Circular-Rotational Motion Mock Exam. Instructions: (92 points) Answer the following questions. SHOW ALL OF YOUR WORK.
|
|
- Magdalen Cannon
- 5 years ago
- Views:
Transcription
1 AP Physics C Sping, 2017 Cicula-Rotational Motion Mock Exam Name: Answe Key M. Leonad Instuctions: (92 points) Answe the following questions. SHOW ALL OF YOUR WORK. ( ) 1. A stuntman dives a motocycle though a vetical loop which has a adius of 2.5 m. The motocycle s engine keeps the motocycle moving at a constant speed as it dives though the loop. (a) (6 pts) Daw a fee-body diagam of the motocycle when bike is at the top of the loop. Solution: Remembe, the centipetal foce is not a foce in and of itself. Theefoe, you should not include it in a fee-body diagam. F g F N (b) (8 pts) What is the minimum speed that the motocycle must maintain to make it though the loop? Solution: When the motocycle eaches the top of the loop, both the gavitational foce and the nomal foce point towads the cente of the cicula path. Theefoe F g + F N = m v 2 /. Fom this fomula we can see that thee is a elationship between F N and v. In geneal, the faste the motocycle tavels though the loop, the lage the nomal foce. Since thee is a minimum value fo the nomal foce (F N = 0 N), thee is likewise a minimum value fo v. Setting F N equal to zeo and solving fo v gives: F g = m v2 m g = m v2 g = v2 v 2 = g v = g = 5 m s (c) (8 pts) Suppose the stuntman dives though the loop at 10 m. If the motocycle and the stuntman s have a combined mass of 150 kg, what was the Nomal Foce exeted by the loop on the motocycle when the bike eached the top of the loop? Solution: Since F g + F N = m v 2 /,
2 AP Physics C/Cicula-Rotational Motion Mock Exam Page 2 of 6 Name: Answe Key F g + F N = m v2 F N = m v2 m g ( ) v 2 = m g = 130 N ( ) 2. A coin is placed 20 cm fom the cente of a tuntable. The tun table stats fom est and begins to otate with an angula acceleation of α = 12 ad.. The coefficient of fiction between the coin and the tuntable is 0.4. (a) (4 pts) What is the tangential acceleation of the coin? s 2 Solution: Recall that tangential acceleation is given by a t = α. Theefoe, a t = (0.2 m)(12ad/s 2 ) = 2.4 m/s 2. (b) (8 pts) What is the maximum centipetal acceleation the coin can withstand without sliding? Solution: The foce esponsible fo holding the coin on the tuntable is the static fictional foce. Since thee is a maximum static fictional foce, thee is a maximum acceleation. We can find the magnitude of the maximum acceleation by setting the fictional foce equal to m a. f = m a µ m g = m a a = µ g Because the tuntable is speeding up as it spins, the coin s acceleation can be boken into two non-zeo components: tangential and centipetal. Because these two components ae othogonal, the magnitude of the coin s total acceleation is a 2 t + a 2 c. Setting this equal to the maximum acceleation we found above, and solving fo a c gives: a 2 t + a 2 c = µ g a 2 t + a 2 c = (µ g) 2 a 2 c = (µ g) 2 a 2 t a c = (µ g) 2 a 2 t ) = 3.2 m s 2 (c) (4 pts) What is the maximum angula velocity the coin can withstand without sliding? Solution: The centipetal acceleation is elated to the angula velocity by a c = ω 2. So, ω = ac / = 4 ad./s. (d) (6 pts) How long afte the tuntable stated spinning was it befoe the coin stated sliding? Solution: Using α = ω/ t gives:
3 AP Physics C/Cicula-Rotational Motion Mock Exam Page 3 of 6 Name: Answe Key α = ω t t = ω α = s ( ) 3. A ginding wheel has a adius of 0.2 m. The wheel stats fom est when a moto begins otating the wheel with an angula acceleation of 1.2 ad s 2. (a) (6 pts) What is the angula velocity of the tuntable the instant it completes one full evolution? Solution: Using ω 2 f = ω α θ gives: 0 ωf 2 = ω 2 f = 2 ω α θ ( 1.2 ad s 2 ω f = 3.88 ad s ) (2π) (b) (4 pts) How fast is a point on the oute edge of the wheel moving the instant it completes one full evolution? Solution: Using v = ω gives v = m/s. (c) (8 pts) What is the magnitude of the acceleation of a point on the oute edge of the wheel? Solution: Recall that when looking at a point on a otationally acceleating object, the acceleation has two components: centipetal, and tangential. The centipetal acceleation is given by: a c = ω 2, while the tangential acceleation is given by a t = α. Since these two components ae othogonal, the magnitude of the acceleation is given by: a = a 2 c + a 2 t = (ω 2 ) 2 + ( α) 2 = ω 4 + α 2 = 15.1 m s 2 (d) (4 pts) Afte the wheel completes one full otation, the moto dies. It takes 10 seconds fo the wheel to come to est. What was the angula acceleation of the wheel afte the moto died? Solution: Using α = ω/ t, we see that α = ad s 2.
