PROBLEM deceleration of the cable attached at B is 2.5 m/s, while that + ] ( )( ) = 2.5 2α. a = rad/s. a 3.25 m/s. = 3.
|
|
- Douglas Buddy Lindsey
- 6 years ago
- Views:
Transcription
1 PROLEM A 5-m steel bem is lowered by mens of two cbles unwinding t the sme speed from overhed crnes. As the bem pproches the ground, the crne opertors pply brkes to slow the unwinding motion. At the instnt considered the decelertion of the cble ttched t is.5 m/s, while tht of the cble ttched t D is 1.5 m/s. Determine () the ngulr ccelertion of the bem, (b) the ccelertion of points A nd E. α = α, ω 0, rd/ = m =.5 m/s, 1.5 m/s D = ( / ) ( / ) = + + D D t D n 1.5 = [.5 ] [α ] ( )( ) ( ) 1.5 =.5 α b A = + ( A / ) + ( A / ) () t n = rd/s = [.5 ] + ( 1.5)( 0.5) ] + [0 ] = 3.5 m/s ( / ) ( / ) = + + E E t E n 3.5 m/s A = = [.5 ] + ( + 1.5)( 0.5) + [0 ] = 0.75 m/s m/s E =
2 PROLEM A 900-mm rod rests on horizontl tble. A force P pplied s shown produces the following ccelertions: A = 3.6 m/s to the right, = 6 rd/s counterclockwise s viewed from bove. Determine the ccelertion () of Point G, (b) of Point. () = + / = [ ] + [( AG) α ] G A GA A G = [3.6 m/s G = [3.6 m/s ] [.7 m/s ] + [(0.45 m)(6 rd/s ) ] + ] (b) = + / = [ ] + [( A) α ] A A A = [3.6 m/s = [3.6 m/s ] [5.4 m/s ] + [(0.9 m)(6 rd/s ) ] + ] = 0.9 m/s G = 1.8 m/s
3 PROLEM Knowing tht t the instnt shown crnk C hs constnt ngulr velocity of 45 rpm clockwise, determine the ccelertion () of Point A, (b) of Point D. Geometry. Velocity nlysis. Let β be ngle AC. 4 in. sin β = β = 30 8 in. ω C = 45 rpm v = rd/s v A = va = ( C) ω = (4)(4.714) = in./s v in./s C v A nd v re prllel; hence, the instntneous center of rottion of rod AD lies t infinity. Accelertion nlysis. α C = 0 Crnk C: ω = 0 v = v = in./s AD A ( ) = ( C) = 0 t Rod A: AD = α AD n = C ωc = ( ) ( ) (4)(4.714) = in./s α A = A = + ( ) + ( ) A A / t A / A = [ A ] = [88.87 ] + [8α AD 30 + ] [8ω AD 60 ] Resolve into components. () : : 0 = cos = 1.81 rd/s A AD = 8 sin 30 = (8)( 1.81)sin 30 = in./s AD AD = 51.3 in./s A
4 PROLEM (Continued) (b) = + ( / ) + ( / ) D D t D t = [88.87 ] + [8α ω 30 + ] [8 [ 60 ] = [88.87 ] + [(8)( 1.81) 30 + ] 0 = [88.87 ] + [ ] = [ ] D = in./s 16.1
5 PROLEM An utomobile trvels to the left t constnt speed of 7 km/h. Knowing tht the dimeter of the wheel is 560 mm, determine the ccelertion () of Point, (b) of Point C, (c) of Point D. v h 1000 m A = 7 km/h 0 m/s 3600 s km = Rolling with no sliding, instntneous center is t C. Accelertion. v = ( AC) ω; 0 m/s = (0.8 m) ω A ω = rd/s Plne motion = Trns. with A + Rottion bout A A / = CA / = DA / = rω = (0.80 m)(71.49 rd/s) = m/s () (b) (c) = + / = m/s A A C= A+ CA / = m/s D= A+ DA / = m/s 60 = 1430 m/s = 1430 m/s C D = 1430 m/s 60
6 PROLEM A hevy crte is being moved short distnce using three identicl cylinders s rollers. Knowing tht t the instnt shown the crte hs velocity of 00 mm/s nd n ccelertion of 400 mm/s, both directed to the right, determine () the ngulr ccelertion of the center cylinder, (b) the ccelertion of point A on the center cylinder. Geometry. Let point C be the center of the center cylinder, its contct point with the crte, nd D its contct point with the ground. Let r be the rdius of the cylinder. r = 100 mm. Velocity nlysis. Accelertion nlysis. Since the contcts t nd D re rolling contcts without slipping, v = 00 mm/s nd v D = 0. Point D is the instntneous center of rottion. v 00 mm/s ω = = = 1 rd/s r 00 mm Point C moves on horizontl line. = [ ] = + ( ) + ( ) = [ ] + [rα D C DC / t DC / n C C C ] + [rω ] Component : 0 = C r (1) = + ( ) + ( ) = [ ] + [rα C C / t C / n C ] + [rω ] Component : Solving (1) nd () simultneously, C = 00 mm/s rα = 00 mm/s () 00 mm/s α = 100 mm 400 mm/s = C + r () =.00 rd/s
7 = + ( ) + ( ) = [ ] + [rα A C AC / t AC / n C PROLEM (Continued) ] + [rω ] = [00 mm/s ] + [00 mm/s ] + [(100 mm) (1 rd/s) ] = [100 mm/s ] + [00 mm/s ] (b) A = 3.6 mm/s A = 0.4 m/s 63.4
8 PROLEM Knowing tht crnk A rottes bout Point A with constnt ngulr velocity of 900 rpm clockwise, determine the ccelertion of the piston P when 10. θ = Lw of sines. sin β sin10 =, β = Velocity nlysis. ω = 900 rpm = 30 p rd/s A v = 0.05ω = 1.5π m/s 60 A v D = v D ω = ω vd/ = 0.15ω v = v + v D D / β [ v D ] = [1.5π 60 + ] [0.15ω β ] Components : 0 = 1.5π cos ω cos β Accelertion nlysis. α A = 0 ω 1.5π cos 60 = = rd/s 0.15 cos β A π = 0.05 ω = (0.05)(30 ) = m/s 30 D = D α = α α D/ = [0.15α A β] + [0.15ω β ] = [6α β ] + [ β ] = + D D / Resolve into components.
