Worked Examples. v max =?



Similar documents
Chapter 30: Magnetic Fields Due to Currents

Solution Derivations for Capa #8

Exam 3: Equation Summary

Experiment 6: Centripetal Force

12. Rolling, Torque, and Angular Momentum

Mechanics 1: Motion in a Central Force Field

Determining solar characteristics using planetary data

7 Circular Motion. 7-1 Centripetal Acceleration and Force. Period, Frequency, and Speed. Vocabulary

Incline and Friction Examples

Chapter 13 Fluids. Use the definition of density to express the mass of the gold sphere: The mass of the copper sphere is given by:

2 r2 θ = r2 t. (3.59) The equal area law is the statement that the term in parentheses,

1240 ev nm 2.5 ev. (4) r 2 or mv 2 = ke2

10. Collisions. Before During After

Gravitation. AP Physics C

PY1052 Problem Set 8 Autumn 2004 Solutions

8.4. Motion of Charged Particles in Magnetic Fields

Phys 2101 Gabriela González. cos. sin. sin

AP Physics Electromagnetic Wrap Up

Magnetic Field and Magnetic Forces. Young and Freedman Chapter 27

FXA Candidates should be able to : Describe how a mass creates a gravitational field in the space around it.

Voltage ( = Electric Potential )

Parameter Identification of DC Motors

Standardized Coefficients

PHYSICS 111 HOMEWORK SOLUTION #13. May 1, 2013

Voltage ( = Electric Potential )

AP Physics Gravity and Circular Motion

The force between electric charges. Comparing gravity and the interaction between charges. Coulomb s Law. Forces between two charges

Chapter 22. Outside a uniformly charged sphere, the field looks like that of a point charge at the center of the sphere.

Mechanics 1: Work, Power and Kinetic Energy

Chapter 11 Relative Velocity

F G r. Don't confuse G with g: "Big G" and "little g" are totally different things.

Description: Conceptual questions about projectile motion and some easy calculations. (uses applets)

PHYSICS 151 Notes for Online Lecture #11

9.5 Amortization. Objectives

Uniform Rectilinear Motion

Multiple choice questions [70 points]

A) When two objects slide against one another, the magnitude of the frictional force is always equal to μ

Solutions to Problems: Chapter 7

Graphs of Equations. A coordinate system is a way to graphically show the relationship between 2 quantities.

Forces & Magnetic Dipoles. r r τ = μ B r

Coordinate Systems L. M. Kalnins, March 2009

Displacement, Velocity And Acceleration

4.4 VOLUME AND SURFACE AREA

1D STEADY STATE HEAT

MECH Statics & Dynamics

(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

Chapter 19: Electric Charges, Forces, and Fields ( ) ( 6 )( 6

Solutions for Physics 1301 Course Review (Problems 10 through 18)

Pearson Physics Level 30 Unit VI Forces and Fields: Chapter 10 Solutions

The Role of Gravity in Orbital Motion

Episode 401: Newton s law of universal gravitation

Gravity. A. Law of Gravity. Gravity. Physics: Mechanics. A. The Law of Gravity. Dr. Bill Pezzaglia. B. Gravitational Field. C.

Gravitation and Kepler s Laws Newton s Law of Universal Gravitation in vectorial. Gm 1 m 2. r 2

6. Friction, Experiment and Theory

Physics Core Topic 9.2 Space

Fluids Lecture 15 Notes

Spirotechnics! September 7, Amanda Zeringue, Michael Spannuth and Amanda Zeringue Dierential Geometry Project

Physics 235 Chapter 5. Chapter 5 Gravitation

TORQUE AND ANGULAR MOMENTUM IN CIRCULAR MOTION

Skills Needed for Success in Calculus 1

(d) False. The orbital period of a planet is independent of the planet s mass.

Figure 2. So it is very likely that the Babylonians attributed 60 units to each side of the hexagon. Its resulting perimeter would then be 360!

Doppler Effect. wavelength

Analytical Proof of Newton's Force Laws

AP Physics B 2012 Free-Response Questions

Effect of Unemployment Insurance Tax On Wages and Employment: A Partial Equilibrium Analysis

Chapter 3 Savings, Present Value and Ricardian Equivalence

4a 4ab b (count number of places from first non-zero digit to

2. Orbital dynamics and tides

Vector Calculus: Are you ready? Vectors in 2D and 3D Space: Review

Experiment MF Magnetic Force

Gauss Law. Physics 231 Lecture 2-1

CHAPTER 5 GRAVITATIONAL FIELD AND POTENTIAL

L-9 Conservation of Energy, Friction and Circular Motion. Kinetic energy. conservation of energy. Potential energy. Up and down the track

Problem Set # 9 Solutions

A r. (Can you see that this just gives the formula we had above?)