4 AP Physics C/Cicula-Rotational Motion Mock Exam Page 4 of 6 Name: Answe Key (12 pts ) 4. A conical pendulum is fomed by attaching a 500 gam ball to the end of a 1.0 m sting. If the mass moves in a cicula path with a adius of 20 cm, how fast is the mass moving? 12 pts 1.0 m 0.2 m Solution: Thee ae two foces acting on the pendulum bob: gavity and tension. T θ F g Because the bob does not move vetically up o down, we can conclude that F y = 0 N. Theefoe, the vetical component of the tension foce must cancel the gavitational foce: T y = F g = 4.9 N. When the bob swings in a cicula path as shown above, a component of the tension points towads the cente of the cicula path. So, T x = m v 2 /. If we could find the x-component of the tension, we could use this elationship to solve fo v. Since we know the vetical component of the tension, we can use the geomety of the poblem to solve fo T x. Fom the fee-body diagam, we can see that tan θ = T x /T y. We can also see fom the fee-body diagam that sin θ = 0.2; theefoe, θ = 11.5 and tan θ = Setting T x /T y = tan θ = and solving fo T x gives: Setting this equal to m v 2 / gives: T x T y = T x = 0.204T y = pts
5 AP Physics C/Cicula-Rotational Motion Mock Exam Page 5 of 6 Name: Answe Key m v2 = T x Tx v = m (1 N)(0.2 m) v = 0.5 kg v = m s (8 pts ) 5. Two satellites, A and B, ae in diffeent cicula obits about the Eath. The obital speed of the satellite A is thee times that of satellite B. Find the atio T A /T B. Solution: Since we ae given the atio of the obital speeds and we ae asked to find the atio of the obital peiods, the easiest way to solve this poblem is using popotions. To do this, we must find a fomula which elates the peiod, T, to the velocity, v. When a satellite is in a cicula obit aound the Eath, the centipetal foce equied to keep the satellite in a cicula path is geneated by gavitational foce. Theefoe: 8 pts F c = F g m v2 = GM E m 2 v 2 = GM E We can elate this to the peiod of the motion, by ecalling that v = 2π. Since we want an expession T elating v to T, we need to solve this equation fo and plug it inso the expession above. Solving fo gives: = vt. Plugging this into the expession above gives: 2π v 2 = GM E ( ) 2π v 2 = GM E vt v 3 = 2πGM E T Theefoe T is popotional to v 3. Theefoe, if v inceases by a facto of 3, T will decease by a facto of 3 3 = 27. Theefoe: T A T B = pts
6 AP Physics C/Cicula-Rotational Motion Mock Exam Page 6 of 6 Name: Answe Key (6 pts ) 6. Planet X obits the sta Omega with an obital peiod of 200 Eath days. Planet Y obits Omega at a distance that is fou times lage than Planet X. How long is a yea on Planet Y? Solution: Since we ae given R Y /R X, and we ae asked about the peiods, we should use Keple s Thid Law. Recall that Keple s Thid Law says: 6 pts O eaanging to solve fo T Y : R 3 X T 2 X = R3 Y T 2 Y RX 3 TX 2 = R3 Y TY 2 ( ( ) 3 TY RY ) 2 = T Y = R X ( ) 3/2 RY R X T Y = ( ) 3/2 RX R Y T Y = ( RX R Y ) 3/2 = 8 (200 days) = 1600 days 6 pts
Circular Motion & Torque Test Review. The period is the amount of time it takes for an object to travel around a circular path once.