9 PROLEM (Continued) : 0 = cos α cos β sin β = rd/s : = sin 30 (0.15)(597.0)sin β cos β D = 96 m/s P = D = 96 m/s P
10 PROLEM 15.1 In the two-cylinder ir compressor shown, the connecting rods nd E re ech 190 mm long nd crnk A rottes bout the fixed Point A with constnt ngulr velocity of 1500 rpm clockwise. Determine the ccelertion of ech piston when θ = 0. Crnk A. v = 0, = 0, ω = 1500 rpm = rd/s, α = 0 A A A v = va + v / A = 0 + [0.05ωA 45 ] = [7.854 m/s 45 ] = + ( ) + ( ) A A / t A / n = 0 + [0.05α A 45 ] + [0.05ω A 45 ] = [(0.05)(157.08) 45 ] = m/s 45 Rod. v D = v D 45 ω = ω v = v + v D / vd 45 = [ ] [0.19ω 45 ] Components 45 : 0 = ω ω = rd/s D = D 45 = + ( ) + ( ) D D / t D / n A [ D 45 ] = [ ] [0.19α 45 + ] [0.19ω 45 ]
11 PROLEM 15.1 (Continued) Components 45 : = (0.19)(41.337) = m/s D Rod E. Since E 0.05 sin γ =, γ = 15.58, β = 45 γ = v E = ve 45 v is prllel to v, ω = 0. E = 1558 m/s 45 D E = E 45 / E n ωe ( ) = 0.19 = 0 ( ) ( ) Drw vector ddition digrm. γ = 45 β = E / t= E / t β E = + ( E / ) t E = tn γ = tnγ = m/s = 337 m/s 45 E
12 PROLEM Knowing tht t the instnt shown br A hs n ngulr velocity of 4 rd/s nd n ngulr ccelertion of rd/s, both clockwise, determine the ngulr ccelertion () of br, (b) of br by using the vector pproch s is done in Smple Problem Reltive position vectors. ra / = (0 in.) i (40 in.) j r = (40 in.) i Velocity nlysis. r A (Rottion bout A): D / r = (0 in.) i (5 in.) j / ω A = 4 rd/s = (4 rd/s) k r = (0 in.) i (40 in.) j v = ω r = ( 4 k) ( 0i 40 j ) A / A A / v = (160 in./s) i+ (80 in./s) j r (Plne motion = Trnsltion with + Rottion bout ): r (Rottion bout E): ω = ω k r = (40 in.) i ω D/ v = v + ω r = v + ( ω k) (40 i) D D/ v = (160 in/s) i+ (40ω + 80 in./s) j D r = (0 in.) i (5in.) j / = ω k D D/ E v = ω r = ( ω k) (0i 5 j) v = 0ω j+ 5ω i D Equting components of the two expression for v D, i: 160 = 5ω ω = 6.4 rd/s j: 40ω + 80 = 0ω 40ω + 80 = 0( 6.4) ω = 5. rd/s E
13 PROLEM (Continued) Summry of ngulr velocities: Accelertion nlysis. r A (Rottion bout A) ω 4 rd/s ω 6.4 rd/s ω 5. rd/s A = = = A = ( rd/s ), =, = k k k = A / A ωa / A r r = ( k) ( 0i 40 j) (4) ( 0i 40 j) = (80 in./s ) i + (40 in./s ) j+ (30 in./s ) i+ (640 in./s ) j = (40 in./s ) i+ (680 in./s ) j r (Trnsltion with + Rottion bout ): D = + α D/ ω D/ α α r r = 40i+ 680 j+ α k (40 i) (5.) (40) i = 40i+ 680j+ 40α j i = i+ ( ) j (1) α r (Rottion bout E): D = α D/ E ω D/ E α r r Equte like components of D expressed by Eqs. (1) nd (). = α k (0i 5 j) (6.4) (0i 5 j) = 0α j+ 5α i 819.0i+ 104j = (5α 819.0) i+ (0α + 104) j () i: j: = = rd/s = (0)( 0.896) = 8.15 rd/s () (b) Angulr ccelertion of br. Angulr ccelertion of br. α = 8.15 rd/s α = rd/s
PROBLEM SOLUTION
PROLEM 15.11 The 18-in.-rdius flywheel is rigidly ttched to 1.5-in.-rdius shft tht cn roll long prllel rils. Knowing tht t the instnt shown the center of the shft hs velocity of 1. in./s nd n ccelertion
More informationKINETICS OF RIGID BODIES PROBLEMS
KINETICS OF RIID ODIES PROLEMS PROLEMS 1. The 6 kg frme C nd the 4 kg uniform slender br of length l slide with negligible friction long the fied horizontl br under the ction of the 80 N force. Clculte
More informationProblems (Motion Relative to Rotating Axes)
1. The disk rolls without slipping on the roblems (Motion Reltie to Rotting xes) horizontl surfce, nd t the instnt represented, the center O hs the elocity nd ccelertion shown in the figure. For this instnt,
More informationKINEMATICS OF RIGID BODIES
KINEMTICS OF RIGID ODIES Introduction In rigid body kinemtics, e use the reltionships governing the displcement, velocity nd ccelertion, but must lso ccount for the rottionl motion of the body. Description
More information1/31/ :33 PM. Chapter 11. Kinematics of Particles. Mohammad Suliman Abuhaiba,Ph.D., P.E.