Moment and couple. In 3-D, because the determination of the distance can be tedious, a vector approach becomes advantageous. r r

Physics HSC Course Stage 6. Space. Part 1: Earth s gravitational field

Motion Control Formulas

AP Physics C: Mechanics 2005 Free-Response Questions

v = x t = x 2 x 1 t 2 t 1 The average speed of the particle is absolute value of the average velocity and is given Distance travelled t

Chapter 10 Velocity, Acceleration, and Calculus

Deflection of Electrons by Electric and Magnetic Fields

Quantity Formula Meaning of variables. 5 C 1 32 F 5 degrees Fahrenheit, 1 bh A 5 area, b 5 base, h 5 height. P 5 2l 1 2w

The Essence of the Electromagnetic Wave is Not Energy

Lab #7: Energy Conservation

Continuous Compounding and Annualization

Lecture L9 - Linear Impulse and Momentum. Collisions

Lecture 14: Transformers. Ideal Transformers

12.4 Problems. Excerpt from "Introduction to Geometry" 2014 AoPS Inc. Copyrighted Material CHAPTER 12. CIRCLES AND ANGLES

Massachusetts Institute of Technology Department of Electrical Engineering and Computer Science


Strength Analysis and Optimization Design about the key parts of the Robot

The Electric Potential, Electric Potential Energy and Energy Conservation. V = U/q 0. V = U/q 0 = -W/q 0 1V [Volt] =1 Nm/C

Purchase and rental subsidies in durable-good oligopolies* 1

Physics 111. Exam #1. January 24, 2014

Chapter 17 The Kepler Problem: Planetary Mechanics and the Bohr Atom

Transcription:

Exaple iction + Unifo Cicula Motion Cicula Hill A ca i diing oe a ei-cicula 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 the Poble If the ca die oe a cicula hill, it tael in a cicle. Appaentl, if the ca tael too fat it cannot eain on a cicula path that follow the hill. One o oe of the foce eponible fo the ca taing on the cicle ut be liited in oe wa. Let ue a fee bod diaga to ee if we can deteine what foce ae eponible fo the ca cicula otion. N Gound on Ca W Eath on Ca he net foce on the ca ut point to the cente of the cicle (becaue it in cicula otion). A the ca tael fate and fate, the hill need to poide le and le noal foce to keep the ca fo penetating it. Eentuall the ca could be going fat enough that the noal foce i zeo if it went an fate that thi citical peed, the weight won t be enough to keep the ca going in a cicle coinciding with the hill and the wheel will lift off of the gound. Since the ca tael in a cicle, we ll alot cetainl ue the elationhip between centipetal acceleation and tangential peed. () epeent the Poble in oal e (Decibe the Phic) We hae been gien o identified the following quantitie: the a of the ca the peed of the ca the adiu of the cicle the ca' following (i.e. the hill' adiu) agnitude N N he noal foce fo the gound on the ca diection up agnitude g W he weight fo the eath on the ca diection down

Exaple iction + Unifo Cicula Motion We alo know the following elationhip that ight be ueful: Ma (Newton' Second Law) a Ca ca net towad cicle cente (Ca' in unifo cicula otion) (3) Plan the Solution We can appl Newton Second Law to the ca at the top of the hill to elate the foce on the ca to it acceleation. We know the fo of the ca acceleation ince it in cicula otion. We can et the noal foce equal to zeo; thi hould happen when the ca i going a fat a it can and till ta on the cicle. (4) Execute the Solution We ll ue the fee bod diaga to wite Newton Second Law fo the ca at the top of the hill. We chooe downwad a poitie becaue we know that the ca i acceleating downwad (i.e. towad the cente of the cicle) Appl Newton Second Law in the -diection. Cicula otion iplie a / towad the cente of the cicle. At the axiu peed, the noal foce will be zeo. he entie downwad foce towad the cente i upplied b the ca weight alone. net, N W Ca ( N ) a + Ca + g ( 0) N Gound on Ca W ax Eath on Ca + g g ax (5) Intepet and Ealuate the Solution A a quick check, let a the hill had a adiu of aound 100. he fatet peed the ca 1 k 3600 k could go would then be ( 9. 81 )( 100 ) 31 11 (about 70 ile pe 1000 1 h h hou). hi ee at leat eaonable fo the peed of a ca lifting off the top of a teep hill.

