Uniform Circular Motion

Similar documents
Chapter 5: Uniform Circular Motion

Chap 5. Circular Motion: Gravitation

ω = θ θ o = θ θ = s r v = rω

Recap. Centripetal acceleration: v r. a = m/s 2 (towards center of curvature)

PHYSICS 220. Lecture 08. Textbook Sections Lecture 8 Purdue University, Physics 220 1

Physics 111 Lecture 5 Circular Motion

Objective Notes Summary

Physics 201 Homework 4

PS113 Chapter 5 Dynamics of Uniform Circular Motion

Physics 101 Lecture 6 Circular Motion

10. Force is inversely proportional to distance between the centers squared. R 4 = F 16 E 11.

1) Consider a particle moving with constant speed that experiences no net force. What path must this particle be taking?

Ch 13 Universal Gravitation

Describing Circular motion

Centripetal Force. Lecture 11. Chapter 8. Course website:

m1 m2 M 2 = M -1 L 3 T -2

Chapter 8. Accelerated Circular Motion

DEVIL PHYSICS THE BADDEST CLASS ON CAMPUS IB PHYSICS

Extra notes for circular motion: Circular motion : v keeps changing, maybe both speed and

OSCILLATIONS AND GRAVITATION

DYNAMICS OF UNIFORM CIRCULAR MOTION

F 12. = G m m 1 2 F 21 = F 12. = G m 1m 2. Review. Physics 201, Lecture 22. Newton s Law Of Universal Gravitation

Uniform Circular Motion

MODULE 5 ADVANCED MECHANICS GRAVITATIONAL FIELD: MOTION OF PLANETS AND SATELLITES VISUAL PHYSICS ONLINE

Uniform Circular Motion

= 4 3 π( m) 3 (5480 kg m 3 ) = kg.

Gravitation. AP/Honors Physics 1 Mr. Velazquez

Circular-Rotational Motion Mock Exam. Instructions: (92 points) Answer the following questions. SHOW ALL OF YOUR WORK.

Quiz 6--Work, Gravitation, Circular Motion, Torque. (60 pts available, 50 points possible)

Physics 2001 Problem Set 5 Solutions

Lecture 1a: Satellite Orbits

Universal Gravitation

Between any two masses, there exists a mutual attractive force.

Chapter 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.

Chapter 13 Gravitation

b) (5) What is the magnitude of the force on the 6.0-kg block due to the contact with the 12.0-kg block?

CHAPTER 5: Circular Motion; Gravitation

c) (6) Assuming the tires do not skid, what coefficient of static friction between tires and pavement is needed?

Radius of the Moon is 1700 km and the mass is 7.3x 10^22 kg Stone. Moon

Physics 4A Chapter 8: Dynamics II Motion in a Plane

Gravitation. Chapter 12. PowerPoint Lectures for University Physics, Twelfth Edition Hugh D. Young and Roger A. Freedman. Lectures by James Pazun

AP Physics 1 - Circular Motion and Gravitation Practice Test (Multiple Choice Section) Answer Section

Physics 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 =

PHYS 1114, Lecture 21, March 6 Contents:

Chapter. s r. check whether your calculator is in all other parts of the body. When a rigid body rotates through a given angle, all

Motion in a Plane Uniform Circular Motion

History of Astronomy - Part II. Tycho Brahe - An Observer. Johannes Kepler - A Theorist

6.4 Period and Frequency for Uniform Circular Motion

Understanding the Concepts

Circular Motion. Mr. Velazquez AP/Honors Physics

Revision Guide for Chapter 11

AST 121S: The origin and evolution of the Universe. Introduction to Mathematical Handout 1

Circular Motion & Torque Test Review. The period is the amount of time it takes for an object to travel around a circular path once.

DEVIL PHYSICS THE BADDEST CLASS ON CAMPUS IB PHYSICS

AH Mechanics Checklist (Unit 2) AH Mechanics Checklist (Unit 2) Circular Motion

Sections and Chapter 10

- 5 - TEST 1R. This is the repeat version of TEST 1, which was held during Session.

Chap13. Universal Gravitation

CIRCULAR MOTION. Particle moving in an arbitrary path. Particle moving in straight line

Physics 1114: Unit 5 Hand-out Homework (Answers)

Paths of planet Mars in sky

Physics 312 Introduction to Astrophysics Lecture 7

Chapter 13: Gravitation

Gaia s Place in Space

Chapter 7. Rotational Motion Angles, Angular Velocity and Angular Acceleration Universal Law of Gravitation Kepler s Laws

Physics 231 Lecture 17

Name. Date. Period. Engage Examine the pictures on the left. 1. What is going on in these pictures?

