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# General Physics PHYS 201

Christopher Newport University

GPA 3.81

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This 77 page Class Notes was uploaded by Bart Thiel Jr. on Monday October 5, 2015. The Class Notes belongs to PHYS 201 at Christopher Newport University taught by Staff in Fall. Since its upload, it has received 20 views. For similar materials see /class/219497/phys-201-christopher-newport-university in Physics 2 at Christopher Newport University.

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Date Created: 10/05/15

Chapter 8 Potential Energy Potential Energy 0 Potential energy is the energy associated with the configuration of a system of objects that exert forces on each other 0 This can be used only with conservative forces 0 When conservative forces act within an isolated system the kinetic energy gained or lost by the system as its members change their relative positions is balanced by an equal loss or gain in potential energy 0 This is Conservation of Mechanical Energy Types of Potential Energy 0 There are many forms of potential energy including o Gravitational o Electromagnetic 0 Chemical 0 Nuclear 0 One form of energy in a system can be converted into another Systems with Multiple Particles e We can extend our definition ofa system to include multiple objects 0 The force can be internal to the system 0 The kinetic energy of the system is the algebraic sum of the kinetic energies ofthe individual objects 0 Sometimes the kinetic energy of one of the objects may be negligible System Example This system consists of Earth and a book Do work on the system by lifting the book through Ay The work done by the gravitational force om the book as it falls from btoamgAr Potential Energy C3130 co 0 O D It CHI 239 o The energy storage mechanism is called potential energy 0 Potential energy is always associated with a system of two or more interacting objects Gravitational Potential Energy cont i W FappAr W mgiyb yaJ W mgyb mng o The quantity mgy is identified as the gravitational potential energy U9 0 U9 mgy 0 Units are joules J Conservation of Mechanical Energy o The mechanical energy of a system is the algebraic sum of the kinetic and potential energies in the system Emech K Ug The statement of Conservation of Mechanical Energy for an isolated system is Kf Uf K UI 0 An isolated system is one for which there are no energy transfers across the boundary Problem 1 A particle of massm 650 kg is released from point A and slides on the frictionless track shown in the figur below hA 730 m a Determine the particle39s speed at points B and C point B ans896 ms 102 ms b Determine the net work done by the gravitational force in moving the particle from A to C ans338 J Elastic Potential Energy Elastic Potential Energy is associated with a spnng The force the spring exerts on a block for example is Fs kx The work done by an external applied force on a springblock system is w kx2 kx2 o The work is equal to the difference between the initial and final values of an expression related to the configuration of the system Conservation of Energy Example 1 Drop a Ball Initial conditions EK umgh u The ball is dropped so K 0 quot l39 The configuration for zero 394 potential energy is the ground Conservation rules applied at some point y above the ground gives u 12 mv2 mgy mgh Conservation of Energy Example 2 Pendulum As the pendulum swings there is a continuous change between potential and kinetic energ39es At A the energy is potential At B all of the potential energy at A is transformed into kinetic energy Let zero potential energy be at B At C the kinetic energy has been transformed back into potential energy 313 0 Conservation of Energy Example 3 Spring Gun 0C 8G 0 Choose point A as the initial point and C as the final point 0 EA I C 39 KA UgAUsAKA UgA UsA o 12 kX2 mgh Conservative Forces 0 The work done by a conservative force on a particle moving between any two points is independent of the path taken by the particle o The work done by a conservative force on a particle moving through any closed path is zero 0 A closed path is one in which the beginning and ending points are the same Conservative Forces cont 0 Examples of conservative forces 0 Gravity 0 Spring force 0 We can associate a potential energy for a system with any conservative force acting between members of the system 0 This can be done only for conservative forces 0 In general WC AU Nonconservative Forces 0 A nonconservative force does not satisfy the conditions of