Section 35 SHM and Circular Motion


 Emmeline Ward
 4 years ago
 Views:
Transcription
1 Section 35 SHM nd Cicul Motion Phsics 204A Clss Notes Wht do objects do? nd Wh do the do it? Objects sometimes oscillte in simple hmonic motion. In the lst section we looed t mss ibting t the end of sping. We pplied the Second Lw to deie the equtions of motion fo SHM. In the pocess, we seemed to be using the ide of ngul fequenc just s we did when we looed t unifom cicul motion. In ddition, the equtions of motion fo SHM loo ee simil to the equtions of motion fo unifom cicul motion. In this section, we will inestigte the connection between SHM nd unifom cicul motion. Net, we ll continue to build ou undestnding of SHM b looing t the oscillto motion of simple pendulum. We ll discoe tht it lso is SHM unde cetin conditions. Section Outline 1. The Connection Between Unifom Cicul Motion nd SHM 2. Eneg in SHM 3. The Simple Pendulum 1. The Connection Between Unifom Cicul Motion nd SHM We he been using the ide of ngul fequenc s we did when we discussed cicul motion. Thee must be some connection, so let s inestigte. At the ight is n object going in cicle on otting tuntble. Just behind the object is sceen whee the shdow of the object cn be seen. The shdow moes bc nd foth s the object goes in cicle. The shdow ppes to be in SHM. At the ight is setch of the object in unifom cicul motion. It hs centipetl cceletion, tngentil elocit, nd position ecto ll shown. If we edw the thee ectos with thei tils t the oigin we cn imgine ll thee spinning s the object ottes. Finding the components of ech the position, elocit, nd cceletion of the shdow, = cos, = sin, nd = cos. The tngentil elocit is elted to this ngul elocit, = ω. Also, the centipetl cceletion is elted to the ngul elocit, = 2 = ( ω )2 = ω 2. Substituting, we get, = cos, = ω sin, nd = ω 2 cos. Notice, just lie SHM we he = ω 2. The ngle chnges with time. We cn wite this using the definition of ngul fequenc, ω d dt d = ωdt = ωt + δ. Now we see nothe w of looing t the phse ngle, δ, s just n integtion constnt. Finll, we cn wite the components of the, position, elocit nd cceletion fo the oscillting shdow s function of time, 351
2 Phsics 204A Clss Notes (t) = cos( ωt + δ ), (t) = ω sin( ωt + δ ), nd (t) = ω 2 cos( ωt + δ ). These e the sme s the SHM equtions of motion with A insted of. The component of the motion of n object in unifom cicul motion is SHM. Tht eplins wh we eep tling bout ngul fequencies! Emple 35.1: A 500g mss ests in equilibium t the end of hoizontl sping with sping constnt 9.80N/m. The mss is gien shp ic esulting in n initil elocit of 0.443m/s to the ight. ()Setch the initil position, elocit, nd cceletion ectos s if the object wee in cicul motion. Find (b)the loction of the equilent object in cicul motion, (c)the phse ngle, nd (d)the eqution fo (t). Gien: = 9.80N/m, m = 0.500g, (0) = 0.443m/s, nd (0) = 0. Find: =?, =?, =?, d =?, nd (t)=? ()We e gien tht the elocit ecto is to the ight nd the initil position is zeo. Theefoe, the component of the elocit must be t mimum nd point to the ight. The  component of the position nd cceletion ectos must be zeo. The cceletion must point towd the cente of the cicle nd the position must point outwd. The nswe then is in the setch t the uppe ight. (b)the equilent object in cicul motion with the ectos pointing the ight diection must be s shown t the ight. (c)looing t the cicle, the phse ngle must be 270 o 3π 2. (d)using the ppopite eqution of motion fo cicul motion, the eqution fo the position s function of time is (t) = o sin(ωt + δ ). The ngul fequenc fo sping is, ω = m = = 4.43 d s. So, (t) = 0.443sin(4.43t + 3π ). 2 Note this esults in (0) = m/s s equied. 352
3 Phsics 204A Clss Notes 2. Eneg in SHM We he led looed t eneg in SHM. The esult ws tht the potentil eneg stoed in sping ws gien b, U s = So, let s loo t n oscillting mss t the end of sping. In the top imge t the ight the mss is t est t the equilibium position of the sping. In the middle, imge the mss hs been pulled to the ight distnce A. The sstem hs totl eneg equl to the potentil eneg in the sping, E o = U s = 1 2 A2. The lowe imge is fte the mss hs been elesed nd it is A heding bc to the left. It hs speed when it eches the position. Thee is still some potentil eneg in the sping plus some inetic eneg, E = K + U s = 1 2 m Appling the Lw of Consetion of Eneg, E o = E 1 2 A2 = 1 2 m , nd soling fo the speed, Recll fo sping, ω = m = ± m (A2 2 )., so the speed cn be witten s, = ±ω (A 2 2 ). This is the eqution of motion fo the speed s function of position. Elie we found this eqution