Phys 331: Ch 9,.6-.7 Noninertial Frames: Centrifugal and Corriolis forces 1. And and

Similar documents
is needed and this can be established by multiplying A, obtained in step 3, by, resulting V = A x y =. = x, located in 1 st quadrant rotated about 2

Work, Energy, and Power. AP Physics C

Electric potential energy Electrostatic force does work on a particle : Potential energy (: i initial state f : final state):

element k Using FEM to Solve Truss Problems

Announcements Candidates Visiting Next Monday 11 12:20 Class 4pm Research Talk Opportunity to learn a little about what physicists do

LEAP FROG TECHNIQUE. Operational Simulation of LC Ladder Filters ECEN 622 (ESS) TAMU-AMSC

Summary chapter 4. Electric field s can distort charge distributions in atoms and molecules by stretching and rotating:

Derivation of the Differential Forms of the Conservation Laws Momentum

Physics 201 Lecture 4

1. A body will remain in a state of rest, or of uniform motion in a straight line unless it

2/4/2012. τ = Reasoning Strategy 1. Select the object to which the equations for equilibrium are to be applied. Ch 9. Rotational Dynamics

Chapter I Matrices, Vectors, & Vector Calculus 1-1, 1-9, 1-10, 1-11, 1-17, 1-18, 1-25, 1-27, 1-36, 1-37, 1-41.

Chapter 4 Motion in Two and Three Dimensions

Physics 11b Lecture #2. Electric Field Electric Flux Gauss s Law

24-2: Electric Potential Energy. 24-1: What is physics

Optimization of the Electron Gun with a Permanent Ion Trap

Section 3: Detailed Solutions of Word Problems Unit 1: Solving Word Problems by Modeling with Formulas

hitt Phy2049: Magnetism 6/10/2011 Magnetic Field Units Force Between Two Parallel Currents Force Between Two Anti-Parallel Currents

Review of Vector Algebra and Vector Calculus Operations

Here is instructions on how to use the simulation program.(the first simulation is used in question 5)

Wp/Lmin. Wn/Lmin 2.5V

_J _J J J J J J J J _. 7 particles in the blue state; 3 particles in the red state: 720 configurations _J J J _J J J J J J J J _

Spring 2002 Lecture #17

Summary 7. ELECTROMAGNETIC JOINT. ROTATING MAGNETIC FIELD. SPACE-PHASOR THEORY... 2

2/24/2014. The point mass. Impulse for a single collision The impulse of a force is a vector. The Center of Mass. System of particles

WYSE Academic Challenge Sectional Mathematics 2006 Solution Set

f = µ mg = kg 9.8m/s = 15.7N. Since this is more than the applied

5.1 Moment of a Force Scalar Formation

Scalars and Vectors Scalar

PHY126 Summer Session I, 2008

V. Principles of Irreversible Thermodynamics. s = S - S 0 (7.3) s = = - g i, k. "Flux": = da i. "Force": = -Â g a ik k = X i. Â J i X i (7.

Set of square-integrable function 2 L : function space F

Introduction of Two Port Network Negative Feedback (Uni lateral Case) Feedback Topology Analysis of feedback applications

Electric Charge. Electric charge is quantized. Electric charge is conserved

AP Physics Kinematic Wrap Up

COLLEGE OF FOUNDATION AND GENERAL STUDIES PUTRAJAYA CAMPUS FINAL EXAMINATION TRIMESTER /2017

Physics 207 Lecture 16

Example 11: The man shown in Figure (a) pulls on the cord with a force of 70

BME 5742 Biosystems Modeling and Control

PHYS 705: Classical Mechanics. Derivation of Lagrange Equations from D Alembert s Principle

Active Load. Reading S&S (5ed): Sec. 7.2 S&S (6ed): Sec. 8.2

Rigid Bodies: Equivalent Systems of Forces

Review for the Mid-Term Exam

WYSE Academic Challenge 2004 Sectional Physics Solution Set

ME 3600 Control Systems Frequency Domain Analysis

Lecture 2 Feedback Amplifier

A) 100 K B) 150 K C) 200 K D) 250 K E) 350 K

Magnetism. Chapter 21

UNIT10 PLANE OF REGRESSION

Chapter Fifiteen. Surfaces Revisited

A) (0.46 î ) N B) (0.17 î ) N

Rotational Kinematics. Rigid Object about a Fixed Axis Western HS AP Physics 1

Physics 2A Chapter 11 - Universal Gravitation Fall 2017

Summary of DLT method for stereo camera calibration, 3D reconstruction and robot-camera integration