Honos Physics Fall, 2016 Cicula Motion & Toque Test Review Name: M. Leonad Instuctions: Complete the following woksheet. SHOW ALL OF YOUR WORK ON A SEPARATE SHEET OF PAPER. 1. Detemine whethe each statement
More informationPhysics 111 Lecture 5 Circular Motion
Physics 111 Lectue 5 Cicula Motion D. Ali ÖVGÜN EMU Physics Depatment www.aovgun.com Multiple Objects q A block of mass m1 on a ough, hoizontal suface is connected to a ball of mass m by a lightweight
More informationPS113 Chapter 5 Dynamics of Uniform Circular Motion
PS113 Chapte 5 Dynamics of Unifom Cicula Motion 1 Unifom cicula motion Unifom cicula motion is the motion of an object taveling at a constant (unifom) speed on a cicula path. The peiod T is the time equied
More informationChapter 5: Uniform Circular Motion
Chapte 5: Unifom Cicula Motion Motion at constant speed in a cicle Centipetal acceleation Banked cuves Obital motion Weightlessness, atificial gavity Vetical cicula motion Centipetal Foce Acceleation towad
More informationPhysics 101 Lecture 6 Circular Motion
Physics 101 Lectue 6 Cicula Motion Assist. Pof. D. Ali ÖVGÜN EMU Physics Depatment www.aovgun.com Equilibium, Example 1 q What is the smallest value of the foce F such that the.0-kg block will not slide
More informationPhysics 4A Chapter 8: Dynamics II Motion in a Plane
Physics 4A Chapte 8: Dynamics II Motion in a Plane Conceptual Questions and Example Poblems fom Chapte 8 Conceptual Question 8.5 The figue below shows two balls of equal mass moving in vetical cicles.
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 informationObjective Notes Summary
Objective Notes Summay An object moving in unifom cicula motion has constant speed but not constant velocity because the diection is changing. The velocity vecto in tangent to the cicle, the acceleation
More informationChapter 5. Uniform Circular Motion. a c =v 2 /r
Chapte 5 Unifom Cicula Motion a c =v 2 / Unifom cicula motion: Motion in a cicula path with constant speed s v 1) Speed and peiod Peiod, T: time fo one evolution Speed is elated to peiod: Path fo one evolution:
More informationPhysics 1114: Unit 5 Hand-out Homework (Answers)
Physics 1114: Unit 5 Hand-out Homewok (Answes) Poblem set 1 1. The flywheel on an expeimental bus is otating at 420 RPM (evolutions pe minute). To find (a) the angula velocity in ad/s (adians/second),
More informationPhysics 2001 Problem Set 5 Solutions
Physics 2001 Poblem Set 5 Solutions Jeff Kissel Octobe 16, 2006 1. A puck attached to a sting undegoes cicula motion on an ai table. If the sting beaks at the point indicated in the figue, which path (A,
More informationPhysics 201 Homework 4
Physics 201 Homewok 4 Jan 30, 2013 1. Thee is a cleve kitchen gadget fo dying lettuce leaves afte you wash them. 19 m/s 2 It consists of a cylindical containe mounted so that it can be otated about its
More informationω = θ θ o = θ θ = s r v = rω
Unifom Cicula Motion Unifom cicula motion is the motion of an object taveling at a constant(unifom) speed in a cicula path. Fist we must define the angula displacement and angula velocity The angula displacement
More informationRecap. Centripetal acceleration: v r. a = m/s 2 (towards center of curvature)
a = c v 2 Recap Centipetal acceleation: m/s 2 (towads cente of cuvatue) A centipetal foce F c is equied to keep a body in cicula motion: This foce poduces centipetal acceleation that continuously changes
More informationChap 5. Circular Motion: Gravitation
Chap 5. Cicula Motion: Gavitation Sec. 5.1 - Unifom Cicula Motion A body moves in unifom cicula motion, if the magnitude of the velocity vecto is constant and the diection changes at evey point and is
More information- 5 - TEST 1R. This is the repeat version of TEST 1, which was held during Session.
- 5 - TEST 1R This is the epeat vesion of TEST 1, which was held duing Session. This epeat test should be attempted by those students who missed Test 1, o who wish to impove thei mak in Test 1. IF YOU
More informationc) (6) Assuming the tires do not skid, what coefficient of static friction between tires and pavement is needed?
Geneal Physics I Exam 2 - Chs. 4,5,6 - Foces, Cicula Motion, Enegy Oct. 10, 2012 Name Rec. Inst. Rec. Time Fo full cedit, make you wok clea to the gade. Show fomulas used, essential steps, and esults with
More informationUniform Circular Motion
Unifom Cicula Motion constant speed Pick a point in the objects motion... What diection is the velocity? HINT Think about what diection the object would tavel if the sting wee cut Unifom Cicula Motion
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 informationSpring 2001 Physics 2048 Test 3 solutions
Sping 001 Physics 048 Test 3 solutions Poblem 1. (Shot Answe: 15 points) a. 1 b. 3 c. 4* d. 9 e. 8 f. 9 *emembe that since KE = ½ mv, KE must be positive Poblem (Estimation Poblem: 15 points) Use momentum-impulse
More informationHoizontal Cicula Motion 1. A paticle of mass m is tied to a light sting and otated with a speed v along a cicula path of adius. If T is tension in the sting and mg is gavitational foce on the paticle then,
More information1) Consider a particle moving with constant speed that experiences no net force. What path must this particle be taking?