conservative forces 0 Nonconservative forces acting in a system cause a change in the mechanical energy of the system 313 0 Mechanical Energy and Nonconservative Forces O 8G 0 In general if friction is acting in a system 0 AEmech AK AU fkd 0 AU is the change in all forms of potential energy 0 If friction is zero this equation becomes the same as Conservation of Mechanical Energy Nonconservative Forces Example AEmech AK AU AEmech Kf Uf f K U 7 If AE lt 0 9 friction is mech reducmg the mechnlcal energy of the system AE 12mvf2 mghfkd mech Problem 0 A crate of mass 106 kg is pulled up a rough incline with an initial speed of 144 ms The pulling force is 108 N parallel to the incline which makes an angle of 203 with the horizontal The coefficient of kinetic friction is 0400 and the crate is pulled 510 m a How much work is done by the gravitational force on the crate Ans 184 J b Determine the increase in internal energy of the crateincline system due friction Ans199 J c How much work is done by the 108 N force on the crate Ans 551 J Chapter 4 Motion in Two Dimensions Position and Displacement o The position of an object is described by its position vector r o The displacement of the object is defined as the change in its position 0 Ar l f l i 2 Path of rf particle X a 2m ThurmanEmir m 0quot QC 000 C 0 Q C O C 0 Motion in Two Dimensions 0 Using or signs is not always sufficient to fully describe motion in more than one dimension o Vectors can be used to more fully describe motion 0 Still interested in displacement velocity and acceleration o Will serve as the basis of multiple types of motion in future chapters General Motion Ideas o In two or threedimensional kinematics everything is the same as as in one dimensional motion except that we must m use full vector notation o Positive and negative signs are no longer sufficient to determine the direction soon 0 9000 cc Average Velocity o The average velocity is the 339 ratio of the displacement to the time interval for the displacement Direction ofv at i Ar V 7 At 0 The direction of the average velocity is the direction of the displacement vector Ar mmmmmmmmmmmm ale Instantaneous Velocity o The instantaneous velocity is the limit of the average velocity as At approaches zero 0 The direction of the instantaneous velocity is along a line that is tangent to the path of the particle s direction of motion 1A 3At It Average Velocity cont o The average velocity between points is independent of the path taken 0 This is because it is dependent on the displacement also independent of the path 4 0 lt30 09 J Average Acceleration 0 The average acceleration of a particle as it moves is defined as the change in the instantaneous velocity vector divided by the time interval during which that change occurs vf Vl AV 5 tf tl At Average Acceleration cont 0 As a particle moves Av can be found in different ways 0 The average acceleration is a vector quantity directed along Av Producing An Acceleration o Various changes in a particle s motion may produce an acceleration o The magnitude of the velocity vector may change o The direction of the velocity vector may change 0 Even if the magnitude remains constant o Both may change simultaneously 3 Instantaneous Acceleration o The instantaneous acceleration is the limit of the average acceleration as AvAt approaches zero AV dV a239 lAltLr At dt Kinematic Equations Component Equations o vf v atbecomes 0 V vX aXtand o vyf vy ayt o r r vz 12az 2 becomes 0 xfxvXl 12axl 2 and o yfyvyl 12ayl 2 2D problem 0 A golf ball is hit off a tee at the edge of a cliff ts x and y coordinates are given by the following expressions x 18 y 4 749 Find position using i and unit vectors Velocity vector v Acceleration vector a Velocity at t 300 seconds Assumptions of Projectile Motion o The freefall acceleration g is constant over the range of motion And is directed downward o The effect of air friction is negligible 0 With these assumptions an object in projectile motion will follow a parabolic path This path is called the trajectory Projectile Motion 0 An object may move in both the Xand y directions simultaneously 0 The form of twodimensional motion we will deal with is called projectile motion Verifying the Parabolic Trajectory 0 Reference frame chosen y is vertical with upward positive 0 Acceleration components ayg and aX0 0 Initial velocity components VX vcos19 and V vsinz9 Verifying the Parabolic Trajectory cont 0 Displacements 1 2 xf vxit Eaxt 2 v1 cos 19t O 1 2 2 yr vi sm 29t gt 1 yf vyt a 000 Projectile Motion Implications 3 Q o The y