Physics 1501 Lecture 19

LINEAR MOMENTUM. product of the mass m and the velocity v r of an object r r

Chapter 23: Electric Potential

Engineering Mechanics. Force resultants, Torques, Scalar Products, Equivalent Force systems

CHAPTER 24 GAUSS LAW

Chapter 7. Systems 7.1 INTRODUCTION 7.2 MATHEMATICAL MODELING OF LIQUID LEVEL SYSTEMS. Steady State Flow. A. Bazoune

PHY2053 Summer 2012 Exam 2 Solutions N F o f k

PHYS 314 HOMEWORK #3

CHAPTER GAUSS'S LAW

ME306 Dynamics, Spring HW1 Solution Key. AB, where θ is the angle between the vectors A and B, the proof

Chapter 8. Linear Momentum, Impulse, and Collisions

Remember: When an object falls due to gravity its potential energy decreases.

β A Constant-G m Biasing

EE 221 Practice Problems for the Final Exam

One-dimensional kinematics

where v means the change in velocity, and t is the

March 15. Induction and Inductance Chapter 31

Example

Equilibrium of Stress

Work, Energy, and Power

1 cos. where v v sin. Range Equations: for an object that lands at the same height at which it starts. v sin 2 i. t g. and. sin g

Hotelling s Rule. Therefore arbitrage forces P(t) = P o e rt.

-' DATE PERIOD DATE PERIOD. Midpoint and Distance Formulas Find the midpoint of each line segment with endpoints at the given coordinates.

ELECTROMAGNETIC INDUCTION PREVIOUS EAMCET BITS

CSU ATS601 Fall Other reading: Vallis 2.1, 2.2; Marshall and Plumb Ch. 6; Holton Ch. 2; Schubert Ch r or v i = v r + r (3.

ALL QUESTIONS ARE WORTH 20 POINTS. WORK OUT FIVE PROBLEMS.

ESCI 342 Atmospheric Dynamics I Lesson 3 Fundamental Forces II

Subjects of this chapter - Propagation harmonic waves - Harmonic waves on an interface (Boundary condition) - Introduction of phasor

gravity r2,1 r2 r1 by m 2,1

10/15/2013. PHY 113 C General Physics I 11 AM-12:15 PM MWF Olin 101

ECEN474/704: (Analog) VLSI Circuit Design Spring 2018

Chapter 6 : Gibbs Free Energy

Solution: (a) C 4 1 AI IC 4. (b) IBC 4

Feedback Principle :-

Section 4.2 Radians, Arc Length, and Area of a Sector

Chapter 3 Kinematics in Two Dimensions; Vectors

4. The material balances for isothermal ideal reactor models

( ) ( ) Review of Force. Review of Force. r = =... Example 1. What is the dot product for F r. Solution: Example 2 ( )

Outline. Steady Heat Transfer with Conduction and Convection. Review Steady, 1-D, Review Heat Generation. Review Heat Generation II

Fri. 10/23 (C14) Linear Dielectrics (read rest at your discretion) Mon. (C 17) , E to B; Lorentz Force Law: fields

The Gradient and Applications This unit is based on Sections 9.5 and 9.6, Chapter 9. All assigned readings and exercises are from the textbook

Physics 201, Lecture 4. Vectors and Scalars. Chapters Covered q Chapter 1: Physics and Measurement.