1/31/18 1:33 PM Chpter 11 Kinemtics of Prticles 1 1/31/18 1:33 PM First Em Sturdy 1//18 3 1/31/18 1:33 PM Introduction Mechnics Mechnics = science which describes nd predicts conditions of rest or motion
More informationTHREE-DIMENSIONAL KINEMATICS OF RIGID BODIES
THREE-DIMENSIONAL KINEMATICS OF RIGID BODIES 1. TRANSLATION Figure shows rigid body trnslting in three-dimensionl spce. Any two points in the body, such s A nd B, will move long prllel stright lines if
More information2/2/ :36 AM. Chapter 11. Kinematics of Particles. Mohammad Suliman Abuhaiba,Ph.D., P.E.
//16 1:36 AM Chpter 11 Kinemtics of Prticles 1 //16 1:36 AM First Em Wednesdy 4//16 3 //16 1:36 AM Introduction Mechnics Mechnics = science which describes nd predicts the conditions of rest or motion
More informationDYNAMICS. Kinematics of Rigid Bodies VECTOR MECHANICS FOR ENGINEERS: Tenth Edition CHAPTER
Tenth E CHTER 15 VECTOR MECHNICS FOR ENGINEERS: YNMICS Ferdinnd. eer E. Russell Johnston, Jr. hillip J. Cornwell Lecture Notes: rin. Self Cliforni olytechnic Stte Uniersity Kinemtics of Rigid odies 013
More informationVersion 001 HW#6 - Circular & Rotational Motion arts (00223) 1
Version 001 HW#6 - Circulr & ottionl Motion rts (00223) 1 This print-out should hve 14 questions. Multiple-choice questions my continue on the next column or pge find ll choices before nswering. Circling
More information2/20/ :21 AM. Chapter 11. Kinematics of Particles. Mohammad Suliman Abuhaiba,Ph.D., P.E.
//15 11:1 M Chpter 11 Kinemtics of Prticles 1 //15 11:1 M Introduction Mechnics Mechnics = science which describes nd predicts the conditions of rest or motion of bodies under the ction of forces It is
More informationThe Spring. Consider a spring, which we apply a force F A to either stretch it or compress it
The Spring Consider spring, which we pply force F A to either stretch it or copress it F A - unstretched -F A 0 F A k k is the spring constnt, units of N/, different for different terils, nuber of coils
More informationKEY. Physics 106 Common Exam 1, Spring, 2004
Physics 106 Common Exm 1, Spring, 2004 Signture Nme (Print): A 4 Digit ID: Section: Instructions: Questions 1 through 10 re multiple-choice questions worth 5 points ech. Answer ech of them on the Scntron
More informationDynamics: Newton s Laws of Motion
Lecture 7 Chpter 4 Physics I 09.25.2013 Dynmics: Newton s Lws of Motion Solving Problems using Newton s lws Course website: http://fculty.uml.edu/andriy_dnylov/teching/physicsi Lecture Cpture: http://echo360.uml.edu/dnylov2013/physics1fll.html
More informationCorrect answer: 0 m/s 2. Explanation: 8 N
Version 001 HW#3 - orces rts (00223) 1 his print-out should hve 15 questions. Multiple-choice questions my continue on the next column or pge find ll choices before nswering. Angled orce on Block 01 001
More informationDynamics and control of mechanical systems. Content
Dynmics nd control of mechnicl systems Dte Dy 1 (01/08) Dy (03/08) Dy 3 (05/08) Dy 4 (07/08) Dy 5 (09/08) Dy 6 (11/08) Content Review of the bsics of mechnics. Kinemtics of rigid bodies plne motion of
More informationA wire. 100 kg. Fig. 1.1
1 Fig. 1.1 shows circulr cylinder of mss 100 kg being rised by light, inextensible verticl wire. There is negligible ir resistnce. wire 100 kg Fig. 1.1 (i) lculte the ccelertion of the cylinder when the
More informationPhysicsAndMathsTutor.com
1. A uniform circulr disc hs mss m, centre O nd rdius. It is free to rotte bout fixed smooth horizontl xis L which lies in the sme plne s the disc nd which is tngentil to the disc t the point A. The disc
More informationFirst, we will find the components of the force of gravity: Perpendicular Forces (using away from the ramp as positive) ma F
1.. In Clss or Homework Eercise 1. An 18.0 kg bo is relesed on 33.0 o incline nd ccelertes t 0.300 m/s. Wht is the coeicient o riction? m 18.0kg 33.0? 0 0.300 m / s irst, we will ind the components o the
More informationSOLUTIONS TO CONCEPTS CHAPTER 6
SOLUIONS O CONCEPS CHAPE 6 1. Let ss of the block ro the freebody digr, 0...(1) velocity Agin 0 (fro (1)) g 4 g 4/g 4/10 0.4 he co-efficient of kinetic friction between the block nd the plne is 0.4. Due
More informationMiscellaneous Problems. pinned to the ground
Miscellneous Problems Problem. Use the mobilit formul to determine the number of degrees of freedom for this sstem. pinned to the ground pinned to the ground Problem. For this mechnism: () Define vectors
More information3. Vectors. Vectors: quantities which indicate both magnitude and direction. Examples: displacemement, velocity, acceleration
