2008 QuarterFinal Exam Solutions


 Brandon Lawrence
 2 years ago
 Views:
Transcription
1 2008 Quatefinal Exam  Solutions QuateFinal Exam Solutions 1 A chaged paticle with chage q and mass m stats with an initial kinetic enegy K at the middle of a unifomly chaged spheical egion of total chage Q and adius R q and Q have opposite signs The spheically chaged egion is not fee to move (a) Find the value of K 0 such that the paticle will just each the bounday of the spheically chaged egion (b) How much time does it take fo the paticle to each the bounday of the egion if it stats with the kinetic enegy K 0 found in pat (a)? Solution: Assume that q is negative and Q is positive (a) Apply Gauss s Law to a spheical shell of adius whee < R Then, whee the chage density ρ = 3Q 4πR 3 E4π 2 = 4πρ3 ɛ 0 Solving fo the Electic Field at a distance fom the cente, we find E = ρ 3ɛ 0 We now find the potential diffeence between the cente of the sphee and the oute bounday of the chaged cloud V = R Substituting in ou expession fo E, we now find 0 Ed Using consevation of enegy, V = ρr2 6ɛ 0 K lost = U gained K 0 = q V = qq 8πRɛ 0 (b) Apply Newton s Second Law to the object of chage q when it is located a distance away fom the cente of the sphee
2 2008 Quatefinal Exam  Solutions 2 F net = ma ( q)e = m d2 dt 2 d 2 dt 2 = qρ 3ɛ 0 m We ecognize that this is the diffeential equation fo simple hamonic motion whee ω = qρ 3ɛ 0 m d 2 dt 2 = ω2 Since the chage has the minimum kinetic enegy needed to each the suface, the tip fom the cente of the sphee to the oute bounday is onefouth of a cycle of SHM Theefoe, Substituting, in fo ho, we find t = T 4 = π 2ω = π 3ɛ0 m 2 qρ t = π 2 4πɛ 0 mr 3 2 A unifom pool ball of adius begins at est on a pool table The ball is given a hoizontal impulse J of fixed magnitude at a distance β above its cente, whee 1 β 1 The coefficient of kinetic fiction between the ball and the pool table is µ You may assume the ball and the table ae pefectly igid Ignoe effects due to defomation (The moment of inetia about the cente of mass of a solid sphee of mass m and adius is I cm = 2 5 m2 ) qq (a) Find an expession fo the final speed of the ball as a function of J, m, and β (b) Fo what value of β does the ball immediately begin to oll without slipping, egadless of the value of µ Solution 1: (a) Conside an axis pependicula to the initial impulse and coplana with the table (Thoughout this solution we conside only toques and angula momenta with espect to this axis) Afte the initial impulse, the toque about this axis is always zeo, so angula momentum is conseved The initial impulse occus a pependicula distance (β + 1) fom the axis, so the angula momentum is L = (β + 1) J Afte the ball has skidded along the table fo a cetain distance, it will begin to oll without slipping At that point, ω = v f
3 2008 Quatefinal Exam  Solutions 3 Meanwhile, its moment of inetia about this axis is (by the paallelaxis theoem) I = I cm + m 2 = 7 5 m2, so that its final angula velocity ω is given by L = Iω (β + 1) J = 7 5 m2 ω Theefoe, (β + 1) J = 7 5 mv f v f = 5J (1 + β) 7m (b) The ball acquies linea momentum J as a esult of the hoizontal impulse, so its initial velocity v is given by mv = J We want the initial angula velocity and initial velocity to satisfy the noslip condition v = ω; thus (β + 1) J = 7 5 J Solution 2: β = 2 5 Conside toques and angula momenta about the cente of mass If the hoizontal impulse is lage compaed with the hoizontal impulse fom fiction duing the time that the cue stick is in contact with the ball, then angula impulse = change in angula momentum becomes: Jβ = 2 5 m2 ω 0 Linea impulse = change in linea momentum yields: J = mv 0 (b) We want the initial angula velocity and initial velocity to satisfy the noslip condition v 0 = ω 0 ; thus Jβ = 2 5 mv 0 = 2 5 J β = 2 5
4 2008 Quatefinal Exam  Solutions 4 (a) In the case that the ball does not immediately begin to oll without slipping, fiction will exet an angula impulse, ft, about the cente of mass of the ball as the ball skids along the suface: ft = 2 5 m2 (ω f ω 0 ) Fiction will also exet a linea impulse ft that will cause a change in linea momentum: Combining the last two equations: ft = m(v f v 0 ) m(v f v 0 ) = 2 5 m(ω f ω 0 ) Afte the ball has skidded along the table fo a cetain distance, it will begin to oll without slipping At that point, ω = v f v f + v 0 = 2 5 v f 2 5 ω 0 Now, using the elationships between the impulse J and the initial angula and linea momentum, v f + J m = 2 5 v f Jβ m J m (1 + β) = 7 5 v f v f = 5J (1 + β) 7m 3 A block of mass m slides on a cicula tack of adius whose wall and floo both have coefficient of kinetic fiction µ with the block The floo lies in a hoizontal plane and the wall is vetical The block is in constant contact with both the wall and the floo The block has initial speed v 0 (a) Let the block have kinetic enegy E afte taveling though an angle θ Deive an expession fo de dθ in tems of g,, µ, m and E (b) Suppose the block cicles the tack exactly once befoe coming to a halt Detemine v 0 in tems of g,, and µ