Electromagnetic Waves

Approach: (Equilibrium) TD analysis, i.e., conservation eqns., state equations Issues: how to deal with

Capítulo. Three Dimensions

OBJECTIVE To investigate the parallel connection of R, L, and C. 1 Electricity & Electronics Constructor EEC470

Transcription:

Phs 331: Ch 9 6-7 Nnnetal Fames: Centfual and Cls fces 1 Mn 1/5 Wed 1/7 Thus F Mn 1/6 96-7 Fctnal Fces: Centfual and Cls 98-9 Fee-Fall Cls Fucault 101- Cente f Mass & Rtatn abut a Fed As 103-4 Rtatn abut an As Ineta Tens Pncple Aes Fnal Just Ch 9 & bennn f 10 F 1/17 6 pm Fnal Eam Equpment: Glbe Ball wth cdnate aes Tuntable wth pape taped t t Ch 30-mn test HW9 101- Cente f Mass & Rtatn abut a Fed As Nn-netal Fames: Rtatn Last tme we leaned that when ne fame s tatn elatve t the the sa the Eath elatve t the fed stas then velct and acceleatn measuements made n the tw fames ae elated b V () f whee ( ) e V ) ˆ V f f ( as And Af ( ) AC ( ) whee Af Acent as ˆ as and A c S S O Of cuse Newtn s nd Law apples nl n an netal fame Fnet m S F net m A ( ) A ( ) f C

Fcttus Inetal / Fame Fce m F net F fame Ffame m Af ( ) AC ( ) Fcentpetal Fcls whee F ma m m ˆ and F cent cent as as c ma c m Tda and tmw we ll lk at sme effects f the centfual fce: and the Cls fce: 96 Centfual Fce: F cf m F m c Suppse a mass has a fed pstn n a tatn cdnate sstem wth anula velct What s the centfual fce n ths case? Label the anle between and as The daam belw wll be helpful as = sn F cf m The se f the fst css pduct s: sn It s tanent t the path f the mass (ccle) as t tates The esult s als pependcula t the anula velct: That means that the se f the centfual fce s: The centfual fce F cf s pependcula t F cf m m sn and pnts awa fm the as f tatn Suppse we ae descbn an bject nea the suface f the eath The se f the centfual fce s lae nea the equat As descbed b an bseve n the eath the dectn f the centfual fce (hw much N-S and hw much In-Out) wll depend n the lcatn We wll

use the anle fm the anula mmentum (the Nth Ple) Ths s knwn as the clattude and s 90 mnus the lattude See the daam belw adal (upwad) F cf m tanental (hntal) In the tatn fame the effectve fce n an bject that s nt mvn (elatve t the tatn fame) s: F eff F av F cf Defne as the avtatnal acceleatn that wuld be felt f thee was n tatn We wll nw use " " f the effectve avtatnal acceleatn n the tatn fame The ndvdual and effectve fces ae shwn belw adal m m" " m " " F cf F cf ad sn tanental The effectve avtatnal acceleatn n the tatn fame s ven b: Fcf " " m We can splt ths nt adal / n-ut and tanental cmpnents / nth-suth The se f the adal cmpnent s: ad ad F sn R sn (sn flp when dppn the abslute value sns snce wll eneall be lae than the centfual cntbutn) cf 3

At the ples 0 ad and at the equat t s less b: R 73 10 5 s 64 10 6 m 0034 m/s The tanental cmpnent s: tan tan F cf cs Rsn cs The anle between and the adal dectn ( ) s alwas small s (n adans): tan 1 tan ad tan ad Rsn cs A plumb lne (a bb n a stn) wll han at ths anle t n equlbum n the tatn fame The anle s laest at 45 whee: 97 Cls Fce: R 0034 m/s 98 m/s 00017 ad 01 Let s cnsde an bject mvn clse t the suface f the eath The Cls fce depends n what dectn an bject s n (velct) elatve t the anula velct Nte that what t means t nth depends n the lcatn n eath! F c v F c v F c v v F c 4

Cmbnn all f the pctues we can eplan wh hucanes tend t tate cunteclckwse n the Nthen Hemsphee If a s mvn nwad twad an aea f lw pessue t s deflected b the Cls fce n the wa shwn belw (vewed fm abve) lw pessue The ppste tatn esults n the Suthen Hemsphee Ths effect s t small t detemne the wa wate tates as t flws dwn the dan (e lke when the Smpsn s vst Austala) Bth Fces: The mtn f a fctnless puck n a hntal tatn tuntable s an nteestn eample F an netal bseve (IGNORE EARTH S ROTATION!) the puck wll smpl mve n a staht lne because thee s n net fce A (nnnetal) tatn bseve ma bseve me cmplcated mtns whch wll be eplaned b the centfual and the Cls fces Eample #: Pb 90 (backund f 94) Suppse a fctnless puck mves n a hntal tuntable tatn cunteclckwse (vewed fm abve) at an anula speed Wte dwn the equatns f mtn f the puck n the tatn sstem f the puck stats at an ntal pstn 0 wth an ntal velct v v v Ine Eath s tatn! In an netal fame thee s n net fce n the puck s 0 and we d see the puck mvn wth cnstant velct / n a staht lne m F net Hw wuld t lk t a lttle bu ddn n the tuntable? In that nnnetal fame that tates wth the tuntable Newtn s secnd law s: m F F m m cf c Takn the anula velct f the tuntable t be 00 the pstn s 0 Calculate the css pducts: ˆ ˆ det 0 0 det ˆ 0 ˆ ˆ ˆ 0 0 0 0 0 5

