 Class Background and Interests Majors Physics? EE? CS? Other? Preparation Assume Assume Assume Assume Math 231 (Vector Calc) Phys 242 (Waves) Math 335 (Diff. Eq) concurrent Phys 333 (E&M) concurrent Year at tech 2nd 3rd 4th 5th Graduate school? Greatest area of interest in mechanics 5 :08 Physics Motivation Physics component Classical mechanics is incredibly useful Applies to everything bigger than an atom and slower than about 100,000 miles/sec Lagrangian method allows “automatic” generation of correct differential equations for complex mechanical systems, in generalized coordinates, with constraints Machines and structures / Electron beams / atmospheric phenomena / stellar-planetary motions / vehicles / fluids in pipes 6 :10 Mathematics Motivation Mathematics component Hamiltonian formulation transfers DIRECTLY to quantum mechanics Matrix approaches also critical for quantum Differential equations and vector calculus completely relevant for advanced E&M and wave propagation classes Functionals, partial derivatives, vector calculus. “Real math”. Good grad-school preparation. 7 :12 About instructor 15 years post-doctoral industry experience Materials studies (tribology) for hard-drives Automated mechanical and magnetic measurements of hard-drives Bringing a 20-million unit/year product to market Likes engineering applications of physics Will endeavor to provide interesting problems that correspond to the real world 8 :16 Course “Mechanics” WebCT / Syllabus and Homework Office hours, Testing and Grading 9 :26 Vectors and Central forces r1 r2 r1 r2 Vectors r1 r2 r2 Many forces are of form F (r1 r2 ) Remove dependence of result on choice of origin r1 Origin 1 Origin 2 10 :30 Vector relationships Vectors dr dx dy dz xˆ yˆ zˆ Allow ready dt dt dt dt representation of 3 (or more!) x r r r r xˆ components at once. x Equations written in r s r s cos( ) vector notation are more compact 3 ri si i 1 11 :33 Vector Relationships -- Problem #L1-1 “The dot-product trick” Given vectors A and B which correspond to symmetry axes of a crystal: B A 2 xˆ B 3xˆ 3 yˆ 3zˆ Calculate: A, B , A Where theta is angle between A and B 12 :38 Vector relationships II – Cross product q r s r s sin( ) xˆ r s det rx sx qi yˆ ry sy zˆ rz s z 3 r s j , k 1 j k ijk ijk 0 For any two indices equal ijk 1 I,j,k even permutation of 1,2,3 ijk 1 I,j,k odd permutation of 1,2,3 Determinant Is a convenient formalism to remember the signs in the cross-product Levi-Civita Density (epsilon) Is a fancy notation worth noting for future reference (and means the same thing) 13 Newton’s Laws 1. A Body at rest remains at rest, while a body in motion at constant velocity remains in motion Unless acted on by an external force dP F dt 2. The rate of change of momentum is directly proportional to the applied force. F12 F21 3. Two bodies exert equal and opposite dP1 dP2 dt dt <--- Using 2 and 3 Together forces on each other 14 :42 Newton’s Laws imply momentum conservation dP1 dP2 dt dt d P1 P2 0 dt P1 P2 C In absence of external force, momentum change is equal and opposite in two-body system. Regroup terms Integrate. Q.E.D. Newton’s laws are valid in all inertial (i.e. constant velocity) reference frames 15 :45 Two types of mass? Gravitational mass mG mG W= mGg Inertial mass mI g F=mIa mI a=0 a>0 “Gravitational forces and acceleration are fundamentally indistinguishable” – A.Einstein 16 :65 Momentum Conservation -- Problem #L1-2 “A car crash” James and Joan were drinking straight tequila while driving two cars of mass 1000 kg and 2000 kg with velocity vectors 30 x m/s and 10 x 60 y m/s Their vehicles collide “perfectly inelastically” (i.e. they stick together) Assume that the resultant wreck slides with velocity vector v final Friction has not had time to work yet. Calculate v final and v final 17 :55 Two types of mass -- Problem #L1-3 a-b “Galileo in an alternate universe” A cannonball (mG = 10 kg) and a golf-ball (mG = 0.1 kg) are simultaneously dropped from a 98 m tall leaning tower in Italy. g 