5 2008 Quatefinal Exam  Solutions 5 Solution Let v be the speed of the block afte it has taveled though an angle θ When we apply Newton s Second Law to the block as it is taveling aound the cicula path of adius at a speed v, we find that the foce of the wall on the block, F w, is and the foce of the floo on the block, F N, is F w = mv2 F N = mg Theefoe, the total foce of fiction on the block is F F ic = µm( v2 + g) Let de denote the loss of kinetic enegy as a esult of the wok W that fiction does on the block when it has taveled though an angle dθ The block will tavel a distance ds = dθ as it tavels though an angle dθ Then, de = W = F F ic ds = F F ic dθ Now, substituting in the expessions fo F F ic, we find de = µm(v 2 + g)dθ de dθ = µ(mv2 + mg) = µ(2e + mg) whee we have simply used the fact that the kinetic enegy E is E = mv2 2 (b) Now, we sepaate vaiables and solve the diffeential equation Exponentiating both sides, yields de 2E + mg = µdθ 1 ln 2E + mg = µθ + C 2 2E + mg = Ce 2µθ Let E 0 denote the initial kinetic enegy Then, using initial conditions, we find that C = 2E 0 + mg Now, using the fact that E = 0 when θ = 2π, we obtain mg = (2E 0 + mg)e 4πµ
6 2008 Quatefinal Exam  Solutions 6 Solving fo E 0, Then, using E 0 = mv2 0 2, and we find that E 0 = 1 2 mg(e4µπ 1) v 0 = g(e 4µπ 1) 4 Two beads, each of mass m, ae fee to slide on a igid, vetical hoop of mass m h The beads ae theaded on the hoop so that they cannot fall off of the hoop They ae eleased with negligible velocity at the top of the hoop and slide down to the bottom in opposite diections What is the maximum value of the atio m/m h such that the hoop always emains in contact with the gound? Neglect fiction Solution 1: Daw a feebody diagam fo each bead; let F N be the (inwad) nomal foce exeted by the hoop on the bead Let θ be the angula position of the bead, measued fom the top of the hoop, and let the hoop have adius We see that F N + mg cos θ = m v2 F N = m v2 mg cos θ The (downwad) vetical component F Ny is given by F Ny = F N cos θ Fom Newton s thid law, the two beads togethe exet an upwads vetical foce on the hoop given by F u = 2F Ny ( ) v 2 F u = 2m cos θ g cos θ noting that the beads clealy each the same position at the same time Meanwhile, when each bead is at a position θ it has moved though a vetical distance (1 cos θ) Thus fom enegy consevation, 1 2 mv2 = mg (1 cos θ) Inseting this into the pevious esult, v 2 = 2g (1 cos θ) F u = 2m cos θ (2g (1 cos θ) g cos θ) F u = 2mg ( 2 cos θ 3 cos 2 θ )
7 2008 Quatefinal Exam  Solutions 7 If the beads eve exet an upwad foce on the hoop geate than m h g, the hoop will leave the gound; ie, the condition fo the hoop to emain in contact with the gound is that fo all θ, F u m h g We can eplace the left side by its maximum value Letting s = cos θ, The deivative is zeo at s = 1 3, whee F u = 2mg ( 2s 3s 2) d ds F u = 2mg (2 6s) F u(max) = 2 3 mg So ou condition is 2 3 mg m hg m 3 m h 2 Solution 2: As befoe, we apply enegy consevation to find the speed of the beads: v = 2g (1 cos θ) The vetical (downwad) component of the beads velocity is thus v y = v sin θ v y = sin θ 2g (1 cos θ) While the hoop is in contact with the gound, the beads ae the only pat of the system in motion, so the momentum of the system is simply the beads momentum: The net (downwad) foce on the system is p y = 2mv y p y = 2m sin θ 2g (1 cos θ) F net = dp y dt = dp y dθ dθ dt = dp y dθ ) F net = 2m 2g d ( sin θ 1 cos θ 1 2g (1 cos θ) dt ( F net = 4mg cos θ ) cos θ + sin θ 2 1 cos θ sin θ cos θ v ( F net = 4mg cos θ cos 2 θ sin2 θ F net = 2mg ( 2 cos θ 3 cos 2 θ + 1 ) )
8 2008 Quatefinal Exam  Solutions 8 This downwad foce is povided by gavity and the nomal foce upwad on the hoop: F net = 2mg + m h g F N If the hoop is to emain in contact with the gound, the nomal foce can neve be negative: F N 0 2mg + m h g F net 0 2mg + m h g 2mg ( 2 cos θ 3 cos 2 θ + 1 ) 0 2mg ( 2 cos θ 3 cos 2 θ ) m h g and we poceed as above
12. Rolling, Torque, and Angular Momentum
12. olling, Toque, and Angula Momentum 1 olling Motion: A motion that is a combination of otational and tanslational motion, e.g. a wheel olling down the oad. Will only conside olling with out slipping.