b ppling the Second Lw nd the definitions of elocit nd cceletion. Now, we see it is just n epession of the Lw of Consetion of Eneg. 3. The Simple Pendulum A mss t the end of sting cn cetinl oscillte. The question is, is it SHM. Recll tht the condition fo SHM is, () = ω 2, whee ω is constnt. Fo mss t the end of sping, Newton s Second Lw ge us, () =. m Fom this eqution we found deduced tht the motion ws SHM with n ngul fequenc equl to the oot of the constnts on the ight hnd side, ω = m. If we ppl the Second Lw to othe sstems nd find tht the cceletion is equl to the negtie of some constnts multiplied b the position, then we cn follow the sme logic to deduce tht the motion will be SHM with n ngul fequenc equl to the oot of the constnts. Let s loo t some oscillto sstems nd see if the e, in fct, SHM. 353
4 Phsics 204A Clss Notes A simple pendulum consists of e light sting with concentted mss t the end. The foces on the mss t the end e git nd the tension. Howee, the tension eets no toque bout the top of the sting. Appling the Second Lw fo Rottion, Στ p = Iα mg sin = m 2 α α = g sin. Since we e looing t ottionl motion, we e checing to see if the ngul cceletion is equl to the negtie of some constnts multiplied b the ngul position. Sdl, this is not the cse fo the simple pendulum becuse we he sin insted of just. Howee, fo smll ngles, sin α g. This is the SHM eqution, the cceletion (ngul, in this cse) is equl to minus some constnts times the position (gin, ngul). This is the sme eqution we got fo the motion of the mss on the end of sping, ecept tht eplces. In othe wods, the equtions of motion fo the ngle,, will be the simple hmonic motion equtions with n ngul fequenc equl to the oot of the constnts so long s the ngle is smll. P F t F g Angul Fequenc of Simple Pendulum ω = g Emple 35.2: The pendulum in gndfthe cloc must he peiod of 2.00s so tht ech swing moes the second hnd twice. Find the length of the pendulum. Gien: T = 2.00s Find: =? The ngul fequenc of simple pendulum is, It is elted to the peiod, Plugging in the numbes, ω = g. ω = 2πf = 2π T T = 2π ω T = 2π g = gt2 4π 2. = (9.80)(2.00)2 4π 2 = 0.993m. This eplins wh ll pendulums in gndfthe clocs e bout this size. 354
5 Phsics 204A Clss Notes Section Summ Wh do objects do wht the do? We he been building ou undestnding of simple hmonic motion. We he lened tht n object with n cceletion tht is equl to minus the poduct of some constnt nd the position is in SHM n obes the SHM equtions of motion, () = ω 2 () = ±ω A 2 2 (t) = A cos( ωt + δ) (t) = ωasin( ωt + δ) (t) = ω 2 Acos ωt + δ ( ) whee A is the mplitude of the motion, ω is the ngul fequenc, nd δ is the phse ngle. The ngul fequenc will be equl to the oot of the constnts. We emined the connection between cicul motion nd SHM. SHM is the motion of the shdow of n object in unifom cicul motion. In othe wods, the equtions of motion fo the component of unifom cicul motion e identicl to the equtions of motion fo SHM. With the nowledge boe, we loo t the oscilltions of simple pendulum nd found tht the e indeed SHM with n ngul fequenc gien b, ω = g. 355
Chapter 4 Kinematics in Two Dimensions
D Kinemtic Quntities Position nd Velocit Acceletion Applictions Pojectile Motion Motion in Cicle Unifom Cicul Motion Chpte 4 Kinemtics in Two Dimensions D Motion Pemble In this chpte, we ll tnsplnt the
More informationChapter 4 TwoDimensional Motion
D Kinemtic Quntities Position nd Velocit Acceletion Applictions Pojectile Motion Motion in Cicle Unifom Cicul Motion Chpte 4 TwoDimensionl Motion D Motion Pemble In this chpte, we ll tnsplnt the conceptul
More informationPhysics 111. Uniform circular motion. Ch 6. v = constant. v constant. Wednesday, 89 pm in NSC 128/119 Sunday, 6:308 pm in CCLIR 468
ics Announcements dy, embe 28, 2004 Ch 6: Cicul Motion  centipetl cceletion Fiction Tension  the mssless sting Help this week: Wednesdy, 89 pm in NSC 128/119 Sundy, 6:308 pm in CCLIR 468 Announcements
More information(a) CounterClockwise (b) Clockwise ()N (c) No rotation (d) Not enough information
m m m00 kg dult, m0 kg bby. he seesw stts fom est. Which diection will it ottes? ( CounteClockwise (b Clockwise ( (c o ottion ti (d ot enough infomtion Effect of Constnt et oque.3 A constnt nonzeo toque
More information( ) ( ) ( ) ( ) ( ) # B x ( ˆ i ) ( ) # B y ( ˆ j ) ( ) # B y ("ˆ ( ) ( ) ( (( ) # ("ˆ ( ) ( ) ( ) # B ˆ z ( k )
Emple 1: A positie chge with elocit is moing though unifom mgnetic field s shown in the figues below. Use the ighthnd ule to detemine the diection of the mgnetic foce on the chge. Emple 1 ˆ i = ˆ ˆ i