9.8m / s 2 Neglect air-resistance How long does each ball take to hit the ground if: a) mI=mG b) mI =mG * mG 18 :65 Lecture #1 Wind-up dP .F dt First homework due in class Thursday 8/29 Office hours today 3-5 Get on WebCT 19 Lecture #2 of 25 Questions on Assignment #1? Homework Expectations /Office hours/Reserve (REA Mechanics Problem Solver, ISBN 0-87891-519-2 --$25) Cross-product problem Impulse and momentum conservation Rocket propulsion Worked problems Saturn V launch I-4 About Newton (http://scienceworld.wolfram.com/biography/Newton.html) Inclined plane 20 :02 Lecture #2 of 25 Questions on Assignment #1? Homework Expectations /Office hours/Reserve (REA Mechanics Problem Solver, ISBN 0-87891-519-2 --$25) Impulse and momentum conservation Rocket propulsion Worked problems Cross-product problem Saturn V launch I-4 21 :10 Homework Bliss GOOD NOT GOOD mr mass of cannon 2.1 103 kg m1 2.1103 mc mass of chicken 2.4 kg m2 2.4 J cannon impulse (1000 Nt s ) xˆ vic , v fc initial , final velocity of chicken impulse 1000 mcannon 2.1 10 kg ; mchicken 2.4kg 3 vi chicken , v f chicken vi , v f Brief description # of points No attempt Just started / confused Well along Nearly perfect Physics Bliss 0 4 7 9 10 22 :17 Definition of Impulse J F (t ) t J t2 F (t ) dt Impulse J is a useful concept in the study of collisions. (e.g. Balls and bats, automobiles, comets and planets) Impulse is the average force acting W F ( r ) r over a time period multiplied by r2 the time period. W F ( r ) dr r1 It may also be written as an integral Note the difference between impulse J and work W. t1 23 :22 Impulse and momentum change t2 J F (t )dt t1 J t2 t1 dp (t ) dt p (t2 ) p (t1 ) p dt Two equivalent impulses with different Fmax 1 Force (N) 0.8 0.6 0.4 0.2 0 0 2 4 6 Time (s) 8 10 Impulse is useful because it directly allows expression of momentum change. 24 :27 Impulse I -- Problem #L2-2 “A car crash” James and Joan were drinking straight tequila while driving two cars of mass 1000 kg and 2000 kg with velocity vectors 30 x m/s and 10 x 60 y m/s We may look at their collision in terms of impulse. Joan’s car applied an impulse to James’ car as follows: 3 J (13.33xˆ 40 yˆ ) 10 Nt s What is the final momentum of James’ car only? What is his final velocity? Calculate pJames final , vJames final and vJames final 25 :32 Rocket Science I F ma Not the whole story dP More general – get used to it F dt Case where m not constant d F m v dt v) (mv m vexhaust defined as " thrust" m marocket thrust Instantaneous acceleration, but rocket keeps getting lighter!! 26 :40 Rocket Science II mr v ve m In absence of other forces v m f dm dv ve Separable ODE. 0 mi m Integrate both sides mf v f ve ln Rocket velocity in mi terms of lost mass m f and exhaust velocity v f 2.3ve log10 mi 27 :45 Rocket Science III mf v f 2.3ve log10 mi If a rocket is 60% fuel, then it limits at ve If a rocket is 90% fuel, then limit at 2.3 ve At 99% fuel, limit is 4.6 ve Achieving escape velocity with Single-Stage to Orbit rocket 70000 Exhaust velocity 3350 mph Exhaust velocity 6700 mph Exhaust velocity 13400 Escape velocity 65000 60000 55000 50000 Velocity (mph) 45000 40000 35000 30000 25000 20000 15000 10000 5000 0 90 80 70 60 50 40 30 Percentage of rocket mass remaining 20 10 1 28 :50 Specific Impulse I sp Engine Type Best Chemical Best Nuclear Xenon Ion Linear Accel. EM Catapult ve g Specific Impulse 500 2,000 3,800 1,000,000 N/A In practice of rocketry, the “specific impulse” is often quoted Exhaust Velocity m/s mph 4,900 10,363 19,600 41,452 37,240 78,759 9,800,000 20,726,119 45,000,000 95,170,955 Rocket Vfinal mph 23,835 95,340 181,146 47,670,074 95,170,955 Travel to Alpha-C (yrs) 106,477 26,619 14,010 53 27 29 :45 :55 Problem T3.7 The first couple of minutes after a space shuttle launch can be described as follows: The initial mass is 2x10E6 kg, the final mass (after 2 minutes) is about 1x10E6 kg, the average exhaust speed is about 3000 m/s. If all this were taking place in outer space, with negligible gravity, what would be the shuttle’s speed at the end of this stage? What is the thrust during this same period and how does it compare with the total initial weight of the shuttle (on earth)? 