Rutgers University Deprtment of Physics & Astronomy 01:750:271 Honors Physics I Lecture 3 Pge 1 of 57 3. Vectors Vectors: quntities which indicte both mgnitude nd direction. Exmples: displcemement, velocity,
More informationSOLUTIONS TO CONCEPTS CHAPTER 10
SOLUTIONS TO CONCEPTS CHPTE 0. 0 0 ; 00 rev/s ; ; 00 rd/s 0 t t (00 )/4 50 rd /s or 5 rev/s 0 t + / t 8 50 400 rd 50 rd/s or 5 rev/s s 400 rd.. 00 ; t 5 sec / t 00 / 5 8 5 40 rd/s 0 rev/s 8 rd/s 4 rev/s
More informationME 141. Lecture 10: Kinetics of particles: Newton s 2 nd Law
ME 141 Engineering Mechnics Lecture 10: Kinetics of prticles: Newton s nd Lw Ahmd Shhedi Shkil Lecturer, Dept. of Mechnicl Engg, BUET E-mil: sshkil@me.buet.c.bd, shkil6791@gmil.com Website: techer.buet.c.bd/sshkil
More informationPhysics 105 Exam 2 10/31/2008 Name A
Physics 105 Exm 2 10/31/2008 Nme_ A As student t NJIT I will conduct myself in professionl mnner nd will comply with the proisions of the NJIT Acdemic Honor Code. I lso understnd tht I must subscribe to
More informationDEFINITION OF ASSOCIATIVE OR DIRECT PRODUCT AND ROTATION OF VECTORS
3 DEFINITION OF ASSOCIATIVE OR DIRECT PRODUCT AND ROTATION OF VECTORS This chpter summrizes few properties of Cli ord Algebr nd describe its usefulness in e ecting vector rottions. 3.1 De nition of Associtive
More informationDYNAMICS VECTOR MECHANICS FOR ENGINEERS: Plane Motion of Rigid Bodies: Forces and Accelerations. Seventh Edition CHAPTER
CHAPTER 16 VECTOR MECHANICS FOR ENGINEERS: DYNAMICS Ferdinnd P. Beer E. Ruell Johnton, Jr. Lecture Note: J. Wlt Oler Tex Tech Univerity Plne Motion of Rigid Bodie: Force nd Accelertion Content Introduction
More informationKINEMATICS OF RIGID BODIES
KINEMTICS OF RIGI OIES Introduction In rigid body kinemtics, e use the reltionships governing the displcement, velocity nd ccelertion, but must lso ccount for the rottionl motion of the body. escription
More informationTrigonometric Functions
Exercise. Degrees nd Rdins Chpter Trigonometric Functions EXERCISE. Degrees nd Rdins 4. Since 45 corresponds to rdin mesure of π/4 rd, we hve: 90 = 45 corresponds to π/4 or π/ rd. 5 = 7 45 corresponds
More informationTime : 3 hours 03 - Mathematics - March 2007 Marks : 100 Pg - 1 S E CT I O N - A
Time : hours 0 - Mthemtics - Mrch 007 Mrks : 100 Pg - 1 Instructions : 1. Answer ll questions.. Write your nswers ccording to the instructions given below with the questions.. Begin ech section on new
More informationE S dition event Vector Mechanics for Engineers: Dynamics h Due, next Wednesday, 07/19/2006! 1-30
Vector Mechnics for Engineers: Dynmics nnouncement Reminders Wednesdy s clss will strt t 1:00PM. Summry of the chpter 11 ws posted on website nd ws sent you by emil. For the students, who needs hrdcopy,
More informationMASTER CLASS PROGRAM UNIT 4 SPECIALIST MATHEMATICS WEEK 11 WRITTEN EXAMINATION 2 SOLUTIONS SECTION 1 MULTIPLE CHOICE QUESTIONS
MASTER CLASS PROGRAM UNIT 4 SPECIALIST MATHEMATICS WEEK WRITTEN EXAMINATION SOLUTIONS FOR ERRORS AND UPDATES, PLEASE VISIT WWW.TSFX.COM.AU/MC-UPDATES SECTION MULTIPLE CHOICE QUESTIONS QUESTION QUESTION
More informationJURONG JUNIOR COLLEGE
JURONG JUNIOR COLLEGE 2010 JC1 H1 8866 hysics utoril : Dynmics Lerning Outcomes Sub-topic utoril Questions Newton's lws of motion 1 1 st Lw, b, e f 2 nd Lw, including drwing FBDs nd solving problems by
More informationSECTION B Circular Motion
SECTION B Circulr Motion 1. When person stnds on rotting merry-go-round, the frictionl force exerted on the person by the merry-go-round is (A) greter in mgnitude thn the frictionl force exerted on the
More information= 40 N. Q = 60 O m s,k
Multiple Choice ( 6 Points Ech ): F pp = 40 N 20 kg Q = 60 O m s,k = 0 1. A 20 kg box is pulled long frictionless floor with n pplied force of 40 N. The pplied force mkes n ngle of 60 degrees with the
More informationForces from Strings Under Tension A string under tension medites force: the mgnitude of the force from section of string is the tension T nd the direc
Physics 170 Summry of Results from Lecture Kinemticl Vribles The position vector ~r(t) cn be resolved into its Crtesin components: ~r(t) =x(t)^i + y(t)^j + z(t)^k. Rtes of Chnge Velocity ~v(t) = d~r(t)=
More informationNarayana IIT Academy
INDIA Sec: Sr. IIT_IZ Jee-Advnced Dte: --7 Time: 09:00 AM to :00 Noon 0_P Model M.Mrks: 0 KEY SHEET CHEMISTRY C D 3 D B 5 A 6 D 7 B 8 AC 9 BC 0 ABD ABD A 3 C D 5 B 6 B 7 9 8 9 0 7 8 3 3 6 PHYSICS B 5 D
More informationPhysics 110. Spring Exam #1. April 16, Name
Physics 110 Spring 010 Exm #1 April 16, 010 Nme Prt Multiple Choice / 10 Problem #1 / 7 Problem # / 7 Problem #3 / 36 Totl / 100 In keeping with the Union College policy on cdemic honesty, it is ssumed