More information10 m, so the distance from the Sun to the Moon during a solar eclipse is. The mass of the Sun, Earth, and Moon are = =
Chpte 1 nivesl Gvittion 11 *P1. () The unth distnce is 1.4 nd the thmoon 8 distnce is.84, so the distnce fom the un to the Moon duing sol eclipse is 11 8 11 1.4.84 = 1.4 The mss of the un, th, nd Moon
More information1 Using Integration to Find Arc Lengths and Surface Areas
Novembe 9, 8 MAT86 Week Justin Ko Using Integtion to Find Ac Lengths nd Sufce Aes. Ac Length Fomul: If f () is continuous on [, b], then the c length of the cuve = f() on the intevl [, b] is given b s
More informationMark Scheme (Results) January 2008
Mk Scheme (Results) Jnuy 00 GCE GCE Mthemtics (6679/0) Edecel Limited. Registeed in Englnd nd Wles No. 4496750 Registeed Office: One90 High Holbon, London WCV 7BH Jnuy 00 6679 Mechnics M Mk Scheme Question
More informationFluids & Bernoulli s Equation. Group Problems 9
Goup Poblems 9 Fluids & Benoulli s Eqution Nme This is moe tutoillike thn poblem nd leds you though conceptul development of Benoulli s eqution using the ides of Newton s 2 nd lw nd enegy. You e going
More informationSolutions to Midterm Physics 201
Solutions to Midtem Physics. We cn conside this sitution s supeposition of unifomly chged sphee of chge density ρ nd dius R, nd second unifomly chged sphee of chge density ρ nd dius R t the position of
More informationr a + r b a + ( r b + r c)
AP Phsics C Unit 2 2.1 Nme Vectos Vectos e used to epesent quntities tht e chcteized b mgnitude ( numeicl vlue with ppopite units) nd diection. The usul emple is the displcement vecto. A quntit with onl
More informationAnswers to test yourself questions
Answes to test youself questions opic Descibing fields Gm Gm Gm Gm he net field t is: g ( d / ) ( 4d / ) d d Gm Gm Gm Gm Gm Gm b he net potentil t is: V d / 4d / d 4d d d V e 4 7 9 49 J kg 7 7 Gm d b E
More information1. Viscosities: μ = ρν. 2. Newton s viscosity law: 3. Infinitesimal surface force df. 4. Moment about the point o, dm
3 Fluid Mecnics Clss Emple 3: Newton s Viscosit Lw nd Se Stess 3 Fluid Mecnics Clss Emple 3: Newton s Viscosit Lw nd Se Stess Motition Gien elocit field o ppoimted elocit field, we wnt to be ble to estimte
More informationCourse Updates. Reminders: 1) Assignment #8 available. 2) Chapter 28 this week.
Couse Updtes http://www.phys.hwii.edu/~vne/phys7sp1/physics7.html Remindes: 1) Assignment #8 vilble ) Chpte 8 this week Lectue 3 iotsvt s Lw (Continued) θ d θ P R R θ R d θ d Mgnetic Fields fom long
More information( ) ( ) Physics 111. Lecture 13 (Walker: Ch ) Connected Objects Circular Motion Centripetal Acceleration Centripetal Force Sept.
Physics Lectue 3 (Wlke: Ch. 6.45) Connected Objects Cicul Motion Centipetl Acceletion Centipetl Foce Sept. 30, 009 Exmple: Connected Blocks Block of mss m slides on fictionless tbletop. It is connected
More informationEnergy Dissipation Gravitational Potential Energy Power
Lectue 4 Chpte 8 Physics I 0.8.03 negy Dissiption Gvittionl Potentil negy Powe Couse wesite: http://fculty.uml.edu/andiy_dnylov/teching/physicsi Lectue Cptue: http://echo360.uml.edu/dnylov03/physicsfll.html
More informationCHAPTER 18: ELECTRIC CHARGE AND ELECTRIC FIELD
ollege Physics Student s Mnul hpte 8 HAPTR 8: LTRI HARG AD LTRI ILD 8. STATI LTRIITY AD HARG: OSRVATIO O HARG. ommon sttic electicity involves chges nging fom nnocoulombs to micocoulombs. () How mny electons
More informationπ,π is the angle FROM a! TO b
Mth 151: 1.2 The Dot Poduct We hve scled vectos (o, multiplied vectos y el nume clled scl) nd dded vectos (in ectngul component fom). Cn we multiply vectos togethe? The nswe is YES! In fct, thee e two
More informationClass Summary. be functions and f( D) , we define the composition of f with g, denoted g f by
Clss Summy.5 Eponentil Functions.6 Invese Functions nd Logithms A function f is ule tht ssigns to ech element D ectly one element, clled f( ), in. Fo emple : function not function Given functions f, g:
More informationOSCILLATIONS AND GRAVITATION
1. SIMPLE HARMONIC MOTION Simple hamonic motion is any motion that is equivalent to a single component of unifom cicula motion. In this situation the velocity is always geatest in the middle of the motion,
More informationU>, and is negative. Electric Potential Energy
Electic Potentil Enegy Think of gvittionl potentil enegy. When the lock is moved veticlly up ginst gvity, the gvittionl foce does negtive wok (you do positive wok), nd the potentil enegy (U) inceses. When
More informationPhysics 1502: Lecture 2 Today s Agenda