30 :65 Impulse II -- Problem #L2-3 “Another car crash” James and Joan are partly recovered from their previous injuries, and haven’t learned from their experience. They are drinking Jack Daniels and not wearing seatbelts. James’ vehicle has velocity vector 30 xˆ m / s Joan’s vehicle has 30 xˆ m / s Both vehicles’ mass=M. Both people’s mass=70 kg. Solve for case of inelastic and elastic collisions of vehicles. Joan has an airbag in her vehicle. It takes her 100 millisec to reduce her velocity relative to her vehicle to zero. James stops 5 millisec after impacting the steering wheel. What impulse does each person experience? What is 31 average force for each? How many “g’s” do they feel.:75 Cross product – Problem #L2-1 q r s r s sin( ) xˆ r s det rx sx qi yˆ ry sy zˆ rz s z Calculate A) Use the determinant method where r (2,3, 4) s (4,3, 2) 3 r s j , k 1 j k ijk q r s B) Use the Levi-Civita density where ijk 0 For any two indices equal ijk 1 I,j,k even permutation of 1,2,3 ijk 1 I,j,k odd permutation of 1,2,3 r (4,3, 2) s (2,3, 4) 32 :Bonus Lecture #2 Wind-up . Impulse J t F (t )dt p t . mv exhaust defined as "thrust " . mf v f ve ln mi Got on WebCT? / Got Books? Office hours Friday 5-6 Homework problems in Taylor, Handout, Galileo problem – check the web. Second homework due in class Thursday 9/5 2 1 (Includes introducing gravity into rocket equation) 33 :72 Lecture #3 of 25 Homework #1 expectations Questions on Assignment #2 Center of Mass Defined Relation to momentum Polar and Spherical Coordinates Worked problems DVD Demonstration on momentum cons. and CM motion 34 :10 Center of Mass Center of Mass and Center of gravity happen to be equivalent For a multi-particle discrete mass-distribution N R CM N m r m r 1 N m 1 M total 1 For a continuous mass-distribution RCM rdm rdm r (r )dV r (r )dA M M dm ( r ) dV total total 35 :15 Linear Momentum and CM N RCM m r 1 M total N M RCM m r 1 N Ptotal m r Ptotal M R CM 1 Ptotal Fexternal , Ptotal M RCM M R CM 0 Fexternal M RCM 36 :20 Spherical Coordinates and Earth Radius on surface of earth is fixed, so two coordinates are sufficient to specify any point on Earth Coordinates are: Latitude (North-south) “Co-latitude” is 90 degrees minus latitude Longitude (East-West) Spherical coordinates “Phi” or “Fee” j – East-west same as longitude “Theta” – North-south, same as Colatitude is 0 at north pole, 180 at south pole, 90 at equator For third dimension, add “r” (radius) 37 :25 Cylindrical and Spherical Coordinates Table of System Area and dA Volume Elements dV Cartesian dxdy dxdydz Cylindrical (rdj )dz (rdj )drdz Spherical (r sin dj )(rd ) (r sin dj )(rd )dr Polar (rdj )dr (rdj )drdz 38 :30 Impulse II -- Problem #L2-3 “Another car crash” James and Joan are partly recovered from their previous injuries, and haven’t learned from their experience. They are drinking Jack Daniels and not wearing seatbelts. James’ vehicle has velocity vector 30 xˆ m / s Joan’s vehicle has 30 xˆ m / s Both vehicles’ mass=M. Both people’s mass=70 kg. Solve for case of inelastic and elastic collisions of vehicles. Joan has an airbag in her vehicle. It takes her 100 millisec to reduce her velocity relative to her vehicle to zero. James stops 5 millisec after impacting the steering wheel. What impulse does each person experience? What is 39 average force for each? How many “g’s” do they feel.:40 Worked Example L3-1 – Discrete masses Given m1 to m10 2 units y O2 y x O1 x m= m m = 3m Calculate 1 unit N R CM N m r m r 1 N m 1 1 M total RCM Given origin O1 For homework given O2 40 :50 Worked Example L3-2 – Continuous mass 2 km Given quarter circle with uniform mass-density and radius 2 km: j O1 r Calculate M total Write r in polar coords Write out double integral, xˆ and yˆ components in terms of r and phi. Solve integral Calculate R CM rdm r (r )dA M total M total RCM Given origin O1 dA (rdj )dr 41 :40 Lecture #3 Wind-up . Fexternal M R CM . rdm rdm . RCM M dm total Office hours Today 2:30-4, tomorrow 4-5:30. Homework problems in Taylor, Handout, Galileo problem – check the web. Second homework due in class Thursday 9/5 42 :72 Lecture #4 of 25 Questions on Assignment #2 Angular Momentum And torque And Central force Moment of Inertia DVD Demonstration on momentum cons. and CM motion 43 :10 Impulse II -- Problem #L2-3 “Another car crash” James and Joan are partly recovered from their previous injuries, and haven’t learned from their experience. They are drinking Jack Daniels and not wearing seatbelts. James’ vehicle has velocity vector 30 xˆ m / s Joan’s vehicle has 30 xˆ m / s Both vehicles’ mass=M. Both people’s mass=70 kg. Solve for case of inelastic and elastic collisions of vehicles. Joan has an airbag in her vehicle. It takes her 100 millisec to reduce her velocity relative to her vehicle to zero. James stops 5 millisec after impacting the steering wheel. What impulse does each person experience? What is 44 :20 average force for each? How many “g’s” do they feel. Worked Example L3-1 – Discrete masses Given m1 to m10 2 units y O2 y x O1 x m= m m = 3m Calculate 1 unit N R CM N m r m r 1 N m 1 1 M total RCM Given origin O1 For homework given O2 45 :30 Worked Example L3-2 – Continuous mass 2 km Given quarter circle with uniform mass-density and radius 2 km: j O1 r Calculate M total Write r in polar coords Write out double integral, xˆ and yˆ components in terms of r and phi. Solve integral Calculate R CM rdm r (r )dA M total M total RCM Given origin O1 dA (rdj )dr 46 :40 Equivalence of Linear and Angular motion equations Fext dPtotal dt Ptotal M cm vcm ext dLtotal dt Ltotal I L r p; r F vtangential r 47 :45 Moment of Inertia vs. Center of Mass For a multi-particle discrete mass-distribution N N I m r r R CM 1 m r 1 M total For a continuous mass-distribution 2 2 2 I r dm r dV r dA . R CM rdm r (r )dV r (r )dA M total M total M total 48 :45 Angular Momentum and Torque dL d Lrp (r p ) dt dt d (r p) (r p) (r p) dt p mr (r p ) (r mr ) 0 d (r p) (r p) (r F ) dt dL (r F ) dt 49 :50 Parallel Axis Theorem d Axis 2 (Parallel to axis 1) CM Axis 1 (through CM) I Parallel ICM Mtotal d 2 50 :55 Angular Momentum and Central Forces Fcentral F ( r )rˆ dL (r F ) (r F (r ) rˆ) 0 because r rˆ 0 dt L constant : Angular momentum conserved by Central forces 51 :55 Moment of Inertia worked problem 2 L 1 Calculate the moment of inertia and kinetic energy of a wire of uniform mass-density lambda, mass M, and length L. A) If rotated about axis at midpoint at angular velocity 1 B) If rotated about axis at endpoint at angular velocity 2 52 :60 Lecture #4 Wind-up 2 . I r dm . dLsystem . external dt I Parallel ICM Mtotal d 2 We are done w/ chapters 1 and 3, read all. Assignment includes Taylor /Supplement / Lecture probs Next assignment is longer – Give it more time. 53 :72 Challenge questions Given a cylinder with rho of r, what function of should rho be to maximize the moment of inertia of the body? 54 Lecture #5 of 25 Questions on Assignment #2 Angular Momentum And torque And Central force Moment of Inertia DVD Demonstration on impulse and collisions 55 Angular Momentum and Central Forces Fcentral F ( r )rˆ dL (r F ) (r F (r ) rˆ) 0 because r rˆ 0 dt L constant : Angular momentum conserved by Central forces Table of Position vectors System Cartesian Cylindrical r xxˆ yyˆ zzˆ rrˆ jjˆ zzˆ rrˆ jjˆ ˆ Spherical 56 :45 Angular Momentum and Central Forces Fcentral F ( r )rˆ dL (r F ) (r F (r ) rˆ) 0 because r rˆ 0 dt L constant : Angular momentum conserved by Central forces Table of Position vectors System Cartesian Cylindrical dr dxxˆ dyyˆ dzzˆ drrˆ rdjjˆ dzzˆ Spherical drrˆ r sin djjˆ rdˆ 57 :45 Velocity Dependent Force F F (r , r , t ) Forces are generally dependent on velocity and time as well as position 2 Fr (r ) br cr Fluid drag force can be approximated with a linear and a quadratic term b k D c kA drag