Chapte 5 Test Cicula Motion and Gavitation 1) Conside a paticle moving with constant speed that expeiences no net foce. What path must this paticle be taking? A) It is moving in a paabola. B) It is moving
More informationCentripetal Force. Lecture 11. Chapter 8. Course website:
Lectue 11 Chapte 8 Centipetal Foce Couse website: http://faculty.uml.edu/andiy_danylov/teaching/physicsi PHYS.1410 Lectue 11 Danylov Depatment of Physics and Applied Physics Today we ae going to discuss:
More informationCircular Motion. Mr. Velazquez AP/Honors Physics
Cicula Motion M. Velazquez AP/Honos Physics Objects in Cicula Motion Accoding to Newton s Laws, if no foce acts on an object, it will move with constant speed in a constant diection. Theefoe, if an object
More informationb) (5) What is the magnitude of the force on the 6.0-kg block due to the contact with the 12.0-kg block?
Geneal Physics I Exam 2 - Chs. 4,5,6 - Foces, Cicula Motion, Enegy Oct. 13, 2010 Name Rec. Inst. Rec. Time Fo full cedit, make you wok clea to the gade. Show fomulas used, essential steps, and esults with
More informationMidterm Exam #2, Part A
Physics 151 Mach 17, 2006 Midtem Exam #2, Pat A Roste No.: Scoe: Exam time limit: 50 minutes. You may use calculatos and both sides of ONE sheet of notes, handwitten only. Closed book; no collaboation.
More informationPHYSICS 220. Lecture 08. Textbook Sections Lecture 8 Purdue University, Physics 220 1
PHYSICS 0 Lectue 08 Cicula Motion Textbook Sections 5.3 5.5 Lectue 8 Pudue Univesity, Physics 0 1 Oveview Last Lectue Cicula Motion θ angula position adians ω angula velocity adians/second α angula acceleation
More informationBetween any two masses, there exists a mutual attractive force.
YEAR 12 PHYSICS: GRAVITATION PAST EXAM QUESTIONS Name: QUESTION 1 (1995 EXAM) (a) State Newton s Univesal Law of Gavitation in wods Between any two masses, thee exists a mutual attactive foce. This foce
More informationCHAPTER 5: Circular Motion; Gravitation
CHAPER 5: Cicula Motion; Gavitation Solution Guide to WebAssign Pobles 5.1 [1] (a) Find the centipetal acceleation fo Eq. 5-1.. a R v ( 1.5 s) 1.10 1.4 s (b) he net hoizontal foce is causing the centipetal
More informationExtra notes for circular motion: Circular motion : v keeps changing, maybe both speed and
Exta notes fo cicula motion: Cicula motion : v keeps changing, maybe both speed and diection ae changing. At least v diection is changing. Hence a 0. Acceleation NEEDED to stay on cicula obit: a cp v /,
More informationAP * PHYSICS B. Circular Motion, Gravity, & Orbits. Teacher Packet
AP * PHYSICS B Cicula Motion, Gavity, & Obits Teache Packet AP* is a tademak of the College Entance Examination Boad. The College Entance Examination Boad was not involved in the poduction of this mateial.
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 information= 4 3 π( m) 3 (5480 kg m 3 ) = kg.
CHAPTER 11 THE GRAVITATIONAL FIELD Newton s Law of Gavitation m 1 m A foce of attaction occus between two masses given by Newton s Law of Gavitation Inetial mass and gavitational mass Gavitational potential
More informationChapter 12. Kinetics of Particles: Newton s Second Law
Chapte 1. Kinetics of Paticles: Newton s Second Law Intoduction Newton s Second Law of Motion Linea Momentum of a Paticle Systems of Units Equations of Motion Dynamic Equilibium Angula Momentum of a Paticle
More informationExam 3: Equation Summary
MAACHUETT INTITUTE OF TECHNOLOGY Depatment of Physics Physics 8. TEAL Fall Tem 4 Momentum: p = mv, F t = p, Fext ave t= t f t = Exam 3: Equation ummay = Impulse: I F( t ) = p Toque: τ =,P dp F P τ =,P
More informationPhysics C Rotational Motion Name: ANSWER KEY_ AP Review Packet
Linea and angula analogs Linea Rotation x position x displacement v velocity a T tangential acceleation Vectos in otational motion Use the ight hand ule to detemine diection of the vecto! Don t foget centipetal
More informationGravitation. AP/Honors Physics 1 Mr. Velazquez
Gavitation AP/Honos Physics 1 M. Velazquez Newton s Law of Gavitation Newton was the fist to make the connection between objects falling on Eath and the motion of the planets To illustate this connection
More informationb) (5) What average force magnitude was applied by the students working together?