1 Lectue 1 Phsics 1502: Lectue 2 Tod s Agend Announcements: Lectues posted on: www.phs.uconn.edu/~cote/ HW ssignments, solutions etc. Homewok #1: On Mstephsics this Fid Homewoks posted on Msteingphsics
More informationAQA Maths M2. Topic Questions from Papers. Circular Motion. Answers
AQA Mths M Topic Questions fom Ppes Cicul Motion Answes PhysicsAndMthsTuto.com PhysicsAndMthsTuto.com Totl 6 () T cos30 = 9.8 Resolving veticlly with two tems Coect eqution 9.8 T = cos30 T =.6 N AG 3 Coect
More informationElectric Potential. and Equipotentials
Electic Potentil nd Euipotentils U Electicl Potentil Review: W wok done y foce in going fom to long pth. l d E dl F W dl F θ Δ l d E W U U U Δ Δ l d E W U U U U potentil enegy electic potentil Potentil
More informationLecture 10. Solution of Nonlinear Equations  II
Fied point Poblems Lectue Solution o Nonline Equtions  II Given unction g : R R, vlue such tht gis clled ied point o the unction g, since is unchnged when g is pplied to it. Whees with nonline eqution
More informationELECTRO  MAGNETIC INDUCTION
NTRODUCTON LCTRO  MAGNTC NDUCTON Whenee mgnetic flu linked with cicuit chnges, n e.m.f. is induced in the cicuit. f the cicuit is closed, cuent is lso induced in it. The e.m.f. nd cuent poduced lsts s
More informationDEPARTMENT OF CIVIL AND ENVIRONMENTAL ENGINEERING FLUID MECHANICS III Solutions to Problem Sheet 3
DEPATMENT OF CIVIL AND ENVIONMENTAL ENGINEEING FLID MECHANICS III Solutions to Poblem Sheet 3 1. An tmospheic vote is moelle s combintion of viscous coe otting s soli boy with ngul velocity Ω n n iottionl
More informationRadial geodesics in Schwarzschild spacetime
Rdil geodesics in Schwzschild spcetime Spheiclly symmetic solutions to the Einstein eqution tke the fom ds dt d dθ sin θdϕ whee is constnt. We lso hve the connection components, which now tke the fom using
More informationAP Calculus AB Exam Review Sheet B  Session 1
AP Clcls AB Em Review Sheet B  Session Nme: AP 998 # Let e the nction given y e.. Find lim nd lim.. Find the solte minimm vle o. Jstiy tht yo nswe is n solte minimm. c. Wht is the nge o? d. Conside the
More informationPhysics 11b Lecture #11
Physics 11b Lectue #11 Mgnetic Fields Souces of the Mgnetic Field S&J Chpte 9, 3 Wht We Did Lst Time Mgnetic fields e simil to electic fields Only diffeence: no single mgnetic pole Loentz foce Moving chge
More informationOn the Eötvös effect
On the Eötvös effect Mugu B. Răuţ The im of this ppe is to popose new theoy bout the Eötvös effect. We develop mthemticl model which loud us bette undestnding of this effect. Fom the eqution of motion
More informationSTD: XI MATHEMATICS Total Marks: 90. I Choose the correct answer: ( 20 x 1 = 20 ) a) x = 1 b) x =2 c) x = 3 d) x = 0
STD: XI MATHEMATICS Totl Mks: 90 Time: ½ Hs I Choose the coect nswe: ( 0 = 0 ). The solution of is ) = b) = c) = d) = 0. Given tht the vlue of thid ode deteminnt is then the vlue of the deteminnt fomed
More informationSchool of Electrical and Computer Engineering, Cornell University. ECE 303: Electromagnetic Fields and Waves. Fall 2007
School of Electicl nd Compute Engineeing, Conell Univesity ECE 303: Electomgnetic Fields nd Wves Fll 007 Homewok 4 Due on Sep. 1, 007 by 5:00 PM Reding Assignments: i) Review the lectue notes. ii) Relevnt
More informationSchool of Electrical and Computer Engineering, Cornell University. ECE 303: Electromagnetic Fields and Waves. Fall 2007
School of Electicl nd Compute Engineeing, Conell Univesity ECE 303: Electomgnetic Fields nd Wves Fll 007 Homewok 3 Due on Sep. 14, 007 by 5:00 PM Reding Assignments: i) Review the lectue notes. ii) Relevnt
More informationAlgebra Based Physics. Gravitational Force. PSI Honors universal gravitation presentation Update Fall 2016.notebookNovember 10, 2016
Newton's Lw of Univesl Gvittion Gvittionl Foce lick on the topic to go to tht section Gvittionl Field lgeb sed Physics Newton's Lw of Univesl Gvittion Sufce Gvity Gvittionl Field in Spce Keple's Thid Lw
More informationHomework 3 MAE 118C Problems 2, 5, 7, 10, 14, 15, 18, 23, 30, 31 from Chapter 5, Lamarsh & Baratta. The flux for a point source is:
. Homewok 3 MAE 8C Poblems, 5, 7, 0, 4, 5, 8, 3, 30, 3 fom Chpte 5, msh & Btt Point souces emit nuetons/sec t points,,, n 3 fin the flux cuent hlf wy between one sie of the tingle (blck ot). The flux fo
More informationNARAYANA I I T / P M T A C A D E M Y. C o m m o n Pr a c t i c e T e s t 0 9 XIIC SPARK Date: PHYSICS CHEMISTRY MATHEMATICS
. (D). (B). (). (). (D). (A) 7. () 8. (B) 9. (B). (). (A). (D). (B). (). (B) NAAYANA I I T / T A A D E Y XISIIITSA (..7) o m m o n c t i c e T e s t 9 XII SA Dte:..7 ANSWE YSIS EISTY ATEATIS. (B).