More informationPY1052 Problem Set 8 Autumn 2004 Solutions
PY052 Poblem Set 8 Autumn 2004 Solutions H h () A solid ball stats fom est at the uppe end of the tack shown and olls without slipping until it olls off the ighthand end. If H 6.0 m and h 2.0 m, what
More informationExam 3: Equation Summary
MASSACHUSETTS INSTITUTE OF TECHNOLOGY Depatment of Physics Physics 8.1 TEAL Fall Tem 4 Momentum: p = mv, F t = p, Fext ave t= t f t= Exam 3: Equation Summay total = Impulse: I F( t ) = p Toque: τ = S S,P
More informationPHYSICS 111 HOMEWORK SOLUTION #5. March 3, 2013
PHYSICS 111 HOMEWORK SOLUTION #5 Mach 3, 2013 0.1 You 3.80kg physics book is placed next to you on the hoizontal seat of you ca. The coefficient of static fiction between the book and the seat is 0.650,
More informationrotation  Conservation of mechanical energy for rotation  Angular momentum  Conservation of angular momentum
Final Exam Duing class (13:55 pm) on 6/7, Mon Room: 41 FMH (classoom) Bing scientific calculatos No smat phone calculatos l ae allowed. Exam coves eveything leaned in this couse. Review session: Thusday
More informationPhys 2101 Gabriela González. cos. sin. sin
1 Phys 101 Gabiela González a m t t ma ma m m T α φ ω φ sin cos α τ α φ τ sin m m α τ I We know all of that aleady!! 3 The figue shows the massive shield doo at a neuton test facility at Lawence Livemoe
More informationRevision Guide for Chapter 11
Revision Guide fo Chapte 11 Contents Student s Checklist Revision Notes Momentum... 4 Newton's laws of motion... 4 Gavitational field... 5 Gavitational potential... 6 Motion in a cicle... 7 Summay Diagams
More informationPHYSICS 111 HOMEWORK SOLUTION #13. May 1, 2013
PHYSICS 111 HOMEWORK SOLUTION #13 May 1, 2013 0.1 In intoductoy physics laboatoies, a typical Cavendish balance fo measuing the gavitational constant G uses lead sphees with masses of 2.10 kg and 21.0
More information14. Gravitation Universal Law of Gravitation (Newton):
14. Gavitation 1 Univesal Law of Gavitation (ewton): The attactive foce between two paticles: F = G m 1m 2 2 whee G = 6.67 10 11 m 2 / kg 2 is the univesal gavitational constant. F m 2 m 1 F Paticle #1
More informationChapter 22. Outside a uniformly charged sphere, the field looks like that of a point charge at the center of the sphere.
Chapte.3 What is the magnitude of a point chage whose electic field 5 cm away has the magnitude of.n/c. E E 5.56 1 11 C.5 An atom of plutonium39 has a nuclea adius of 6.64 fm and atomic numbe Z94. Assuming
More informationGauss Law. Physics 231 Lecture 21
Gauss Law Physics 31 Lectue 1 lectic Field Lines The numbe of field lines, also known as lines of foce, ae elated to stength of the electic field Moe appopiately it is the numbe of field lines cossing
More informationProblems on Force Exerted by a Magnetic Fields from Ch 26 T&M
Poblems on oce Exeted by a Magnetic ields fom Ch 6 TM Poblem 6.7 A cuentcaying wie is bent into a semicicula loop of adius that lies in the xy plane. Thee is a unifom magnetic field B Bk pependicula to
More informationProblem Set 6: Solutions
UNIVESITY OF ALABAMA Depatment of Physics and Astonomy PH 164 / LeClai Fall 28 Poblem Set 6: Solutions 1. Seway 29.55 Potons having a kinetic enegy of 5. MeV ae moving in the positive x diection and ente
More informationChapter 13 Gravitation. Problems: 1, 4, 5, 7, 18, 19, 25, 29, 31, 33, 43
Chapte 13 Gavitation Poblems: 1, 4, 5, 7, 18, 19, 5, 9, 31, 33, 43 Evey object in the univese attacts evey othe object. This is called gavitation. We e use to dealing with falling bodies nea the Eath.