coefficient b = Linear (Stokes Law, Viscous or “skin” drag) c = Quadratic drag coefficient ( Newton’s Law, Inertial or “form” drag) 58 :15 Lecture #5 of 25 Moment of inertia Retarding forces Stokes Law (viscous drag) Newton’s Law (inertial drag) Reynolds number Plausibility of Stokes law Projectile motions with viscous drag Plausibility of Newton’s Law Projectile motions with inertial drag 59 :10 Moment of inertia L5-1 R Given a solid quarter disk with uniform massdensity and radius R: j O1 r Calculate I total Write r in polar coords Write out double integral, both r and phi components Solve integral Calculate 2 I CM r 2 dm r (r )dA IO1 Given that CM is located at (2R/3, p/4 Calculate ICM 60 :10 Velocity Dependent Force F F (r , r , t ) Forces are generally dependent on velocity and time as well as position 2 Fr (r ) br cr Fluid drag force can be approximated with a linear and a quadratic term Ratio f quad flin is important drag factor b = Linear (Stokes Law, Viscous or “skin” drag) c = Quadratic drag factor ( Newton’s Law, Inertial or “form” drag) 61 :15 The Reynolds Number density viscosity R R < 10 – Linear drag 1000< R < 300,000 – Quadratic R > 300,000 – Turbulent D vD R inertial (quad ) drag viscous (linear ) drag v 62 :20 vD R The Reynolds Number II density viscosity D vD R R < 10 – Linear drag Fd 1 2 v 2 CD 1000< R < 300,000 – Quadratic R > 300,000 – Turbulent Linear Regime kD v CD 1 2 v A 2 v 1 D 2 vD Re Quadratic Regime CD kA 12 v 2 1 2 v A 2 k 63 :25 Defining Viscosity Fdrag y A u xˆ y Fdrag u A y x Two planes of Area “A” separated by gap y Top plane moves at relative velocity u xˆ u defines viscosity (“eta”) F A y 2 N s / m MKS Units of are Pascal-seconds Only CGS units (poise) are actually used 2 1 poise=0.1 N s / m 64 :30 Viscous Drag I Fdrag A u xˆ Fdrag du A xˆ dy An object moved through a fluid is surrounded by a “flow-field” (red). Fluid at the surface of the object moves along with the object. Fluid a large distance away does not move at all. We say there is a “velocity gradient” or “shear field” near the object. We are changing the momentum of the nearby fluid. This dp/dt creates a force which we call the viscous drag. 65 :35 Viscous Drag II Fdrag D u xˆ Fdrag kD u xˆ “k” is a “form-factor” which depends on the shape of the object and how that affects the gradient field of the fluid. “D” is a “characteristic length” of the object The higher the velocity of the object, the larger the velocity gradient around it. Thus drag is proportional to velocity 66 :40 Viscous Drag III – Stokes Law Fdrag D u xˆ Fdrag b r Fdrag 3p D u xˆ Form-factor k becomes 3p “D” is diameter of sphere Viscous drag on walls of sphere is responsible for retarding force. George Stokes [1819-1903] (Navier-Stokes equations/ Stokes’ theorem) 67 :45 Falling raindrops L5-2 A small raindrop falls through a cloud. It has a 10 mm radius. The density of water is 1 g/cc. The viscosity of air is 180 mPoise. a) Draw the free-body diagram. b) Quantify the force on the drop for a velocity of 10 mm/sec. c) What is the Reynolds number of this raindrop d) What should be the terminal velocity of the raindrop? Work the same problem with a 100 mm drop. 68 :50 Stokes Dynamics 69 :10 Lecture #5 Wind-up . R vD . g t . v vz v 1 e mg vt 3pD Read sections 2.1-2.4 70 :72 Class #6 of 30 Homework #2 – “HAP” Rocket with gravity Retarding forces Projectile motions with viscous drag Plausibility of Newton’s Law Projectile motions with inertial drag Homework #3 – CM of a sphere Demo – High Speed photography 71 :15 Falling raindrops I Fdrag , mg Problems: A small raindrop falls through a cloud. At time t=0 its velocity v=0. Describe it’s velocity vs. time. Raindrop is 10 mm diameter, density is 1 g/cc, viscosity of air is 180 mPoise z x 72 :18 Falling raindrops II mr m gzˆ br Assume vertical motion mz m g bz dvz b g vz dt m Fdrag mg z x b b Define u g vz u vz m m b t b u u u u (0)e m m 1) Newton 2) On z-axis 3) Rewrite in terms of v 4) Variable substitution 5) Solve by inspection 73 :23 