Exaple iction + Unifo Cicula Motion Plagound Slipping Soe childen ae plaing with a lage ock on the plagound. he keep putting it on a e-go-ound pinning at 1 eolution ee econd. he e ting to find out how fa awa fo the cente the can place the ock befoe it lide to the outide. You look up the coefficient of tatic fiction fo ock on etal and find it aound 0.3. What i the fathet ditance fo the cente of the e-go-ound that the ock can it and till not lide? (1) Copehend the Poble Let tat with a ketch to ee what going on. he ock i itting on a e-go-ound (a flat, pinning dik) pinning at a known ate. 1 e ec op View Side View We know that the ock i taelling in a cicle. he quetion i how lage of a adiu can the cicle hae befoe the ock ut lip? Since it taelling in a cicle at contant peed, we know that the acceleation of the ock i elated to it elocit (a /). In addition, we know how long it take the ock to ake a coplete eolution ( econd). We need to know what foce poduce the ock acceleation towad the cente of the cicle. Let figue thi out b dawing the fee bod diaga fo the ock (we ll ue the ide iew ). f N loo on ock loo on ock W Eath on ock hee ae thee foce on the ock, the noal foce fo the floo of the e-go-ound, the weight fo the eath, and the tatic fiction foce fo the floo of the e-go-ound. Since the tatic fiction i the onl foce in the hoizontal diection, we know it ut point towad the cente of the cicle in ode to pull the ock towad the cente (a deanded b the ock cente-pointing acceleation). he tatic fiction foce ha a axiu tength it can poide, gien b f tatic, ax µ N. A we oe the ock outwad, the tatic fiction foce equied to keep it going aound the cicle inceae. hee will appaentl be a axiu adiu at which the tatic fiction foce hit thi axiu alue. If the ock i placed an fathe awa, it will lip.

Exaple iction + Unifo Cicula Motion () epeent the Poble in oal e (Decibe the Phic) We ae gien o hae identified the following quantitie a ueful: the tie it take the e-go-ound to ake one eolution the a of the ock agnitude N N the noal foce fo the floo on the ock diection up agnitude g W the weight foce fo the eath on the ock diection down agnitude the tatic fictional foce fo the floo on the ock diection towad cente the ditance fo the cicle cente to the ock (the adiu) We alo hae the following elationhip that ight be ipotant. Ma Newton' Second Law f net tatic, ax a c µ N tatic fiction axiu centipetal acceleation (3) Plan the Solution Since it in unifo cicula otion, we know the ock acceleation in te of it elocit and adiu. We can ue Newton Second Law to elate thi acceleation to the foce on the ock. B ineting the axiu alue fo the tatic fictional foce, we hould be able to find the laget adiu at which the ock won t lide. (4) Execute the Solution Daw the fee bod diaga fo the ock. We chooe towad the cente of the cicle a poitie, a thi i the diection the ock actuall acceleate. Appl Newton Second Law in the x- diection. Since the ock i in unifo cicula otion, we know how it acceleation i elated to and. f N net, x a f + N + W a x, x x x x + 0 + 0 W x

Exaple iction + Unifo Cicula Motion Since we don t know the ock elocit, we need to wite it in te of the adiu and the eolution tie (a.k.a. the peiod). he ock tael one cicufeence in tie. eplace the elocit in ou Newton Second Law x-equation with thi expeion fo. Now we hae the adiu in te of thing we know (peiod and a ) and the tatic fiction foce. he axiu adiu i whee the tatic fiction foce i at it axiu. We need the noal foce fo the floo on the ock. We can find thi fo appling Newton Second Law in the -diection. We know the acceleation of the ock in the -diection i zeo. We can inet thi noal foce alue into the expeion fo the axiu adiu to get ou anwe. (Note that the a cancel out) Now we can inet the nueical alue fo the poble tateent. ditance π tie π f ax f, ax ( µ N ) net, f + N + W ( 0), ( g ) a 0 + N + 0 ax ax µ µ g ( ) ( )( ) (. ) N g ( g ) 0. 3 9. 81 100 c ax 0. 30 30 c 4 3 1416 1 he ock will lide if it i placed oe than 30 c (about 1 ft) fo the cente of the e-goound. If we place the ock anwhee cloe to the cente than thi, it will not lide on the e-goound. he ock would then jut tael aound the cente in a cicle with the et of the ego-ound.