component of the velocity is zero at the maximum height of the trajectory o The accleration stays the same throughout the trajectory Projectile Motion Diagram 00 0 KW o o o o e o mmmmmmmmmmmm le Range and Maximum Height a Projectile c When analyzing projec ri39 y motion two characteristics are of special interest 0 The range R is the horizontal distance of ti projectile o The maximum height tiquot projectile reaches is h mmmmm awnBrooks Cain Projectile Motion Problem Solving Hints More About the Range of a Projectile 73163 Select a coordinate system y111 Resolve the initial velocity into X and y components Analyze the horizontal motion using constant 150 velocity techniques 11 501115 o Analyze the vertical motion using constant acceleration techniques 0 Remember that both directions share the same time x011 NonSymmetric Projectile Motion Projectile Problem 1 11 2001115 1 1 201ll111s 0 Follow the general rules for projectile motion o Break the y direction into x j parts o How lon gxdoes it take the rock to reach a up and down or the ground 0 zmwteg gilhgicgteorzsi iie r 0 With what velocity Will the rock hit the of t e height ground 4150111 Y quot 39 391 if 3 7 450111 39 7 Agdvii b V 1 g X Z 1 4 1 1 l quotl l 397 11e4f11l 111 391 w39homnnEmolsW e LY 39 a l l 1 M 1 Chapter 2 Motion in One Dimension Problem 1 A particle moves along the X axis according to the equation X 2 01 2 97tr t2 where XiS in meters and tis in seconds a What is its velocity at t 2 80 s b What is its acceleration at t 2 80 s Kinematic Equations 0 The kinematic equations may be used to solve any problem involving onedimensional motion with a constant acceleration 0 You may need to use two of the equations to solve one pro em 0 Many times there is more than one way to solve a problem Kinematic Equations specific o For constant a fo V 11 0 Can determine an object s velocity at any time twhen we know its initial velocity and its acceleration 0 Does not give any information about displacement Kinematic Equations specific 0 For constant acceleration V vx v3 x 2 o The average velocity can be expressed as the arithmetic mean of the initial and final velocities Kinematic Equations specific o For constant acceleration xf o This gives you the position of the particle in terms of time and velocities 0 Doesn t give you the acceleration 7 1 xivtxi5vxf v t aw Kinematic Equations specific 039 o For constant acceleration x 1 2 f xlvxit axt o Gives final position in terms of velocity and acceleration 0 Doesn t tell you about final velocity Kinematic Equations specific 0 For constant a 2 2 VJ vi 2axf xl o Gives final velocity in terms of acceleration and displacement 0 Does not give any information about the time SUI DO 0 Problem 2 The driver of a car slams on the brakes when he sees a tree blocking the road The car slows uniformly with acceleration 580 ms2 for 400 s making straight skid marks 636 m long ending at the tree With what speed does the car then strike the tree 90 0 Graphical Look at Motion displacement time curve 9000 0000 o The slope of the curve is the velocity x o The curved line Slope vxfygf indicates the velocity is changing I 0 Therefore there IS an I acceleration Slope vm 0 i 5 a Graphical Look at Motion velocity time curve o The slope gives the acceleration W o The straight line Skqxz ax 6 VV Indicates a constant as I I t39 y I axt acce era Ion U i m r w 0 It i t b Graphical Look at Motion acceleration time curve 0 The zero slope indicates a constant a acceleration Slope 0 Problem 3 The position versus time for a certain particle moving along the xaxis is shown in the figure below Find the average velocity in the following time intervals a 0 to 2 s 39 Problem 4 A positiontime graph for a particle moving along the xaxis is shown in the figure The divisions along the horizontal axis represent 150 s and the divisions along the vertical axis represent 50 m Determine the instantaneous velocity at f 600 s where the tangent line touches the curve by measuring the slope of the tangent line shown in the graph r trmr Freely Falling Objects o A freely falling object is any object moving freely under the influence of gravity alone 0 It does not depend upon the initial motion of the object o Dropped released from rest o Thrown downward o Thrown upward Acceleration of Freely Falling Object o The acceleration of an object in free fall is directed downward regardless of the initial motion o The magnitude of free fall acceleration is g 980 ms2 g decreases with increasing altitude gvaries with latitude 980 ms2 is the average at the Earth