More information10. Collisions. Before During After
10. Collisions Use conseation of momentum and enegy and the cente of mass to undestand collisions between two objects. Duing a collision, two o moe objects exet a foce on one anothe fo a shot time: F(t)
More informationDeflection of Electrons by Electric and Magnetic Fields
Physics 233 Expeiment 42 Deflection of Electons by Electic and Magnetic Fields Refeences Loain, P. and D.R. Coson, Electomagnetism, Pinciples and Applications, 2nd ed., W.H. Feeman, 199. Intoduction An
More informationMagnetic Field and Magnetic Forces. Young and Freedman Chapter 27
Magnetic Field and Magnetic Foces Young and Feedman Chapte 27 Intoduction Reiew  electic fields 1) A chage (o collection of chages) poduces an electic field in the space aound it. 2) The electic field
More informationHour Exam No.1. p 1 v. p = e 0 + v^b. Note that the probe is moving in the direction of the unit vector ^b so the velocity vector is just ~v = v^b and
Hou Exam No. Please attempt all of the following poblems befoe the due date. All poblems count the same even though some ae moe complex than othes. Assume that c units ae used thoughout. Poblem A photon
More information2 r2 θ = r2 t. (3.59) The equal area law is the statement that the term in parentheses,
3.4. KEPLER S LAWS 145 3.4 Keple s laws You ae familia with the idea that one can solve some mechanics poblems using only consevation of enegy and (linea) momentum. Thus, some of what we see as objects
More informationTORQUE AND ANGULAR MOMENTUM IN CIRCULAR MOTION
MISN034 TORQUE AND ANGULAR MOMENTUM IN CIRCULAR MOTION shaft TORQUE AND ANGULAR MOMENTUM IN CIRCULAR MOTION by Kiby Mogan, Chalotte, Michigan 1. Intoduction..............................................
More informationFXA 2008. Candidates should be able to : Describe how a mass creates a gravitational field in the space around it.
Candidates should be able to : Descibe how a mass ceates a gavitational field in the space aound it. Define gavitational field stength as foce pe unit mass. Define and use the peiod of an object descibing
More informationAlgebra and Trig. I. A point is a location or position that has no size or dimension.
Algeba and Tig. I 4.1 Angles and Radian Measues A Point A A B Line AB AB A point is a location o position that has no size o dimension. A line extends indefinitely in both diections and contains an infinite
More informationGravitation. AP Physics C
Gavitation AP Physics C Newton s Law of Gavitation What causes YOU to be pulled down? THE EARTH.o moe specifically the EARTH S MASS. Anything that has MASS has a gavitational pull towads it. F α Mm g What
More informationGravitation and Kepler s Laws Newton s Law of Universal Gravitation in vectorial. Gm 1 m 2. r 2
F Gm Gavitation and Keple s Laws Newton s Law of Univesal Gavitation in vectoial fom: F 12 21 Gm 1 m 2 12 2 ˆ 12 whee the hat (ˆ) denotes a unit vecto as usual. Gavity obeys the supeposition pinciple,
More informationThe force between electric charges. Comparing gravity and the interaction between charges. Coulomb s Law. Forces between two charges
The foce between electic chages Coulomb s Law Two chaged objects, of chage q and Q, sepaated by a distance, exet a foce on one anothe. The magnitude of this foce is given by: kqq Coulomb s Law: F whee
More informationVoltage ( = Electric Potential )
V1 of 9 Voltage ( = lectic Potential ) An electic chage altes the space aound it. Thoughout the space aound evey chage is a vecto thing called the electic field. Also filling the space aound evey chage
More informationThe Electric Potential, Electric Potential Energy and Energy Conservation. V = U/q 0. V = U/q 0 = W/q 0 1V [Volt] =1 Nm/C
Geneal Physics  PH Winte 6 Bjoen Seipel The Electic Potential, Electic Potential Enegy and Enegy Consevation Electic Potential Enegy U is the enegy of a chaged object in an extenal electic field (Unit
More informationMultiple choice questions [60 points]
1 Multiple choice questions [60 points] Answe all o the ollowing questions. Read each question caeully. Fill the coect bubble on you scanton sheet. Each question has exactly one coect answe. All questions
More informationExperiment 6: Centripetal Force
Name Section Date Intoduction Expeiment 6: Centipetal oce This expeiment is concened with the foce necessay to keep an object moving in a constant cicula path. Accoding to Newton s fist law of motion thee
More informationMechanics 1: Work, Power and Kinetic Energy
Mechanics 1: Wok, Powe and Kinetic Eneg We fist intoduce the ideas of wok and powe. The notion of wok can be viewed as the bidge between Newton s second law, and eneg (which we have et to define and discuss).