More information1. The sphere P travels in a straight line with speed
1. The sphee P tels in stight line with speed = 10 m/s. Fo the instnt depicted, detemine the coesponding lues of,,,,, s mesued eltie to the fixed Oxy coodinte system. (/134) + 38.66 1.34 51.34 10sin 3.639
More informationChapter 2: Electric Field
P 6 Genel Phsics II Lectue Outline. The Definition of lectic ield. lectic ield Lines 3. The lectic ield Due to Point Chges 4. The lectic ield Due to Continuous Chge Distibutions 5. The oce on Chges in
More informationN for static friction and N
Fiction: Epeimentll the following fetues e obseed to be tue of the foce of fiction: ) Fiction lws opposes the motion. The foce is dissiptie nd its diection is pllel to the sufce of the object in motion.
More informationTopics for Review for Final Exam in Calculus 16A
Topics fo Review fo Finl Em in Clculus 16A Instucto: Zvezdelin Stnkov Contents 1. Definitions 1. Theoems nd Poblem Solving Techniques 1 3. Eecises to Review 5 4. Chet Sheet 5 1. Definitions Undestnd the
More informationGeneral Physics II. number of field lines/area. for whole surface: for continuous surface is a whole surface
Genel Physics II Chpte 3: Guss w We now wnt to quickly discuss one of the moe useful tools fo clculting the electic field, nmely Guss lw. In ode to undestnd Guss s lw, it seems we need to know the concept
More informationChapter 28 Sources of Magnetic Field
Chpte 8 Souces of Mgnetic Field  Mgnetic Field of Moving Chge  Mgnetic Field of Cuent Element  Mgnetic Field of Stight CuentCying Conducto  Foce Between Pllel Conductos  Mgnetic Field of Cicul Cuent
More informationChapter 21: Electric Charge and Electric Field
Chpte 1: Electic Chge nd Electic Field Electic Chge Ancient Gees ~ 600 BC Sttic electicit: electic chge vi fiction (see lso fig 1.1) (Attempted) pith bll demonsttion: inds of popeties objects with sme
More informationSimple Harmonic Motion I Sem
Simple Hrmonic Motion I Sem Sllus: Differentil eqution of liner SHM. Energ of prticle, potentil energ nd kinetic energ (derivtion), Composition of two rectngulr SHM s hving sme periods, Lissjous figures.
More informationSatellite Orbits. Orbital Mechanics. Circular Satellite Orbits
Obitl Mechnic tellite Obit Let u tt by king the quetion, Wht keep tellite in n obit ound eth?. Why doen t tellite go diectly towd th, nd why doen t it ecpe th? The nwe i tht thee e two min foce tht ct
More informationPicking Coordinate Axes
Picing Coodinte Axes If the object you e inteested in Is cceleting Choose one xis long the cceletion Su of Foce coponents long tht xis equls Su of Foce coponents long ny othe xis equls 0 Clcultions e esie
More informationMAGNETIC EFFECT OF CURRENT & MAGNETISM
TODUCTO MAGETC EFFECT OF CUET & MAGETM The molecul theo of mgnetism ws given b Webe nd modified lte b Ewing. Oested, in 18 obseved tht mgnetic field is ssocited with n electic cuent. ince, cuent is due
More informationWeek 8. Topic 2 Properties of Logarithms
Week 8 Topic 2 Popeties of Logithms 1 Week 8 Topic 2 Popeties of Logithms Intoduction Since the esult of ithm is n eponent, we hve mny popeties of ithms tht e elted to the popeties of eponents. They e
More informationThe Area of a Triangle
The e of Tingle tkhlid June 1, 015 1 Intodution In this tile we will e disussing the vious methods used fo detemining the e of tingle. Let [X] denote the e of X. Using se nd Height To stt off, the simplest
More informationThis immediately suggests an inversesquare law for a "piece" of current along the line.
Electomgnetic Theoy (EMT) Pof Rui, UNC Asheville, doctophys on YouTube Chpte T Notes The iotsvt Lw T nvesesque Lw fo Mgnetism Compe the mgnitude of the electic field t distnce wy fom n infinite line
More informationSPA7010U/SPA7010P: THE GALAXY. Solutions for Coursework 1. Questions distributed on: 25 January 2018.