s surface Acceleration of Free Fall cont c We will neglect air resistance 0 Free fall motion is constantly accelerated motion in one dimension 0 Let upward be positive 0 Use the kinematic equations with ay g 980 ms2 Free Fall Example A ball is thrown vertically upward from the giround with an initial speed of 150 ms a w Ion oes it ta et e ball to reach its maximum altitude b What is its maximum altitude c Determine the velocity and acceleration of the ball at t 200 5 Answer a 153 s b 115 m c v 460 ms a 980 ms2 General Problem Solving Strategy o Conceptualize o Categorize o Analyze 0 Finalize Problem Solving Conceptualize Think about and understand the situation Make a quick drawing of the situation Gatherthe numerical information 0 Include algebraic meanings of phrases Focus on the expected result 0 Think aboutunits Think about what a reasonable answer should be Problem Solving Categorize o Simplify the problem Can you ignore air resistance Model objects as particles 0 Classify the type of problem 0 Try to identify similar problems you have already solve Problem Solving Analyze 0 Select the relevant equations to apply 0 Solve for the unknown variable 0 Substitute appropriate numbers 0 Calculate the results Include units 0 Round the result to the appropriate number of significant figures Problem Solving Finalize 0 Check your result Does it have the correct units Does it agree with your conceptualized ideas 0 Look at limiting situations to be sure the results are reasonable 0 Compare the result with those of similar problems Problem Solving Some Final Ideas c When solving complex problems you may need to identify subproblems and apply the problemsolving strategy to each subpart 0 These steps can be a guide for solving problems in this course Physics for Scientists and Engineers Introduction and Chapter 1 o e 0 Classical Physics 0 Mechanics and electromagnetism are basic to all other branches of classical and modern physics 0 Classical physics Developed before 1900 Our study will start with Classical Mechanics 0 Also called Newtonian Mechanics or Mechanics 0 Modern physics From about 1900 to the present Physics Fundamental Science 0 Concerned with the fundamental principles of the Universe 0 Foundation of other physical sciences 0 Has simplicity of fundamental concepts Divided into five major areas Classical Mechanics Relativi Thermodynamics Electromagnetism ptics Quantum Mechanics Objectives of Physics 0 To find the limited number of fundamental laws that govern natural phenomena 0 To use these laws to develop theories 0 Express the laws in the language of mathematics Mathematics provides the bridge between theory and experiment Theory and Experiments 0 Should complement each other 0 When a discrepancy occurs theory may be modified Theory may apply to limited conditions 0 Example Newtonian Mechanics is confined to objects traveling slowly with respect to the speed of light Try to develop a more general theory Modern Physics 0 Began near the end of the 19 h century 0 Phenomena that could not be explained by classical physics 0 Includes theories of relativity and quantum mechanics Classical Physics Overview Classical physics includes principles in many branches developed before 1900 Mechanics 0 Major developments by Newton and continuing through the 18 h century Thermodynamics optics and electromagnetism o Developed in the latter part of the 19 h century Quantum Mechanics o Formulated to describe physical phenomena at the atomic level 0 Led to the development of many practical devices Fundamental Quantities and Their Units Quantity SI Unit Length meter Mass kilogram Time second Temperature Kelvin Electric Current Ampere Luminous Intensity Candela Amount of Substance mole Length 0 Length is the distance between two points in space 0 Units SI meter m 0 Defined in terms of a meter the distance traveled by light in a vacuum during a given time 0 See Table 11 for some examples of lengths Quantities Used in Mechanics o In mechanics three basic quantifies are used Length Mass Time 0 Will also use derived quantifies These are other quantities that can be expressed in terms of the basic quantities 0 Example Area is the product of two lengths Area is aderived quanlil Length is the fundamental quantity Mass 0 Units SI kilogram kg 0 Defined in terms of a kilogram based on a specific cylinder kept at the International Bureau of Standards 0 See Table 12 for masses of various objects Time 0 Units 0 secondss 0 Defined in terms of the oscillation of radiation from a cesium atom 0 See Table 13 for some