More informationVoltage ( = Electric Potential )
V1 Voltage ( = Electic Potential ) An electic chage altes the space aound it. Thoughout the space aound evey chage is a vecto thing called the electic field. Also filling the space aound evey chage is
More informationVector Calculus: Are you ready? Vectors in 2D and 3D Space: Review
Vecto Calculus: Ae you eady? Vectos in D and 3D Space: Review Pupose: Make cetain that you can define, and use in context, vecto tems, concepts and fomulas listed below: Section 7.7. find the vecto defined
More informationSolutions for Physics 1301 Course Review (Problems 10 through 18)
Solutions fo Physics 1301 Couse Review (Poblems 10 though 18) 10) a) When the bicycle wheel comes into contact with the step, thee ae fou foces acting on it at that moment: its own weight, Mg ; the nomal
More informationSolution Derivations for Capa #8
Solution Deivations fo Capa #8 1) A ass spectoete applies a voltage of 2.00 kv to acceleate a singly chaged ion (+e). A 0.400 T field then bends the ion into a cicula path of adius 0.305. What is the ass
More informationDetermining solar characteristics using planetary data
Detemining sola chaacteistics using planetay data Intoduction The Sun is a G type main sequence sta at the cente of the Sola System aound which the planets, including ou Eath, obit. In this inestigation
More information7 Circular Motion. 71 Centripetal Acceleration and Force. Period, Frequency, and Speed. Vocabulary
7 Cicula Motion 71 Centipetal Acceleation and Foce Peiod, Fequency, and Speed Vocabulay Vocabulay Peiod: he time it takes fo one full otation o evolution of an object. Fequency: he numbe of otations o
More informationProblems of the 2 nd International Physics Olympiads (Budapest, Hungary, 1968)
Poblems of the nd ntenational Physics Olympiads (Budapest Hungay 968) Péte Vankó nstitute of Physics Budapest Univesity of Technical Engineeing Budapest Hungay Abstact Afte a shot intoduction the poblems
More information1240 ev nm 2.5 ev. (4) r 2 or mv 2 = ke2
Chapte 5 Example The helium atom has 2 electonic enegy levels: E 3p = 23.1 ev and E 2s = 20.6 ev whee the gound state is E = 0. If an electon makes a tansition fom 3p to 2s, what is the wavelength of the
More informationAP Physics Electromagnetic Wrap Up
AP Physics Electomagnetic Wap Up Hee ae the gloious equations fo this wondeful section. F qsin This is the equation fo the magnetic foce acting on a moing chaged paticle in a magnetic field. The angle
More information81 Newton s Law of Universal Gravitation
81 Newton s Law of Univesal Gavitation One of the most famous stoies of all time is the stoy of Isaac Newton sitting unde an apple tee and being hit on the head by a falling apple. It was this event,
More informationLINES AND TANGENTS IN POLAR COORDINATES
LINES AND TANGENTS IN POLAR COORDINATES ROGER ALEXANDER DEPARTMENT OF MATHEMATICS 1. Polacoodinate equations fo lines A pola coodinate system in the plane is detemined by a point P, called the pole, and
More informationIntroduction to Fluid Mechanics
Chapte 1 1 1.6. Solved Examples Example 1.1 Dimensions and Units A body weighs 1 Ibf when exposed to a standad eath gavity g = 3.174 ft/s. (a) What is its mass in kg? (b) What will the weight of this body
More informationDisplacement, Velocity And Acceleration
Displacement, Velocity And Acceleation Vectos and Scalas Position Vectos Displacement Speed and Velocity Acceleation Complete Motion Diagams Outline Scala vs. Vecto Scalas vs. vectos Scala : a eal numbe,
More informationPhysics 235 Chapter 5. Chapter 5 Gravitation
Chapte 5 Gavitation In this Chapte we will eview the popeties of the gavitational foce. The gavitational foce has been discussed in geat detail in you intoductoy physics couses, and we will pimaily focus
More informationFluids Lecture 15 Notes
Fluids Lectue 15 Notes 1. Unifom flow, Souces, Sinks, Doublets Reading: Andeson 3.9 3.12 Unifom Flow Definition A unifom flow consists of a velocit field whee V = uî + vĵ is a constant. In 2D, this velocit
More informationLab #7: Energy Conservation
Lab #7: Enegy Consevation Photo by Kallin http://www.bungeezone.com/pics/kallin.shtml Reading Assignment: Chapte 7 Sections 1,, 3, 5, 6 Chapte 8 Sections 14 Intoduction: Pehaps one of the most unusual
More informationGauss s law relates to total electric flux through a closed surface to the total enclosed charge.