SPA7U/SPA7P: THE GALAXY Solutions fo Cousewok Questions distibuted on: 25 Jnuy 28. Solution. Assessed question] We e told tht this is fint glxy, so essentilly we hve to ty to clssify it bsed on its spectl
More information6. Gravitation. 6.1 Newton's law of Gravitation
Gvittion / 1 6.1 Newton's lw of Gvittion 6. Gvittion Newton's lw of gvittion sttes tht evey body in this univese ttcts evey othe body with foce, which is diectly popotionl to the poduct of thei msses nd
More informationUCSD Phys 4A Intro Mechanics Winter 2016 Ch 5 Solutions
UCSD Phs 4 Into Mechanics Winte 016 Ch 5 Solutions 0. Since the uppe bloc has a highe coefficient of iction, that bloc will dag behind the lowe bloc. Thus thee will be tension in the cod, and the blocs
More informationApplied Physics Introduction to Vibrations and Waves (with a focus on elastic waves) Course Outline
Applied Physics Introduction to Vibrtions nd Wves (with focus on elstic wves) Course Outline Simple Hrmonic Motion && + ω 0 ω k /m k elstic property of the oscilltor Elstic properties of terils Stretching,
More informationAP Physics 1  Circular Motion and Gravitation Practice Test (Multiple Choice Section) Answer Section
AP Physics 1  Cicula Motion and Gaitation Pactice est (Multiple Choice Section) Answe Section MULIPLE CHOICE 1. B he centipetal foce must be fiction since, lacking any fiction, the coin would slip off.
More information1. A man pulls himself up the 15 incline by the method shown. If the combined mass of the man and cart is 100 kg, determine the acceleration of the
1. n pulls hiself up the 15 incline b the ethod shown. If the cobined ss of the n nd ct is 100 g deteine the cceletion of the ct if the n eets pull of 50 on the ope. eglect ll fiction nd the ss of the
More informationImportant design issues and engineering applications of SDOF system Frequency response Functions
Impotnt design issues nd engineeing pplictions of SDOF system Fequency esponse Functions The following desciptions show typicl questions elted to the design nd dynmic pefomnce of secondode mechnicl system
More informationChapter 8. Ch.8, Potential flow
Ch.8, Voticit (epetition) Velocit potentil Stem function Supeposition Cicultion dimensionl bodies KuttJoukovskis lift theoem Comple potentil Aismmetic potentil flow Rotting fluid element Chpte 4 Angul
More informationExample 2: ( ) 2. $ s ' 9.11" 10 *31 kg ( )( 1" 10 *10 m) ( e)
Emple 1: Two point chge e locted on the i, q 1 = e t = 0 nd q 2 = e t =.. Find the wok tht mut be done b n etenl foce to bing thid point chge q 3 = e fom infinit to = 2. b. Find the totl potentil eneg
More informationElectric Field F E. q Q R Q. ˆ 4 r r   Electric field intensity depends on the medium! origin
1 1 Electic Field + + q F Q R oigin E 0 0 F E ˆ E 4 4 R q Q R Q   Electic field intensity depends on the medium! Electic Flux Density We intoduce new vecto field D independent of medium. D E So, electic
More informationOptimization. x = 22 corresponds to local maximum by second derivative test
Optimiztion Lectue 17 discussed the exteme vlues of functions. This lectue will pply the lesson fom Lectue 17 to wod poblems. In this section, it is impotnt to emembe we e in Clculus I nd e deling onevible
More informationELECTROSTATICS. 4πε0. E dr. The electric field is along the direction where the potential decreases at the maximum rate. 5. Electric Potential Energy:
LCTROSTATICS. Quntiztion of Chge: Any chged body, big o smll, hs totl chge which is n integl multile of e, i.e. = ± ne, whee n is n intege hving vlues,, etc, e is the chge of electon which is eul to.6
More informationElectricity & Magnetism Lecture 6: Electric Potential
Electicity & Mgnetism Lectue 6: Electic Potentil Tody s Concept: Electic Potenl (Defined in tems of Pth Integl of Electic Field) Electicity & Mgnesm Lectue 6, Slide Stuff you sked bout:! Explin moe why
More informationChapter 7. Kleene s Theorem. 7.1 Kleene s Theorem. The following theorem is the most important and fundamental result in the theory of FA s:
Chpte 7 Kleene s Theoem 7.1 Kleene s Theoem The following theoem is the most impotnt nd fundmentl esult in the theoy of FA s: Theoem 6 Any lnguge tht cn e defined y eithe egul expession, o finite utomt,
More informationUniform Circular Motion
Unfom Ccul Moton Unfom ccul Moton An object mong t constnt sped n ccle The ntude of the eloct emns constnt The decton of the eloct chnges contnuousl!!!! Snce cceleton s te of chnge of eloct:!! Δ Δt The
More information1.4 Using Newton s laws, show that r satisfies the differential equation 2 2
EN40: Dnmics nd Vibtions Homewok 3: Solving equtions of motion fo pticles School of Engineeing Bown Univesit. The figue shows smll mss m on igid od. The sstem stts t est with 0 nd =0, nd then the od begins
More information9.4 The response of equilibrium to temperature (continued)
9.4 The esponse of equilibium to tempetue (continued) In the lst lectue, we studied how the chemicl equilibium esponds to the vition of pessue nd tempetue. At the end, we deived the vn t off eqution: d
More informationDYNAMICS. Kinetics of Particles: Newton s Second Law VECTOR MECHANICS FOR ENGINEERS: Ninth Edition CHAPTER. Ferdinand P. Beer E. Russell Johnston, Jr.