Geneal Physics I Exam 3 - Chs. 7,8,9 - Momentum, Rotation, Equilibium Nov. 3, 2010 Name Rec. Inst. Rec. Time Fo full cedit, make you wok clea to the gade. Show fomulas used, essential steps, and esults
More information10. Force is inversely proportional to distance between the centers squared. R 4 = F 16 E 11.
NSWRS - P Physics Multiple hoice Pactice Gavitation Solution nswe 1. m mv Obital speed is found fom setting which gives v whee M is the object being obited. Notice that satellite mass does not affect obital
More informationCIRCULAR MOTION. Particle moving in an arbitrary path. Particle moving in straight line
1 CIRCULAR MOTION 1. ANGULAR DISPLACEMENT Intoduction: Angle subtended by position vecto of a paticle moving along any abitay path w..t. some fixed point is called angula displacement. (a) Paticle moving
More informationMotion in a Plane Uniform Circular Motion
Lectue 11 Chapte 8 Physics I Motion in a Plane Unifom Cicula Motion Couse website: http://faculty.uml.edu/andiy_danylo/teaching/physicsi PHYS.1410 Lectue 11 Danylo Depatment of Physics and Applied Physics
More informationChapter 8. Accelerated Circular Motion
Chapte 8 Acceleated Cicula Motion 8.1 Rotational Motion and Angula Displacement A new unit, adians, is eally useful fo angles. Radian measue θ(adians) = s = θ s (ac length) (adius) (s in same units as
More informationCircular motion. Objectives. Physics terms. Assessment. Equations 5/22/14. Describe the accelerated motion of objects moving in circles.
Cicula motion Objectives Descibe the acceleated motion of objects moving in cicles. Use equations to analyze the acceleated motion of objects moving in cicles.. Descibe in you own wods what this equation
More informationName. Date. Period. Engage Examine the pictures on the left. 1. What is going on in these pictures?
AP Physics 1 Lesson 9.a Unifom Cicula Motion Outcomes 1. Define unifom cicula motion. 2. Detemine the tangential velocity of an object moving with unifom cicula motion. 3. Detemine the centipetal acceleation
More informationPHYS 1114, Lecture 21, March 6 Contents:
PHYS 1114, Lectue 21, Mach 6 Contents: 1 This class is o cially cancelled, being eplaced by the common exam Tuesday, Mach 7, 5:30 PM. A eview and Q&A session is scheduled instead duing class time. 2 Exam
More informationQuiz 6--Work, Gravitation, Circular Motion, Torque. (60 pts available, 50 points possible)
Name: Class: Date: ID: A Quiz 6--Wok, Gavitation, Cicula Motion, Toque. (60 pts available, 50 points possible) Multiple Choice, 2 point each Identify the choice that best completes the statement o answes
More informationChapter 13 Gravitation
Chapte 13 Gavitation In this chapte we will exploe the following topics: -Newton s law of gavitation, which descibes the attactive foce between two point masses and its application to extended objects
More informationAnswers to test yourself questions
Answes to test youself questions opic. Cicula motion π π a he angula speed is just ω 5. 7 ad s. he linea speed is ω 5. 7 3. 5 7. 7 m s.. 4 b he fequency is f. 8 s.. 4 3 a f. 45 ( 3. 5). m s. 3 a he aeage
More informationPhys 201A. Homework 6 Solutions. F A and F r. B. According to Newton s second law, ( ) ( )2. j = ( 6.0 m / s 2 )ˆ i ( 10.4m / s 2 )ˆ j.