Chapte : Gauss s Law Gauss s Law is an altenative fomulation of the elation between an electic field and the souces of that field in tems of electic flu. lectic Flu Φ though an aea ~ Numbe of Field Lines
More informationThe Role of Gravity in Orbital Motion
! The Role of Gavity in Obital Motion Pat of: Inquiy Science with Datmouth Developed by: Chistophe Caoll, Depatment of Physics & Astonomy, Datmouth College Adapted fom: How Gavity Affects Obits (Ohio State
More information2  ELECTROSTATIC POTENTIAL AND CAPACITANCE Page 1
 ELECTROSTATIC POTENTIAL AND CAPACITANCE Page. Line Integal of Electic Field If a unit positive chage is displaced by `given by dw E. dl dl in an electic field of intensity E, wok done is Line integation
More informationForces & Magnetic Dipoles. r r τ = μ B r
Foces & Magnetic Dipoles x θ F θ F. = AI τ = U = Fist electic moto invented by Faaday, 1821 Wie with cuent flow (in cup of Hg) otates aound a a magnet Faaday s moto Wie with cuent otates aound a Pemanent
More informationNewton s Shell Theorem
Newton Shell Theoem Abtact One of the pincipal eaon Iaac Newton wa motivated to invent the Calculu wa to how that in applying hi Law of Univeal Gavitation to pheicallyymmetic maive bodie (like planet,
More informationIntroduction to Electric Potential
Univesiti Teknologi MARA Fakulti Sains Gunaan Intoduction to Electic Potential : A Physical Science Activity Name: HP: Lab # 3: The goal of today s activity is fo you to exploe and descibe the electic
More informationCh. 8 Universal Gravitation. Part 1: Kepler s Laws. Johannes Kepler. Tycho Brahe. Brahe. Objectives: Section 8.1 Motion in the Heavens and on Earth
Ch. 8 Univesal Gavitation Pat 1: Keple s Laws Objectives: Section 8.1 Motion in the Heavens and on Eath Objectives Relate Keple s laws of planetay motion to Newton s law of univesal gavitation. Calculate
More informationMechanics 1: Motion in a Central Force Field
Mechanics : Motion in a Cental Foce Field We now stud the popeties of a paticle of (constant) ass oving in a paticula tpe of foce field, a cental foce field. Cental foces ae ve ipotant in phsics and engineeing.
More informationMoment and couple. In 3D, because the determination of the distance can be tedious, a vector approach becomes advantageous. r r
Moment and couple In 3D, because the detemination of the distance can be tedious, a vecto appoach becomes advantageous. o k j i M k j i M o ) ( ) ( ) ( + + M o M + + + + M M + O A Moment about an abita
More information(a) The centripetal acceleration of a point on the equator of the Earth is given by v2. The velocity of the earth can be found by taking the ratio of
Homewok VI Ch. 7  Poblems 15, 19, 22, 25, 35, 43, 51. Poblem 15 (a) The centipetal acceleation of a point on the equato of the Eath is given by v2. The velocity of the eath can be found by taking the
More informationEpisode 401: Newton s law of universal gravitation
Episode 401: Newton s law of univesal gavitation This episode intoduces Newton s law of univesal gavitation fo point masses, and fo spheical masses, and gets students pactising calculations of the foce
More informationNuclear models: FermiGas Model Shell Model
Lectue Nuclea mode: FemiGas Model Shell Model WS/: Intoduction to Nuclea and Paticle Physics The basic concept of the Femigas model The theoetical concept of a Femigas may be applied fo systems of weakly
More informationProblems of the 2 nd and 9 th International Physics Olympiads (Budapest, Hungary, 1968 and 1976)
Poblems of the nd and 9 th Intenational Physics Olympiads (Budapest Hungay 968 and 976) Péte Vankó Institute of Physics Budapest Univesity of Technology and Economics Budapest Hungay Abstact Afte a shot
More informationMultiple choice questions [70 points]
Multiple choice questions [70 points] Answe all of the following questions. Read each question caefull. Fill the coect bubble on ou scanton sheet. Each question has exactl one coect answe. All questions
More informationGravity. A. Law of Gravity. Gravity. Physics: Mechanics. A. The Law of Gravity. Dr. Bill Pezzaglia. B. Gravitational Field. C.
Physics: Mechanics 1 Gavity D. Bill Pezzaglia A. The Law of Gavity Gavity B. Gavitational Field C. Tides Updated: 01Jul09 A. Law of Gavity 3 1a. Invese Squae Law 4 1. Invese Squae Law. Newton s 4 th law
More informationCoordinate Systems L. M. Kalnins, March 2009
Coodinate Sstems L. M. Kalnins, Mach 2009 Pupose of a Coodinate Sstem The pupose of a coodinate sstem is to uniquel detemine the position of an object o data point in space. B space we ma liteall mean
More informationCarterPenrose diagrams and black holes
CatePenose diagams and black holes Ewa Felinska The basic intoduction to the method of building Penose diagams has been pesented, stating with obtaining a Penose diagam fom Minkowski space. An example
More informationChapter 2. Electrostatics
Chapte. Electostatics.. The Electostatic Field To calculate the foce exeted by some electic chages,,, 3,... (the souce chages) on anothe chage Q (the test chage) we can use the pinciple of supeposition.
More informationFigure 1.1: Fluid flow in a plane narrow slit.