Ninth E CHPTER VECTOR MECHNICS OR ENGINEERS: DYNMICS edinnd P. ee E. Russell Johnston, J. Lectue Notes: J. Wlt Ole Texs Tech Univesity Kinetics of Pticles: Newton s Second Lw The McGwHill Copnies, Inc.
More informationPhysics 201 Lecture 18
Phsics 0 ectue 8 ectue 8 Goals: Define and anale toque ntoduce the coss poduct Relate otational dnamics to toque Discuss wok and wok eneg theoem with espect to otational motion Specif olling motion (cente
More information+ r Position Velocity
1. The phee P tel in tight line with contnt peed of =100 m/. Fo the intnt hown, detemine the coeponding lue of,,,,, eltie to the fixed Ox coodinte tem. meued + + Poition Velocit e 80 e 45 o 113. 137 d
More informationEECE 260 Electrical Circuits Prof. Mark Fowler
EECE 60 Electicl Cicuits Pof. Mk Fowle Complex Numbe Review /6 Complex Numbes Complex numbes ise s oots of polynomils. Definition of imginy # nd some esulting popeties: ( ( )( ) )( ) Recll tht the solution
More information(A) 6.32 (B) 9.49 (C) (D) (E) 18.97
Univesity of Bhin Physics 10 Finl Exm Key Fll 004 Deptment of Physics 13/1/005 8:30 10:30 e =1.610 19 C, m e =9.1110 31 Kg, m p =1.6710 7 Kg k=910 9 Nm /C, ε 0 =8.8410 1 C /Nm, µ 0 =4π10 7 T.m/A Pt : 10
More informationMultiplying and Dividing Rational Expressions
Lesson Peview Pt  Wht You ll Len To multipl tionl epessions To divide tionl epessions nd Wh To find lon pments, s in Eecises 0 Multipling nd Dividing Rtionl Epessions Multipling Rtionl Epessions Check
More informationSection 26 The Laws of Rotational Motion
Physics 24A Class Notes Section 26 The Laws of otational Motion What do objects do and why do they do it? They otate and we have established the quantities needed to descibe this motion. We now need to
More informationJEE(Advanced) 2018 TEST PAPER WITH SOLUTION PHYSICS. (HELD ON SUNDAY 20 th MAY, 2018) PART1 : PHYSICS. (C) L = mkr ALLEN
JEE(Advnced) 08 TEST PAPE WITH SOUTION (HED ON SUNDAY 0 th MAY, 08) PAT : JEE(Advnced) 08/Ppe. The potentil enegy of pticle of mss m t distnce fom fixed point O is given by V () k /, whee k is positive
More informationPX3008 Problem Sheet 1
PX38 Poblem Sheet 1 1) A sphee of dius (m) contins chge of unifom density ρ (Cm 3 ). Using Guss' theoem, obtin expessions fo the mgnitude of the electic field (t distnce fom the cente of the sphee) in
More informationB 20 kg. 60 kg A. m s, m k
1. he sste is elesed o est with the cble tut. o the iction coeicients s =.5 nd =. clculte the cceletion o ech bod nd the tension in the cble. eglect the sll ss nd iction o the pulles.(3/9) s 6 g 3 g W
More informationChapter 23 Electrical Potential
hpte Electicl Potentil onceptul Polems [SSM] A poton is moved to the left in unifom electic field tht points to the ight. Is the poton moving in the diection of incesing o decesing electic potentil? Is
More informationPhysics 604 Problem Set 1 Due Sept 16, 2010
Physics 64 Polem et 1 Due ept 16 1 1) ) Inside good conducto the electic field is eo (electons in the conducto ecuse they e fee to move move in wy to cncel ny electic field impessed on the conducto inside
More informationTopic 1 Notes Jeremy Orloff
Topic 1 Notes Jerem Orloff 1 Introduction to differentil equtions 1.1 Gols 1. Know the definition of differentil eqution. 2. Know our first nd second most importnt equtions nd their solutions. 3. Be ble
More information7.5Determinants in Two Variables
7.eteminnts in Two Vibles efinition of eteminnt The deteminnt of sque mti is el numbe ssocited with the mti. Eve sque mti hs deteminnt. The deteminnt of mti is the single ent of the mti. The deteminnt
More informationThe Wave Equation I. MA 436 Kurt Bryan
1 Introduction The Wve Eqution I MA 436 Kurt Bryn Consider string stretching long the x xis, of indeterminte (or even infinite!) length. We wnt to derive n eqution which models the motion of the string
More informationPhysics 505 Fall 2005 Midterm Solutions. This midterm is a two hour open book, open notes exam. Do all three problems.
Physics 55 Fll 5 Midtem Solutions This midtem is two hou open ook, open notes exm. Do ll thee polems. [35 pts] 1. A ectngul ox hs sides of lengths, nd c z x c [1] ) Fo the Diichlet polem in the inteio
More informationFULL MECHANICS SOLUTION
FULL MECHANICS SOLUION. m 3 3 3 f For long the tngentil direction m 3g cos 3 sin 3 f N m 3g sin 3 cos3 from soling 3. ( N 4) ( N 8) N gsin 3. = ut + t = ut g sin cos t u t = gsin cos = 4 5 5 = s] 3 4 o
More information1.2. Linear Variable Coefficient Equations. y + b "! = a y + b " Remark: The case b = 0 and a nonconstant can be solved with the same idea as above.