approximate time intervals Prefixes o Prefixes correspond to powers of 10 0 Each prefix has a specific name 0 Each prefix has a specific abbreviation Reasonableness of Results 0 When solving a problem you need to check your answer to see if it seems reasonable Prefixes cont 0 The prefixes can be used with any basic units 0 They are multipliers of the basic unit 0 Examples o 1 mm 10393 m 1 1 3 TA BLE 39 AL mg 0 g Prefixes for Powers of Ten Power Prefix Abbreviation Power Prefix Abbreviation 1041 yocto y 10395 kilo k 10 zepto 7 106 mega M 10 8 alto a inquot gigzi G 10 15 femto f If era T 10le pico 1 ii15 peta P 10 9 1121110 11 Ni18 le E 10396 micro u 10239 Zena Z 1l 3 milli m 102quot yotta 391 10 csnti 10 1 led l nnnnnnnnnnnnnnnnnnnnnnn an Basic Quantities and Their Dimension o Dimension has a specific meaning it denotes the physical nature of a quantity o Dimensions are denoted with square brackets 0 Length L 0 Mass M 0 Time T Dimensional Analysis example o Given the equation X 12 at2 o Check dimensions on each side Dimensions and Units 0 Each dimension can have many actual units 0 Table 15 for the dimensions and units of some derived quantities TABLE 39l 5 Dimensions and Units of Four Derived Quantities Quantity Area Volume Speed Acceleration Dimensions L2 L3 LT LT2 SI units 111 2 1113 ms 11152 USi customary unit39s ft ftg fts ftsg Significant Figures 0 A significant figure is one that is reliably known 0 Zeros may or may not be significant 0 Those used to position the decimal point are not significant 0 To remove ambiguity use scientific notation o In a measurement the significant figures include the first estimated digit Chapter 10 Rotation of a Rigid Object about a Fixed Axis Angular Position 2 Point Pwill rotate about the origin in a circle of radius r Every particle on the disc undergoes circular motion about the origin 0 P is located at r 639 where ris the distance from the origin to P and 6is the measured counterclockwise from the reference line Angular Position Axis of rotation is the center of the disc Choose a fixed reference line T 0 Point Pis at afixed 0 13 R f distance rfrom the v e 53 origin M a 0 O 0 3 Angular POSItIon 3 As the particle moves the only coordinate that changes is 6 p As the particle moves a 139 5 through 6 it moves 0 though an arc length 0 Reference 8 339 line The arc length and r are related b o s 0r W Radian 0 This can also be exoressed as 6 quot 0 Bis a pure number but commonly is given the artificial unit radian 0 One radian is the angle subtended by an arc length equal to the radius of the arc Angular Displacement o The angular displacement is V defined as the angle the object rotates through during some time interval 0 Aha 491 Conversions 0 Comparing degrees and radians 1 rad 3600 27 0 Converting from degrees to radians 6rad 7T 9 degrees 180 Average Angular Speed 0 The average angular speed a of a rotating rigid object is the ratio of the angular displacement to the time interval QA6 a7 tf tl At Average Angular Acceleration 0 The average angular acceleration or of an object is defined as the ratio of the change in the angular speed to the time it takes for the object to undergo the change 07 t tl At f Comparison Between Rotation and Linear Equations Table 101 Kinematic Equations for Rotational and Linear Motion Under Constant Acceleration Rotational Motion About Fixed Axis Linear Motion of curlat of viat l 9 l c 0 0 l all ECU x Uil Eatz w2 02 2aaf 0i 39 1 waymww 112 11 2a9r xi 99 Xi av 11f mam Yhomsn mmmmmm is Rotational Kinematic Equa ons wfzwiat 6f 6i aitlart2 2 a wf 2a6f 6 6f 6iaiaft Relationship Between Angular 5555 n n n O and Linear Quantities 0 Displacements o Every point on the S 6r rotating object has the speeds same angular motion V Z or 0 Every point on the o Accelerations 6105 rotating object does not have the same linear motion 0 O 00 23 Speed Comparison 0 The linear velocity is y always tangent to the w circular path v V 0 called the tangential velocity P o The magnitude is r 5 defined by the 0 l tangential speed 0 l x I ds d6 V r r60 mewwm dt dt 0 O 0 33 Centripetal Acceleration 0 An object traveling in a circle even though it moves with a constant speed will have an acceleration o Therefore each point on a rotating rigid object will experience a centripetal acceleration 2 2 V M I I Acceleration Comparison 0 The tangential acceleration is the derivative of the tangential velocity dv day a r E E Rotational Motion Example 0 For a compact disc player to read a CD the angular speed must vary to keep the tangential speed M vt ar 0 At the