Poblem Conside a fluid of density ρ in incompessible, lamina flow in a plane naow slit of length L and width W fomed by two flat paallel walls that ae a distance 2B apat. End effects may be neglected because
More informationChapter 9 Rotation. Conceptual Problems
Chapte 9 otation Conceptual Poblems 7 Duing a baseball game, the pitche has a blazing astball. You have not been able to swing the bat in time to hit the ball. You ae now just tying to make the bat contact
More informationTheory and measurement
Gavity: Theoy and measuement Reading: Today: p11  Theoy of gavity Use two of Newton s laws: 1) Univesal law of gavitation: ) Second law of motion: Gm1m F = F = mg We can combine them to obtain the gavitational
More informationEXPERIMENT 16 THE MAGNETIC MOMENT OF A BAR MAGNET AND THE HORIZONTAL COMPONENT OF THE EARTH S MAGNETIC FIELD
260 161. THEORY EXPERMENT 16 THE MAGNETC MOMENT OF A BAR MAGNET AND THE HORZONTAL COMPONENT OF THE EARTH S MAGNETC FELD The uose of this exeiment is to measue the magnetic moment μ of a ba magnet and
More informationChapter 4: Fluid Kinematics
Oveview Fluid kinematics deals with the motion of fluids without consideing the foces and moments which ceate the motion. Items discussed in this Chapte. Mateial deivative and its elationship to Lagangian
More informationMotion Control Formulas
ems: A = acceleation ate {in/sec } C = caiage thust foce {oz} D = deceleation ate {in/sec } d = lead of scew {in/ev} e = lead scew efficiency ball scew 90% F = total fictional foce {oz} GR = gea atio J
More informationUNIT 21: ELECTRICAL AND GRAVITATIONAL POTENTIAL Approximate time two 100minute sessions
Name St.No.  Date(YY/MM/DD) / / Section Goup# UNIT 21: ELECTRICAL AND GRAVITATIONAL POTENTIAL Appoximate time two 100minute sessions OBJECTIVES I began to think of gavity extending to the ob of the moon,
More informationExperiment MF Magnetic Force
Expeiment MF Magnetic Foce Intoduction The magnetic foce on a cuentcaying conducto is basic to evey electic moto  tuning the hands of electic watches and clocks, tanspoting tape in Walkmans, stating
More informationUNIT CIRCLE TRIGONOMETRY
UNIT CIRCLE TRIGONOMETRY The Unit Cicle is the cicle centeed at the oigin with adius unit (hence, the unit cicle. The equation of this cicle is + =. A diagam of the unit cicle is shown below: + =   
More informationF G r. Don't confuse G with g: "Big G" and "little g" are totally different things.
G1 Gavity Newton's Univesal Law of Gavitation (fist stated by Newton): any two masses m 1 and m exet an attactive gavitational foce on each othe accoding to m m G 1 This applies to all masses, not just
More informationChapter 10 Angular Momentum
Chapte 0 Angula Momentum Conceptual Poblems 5 A paticle tavels in a cicula path and point P is at the cente o the cicle. (a) the paticle s linea momentum p is doubled without changing the adius o the cicle,
More informationStructure and evolution of circumstellar disks during the early phase of accretion from a parent cloud
Cente fo Tubulence Reseach Annual Reseach Biefs 2001 209 Stuctue and evolution of cicumstella disks duing the ealy phase of accetion fom a paent cloud By Olusola C. Idowu 1. Motivation and Backgound The
More informationPhysics HSC Course Stage 6. Space. Part 1: Earth s gravitational field
Physics HSC Couse Stage 6 Space Pat 1: Eath s gavitational field Contents Intoduction... Weight... 4 The value of g... 7 Measuing g...8 Vaiations in g...11 Calculating g and W...13 You weight on othe
More informationChapter 4. Electric Potential
Chapte 4 Electic Potential 4.1 Potential and Potential Enegy... 43 4.2 Electic Potential in a Unifom Field... 47 4.3 Electic Potential due to Point Chages... 48 4.3.1 Potential Enegy in a System of
More informationOrbital Motion & Gravity
Astonomy: Planetay Motion 1 Obital Motion D. Bill Pezzaglia A. Galileo & Fee Fall Obital Motion & Gavity B. Obits C. Newton s Laws Updated: 013Ma05 D. Einstein A. Galileo & Fee Fall 3 1. Pojectile Motion
More informationESCAPE VELOCITY EXAMPLES
ESCAPE VELOCITY EXAMPLES 1. Escape velocity is the speed that an object needs to be taveling to beak fee of planet o moon's gavity and ente obit. Fo example, a spacecaft leaving the suface of Eath needs
More informationVIII. Magnetic Fields  Worked Examples
MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics 8.0 Spring 003 VIII. Magnetic Fields  Worked Examples Example : Rolling rod A rod with a mass m and a radius R is mounted on two parallel rails
More informationClassical Mechanics (CM):
Classical Mechanics (CM): We ought to have some backgound to aeciate that QM eally does just use CM and makes one slight modification that then changes the natue of the oblem we need to solve but much
More informationBead moving along a thin, rigid, wire.