1. 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

Circular Orbits. and g =

Chapter 5. Uniform Circular Motion. a c =v 2 /r

Escape Velocity. GMm ] B

PHYSICS NOTES GRAVITATION

Class 6 - Circular Motion and Gravitation

3.2 Centripetal Acceleration

Chapter 12. Kinetics of Particles: Newton s Second Law

Introduction to Mechanics Centripetal Force

Physics 111. Lecture 14 (Walker: Ch. 6.5) Circular Motion Centripetal Acceleration Centripetal Force February 27, 2009


AP * PHYSICS B. Circular Motion, Gravity, & Orbits. Teacher Packet

Unit 6 Practice Test. Which vector diagram correctly shows the change in velocity Δv of the mass during this time? (1) (1) A. Energy KE.

Newton s Laws, Kepler s Laws, and Planetary Orbits

Spring 2001 Physics 2048 Test 3 solutions

Midterm Exam #2, Part A

b) (5) What average force magnitude was applied by the students working together?

Lecture 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:

r cos, and y r sin with the origin of coordinate system located at

Chapter 5. Applying Newton s Laws. Newton s Laws. r r. 1 st Law: An object at rest or traveling in uniform. 2 nd Law:

Unit 6 Practice Test. Which vector diagram correctly shows the change in velocity Δv of the mass during this time? (1) (1) A. Energy KE.

Physics C Rotational Motion Name: ANSWER KEY_ AP Review Packet

Lab #9: The Kinematics & Dynamics of. Circular Motion & Rotational Motion

KEPLER S LAWS OF PLANETARY MOTION

Section 26 The Laws of Rotational Motion

Determining solar characteristics using planetary data

Answers to test yourself questions

10. Universal Gravitation

Kinematics in 2-D (II)

ISSUED BY K V - DOWNLOADED FROM CIRCULAR MOTION

F g. = G mm. m 1. = 7.0 kg m 2. = 5.5 kg r = 0.60 m G = N m 2 kg 2 = = N

HW Solutions # MIT - Prof. Please study example 12.5 "from the earth to the moon". 2GmA v esc

kg 2 ) 1.9!10 27 kg = Gm 1

Transcription:

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 constant speed v v always tangential to the cicle Now pick anothe point in the objects motion. What diection is the velocity? Compae it to the pevious point Is the object acceleating?

stat hee What diection is the acceleation? v 1 v 2 vecto pat of a a = Δv = v f v i = v 2 v 1 t t t acceleation points in the diection of Δv!!! Δv v 1 v 2 Δv v 2 v 1 *emembe constant speed > v 1 = v 2 = 45 o Δv and theefoe a point towad the cente of the cicle!!! This is called centipetal acceleation, a c d = vδt These ae 2 simila isosceles tiangles Can compae coesponding sides!!! Δv v 1 v 2 Δv = vδt v Δv v = 2 Δt a c = v2

Definitions fo Cicula Motion Cycle in geneal: an object coming back to its initial position afte leaving cicula motion one evolution Peiod the time fo an object to make one cycle T = time cycle Fequency the # of cycle made in a cetain amount of time (invese of the peiod) f = # cycles time **** T = 1 f = 1 **** f T Now we know thee is an acceleation associated with cicula motion which means?... Thee must be a FORCE associated with cicula motion!! F net = ma

Centipetal Foce, F c F net = ma F c = ma c a c = v 2 F c = mv 2 v c = d = 2π F c = 4π 2 m t T T 2 **** F c is always the net foce!! **** must be povided by some othe foce Tension, Gavity, Nomal etc. Centipetal Foce, F c What diection is F c? ***Remembe, F net and acceleation always point in the same diection! a c F c Δv v 2 v 1 **F c always points towad the cente of the cicle!

Find v Find v

Rounding a Cone A 1200 kg ca ounds a FLAT cicula cone o adius, = 45m. If the μs between the ties and the oad is, μs = 0.82. What is the geatest speed the ca can have without skidding? (not slipping off of its cicula path) Top View Side View No matte how massive the ca they all have the same maximum speed!! What would the adius of the cicle be in which the maximum speed, without skidding, is doubled? (same fictional foce exeted) f = m v 2 α v 2 f = m v 2 constants 4 α (2v) 2 quadupled? α v 2 16 α (4v) 2

AIRPLANES An aiplane is able to get off the gound because of a foce called "lift". The wings ae designed to push ai down and back, "pushing" the plane fowad and up. (3 d Law) Lift Lift is always pependicula to the wing suface Because of Newton's 3d Law!!! AIRPLANES Once in the ai, How do planes tun? The plane must tilt o "bank" its wings and in the pocess ceate some F c What povides that F c?