7. We denote the two foces F A + F B = ma,sof B = ma F A. (a) In unit vecto notation F A = ( 20.0 N)ˆ i and Theefoe, Phys 201A Homewok 6 Solutions F A and F B. Accoding to Newton s second law, a = [ (
More informationPhysics: Work & Energy Beyond Earth Guided Inquiry
Physics: Wok & Enegy Beyond Eath Guided Inquiy Elliptical Obits Keple s Fist Law states that all planets move in an elliptical path aound the Sun. This concept can be extended to celestial bodies beyond
More informationSections and Chapter 10
Cicula and Rotational Motion Sections 5.-5.5 and Chapte 10 Basic Definitions Unifom Cicula Motion Unifom cicula motion efes to the motion of a paticle in a cicula path at constant speed. The instantaneous
More informationUniversal Gravitation
Chapte 1 Univesal Gavitation Pactice Poblem Solutions Student Textbook page 580 1. Conceptualize the Poblem - The law of univesal gavitation applies to this poblem. The gavitational foce, F g, between
More informationUniform Circular Motion
Unifom Cicula Motion Intoduction Ealie we defined acceleation as being the change in velocity with time: a = v t Until now we have only talked about changes in the magnitude of the acceleation: the speeding
More informationExperiment 09: Angular momentum
Expeiment 09: Angula momentum Goals Investigate consevation of angula momentum and kinetic enegy in otational collisions. Measue and calculate moments of inetia. Measue and calculate non-consevative wok
More information10. Universal Gravitation
10. Univesal Gavitation Hee it is folks, the end of the echanics section of the couse! This is an appopiate place to complete the study of mechanics, because with his Law of Univesal Gavitation, Newton
More informationChap13. Universal Gravitation
Chap13. Uniesal Gaitation Leel : AP Physics Instucto : Kim 13.1 Newton s Law of Uniesal Gaitation - Fomula fo Newton s Law of Gaitation F g = G m 1m 2 2 F21 m1 F12 12 m2 - m 1, m 2 is the mass of the object,
More informationDescribing Circular motion
Unifom Cicula Motion Descibing Cicula motion In ode to undestand cicula motion, we fist need to discuss how to subtact vectos. The easiest way to explain subtacting vectos is to descibe it as adding a
More information3.2 Centripetal Acceleration
unifom cicula motion the motion of an object with onstant speed along a cicula path of constant adius 3.2 Centipetal Acceleation The hamme thow is a tack-and-field event in which an athlete thows a hamme
More informationUnderstanding the Concepts
Chistian Bache Phsics Depatment Bn Maw College Undestanding the Concepts PHYSICS 101-10 Homewok Assignment #5 - Solutions 5.7. A cclist making a tun must make use of a centipetal foce, one that is pependicula
More informationCircular Orbits. and g =
using analyse planetay and satellite motion modelled as unifom cicula motion in a univesal gavitation field, a = v = 4π and g = T GM1 GM and F = 1M SATELLITES IN OBIT A satellite is any object that is
More informationPhysics 111. Ch 12: Gravity. Newton s Universal Gravity. R - hat. the equation. = Gm 1 m 2. F g 2 1. ˆr 2 1. Gravity G =
ics Announcements day, embe 9, 004 Ch 1: Gavity Univesal Law Potential Enegy Keple s Laws Ch 15: Fluids density hydostatic equilibium Pascal s Pinciple This week s lab will be anothe physics wokshop -
More informationChapter 4. Newton s Laws of Motion
Chapte 4 Newton s Laws of Motion 4.1 Foces and Inteactions A foce is a push o a pull. It is that which causes an object to acceleate. The unit of foce in the metic system is the Newton. Foce is a vecto
More informationCh 13 Universal Gravitation
Ch 13 Univesal Gavitation Ch 13 Univesal Gavitation Why do celestial objects move the way they do? Keple (1561-1630) Tycho Bahe s assistant, analyzed celestial motion mathematically Galileo (1564-1642)
More informationRotational Motion. Every quantity that we have studied with translational motion has a rotational counterpart
Rotational Motion & Angula Momentum Rotational Motion Evey quantity that we have studied with tanslational motion has a otational countepat TRANSLATIONAL ROTATIONAL Displacement x Angula Position Velocity
More informationGravitation. Chapter 12. PowerPoint Lectures for University Physics, Twelfth Edition Hugh D. Young and Roger A. Freedman. Lectures by James Pazun
Chapte 12 Gavitation PowePoint Lectues fo Univesity Physics, Twelfth Edition Hugh D. Young and Roge A. Feedman Lectues by James Pazun Modified by P. Lam 5_31_2012 Goals fo Chapte 12 To study Newton s Law
More informationChapter 5. really hard to start the object moving and then, once it starts moving, you don t have to push as hard to keep it moving.
Chapte 5 Fiction When an object is in motion it is usually in contact with a viscous mateial (wate o ai) o some othe suface. So fa, we have assumed that moving objects don t inteact with thei suoundings
More information06 - ROTATIONAL MOTION Page 1 ( Answers at the end of all questions )
06 - ROTATIONAL MOTION Page ) A body A of mass M while falling vetically downwads unde gavity beaks into two pats, a body B of mass ( / ) M and a body C of mass ( / ) M. The cente of mass of bodies B and
More informationLecture 22. PE = GMm r TE = GMm 2a. T 2 = 4π 2 GM. Main points of today s lecture: Gravitational potential energy: Total energy of orbit:
Lectue Main points of today s lectue: Gavitational potential enegy: Total enegy of obit: PE = GMm TE = GMm a Keple s laws and the elation between the obital peiod and obital adius. T = 4π GM a3 Midtem
More information2013 Checkpoints Chapter 7 GRAVITY
0 Checkpoints Chapte 7 GAVIY Question 64 o do this question you must et an equation that has both and, whee is the obital adius and is the peiod. You can use Keple s Law, which is; constant. his is a vey
More informationUnit 6 Practice Test. Which vector diagram correctly shows the change in velocity Δv of the mass during this time? (1) (1) A. Energy KE.