Bead moving along a thin, rigid, wire. odolfo. osales, Department of Mathematics, Massachusetts Inst. of Technology, Cambridge, Massachusetts, MA 02139 October 17, 2004 Abstract An equation describing
More informationCHAPTER 5 GRAVITATIONAL FIELD AND POTENTIAL
CHATER 5 GRAVITATIONAL FIELD AND OTENTIAL 5. Intoduction. This chapte deals with the calculation of gavitational fields and potentials in the vicinity of vaious shapes and sizes of massive bodies. The
More information4a 4ab b 4 2 4 2 5 5 16 40 25. 5.6 10 6 (count number of places from first nonzero digit to
. Simplify: 0 4 ( 8) 0 64 ( 8) 0 ( 8) = (Ode of opeations fom left to ight: Paenthesis, Exponents, Multiplication, Division, Addition Subtaction). Simplify: (a 4) + (a ) (a+) = a 4 + a 0 a = a 7. Evaluate
More informationPoynting Vector and Energy Flow in a Capacitor Challenge Problem Solutions
Poynting Vecto an Enegy Flow in a Capacito Challenge Poblem Solutions Poblem 1: A paallelplate capacito consists of two cicula plates, each with aius R, sepaate by a istance. A steay cuent I is flowing
More informationdz + η 1 r r 2 + c 1 ln r + c 2 subject to the boundary conditions of noslip side walls and finite force over the fluid length u z at r = 0
Poiseuille Flow Jean Louis Maie Poiseuille, a Fench physicist and physiologist, was inteested in human blood flow and aound 1840 he expeimentally deived a law fo flow though cylindical pipes. It s extemely
More informationChapter 13. VectorValued Functions and Motion in Space 13.6. Velocity and Acceleration in Polar Coordinates
13.6 Velocity and Acceleation in Pola Coodinates 1 Chapte 13. VectoValued Functions and Motion in Space 13.6. Velocity and Acceleation in Pola Coodinates Definition. When a paticle P(, θ) moves along
More informationLab M4: The Torsional Pendulum and Moment of Inertia
M4.1 Lab M4: The Tosional Pendulum and Moment of netia ntoduction A tosional pendulum, o tosional oscillato, consists of a disklike mass suspended fom a thin od o wie. When the mass is twisted about the
More informationChapter 19: Electric Charges, Forces, and Fields ( ) ( 6 )( 6
Chapte 9 lectic Chages, Foces, an Fiels 6 9. One in a million (0 ) ogen molecules in a containe has lost an electon. We assume that the lost electons have been emove fom the gas altogethe. Fin the numbe
More informationNotes on Electric Fields of Continuous Charge Distributions
Notes on Electic Fields of Continuous Chage Distibutions Fo discete pointlike electic chages, the net electic field is a vecto sum of the fields due to individual chages. Fo a continuous chage distibution
More informationFunctions of a Random Variable: Density. Math 425 Intro to Probability Lecture 30. Definition Nice Transformations. Problem
Intoduction One Function of Random Vaiables Functions of a Random Vaiable: Density Math 45 Into to Pobability Lectue 30 Let gx) = y be a onetoone function whose deiatie is nonzeo on some egion A of the
More informationIn the lecture on double integrals over nonrectangular domains we used to demonstrate the basic idea
Double Integals in Pola Coodinates In the lectue on double integals ove nonectangula domains we used to demonstate the basic idea with gaphics and animations the following: Howeve this paticula example
More informationUniform Rectilinear Motion
Engineeing Mechanics : Dynamics Unifom Rectilinea Motion Fo paticle in unifom ectilinea motion, the acceleation is zeo and the elocity is constant. d d t constant t t 111 Engineeing Mechanics : Dynamics
More informationProblem Set # 9 Solutions
Poblem Set # 9 Solutions Chapte 12 #2 a. The invention of the new highspeed chip inceases investment demand, which shifts the cuve out. That is, at evey inteest ate, fims want to invest moe. The incease
More informationChapter 4: Fluid Kinematics
41 Lagangian g and Euleian Desciptions 42 Fundamentals of Flow Visualization 43 Kinematic Desciption 44 Reynolds Tanspot Theoem (RTT) 41 Lagangian and Euleian Desciptions (1) Lagangian desciption
More informationVISCOSITY OF BIODIESEL FUELS
VISCOSITY OF BIODIESEL FUELS One of the key assumptions fo ideal gases is that the motion of a given paticle is independent of any othe paticles in the system. With this assumption in place, one can use
More informationWorked Examples. v max =?
Exaple iction + Unifo Cicula Motion Cicula Hill A ca i diing oe a eicicula hill of adiu. What i the fatet the ca can die oe the top of the hill without it tie lifting off of the gound? ax? (1) Copehend
More informationRotational inertia (moment of inertia)
Rotational inertia (moment of inertia) Define rotational inertia (moment of inertia) to be I = Σ m i r i 2 or r i : the perpendicular distance between m i and the given rotation axis m 1 m 2 x 1 x 2 Moment
More informationGraphs of Equations. A coordinate system is a way to graphically show the relationship between 2 quantities.
Gaphs of Equations CHAT PeCalculus A coodinate sstem is a wa to gaphicall show the elationship between quantities. Definition: A solution of an equation in two vaiables and is an odeed pai (a, b) such
More information