ˆ det 0 ˆ 0 ˆ 0 0 S the equatn f mtn ves (dvdn ut the mass): 0 0 0 the equatns f the and cmpnents ae: and A tck f slvn bth f these cupled dffeental equatns at the same tme (see Sect 7) s t defne If we add tmes the -equatn t the -equatn we et: Ths lks an awful lt lke the damped hamnc scllat (asde fm that fact f and the lack f a neatve sn n the lnea tem) S we can uess the basc fm f the slutn that we d uessed n that case Snce ths s a lnea dffeental equatn uess the slutn aula equatn: 0 e t whch ves the Ths mples that Thee s nl ne slutn f s we need a secnd slutn (the dffeental equatn s secnd de) Ths sunds a lt lke the pblem wth the ctcall damped smple hamnc scllat S we ve t a d chance that a smla slutn wll wk Yu can check that n addtn t e t te t s a slutn s the eneal slutn s: t e t C 1 C t whee the ceffcents ma be cmple The ntal cndtns ae 0 0 v v 0 and v v The fst cndtn mples that C 1 and the devatve s: s: C t t t e C1 Ct Ce 0 C C1 v v v v C1 v v and Ths ves: t e t v t v t cs t sn t v t v t 6

The eal pat f s t and the mana pat s t whch ves (Eq 97): t t v t cs t v t sn t v t sn t v t sn t Yu wll eple (cmputatnall) the behav f the mtn f dffeent ntal velctes n the hmewk (Pb 94) Hw can we bseve an effect f the Cls fce n the mtn f an bject nea the eath? (We alead cnsdeed hucanes but the ae cmplcated sstems f patcles) Ou calculatns wll be dne f the nthen hemsphee Fee Fall: We wll use the cdnate aes and wth the n n the suface f the eath at the clattude (belw n the left) Thse cdnates pnt n the same dectns as tatn cdnate aes and wth the n at the cente f the eath (belw n the ht) (nth) (east) (up) ' ' R ' R The pstn f the patcle can be wtten as R whee R s a vect fm the cente f the eath t a pnt n the suface at clattude and s the pstn elatve the pnt n the suface We ll assume that the epement takes place nea the suface f the eath s R and R R The centfual fce s appmatel: F cf m R A plumb lne wll pnt aln the bseved whch s R (as dscussed last tme) The dectn f defnes the dectn f the as We wll use the clattude as the anle between the as and the anula mmentum vect nn the slht cectn dscussed last tme Newtn s secnd law n the tatn fame ves: 7

m m m m R The anula velct n the tatn cdnate sstem s: 0 sn cs s the css pduct s: ˆ det 0 ˆ sn ˆ cs cs sn cs sn The cmpnents f the equatn f mtn ae: cs cs sn sn Suppse that an bject s dpped fm est at 0 and h As a eeth de appmatn we can dp all tems cntann s nteatn twce ves: The bject wll land ( 0 and 1 0 t and h t 0) at abut the tme: t h Ths ves: T et a fst de appmatn put the eeth de appmatn f the cmpnent f the velct n the nal equatns f mtn t et: The nteatn the equatn twce ves: 0 t sn t sn When the bject lands: 1 3 t sn 3 1 3 h 3 sn 3 h 3 sn 8

At a clattude (& lattude) f be: 45 and heht f 100 metes the eastwad deflectn wuld 3 73 10 5 ad/s 100 m 3 98 m/s sn45 00155 m 155 cm 9