More informationPhysics 2135 Exam 3 April 21, 2015
Em Totl hysics 2135 Em 3 April 21, 2015 Key rinted Nme: 200 / 200 N/A Rec. Sec. Letter: Five multiple choice questions, 8 points ech. Choose the best or most nerly correct nswer. 1. C Two long stright
More informationUCSD Phys 4A Intro Mechanics Winter 2016 Ch 4 Solutions
USD Phys 4 Intro Mechnics Winter 06 h 4 Solutions 0. () he 0.0 k box restin on the tble hs the free-body dir shown. Its weiht 0.0 k 9.80 s 96 N. Since the box is t rest, the net force on is the box ust
More information- 5 - TEST 2. This test is on the final sections of this session's syllabus and. should be attempted by all students.
- 5 - TEST 2 This test is on the finl sections of this session's syllbus nd should be ttempted by ll students. Anything written here will not be mrked. - 6 - QUESTION 1 [Mrks 22] A thin non-conducting
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 informationNavigation Mathematics: Angular and Linear Velocity EE 570: Location and Navigation
Lecture Nvigtion Mthemtics: Angulr n Liner Velocity EE 57: Loction n Nvigtion Lecture Notes Upte on Februry, 26 Kevin Weewr n Aly El-Osery, Electricl Engineering Dept., New Mexico Tech In collbortion with
More informationPROBLEM Copyright McGraw-Hill Education. Permission required for reproduction or display. SOLUTION
PROLEM 15.10 The bent rod E rotates about a line joining Points and E with a constant angular elocity of 9 rad/s. Knowing that the rotation is clockwise as iewed from E, determine the elocity and acceleration
More informationPhysics 207 Lecture 7
Phsics 07 Lecture 7 Agend: Phsics 07, Lecture 7, Sept. 6 hpter 6: Motion in (nd 3) dimensions, Dnmics II Recll instntneous velocit nd ccelertion hpter 6 (Dnmics II) Motion in two (or three dimensions)
More informationRolling Contact Bearings (pg 599)
Bering V9.xmcd [Pg / 6] Title [234] The Units used s stndrd: m, kg, N, P, sec, wtts N, kg, m, P, sec/min, wtts/kw Rolling Contct Berings (pg 599) This note is only guideline for using the text book. Detiled
More informationMathematics Extension Two
Student Number 04 HSC TRIAL EXAMINATION Mthemtics Etension Two Generl Instructions Reding time 5 minutes Working time - hours Write using blck or blue pen Bord-pproved clcultors my be used Write your Student
More informationYear 12 Mathematics Extension 2 HSC Trial Examination 2014
Yer Mthemtics Etension HSC Tril Emintion 04 Generl Instructions. Reding time 5 minutes Working time hours Write using blck or blue pen. Blck pen is preferred. Bord-pproved clcultors my be used A tble of
More informationMathematics of Motion II Projectiles
Chmp+ Fll 2001 Dn Stump 1 Mthemtics of Motion II Projectiles Tble of vribles t time v velocity, v 0 initil velocity ccelertion D distnce x position coordinte, x 0 initil position x horizontl coordinte
More informationCoimisiún na Scrúduithe Stáit State Examinations Commission LEAVING CERTIFICATE 2010 MARKING SCHEME APPLIED MATHEMATICS HIGHER LEVEL
Coimisiún n Scrúduithe Stáit Stte Emintions Commission LEAVING CERTIFICATE 00 MARKING SCHEME APPLIED MATHEMATICS HIGHER LEVEL Generl Guidelines Penlties of three types re pplied to cndidtes' work s follows:
More informationStudy Guide Final Exam. Part A: Kinetic Theory, First Law of Thermodynamics, Heat Engines
Msschusetts Institute of Technology Deprtment of Physics 8.0T Fll 004 Study Guide Finl Exm The finl exm will consist of two sections. Section : multiple choice concept questions. There my be few concept
More informationAbstract. Introduction
Apprent chnge of position of light source reltive to detector/observer due to rottion nd ccelertion - new interprettion nd nlysis of Michelson-Morley nd gnc experiments Abstrct Henok Tdesse, Electricl
More information200 points 5 Problems on 4 Pages and 20 Multiple Choice/Short Answer Questions on 5 pages 1 hour, 48 minutes
PHYSICS 132 Smple Finl 200 points 5 Problems on 4 Pges nd 20 Multiple Choice/Short Answer Questions on 5 pges 1 hour, 48 minutes Student Nme: Recittion Instructor (circle one): nme1 nme2 nme3 nme4 Write
More informationAP Physics 1. Slide 1 / 71. Slide 2 / 71. Slide 3 / 71. Circular Motion. Topics of Uniform Circular Motion (UCM)
Slide 1 / 71 Slide 2 / 71 P Physics 1 irculr Motion 2015-12-02 www.njctl.org Topics of Uniform irculr Motion (UM) Slide 3 / 71 Kinemtics of UM lick on the topic to go to tht section Period, Frequency,
More information2A1A Vector Algebra and Calculus I
Vector Algebr nd Clculus I (23) 2AA 2AA Vector Algebr nd Clculus I Bugs/queries to sjrob@robots.ox.c.uk Michelms 23. The tetrhedron in the figure hs vertices A, B, C, D t positions, b, c, d, respectively.