AIRPLANES F L sin F L cos F L What povides that F c? mg AIRPLANES The hoizontal component of the lift!! (in geneal the pat of the lift that is in the same plane as the cicle) F L sin F c F L F L cos mg Find an expession fo the speed of the plane as function of g, and

AIRPLANES F L sin F L F L cos F c mg The hoizontal component of the lift!! (in geneal the pat of the lift that is in the same plane as the cicle) AIRPLANES F L sin F L F L cos F c mg

Banked Cuves The Nomal Foce povides F c!!! Nsin N Ncos notice cicula path is hoizontal NOT along incline mg Find an expession fo the speed of the ca as function of g, and Banked Cuves Nsin N Ncos mg

At what angle should a cuve of adius 150 m be banked so cas can tavel safely at 25 m/s without elying on fiction? Univesal Law of Gavitation At the suface of the Eath: F g = mg in geneal F g depends on how fa away objects ae fom one anothe F Gm 1 m **** 2 g = **** Newton's Constant G = 6.67 x 10 * do NOT have to memoize! 2 11 Nm2 kg 2

Univesal Law of Gavitation At the suface of the Eath: F g = mg in geneal F g depends on how fa away objects ae fom one anothe F 1 F 2 d F 1 = F 2 F Gm 1 m **** 2 g = **** 2 F g α 1 2 F g e What is the gavitational acceleation at the suface of the Eath? (conside an object in feefall) G = 6.67 x 10 11 Nm 2 /kg 2 M e = 5.98 x 10 24 kg e = 6.38 x 10 6 m

e What is the gavitational acceleation at the suface of the Eath? mg (conside an object in feefall) ƩF = ma G M e m e 2 = ma G M e (6.67 x 10 = a = 11 )(5.98 x 10 24 ) = 9.7991 m 2 e (6.38 x 10 6 ) 2 s 2 Keple's Laws: Johannes Keple (1571-1630) Using geomety & mathematics developed " The Kinematics of Obits" I. Law of Obits All planets move in elliptical obits with the Sun at one of the focal points. * 1 AU = distance Eath to Sun (93 million miles- aveage) * Cicula obits ae just a special case of an elliptical obit

II. Law of Equal Aeas A line dawn fom the Sun to any planet sweeps out equal aeas in equal time intevals. t 1 2 t3 4 t 1 2 t 3 4 A 1 2 A 3 4 *close->faste! III Law of Peiods The squae of the obital peiod of any planet is popotional to the cube of the aveage distance fom the planet to the Sun! CONSTANT

Obit and Gavity Fo an object in obit, gavity povides the centipetal foce! ƩF = F g = F c An object of mass m obiting a cental mass M o at a sepeation of Let's compae thei sepaation,, to the peiod of the objects obit, T. GM o m mv 2 ecall: = 2 v = 2π T GM o ( 2π T ) 2 GM = ==> o 4π 2 2 2 = T 2 4π T 2 = 2 GM 3 o 4π 2 GM o = k (constant)!! T 2 3 = k Keple's 3 d Law Obit and Gavity Let's get fomulas fo the obital speed, v, and the obital peiod, T. GM o m = mv 2 2 ƩF = F g = F c v 2 = GM o ==> Using ou deivation of Keple's 3 d : GM o 2 = 4π 2 T T 2 2 = 4π2 3 v = GM o ==> T = 2π 3 GM o GMo **Neithe obital peiod o speed depend on the mass of the object, only the cental mass.**

Definition: Astonomical Unit (AU) aveage distance fom the Eath to the Sun R E = 1 AU 1 AU 1 AU = 1.5 x 10 11 m 93 x 10 6 miles Jupite obits the Sun at 5.2 times the distance Eath does, what is it's peiod?

Astonomes have discoveed a supemassive black hole in the cente of the galaxy M87. They have obseved matte at distance of 5.7x10 17 m fom the cente obiting at a speed of 7.5x10 5 m/s. Find the mass of the supemassive black hole. How much moe massive is it than Eath? The Sun? F g = weight = mg we also know F g deceases as you get futhe away fom the Eath So... as you get futhe away fom the suface of the Eath you actually weigh less!!!