Unit 6 actice Test 1. Which one of the following gaphs best epesents the aiation of the kinetic enegy, KE, and of the gaitational potential enegy, GE, of an obiting satellite with its distance fom the
More informationLab #9: The Kinematics & Dynamics of. Circular Motion & Rotational Motion
Reading Assignment: Lab #9: The Kinematics & Dynamics of Cicula Motion & Rotational Motion Chapte 6 Section 4 Chapte 11 Section 1 though Section 5 Intoduction: When discussing motion, it is impotant to
More informationUnit 6 Practice Test. Which vector diagram correctly shows the change in velocity Δv of the mass during this time? (1) (1) A. Energy KE.
Unit 6 actice Test 1. Which one of the following gaphs best epesents the aiation of the kinetic enegy, KE, and of the gaitational potential enegy, GE, of an obiting satellite with its distance fom the
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 NON-UNIFORM 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 information6.4 Period and Frequency for Uniform Circular Motion
6.4 Peiod and Fequency fo Unifom Cicula Motion If the object is constained to move in a cicle and the total tangential foce acting on the total object is zeo, F θ = 0, then (Newton s Second Law), the tangential
More informationROTATORY MOTION HORIZONTAL AND VERTICAL CIRCULAR MOTION
ROTATORY MOTION HORIZONTAL AND VERTICAL CIRCULAR MOTION POINTS TO REMEMBER 1. Tanslatoy motion: Evey point in the body follows the path of its peceding one with same velocity including the cente of mass..
More informationMODULE 5 ADVANCED MECHANICS GRAVITATIONAL FIELD: MOTION OF PLANETS AND SATELLITES VISUAL PHYSICS ONLINE
VISUAL PHYSICS ONLIN MODUL 5 ADVANCD MCHANICS GRAVITATIONAL FILD: MOTION OF PLANTS AND SATLLITS SATLLITS: Obital motion of object of mass m about a massive object of mass M (m
More informationAH Mechanics Checklist (Unit 2) AH Mechanics Checklist (Unit 2) Circular Motion
AH Mechanics Checklist (Unit ) AH Mechanics Checklist (Unit ) Cicula Motion No. kill Done 1 Know that cicula motion efes to motion in a cicle of constant adius Know that cicula motion is conveniently descibed
More informationChapter 7. Rotational Motion Angles, Angular Velocity and Angular Acceleration Universal Law of Gravitation Kepler s Laws
Chapte 7 Rotational Motion Angles, Angula Velocity and Angula Acceleation Univesal Law of Gavitation Keple s Laws Angula Displacement Cicula motion about AXIS Thee diffeent measues of angles: 1. Degees.
More informationChapter 5 Force and Motion
Chapte 5 Foce and Motion In Chaptes 2 and 4 we have studied kinematics, i.e., we descibed the motion of objects using paametes such as the position vecto, velocity, and acceleation without any insights
More informationFrom Newton to Einstein. Mid-Term 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 Mid-Tem 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 informationChapter 5 Force and Motion
Chapte 5 Foce and Motion In chaptes 2 and 4 we have studied kinematics i.e. descibed the motion of objects using paametes such as the position vecto, velocity and acceleation without any insights as to
More informationr cos, and y r sin with the origin of coordinate system located at
Lectue 3-3 Kinematics of Rotation Duing ou peious lectues we hae consideed diffeent examples of motion in one and seeal dimensions. But in each case the moing object was consideed as a paticle-like object,
More informationTranslation and Rotation Kinematics
Tanslation and Rotation Kinematics Oveview: Rotation and Tanslation of Rigid Body Thown Rigid Rod Tanslational Motion: the gavitational extenal foce acts on cente-of-mass F ext = dp sy s dt dv total cm
More informationRotational Motion. Lecture 6. Chapter 4. Physics I. Course website:
Lectue 6 Chapte 4 Physics I Rotational Motion Couse website: http://faculty.uml.edu/andiy_danylov/teaching/physicsi Today we ae going to discuss: Chapte 4: Unifom Cicula Motion: Section 4.4 Nonunifom Cicula
More informationHW Solutions # MIT - Prof. Please study example 12.5 "from the earth to the moon". 2GmA v esc
HW Solutions # 11-8.01 MIT - Pof. Kowalski Univesal Gavity. 1) 12.23 Escaping Fom Asteoid Please study example 12.5 "fom the eath to the moon". a) The escape velocity deived in the example (fom enegy consevation)