More informationSOLUTIONS TO CONCEPTS CHAPTER
1. m = kg S = 10m Let, ccelertion =, Initil velocity u = 0. S= ut + 1/ t 10 = ½ ( ) 10 = = 5 m/s orce: = = 5 = 10N (ns) SOLUIONS O CONCEPS CHPE 5 40000. u = 40 km/hr = = 11.11 m/s. 3600 m = 000 kg ; v
More informationC D o F. 30 o F. Wall String. 53 o. F y A B C D E. m 2. m 1. m a. v Merry-go round. Phy 231 Sp 03 Homework #8 Page 1 of 4
Phy 231 Sp 3 Hoework #8 Pge 1 of 4 8-1) rigid squre object of negligible weight is cted upon by the forces 1 nd 2 shown t the right, which pull on its corners The forces re drwn to scle in ters of the
More informationStudent Session Topic: Particle Motion
Student Session Topic: Prticle Motion Prticle motion nd similr problems re on the AP Clculus exms lmost every yer. The prticle my be prticle, person, cr, etc. The position, velocity or ccelertion my be
More informationChapter 5 Exercise 5A
Chpter Exercise Q. 1. (i) 00 N,00 N F =,00 00 =,000 F = m,000 = 1,000 = m/s (ii) =, u = 0, t = 0, s =? s = ut + 1 t = 0(0) + 1 ()(00) = 00 m Q.. 0 N 100 N F = 100 0 = 60 F = m 60 = 10 = 1 m/s F = m 60
More informationIn-Class Problems 2 and 3: Projectile Motion Solutions. In-Class Problem 2: Throwing a Stone Down a Hill
MASSACHUSETTS INSTITUTE OF TECHNOLOGY Deprtment of Physics Physics 8T Fll Term 4 In-Clss Problems nd 3: Projectile Motion Solutions We would like ech group to pply the problem solving strtegy with the
More informationHW Solutions # MIT - Prof. Kowalski. Friction, circular dynamics, and Work-Kinetic Energy.
HW Solutions # 5-8.01 MIT - Prof. Kowlski Friction, circulr dynmics, nd Work-Kinetic Energy. 1) 5.80 If the block were to remin t rest reltive to the truck, the friction force would need to cuse n ccelertion
More informationChapter 10 Rotation of a Rigid Object About a Fixed Axis
Chter ottion o igid Object About Fixed Axis P. () ω ωi. rd s α t. s 4. rd s it+ 4. rd s. s 8. rd (b) θ ω αt P.5 rev min π rd π ωi rd s, ω. min 6. s. rev () ω ωi π / t s α. 5.4 s (b) ω + ωi π π θ ωt t rd
More informationPhysics 207 Lecture 5
Phsics 07 Lecture 5 Agend Phsics 07, Lecture 5, Sept. 0 Chpter 4 Kinemtics in or 3 dimensions Independence of, nd/or z components Circulr motion Cured pths nd projectile motion Frmes of reference dil nd
More informationIMPOSSIBLE NAVIGATION
Sclrs versus Vectors IMPOSSIBLE NAVIGATION The need for mgnitude AND direction Sclr: A quntity tht hs mgnitude (numer with units) ut no direction. Vector: A quntity tht hs oth mgnitude (displcement) nd
More informationFirst Semester Review Calculus BC
First Semester Review lculus. Wht is the coordinte of the point of inflection on the grph of Multiple hoice: No lcultor y 3 3 5 4? 5 0 0 3 5 0. The grph of piecewise-liner function f, for 4, is shown below.
More informationMathematics Extension 2
S Y D N E Y B O Y S H I G H S C H O O L M O O R E P A R K, S U R R Y H I L L S 005 HIGHER SCHOOL CERTIFICATE TRIAL PAPER Mthemtics Extension Generl Instructions Totl Mrks 0 Reding Time 5 Minutes Attempt
More informationPhys101 Lecture 4,5 Dynamics: Newton s Laws of Motion
Phys101 Lecture 4,5 Dynics: ewton s Lws of Motion Key points: ewton s second lw is vector eqution ction nd rection re cting on different objects ree-ody Digrs riction Inclines Ref: 4-1,2,3,4,5,6,7,8,9.