T et a secnd de appmatn f the cmpnent f the acceleatn (t s e n the fst de) substtute the fst de appmatn f the cmpnent f the velct nt the nal equatn f : t sn cs T keep tems f de f we ll ne the small cectn t the cmpnent Inteate the equatn twce t et: When the bject lands: 1 6 At a clattude (& lattude) f be: h Ý 3 1 6 sn cs t 3 sn cs t 4 sn cs h sn cs 3 45 and heht f 100 metes the suthwad deflectn wuld 73 10 5 ad/s 100 m 3 98 m/s sn45 cs45 18 10 6 m 18 10 4 cm Gadents n the avtatnal feld f the eath can als cntbute t the suthwad deflectn f a falln bject but ths calculatn ves the ht de f mantude Eastwad deflectns wee measued b seveal epements between abut 1800 and 1900 The fllwn s a summa f epements fm MS Testen and H Sdak Am J Phs 68 () 19-14 (000) In the ntatn s eastwad and s suthwad The suthwad deflectn was t small t measue n the epements An altenatve appach t the calculatn abve s t thnk f the path f the bject as an bt n an netal efeence fame The tatn f the Eath has t be taken nt accunt afte detemnn the bt t fnd the path seen b a tatn bseve (The btal mtn f the Eath abut the Sun wll be nsnfcant dun ths tpe f epement) 10

The Fucault Pendulum: What s specal abut t? Fucault s wndeful dscve was the ealatn that the small effects f the Cls fce culd be eatl multpled b usn a pendulum What a wndeful da t must have been f Fucault when he ntced that the htwad deflectn f ne swn wuld nt be undne n the etun swn; the effects wuld accumulate! RH Rme Am J Phs 51 (8) 683 (1983) Thus the pendulum has the advantae that the effects [f the Cls fce] accumulate and thus the effect mves fm the dman f the t that f bsevatn Lén Fucault A ln sphecal pendulum (nt cnstaned t mve n a plane) that has a ve small ampltude f scllatn mves appmatel n a hntal plane We ll use the same cdnates that we dd f fee fall We can ne the dsplacement ( 0) and velct ( Ý 0) n the vetcal dectn F small scllatns T T m because the acceleatn f the bb s ve small and L The and cmpnents f the tensn ae pptnal t the dsplacements: T L T m L and T L T m L The equatns f mtn (wth an m facted ut) ae the same as f fee fall ecept f the addtn f the tensn: T T m m cs cs sn L L cs cs The hntal cmpnents f the Cls fce when a pendulum s swnn n the Nthen hemsphee ae shwn belw (use the cmpnent equatns abve and ecall the dscussn f hucanes esteda) 11

v F c Defne the natual fequenc f the pendulum L and the cmpnent f the eath s anula mmentum cs t et: The slutn can be fund b defnn the cmple functn Add the fst equatn and tmes the secnd ne t et: 0 0 0 0 Guess that the slutn wll be f the fm e t whch ves the aula equatn: 0 The slutn t the quadatc equatn s: 0 4 1 4 1 because The eneal slutn s: t C 1 e t C e t e t C 1 e t C e t s: t t t C1e Ce If we chse the ntal cndtns A 0 and v v 0 then the ntal cndtns f ae: These ae appmatel satsfed ( ) f: 0 A and 0 0 C 1 C A and C 1 C 0 1

The ceffcents ae C 1 C A and the slutn s: e t Acs t cs t sn t Acs t t t The ampltude f the scllatn s A the fequenc f the scllatn s and cs s the fequenc f the tatn f the dectn the pendulum s swn The anle between the dectn f swn and the as s t The anula speed f the eath s tatn s 360 da s at the Nth Ple ( 0) the pendulum tates nce a da At a lattude f 4 (clattude 48 ): cs48 3 360 da 40 da 10 hu Dem: Pendulum n a tatn platfm The pendulum cntnues t spn n the same as n an netal fame but chanes dectn n the tatn fame The anal wth the Fucault pendulum s nt pefect because the ate f chane n the dectn f the pendulum s swn (n the tatn fame) des nt depend n ts lcatn n the tuntable Als the fce n the pendulum s alwas n the same dectn n ths case but t chanes as a Fucault pendulum es aund the eath 13