More informationPhysics 111. Lecture 14 (Walker: Ch. 6.5) Circular Motion Centripetal Acceleration Centripetal Force February 27, 2009
Physics 111 Lectue 14 (Walke: Ch. 6.5) Cicula Motion Centipetal Acceleation Centipetal Foce Febuay 7, 009 Midtem Exam 1 on Wed. Mach 4 (Chaptes 1-6) Lectue 14 1/8 Connected Objects If thee is a pulley,
More informationChapters 5-8. Dynamics: Applying Newton s Laws
Chaptes 5-8 Dynamics: Applying Newton s Laws Systems of Inteacting Objects The Fee Body Diagam Technique Examples: Masses Inteacting ia Nomal Foces Masses Inteacting ia Tensions in Ropes. Ideal Pulleys
More informationUniform Circular Motion
Unifom Cicula Motion Have you eve idden on the amusement pak ide shown below? As it spins you feel as though you ae being pessed tightly against the wall. The ide then begins to tilt but you emain glued
More informationPhysics 107 TUTORIAL ASSIGNMENT #8
Physics 07 TUTORIAL ASSIGNMENT #8 Cutnell & Johnson, 7 th edition Chapte 8: Poblems 5,, 3, 39, 76 Chapte 9: Poblems 9, 0, 4, 5, 6 Chapte 8 5 Inteactive Solution 8.5 povides a model fo solving this type
More informationPhysics 181. Assignment 4
Physics 181 Assignment 4 Solutions 1. A sphee has within it a gavitational field given by g = g, whee g is constant and is the position vecto of the field point elative to the cente of the sphee. This
More informationPhysics 231 Lecture 17
Physics 31 Lectue 17 Main points of today s lectue: Centipetal acceleation: a c = a c t Rotational motion definitions: Δω Δω α =, α = limδ t 0 Δt Δt Δ s= Δ θ;t = ω;at = α Rotational kinematics equations:
More informationExam 3: Equation Summary
MAACHUETT INTITUTE OF TECHNOLOGY Depatment of Physics Physics 8. TEAL Fall Tem 4 Momentum: p = mv, F t = p, Fext ave t= t f t = Exam 3: Equation ummay = Impulse: I F( t ) = p Toque: τ =,P dp F P τ =,P
More information1. A stone falls from a platform 18 m high. When will it hit the ground? (a) 1.74 s (b) 1.83 s (c) 1.92 s (d) 2.01 s
1. A stone falls fom a platfom 18 m high. When will it hit the gound? (a) 1.74 s (b) 1.83 s (c) 1.9 s (d).01 s Constant acceleation D = v 0 t + ½ a t. Which, if any, of these foces causes the otation of
More information1121 T Question 1
1121 T1 2008 Question 1 ( aks) You ae cycling, on a long staight path, at a constant speed of 6.0.s 1. Anothe cyclist passes you, tavelling on the sae path in the sae diection as you, at a constant speed
More informationCircular Motion. x-y coordinate systems. Other coordinates... PHY circular-motion - J. Hedberg
Cicula Motion PHY 207 - cicula-motion - J. Hedbeg - 2017 x-y coodinate systems Fo many situations, an x-y coodinate system is a geat idea. Hee is a map on Manhattan. The steets ae laid out in a ectangula
More informationrt () is constant. We know how to find the length of the radius vector by r( t) r( t) r( t)
Cicula Motion Fom ancient times cicula tajectoies hae occupied a special place in ou model of the Uniese. Although these obits hae been eplaced by the moe geneal elliptical geomety, cicula motion is still
More informationSAMPLE QUESTION PAPER CLASS NAME & LOGO XII-JEE (MAINS)-YEAR Topic Names: Cicula motion Test Numbe Test Booklet No. 000001 110001 Wite/Check this Code on you Answe Sheet : IMPORTANT INSTRUCTIONS : Wite
More informationDEVIL PHYSICS THE BADDEST CLASS ON CAMPUS IB PHYSICS
DEVIL PHYSICS THE BADDEST CLASS ON CAMPUS IB PHYSICS TSOKOS LESSON 6- THE LAW OF GRAVITATION Essential Idea: The Newtonian idea of gavitational foce acting between two spheical bodies and the laws of mechanics
More informationm1 m2 M 2 = M -1 L 3 T -2
GAVITATION Newton s Univesal law of gavitation. Evey paticle of matte in this univese attacts evey othe paticle with a foce which vaies diectly as the poduct of thei masses and invesely as the squae of
More informationMultiple choice questions [100 points] As shown in the figure, a mass M is hanging by three massless strings from the ceiling of a room.
Multiple choice questions [00 points] Answe all of the following questions. Read each question caefully. Fill the coect ule on you scanton sheet. Each coect answe is woth 4 points. Each question has exactly
More information