1 12 Liner Vrible Coefficient Equtions Section Objective(s): Review: Constnt Coefficient Equtions Solving Vrible Coefficient Equtions The Integrting Fctor Method The Bernoulli Eqution 121 Review: Constnt
More informationFriedmannien equations
..6 Fiedmnnien equtions FLRW metic is : ds c The metic intevl is: dt ( t) d ( ) hee f ( ) is function which detemines globl geometic l popety of D spce. f d sin d One cn put it in the Einstein equtions
More informationLecture 13  Linking E, ϕ, and ρ
Lecture 13  Linking E, ϕ, nd ρ A Puzzle... InnerSurfce Chrge Density A positive point chrge q is locted offcenter inside neutrl conducting sphericl shell. We know from Guss s lw tht the totl chrge on
More informationEinstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi , Ph. : ,
PE ELECTOSTATICS C Popeties of chges : (i) (ii) (iii) (iv) (v) (vi) Two kinds of chges eist in ntue, positive nd negtive with the popety tht unlike chges ttct ech othe nd like chges epel ech othe. Ecess
More informationRadian Measure CHAPTER 5 MODELLING PERIODIC FUNCTIONS
5.4 Radian Measue So fa, ou hae measued angles in degees, with 60 being one eolution aound a cicle. Thee is anothe wa to measue angles called adian measue. With adian measue, the ac length of a cicle is
More informationRELATIVE KINEMATICS. q 2 R 12. u 1 O 2 S 2 S 1. r 1 O 1. Figure 1
RELAIVE KINEMAICS he equtions of motion fo point P will be nlyzed in two diffeent efeence systems. One efeence system is inetil, fixed to the gound, the second system is moving in the physicl spce nd the
More informationGet Solution of These Packages & Learn by Video Tutorials on EXERCISE1
FEE Downlod Study Pckge fom website: www.tekoclsses.com & www.mthsbysuhg.com Get Solution of These Pckges & Len by Video Tutoils on www.mthsbysuhg.com EXECISE * MAK IS MOE THAN ONE COECT QUESTIONS. SECTION
More informationContinuous Charge Distributions
Continuous Chge Distibutions Review Wht if we hve distibution of chge? ˆ Q chge of distibution. Q dq element of chge. d contibution to due to dq. Cn wite dq = ρ dv; ρ is the chge density. = 1 4πε 0 qi
More informationSEE LAST PAGE FOR SOME POTENTIALLY USEFUL FORMULAE AND CONSTANTS
Cicle instucto: Moow o Yethiaj Name: MEMORIL UNIVERSITY OF NEWFOUNDLND DEPRTMENT OF PHYSICS ND PHYSICL OCENOGRPHY Final Eam Phsics 5 Winte 3:5: pil, INSTRUCTIONS:. Do all SIX (6) questions in section
More informationPhysics Honors. Final Exam Review Free Response Problems
Physics Honors inl Exm Review ree Response Problems m t m h 1. A 40 kg mss is pulled cross frictionless tble by string which goes over the pulley nd is connected to 20 kg mss.. Drw free body digrm, indicting
More informationFrom Newton to Einstein. MidTerm Test, 12a.m. Thur. 13 th Nov Duration: 50 minutes. There are 20 marks in Section A and 30 in Section B.
Fom Newton to Einstein MidTem Test, a.m. Thu. 3 th Nov. 008 Duation: 50 minutes. Thee ae 0 maks in Section A and 30 in Section B. Use g = 0 ms in numeical calculations. You ma use the following epessions
More informationAssistant Professor: Zhou Yufeng. N , ,
Aitnt Pofeo: Zhou Yufeng N3.05, 6790448, yfzhou@ntu.edu.g http://www3.ntu.edu.g/home/yfzhou/coue.html . A pojectile i fied t flling tget hown. The pojectile lee the gun t the me intnt tht the tget dopped
More informationconstant t [rad.s 1 ] v / r r [m.s 2 ] (direction: towards centre of circle / perpendicular to circle)
VISUAL PHYSICS ONLINE MODULE 5 ADVANCED MECHANICS NONUNIFORM CIRCULAR MOTION Equation of a cicle x y Angula displacement [ad] Angula speed d constant t [ad.s 1 ] dt Tangential velocity v v [m.s 1 ]
More information13.5. Torsion of a curve Tangential and Normal Components of Acceleration
13.5 osion of cuve ngentil nd oml Components of Acceletion Recll: Length of cuve '( t) Ac length function s( t) b t u du '( t) Ac length pmetiztion ( s) with '( s) 1 '( t) Unit tngent vecto '( t) Cuvtue:
More informationDYNAMICS OF UNIFORM CIRCULAR MOTION
Chapte 5 Dynamics of Unifom Cicula Motion Chapte 5 DYNAMICS OF UNIFOM CICULA MOTION PEVIEW An object which is moing in a cicula path with a constant speed is said to be in unifom cicula motion. Fo an object
More information