Falling raindrops III b t m u u (0)e b mg m u g vz vz u m b b b t mg m vz u (0)e m b b b mg t mg m v vz 1 e 3p D b 3pD t mg 1 e m vz 3pD g t v vz v 1 e 1) Our solution 2) Substitute original variable 3) Apply boundary conditions 4) Expand “b” 5) Define vterminal 74 :30 Inertial Drag I An v t 2 Fdrag cr Fdrag kAn v 2 Plate with area “An” moves a distance v t through fluid with density The mass of the fluid displaced is M An v t Mass “M” must acquire a velocity “v” to move out of the way of the plate. The moving plate is causing p Mv ( An v t ) v Rearranging we get p 2 t An v 75 :35 Inertial Drag II – A sphere 2 Fdrag kAnv 1 2 An p D 4 1 k 4 p 2 2 Fdrag D v vˆ 16 Previously demonstrated “An” means “A normal to velocity” Form factor for sphere Plug ‘n’ play 76 :40 Falling raindrops redux Fdrag mg z Problems: A small raindrop falls through a cloud. At time t=0 its velocity is zero. Describe it’s velocity vs. time. Raindrop is 1000 mm diameter, density is 1 g/cc. x 77 Falling raindrops redux II mr mgzˆ cr 2 Assume vertical motion mz mg cz 2 Fdrag mg 1) Newton 2) On z-axis dvz c 2 z 3) Rewrite in terms g vz dt m of v x dvz vz 2 mg Define v g (1 2 ) 4) Rearrange terms c dt v dvz vz 2 g dt (1 2 ) v 5) Separate variables 78 :45 Falling raindrops redux III v (t ) t dvz dvz g dt gt 2 vz 2 g dt 0 0 vz (1 (1 2 ) 2) vt v dx vz u dvz v du (1 x2 ) arctanh( x) v v du 0 (1 u 2 ) v arctanh(v(t ) / v ) gt v(t ) v tanh( gt / v ) v ( t ) / v 79 :50 Tanh and sinh and cosh, oh my ix e e cos x 2 ix ix e e sin x 2 eix e ix tan x ix ix e e sin, sinh and tanh 5 sin(x) sinh(x) tanh(x) 4.5 4 3.5 3 f(x) ix 2.5 2 1.5 x e e cosh x cos ix 2 x x e e tanh x tan ix x x e e x 1 0.5 0 0 50 100 150 200 x [degrees] 250 300 80 :55 350 L6-1 Liquid water drops on Io A small water droplet is traveling sideways on a cloud on Jupiter’s moon Io. At time t=0 its velocity is purely horizontal v 3xˆ m / s 2 0 Io is tiny. Neglect gravity. Draw forces on the droplet. Write down differential equation for velocity assuming Stokes-law force. Solve to get raindrop’s velocity vs. time. Calculate the “decay time” for droplet velocity assuming R=100 um and for Io’s atmosphere =20 m-poise. 81 :60 Falling raindrops L6-2 A small raindrop falls through a cloud. It has a 1000 mm radius. The density of water is 1 g/cc. The viscosity of air is 180 mPoise. The density of air is 1.3 g/liter at STP. a) Draw the free-body diagram. b) What should be the terminal velocity of the raindrop, using quadratic drag? c) What should be the terminal velocity of the raindrop, using linear drag? d) Which of the previous of two answers should we use?? e) What is the Reynolds number of this 82 raindrop? :70 Integration in Different Coordinates Table of Position vectors System Cartesian r xxˆ yyˆ zzˆ r cos(j ) xˆ r sin(j ) yˆ zzˆ r cos(j ) sin( ) xˆ r sin(j ) sin( ) yˆ r cos( ) zˆ Cylindrical Spherical 83 :45 Lecture #6 Wind-up . Fdrag 3p D u x ˆ Linear Drag g t v vz v 1 e p 2 2 Fdrag D v vˆ Quadratic Drag 16 v(t ) v tanh( gt / v ) Read sections 2.4-2.7 Matlab Basics 84 :72 Class #7 of 30 Integration of vector quantities CM problems revisited Energy and Conservative forces Stokes Theorem and Gauss’s Theorem Line Integrals Curl Work done by a force over a path Angular momentum demo 85 :03 Test #1 of 4 Thurs. 9/26/02 – in class Four problems Bring an index card 3”x5”. Use both sides. Write anything you want that will help. All calculations to be written out and numbers plugged in BEFORE solving with a calculator. Full credit requires a units check. Linear and Angular momentum / Impulse Moment of Inertia / Center of Mass Retarding forces (Stokes / Newton etc.) Conservative forces / Line integrals / curls / energy conservation 86 :08 Integration in Different Coordinates System Cartesian Position vector r xxˆ yyˆ zzˆ Cylindrical r cos(j ) xˆ r sin(j ) yˆ zzˆ Spherical r cos(j )sin( ) xˆ r sin(j )sin( ) yˆ r cos( ) zˆ 87 :12 Worked Example L7-1 – Continuous mass Given hemisphere with uniform mass-density and radius 5 m: Calculate M total Write r in polar coords Write out triple integral, x.