More informationMathematics Extension 2
00 HIGHER SCHOOL CERTIFICATE EXAMINATION Mthemtics Extension Generl Instructions Reding time 5 minutes Working time hours Write using blck or blue pen Bord-pproved clcultors m be used A tble of stndrd
More informationMotion. Acceleration. Part 2: Constant Acceleration. October Lab Phyiscs. Ms. Levine 1. Acceleration. Acceleration. Units for Acceleration.
Motion ccelertion Prt : Constnt ccelertion ccelertion ccelertion ccelertion is the rte of chnge of elocity. = - o t = Δ Δt ccelertion = = - o t chnge of elocity elpsed time ccelertion is ector, lthough
More informationX Fx = F A. If applied force is small, book does not move (static), a x =0, then f=f s
A Appl ewton s nd Lw X 0 X A I pplied orce is sll, boo does not ove sttic, 0, then s A Increse pplied orce, boo still does not ove Increse A ore, now boo oves, 0 > A A here is soe iu sttic rictionl orce,
More informationPROBLEM 11.3 SOLUTION
PROBLEM.3 The verticl motion of mss A is defined by the reltion x= 0 sin t+ 5cost+ 00, where x nd t re expressed in mm nd seconds, respectively. Determine () the position, velocity nd ccelertion of A when
More informationES.182A Topic 32 Notes Jeremy Orloff
ES.8A Topic 3 Notes Jerem Orloff 3 Polr coordintes nd double integrls 3. Polr Coordintes (, ) = (r cos(θ), r sin(θ)) r θ Stndrd,, r, θ tringle Polr coordintes re just stndrd trigonometric reltions. In
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 informationLinear Motion. Kinematics Quantities
Liner Motion Physics 101 Eyres Kinemtics Quntities Time Instnt t Fundmentl Time Interl Defined Position x Fundmentl Displcement Defined Aerge Velocity g Defined Aerge Accelertion g Defined 1 Kinemtics
More informationJUST THE MATHS SLIDES NUMBER INTEGRATION APPLICATIONS 12 (Second moments of an area (B)) A.J.Hobson
JUST THE MATHS SLIDES NUMBER 13.12 INTEGRATION APPLICATIONS 12 (Second moments of n re (B)) b A.J.Hobson 13.12.1 The prllel xis theorem 13.12.2 The perpendiculr xis theorem 13.12.3 The rdius of grtion
More informationCalculus AB Section I Part A A CALCULATOR MAY NOT BE USED ON THIS PART OF THE EXAMINATION
lculus Section I Prt LULTOR MY NOT US ON THIS PRT OF TH XMINTION In this test: Unless otherwise specified, the domin of function f is ssumed to e the set of ll rel numers for which f () is rel numer..
More information13.4 Work done by Constant Forces
13.4 Work done by Constnt Forces We will begin our discussion of the concept of work by nlyzing the motion of n object in one dimension cted on by constnt forces. Let s consider the following exmple: push
More informationAnalytically, vectors will be represented by lowercase bold-face Latin letters, e.g. a, r, q.
1.1 Vector Alger 1.1.1 Sclrs A physicl quntity which is completely descried y single rel numer is clled sclr. Physiclly, it is something which hs mgnitude, nd is completely descried y this mgnitude. Exmples
More informationYour Thoughts. Mechanics Lecture 16, Slide 1
Your Thoughts I get dizzy with ll the equtions being shifted, spun nd switched so much in the pre-lectures. If the pre-lectures for, 3 nd 4 re like tht, I m pretty worried. Are we going to be rcing spheres,
More informationMathematics Extension 2
00 HIGHER SCHOOL CERTIFICATE EXAMINATION Mthemtics Etension Generl Instructions Reding time 5 minutes Working time hours Write using blck or blue pen Bord-pproved clcultors my be used A tble of stndrd
More informationChapter 5 Bending Moments and Shear Force Diagrams for Beams
Chpter 5 ending Moments nd Sher Force Digrms for ems n ddition to illy loded brs/rods (e.g. truss) nd torsionl shfts, the structurl members my eperience some lods perpendiculr to the is of the bem nd will
More informationHigher Checklist (Unit 3) Higher Checklist (Unit 3) Vectors
Vectors Skill Achieved? Know tht sclr is quntity tht hs only size (no direction) Identify rel-life exmples of sclrs such s, temperture, mss, distnce, time, speed, energy nd electric chrge Know tht vector
More informationMethod of Localisation and Controlled Ejection of Swarms of Likely Charged Particles
Method of Loclistion nd Controlled Ejection of Swrms of Likely Chrged Prticles I. N. Tukev July 3, 17 Astrct This work considers Coulom forces cting on chrged point prticle locted etween the two coxil,
More informationWhen a force f(t) is applied to a mass in a system, we recall that Newton s law says that. f(t) = ma = m d dt v,
Impulse Functions In mny ppliction problems, n externl force f(t) is pplied over very short period of time. For exmple, if mss in spring nd dshpot system is struck by hmmer, the ppliction of the force
More information8A Review Solutions. Roger Mong. February 24, 2007
8A Review Solutions Roer Mon Ferury 24, 2007 Question We ein y doin Free Body Dirm on the mss m. Since the rope runs throuh the lock 3 times, the upwrd force on the lock is 3T. (Not ecuse there re 3 pulleys!)
More information/ 3, then (A) 3(a 2 m 2 + b 2 ) = 4c 2 (B) 3(a 2 + b 2 m 2 ) = 4c 2 (C) a 2 m 2 + b 2 = 4c 2 (D) a 2 + b 2 m 2 = 4c 2
SET I. If the locus of the point of intersection of perpendiculr tngents to the ellipse x circle with centre t (0, 0), then the rdius of the circle would e + / ( ) is. There re exctl two points on the
More informationPHYSICS 211 MIDTERM I 21 April 2004
PHYSICS MIDERM I April 004 Exm is closed book, closed notes. Use only your formul sheet. Write ll work nd nswers in exm booklets. he bcks of pges will not be grded unless you so request on the front of
More informationINTRODUCTION. The three general approaches to the solution of kinetics problems are:
INTRODUCTION According to Newton s lw, prticle will ccelerte when it is subjected to unblnced forces. Kinetics is the study of the reltions between unblnced forces nd the resulting chnges in motion. The
More informationDate Lesson Text TOPIC Homework. Solving for Obtuse Angles QUIZ ( ) More Trig Word Problems QUIZ ( )
UNIT 5 TRIGONOMETRI RTIOS Dte Lesson Text TOPI Homework pr. 4 5.1 (48) Trigonometry Review WS 5.1 # 3 5, 9 11, (1, 13)doso pr. 6 5. (49) Relted ngles omplete lesson shell & WS 5. pr. 30 5.3 (50) 5.3 5.4
More informationLevel I MAML Olympiad 2001 Page 1 of 6 (A) 90 (B) 92 (C) 94 (D) 96 (E) 98 (A) 48 (B) 54 (C) 60 (D) 66 (E) 72 (A) 9 (B) 13 (C) 17 (D) 25 (E) 38
Level I MAML Olympid 00 Pge of 6. Si students in smll clss took n em on the scheduled dte. The verge of their grdes ws 75. The seventh student in the clss ws ill tht dy nd took the em lte. When her score
More informationPHYSICS ASSIGNMENT-9
MPS/PHY-XII-11/A9 PHYSICS ASSIGNMENT-9 *********************************************************************************************************** 1. A wire kept long the north-south direction is llowed
More informationMath 33A Discussion Example Austin Christian October 23, Example 1. Consider tiling the plane by equilateral triangles, as below.
Mth 33A Discussion Exmple Austin Christin October 3 6 Exmple Consider tiling the plne by equilterl tringles s below Let v nd w be the ornge nd green vectors in this figure respectively nd let {v w} be
More informationSolutions to Physics: Principles with Applications, 5/E, Giancoli Chapter 16 CHAPTER 16
CHAPTER 16 1. The number of electrons is N = Q/e = ( 30.0 10 6 C)/( 1.60 10 19 C/electrons) = 1.88 10 14 electrons.. The mgnitude of the Coulomb force is Q /r. If we divide the epressions for the two forces,
More informationEdexcel GCE Core Mathematics (C2) Required Knowledge Information Sheet. Daniel Hammocks
Edexcel GCE Core Mthemtics (C) Required Knowledge Informtion Sheet C Formule Given in Mthemticl Formule nd Sttisticl Tles Booklet Cosine Rule o = + c c cosine (A) Binomil Series o ( + ) n = n + n 1 n 1
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 informationDistance And Velocity
Unit #8 - The Integrl Some problems nd solutions selected or dpted from Hughes-Hllett Clculus. Distnce And Velocity. The grph below shows the velocity, v, of n object (in meters/sec). Estimte the totl
More informationMultiple Integrals. Review of Single Integrals. Planar Area. Volume of Solid of Revolution
Multiple Integrls eview of Single Integrls eding Trim 7.1 eview Appliction of Integrls: Are 7. eview Appliction of Integrls: olumes 7.3 eview Appliction of Integrls: Lengths of Curves Assignment web pge
More information( β ) touches the x-axis if = 1
Generl Certificte of Eduction (dv. Level) Emintion, ugust Comined Mthemtics I - Prt B Model nswers. () Let f k k, where k is rel constnt. i. Epress f in the form( ) Find the turning point of f without
More information13.3 CLASSICAL STRAIGHTEDGE AND COMPASS CONSTRUCTIONS
33 CLASSICAL STRAIGHTEDGE AND COMPASS CONSTRUCTIONS As simple ppliction of the results we hve obtined on lgebric extensions, nd in prticulr on the multiplictivity of extension degrees, we cn nswer (in
More informationat its center, then the measure of this angle in radians (abbreviated rad) is the length of the arc that subtends the angle.
Notes 6 ngle Mesure Definition of Rdin If circle of rdius is drwn with the vertex of n ngle Mesure: t its center, then the mesure of this ngle in rdins (revited rd) is the length of the rc tht sutends
More informationLesson 8.1 Graphing Parametric Equations
Lesson 8.1 Grphing Prmetric Equtions 1. rete tle for ech pir of prmetric equtions with the given vlues of t.. x t 5. x t 3 c. x t 1 y t 1 y t 3 y t t t {, 1, 0, 1, } t {4,, 0,, 4} t {4, 0,, 4, 8}. Find
More information