ˆ,yˆ and zˆ components in terms of r and phi Solve integral Calculate R CM rdm r (r )dV M total M total RCM Given origin O1 dV (rd )(r sin dj )dr 88 :20 Worked Example L3-2 – Continuous mass 2 km Given quarter disk with uniform mass-density and radius 2 km: j O1 r Calculate M total Write r in polar coords Write out double integral, xˆ and yˆ components in terms of r and phi. Solve integral Calculate R CM rdm r (r )dA M total M total RCM Given origin O1 dA (rdj )dr 89 :30 Conservative Forces A force is conservative iff: F F (r ) F F (r , r , t ) P2 P1 F dr Const ( over all paths ) P2 P1 1. The force depends only on r Dependence on r , t not allowed. 2. For any two points P1, P2 the work done by the force is independent of the path taken between P1 and P2. F dr Work ( P1 P 2) r U (r ) W (r0 r ) F (r ) dr r0 90 :35 Line integral and Closed loop integral Conservative force P1 F dr Const F dr 0 P2 a) P1 and P2 with two possible integration paths. b) and c) P1 and P2 are brought closer together. d) P1 and P2 brought together to an arbitrarily small distance . Geometric argument that conservative force implies zero closed-loop path integral. 91 :40 Integration by eyeball a b Conservative force F dr 0 c d e f 92 :50 y L7-2 – Path integrals Q(0,1) Taylor 4.3 c F (r ) yxˆ xyˆ b O Calculate, along (a) Calculate, along (b) Calculate, along (c) Calculate x P(1,0) a OQP F dr (modified) Q F dr F dr F dr P P Q P Q 93 :65 Angular Momentum and Central Forces Fcentral F ( r )rˆ dL (r F ) (r F (r ) rˆ) 0 because r rˆ 0 dt L constant : Angular momentum conserved by Central forces Taylor 3-25 Given m, 0 , r0 and central force Calculate given new shorter r Fcentral F ( r )rˆ 94 :70 Lecture #7 Wind-up r . (r ) F (r) dr U r0 P1 P2 For conservative forces F dr C F dr 0 Read Chapter 4 First test 9/26 95 :72 Lecture #8 of 24 Homework #3 – Moment of Inertia of disk w/ hole Buoyancy Energy and Conservative forces Force as gradient of potential Curl Stokes Theorem and Gauss’s Theorem Line Integrals Energy conservation problems Demo – Energy Conservation 96 :02 Disk with Hole R, M d Axis 2 (Parallel to axis 1) Axis 1 (through CM) I Parallel ICM Mtotal d 2 :10 97 Buoyancy Archimedes Impure crown? King Hiero commissions ”Mission oriented Research” Taking bath Noticing water displaced by his body as he got in the tub Running starkers thru the streets shouting “Eureka” The buoyant force on an object is equal to the WEIGHT of the fluid displaced by that object Fb Vg 98 :15 Force as the gradient of potential Scalar Vector x r 0 r0 F ( x) f ( x) dx U (r ) F (r ) dr dF f ( x) dx x x F (r ) U r r xˆ yˆ zˆ x y z 1 ˆ 1 rˆ jˆ r r r sin j U Axy B sin Cz 3 99 :20 Gravitational Potential GMm U (r ) r 1 F U GMm r 1 GMm Fr GMm 2 r r r F Fj 0 xˆ yˆ zˆ x y z 1 ˆ 1 rˆ jˆ r r r sin j 100 :25 Curl as limit of tiny line-integrals F dr (curl F ) nˆ lim ai 0 ai 101 :25 Stokes and Gauss’s theorem’s Gauss – Integrating divergence over a volume is equivalent to integrating function over a surface enclosing that volume. Stokes – Integrating curl over an area is equivalent to integrating function around a path enclosing that area. 102 :30 The curl-o-meter (by Ronco®) a b Conservative force F dr 0 F 0 c xˆ e f det x f x d f yˆ y fy zˆ z f z 103 :35 Maxwell’s Equations m0 B I A Curl is zero EXCEPT Where there is a current I. E ; B 0 0 B E t E B m0 J 0 m0 t 104 :35 y L8-1 – Area integral of curl Q(0,1) Taylor 4.3 c F (r ) yxˆ xyˆ b O a x P(1,0) Calculate, along a,c Calculate, along a,b Calculate, inside a,c Calculate, inside a,b (modified) OQP F dr F dA OQP 105 :50 L8-2 Energy problems y h y O x x A block of mass “m” starts from rest and slides down a ramp of height “h” and angle “theta”. a) Calculate velocity “v” at bottom of ramp b) Do the same for a rolling disk (mass “m”, radius r) c) Do part “a” again, in the presence of friction “m” 106 :65 Retarding forces summary Fdrag 3p Duxˆ buxˆ g t vz v 1 e v Fdrag p Linear Drag on a sphere (Stokes) mg ; v b Falling from rest w/ gravity D v vˆ cv vˆ Quadratic Drag on a 16 2 2 v 2 2 mg c v(t ) v tanh( gt / v ) v(t ) v0 cv0 1 t m Sphere (Newton) Falling from rest w/ gravity Decelerating from v without gravity 107 :72 Lecture #8 Wind-up . F dr Path Surface ( F ) dA Stokes’ Theorem F 0 Read Chapter 4 9/24 Office hours 4-6 pm First test 9/26 HW posted midafternoon For conservative forces 108 :72 Lecture #9 of 24 Test advice Review problems Moment of Inertia of disk w/ hole Line Integrals Energy conservation problems Others of interest Energy Pendulum / Simple and solid 2nd derivative as a spring constant 109 :02 Assumed to know Resolving a vector into two components Drawing a free body diagram and using it to setup differential equation of motion Selecting an appropriate coordinate system and reference frame Solving a separable differential equation Applying energy conservation, momentum conservation and understanding basic laws of the trajectory of flying objects Integrating in polar / cylindrical / spherical coords. Doing line integrals :05 110 Best Advice Draw a picture!!! Check that signs and units make sense Check that magnitude of answer makes sense Try limiting conditions :08 111 Richard’s index card (front) I Parallel ICM Mtotal d r s r s cos( ) 2 3 rs i 1 i Path F dr Surface ( F ) dA i xˆ f det x f x yˆ y fy zˆ z f z xˆ yˆ zˆ x y z Impulse J F (t )dt p t1 1 ˆ 1 Conservative force F dr 0 F 0 ˆ r jˆ r r r r sin j mf U ( r ) F ( r ) dr v f ve ln t2 mi r0 mvexhaust "thrust " N R CM m r 1 M total Fexternal M R CM RCM F (r ) U rr rdm dm L r p dL (r F ) Torques and forces on an Extended object operate on its center of mass as if it were a point object. dt Ltotal ri pi i dLtotal ext dt 112 :12 Richard’s Index card (back side) Fdrag 3p Duxˆ buxˆ g t vz v 1 e v Fdrag p 16 Linear Drag on a sphere (Stokes) mg ; v b Falling from rest w/ gravity D v vˆ cv vˆ Quadratic Drag on a 2 2 1 poise 0.1N s / m2 v 2 2 mg c v(t ) v tanh( gt / v ) v(t ) v0 cv0 1 t m Sphere (Newton) Falling from rest w/ gravity Decelerating from v without gravity 113 :15 Analogy of 1-D system to roller coaster E1 E2 E3 E4 x T(x)=E-U(x) <- General 1-D system T(x)=E-mgx <- Roller Coaster 114 :20 Potential Wells K<0 mx kx x(t ) A sin k t m mx kx x(t ) Ae k m t Be K>0 k t m Mass m Spring constant k 115 :25 Taylor Series Expansion Taylor series -- Generic f ( x x0 ) f ( x0 ) f ( x0 )( x x0 ) 1 2! f ( x0 )( x x0 ) 2 ... Taylor series -- Potential 1 U ( x x0 ) U ( x0 ) U ( x0 )( x x0 ) U ( x0 )( x x0 ) 2 ... 2! Can be ignored or set to zero … “gauge invariance” 1 Is already zero for potential evaluated about a critical point x-0 Thus U ( x x0 ) U ( x0 )( x x0 ) 2 ... 2 1 keffective ( x x0 )2 2 keffective U ( x0 ) 116 :35 L9-1 Potential Wells Equivalent curvatures 30 U ( x) 20 10x x2 x3 x4 x5 20 Y 10 Evaluate keffective 0 -10 What is equation of parabola of equivalent curvature? -20 -30 -2 -1 0 X 1 2 keffective U ( x0 ) 10 0 -10 -20 -30 -40 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 117 :35 L 9-2 Potential Wells -21 x 10 Lennard-Jones potential Van Der Waals attraction (R-6) -12 Pauli repulsion (R ) 1.5 1 Energy (joules) 12 6 U (r ) 4 r r 1.05 1021 joule / molecule 0.316 nm Lennard-Jones potential for water 0.5 Calculate the Equilibrium separation of two water molecules 0 -0.5 -1 -1.5 0 0.5 1 1.5 Distance (nm) 118 :45 y L8-1 – Area integral of curl Q(0,1) Taylor 4.3 c F (r ) yxˆ xyˆ b O a x P(1,0) Calculate, along a,c Calculate, along a,b Calculate, inside a,c Calculate, inside a,b (modified) OQP F dr F dA OQP 119 :55 L8-2 Energy problems y h y O x x A block of mass “m” starts from rest and slides down a ramp of height “h” and angle “theta”. a) Calculate velocity “v” at bottom of ramp b) Do the same for a rolling disk (mass “m”, radius r) c) Do part “a” again, in the presence of friction “m” 120 :65 Lecture #9 Wind-up d2 f keffective 2 dx critical po int. Office hours today 4-6 Wed 4-5:30 Thursday Hand-written index card Official physics league 3’x5’ size 121 :72