inner sections the angular speed is faster than at the outer sec ons mmmmmmmmmmmmm l s Rotational Kinetic Energy n An object rotating about some axis with an angular speed a has rotational kinetic energy even though it may not have any translational kinetic energy Each particle has a kinetic energy of e K V2 my2 Since the tangential velocity depends on the distance r from the axis of rotation we can substitute v a r Rotational Kinetic Energy final There is an analogy between the kinetic energies associated with linear motion K 2 mv2 and the kinetic energy associated with rotational motion Kg 2 I602 Rotational kinetic energy is not a new type of energy the form is different because it is applied to a rotating object The units of rotational kinetic energy are Joules J Rotational Kinetic Energy cont o The total rotational kinetic energy of the rigid object is the sum of the energies of all its particles KR 2K Z m wz 1 7 7 1 7 K 7 a 7101 R 2 0 Where is called the moment of inertia Moment of Inertia o The definition of moment of inertia is I cm 0 The dimensions of moment of inertia are ML2 and its SI units are kgm2 c We can calculate the moment of inertia of an object more easily by assuming it is divided into many small volume elements each of mass Am Moments of Inertia of Various Rigid Objects Torque cont The moment arm d is the perpendicular distance from the axis of rotation to a line drawn along the direction of the force o d rsin D Fsin g F 1 u 39 Fcos 0 k r a y l Llne of d acuou Torque o Torque T is the tendency of a force to rotate an object about some axis 0 Torque is a vector 0 rrFsin Fd o Fistheforce o is the angle the force makes with the horizontal o dis the moment arm or lever arm Torque final o The horizontal component of F Fcos has no tendency to produce a rotation o Torque will have direction o If the turning tendency of the force is counterclockwise the torque will be positive a If the turning tendency is clockwise the torque will be negative 0 O 0 Net Torque 0 The force F1 will tend to cause a counterclockwise rotation about 0 o The force F2 will tend to cause a clockwise rotation about 0 o ZIqz F L f ag Torque and Angular Acceleration Particle cont 0 Since mr2 is the moment of inertia of the particle o 239 la a The torque is directly proportional to the angular acceleration and the constant of proportionality is the moment of inertia 6 V Vork llquot T If quoti Rotational kinetic energy KR 4le Power 97 Ta Angular momentum L 2 la Net torque 2739 ILll Torque and Angular 3 I Acceleration Consider a particle of mass m rotating in a circle of radius runder the influence of tangential force Ft The tangential force provides a tangential acceleration l39 0 Ft mat T 33 Summary of Useful Equations Rotational Motion About a Fixed Axis Linear Motion Angular speed to Hir1 Linear speed 1 I Ix1 Angular acceleration a wIl Linear acceleration I Ivll Net torque ET for Net force 2 mu ll w w a ll zif 11 a a constant 6 6i 1 Eat2 a constant X xi 11 01 w2 02 2a0 6 112 vi 2rx xi 1 Nork Hquot I R 13 x Kinetic energy K 11113quot Power Fr Linear momentum 2 nm Net force 2F ripll Total Kinetic Energy of a Rolling Object 0 The total kinetic energy of a rolling object is the sum of the translational energy of its center of mass and the rotational kinetic energy about its center of mass K12ICM w2 VszCM2 Total Kinetic Energy Example cont Despite the friction no loss of mechanical energy occurs because the contact point is at rest relative to the surface at any instant Let U 0 at the bottom of the plane Kf Uf K K 12 ICM Ff 2 VCM2 12 MVCM2 U Mgh 39 Ur 0 Total Kinetic Energy Example 0 Accelerated rolling motion is possible only M if friction is present between the sphere and the incline I o The friction producesthe x aquot net torque required for 7quot rotation Chapter 11 Angular Momentum The Vector Product and Torque o The torque vector lies in a direction perpendicular to the plane formed by the position vector and the force vector 0 C I X F o The torque is the vector or cross product of the position vector and the force vector The Vector Product 0 There are instances where the product of two vectors is another vector 0 Earlier we saw where the product of two vectors was a scalar c This was called the dot product 0 The vector product of two vectors is also called the cross product The Vector Product Defined 0 Given two vectors A and B c The vector cross product of A and B is defined as a third vector C c C is read as A cross B c The magnitude of C is AB sin 6 0 His the angle between A and B More About the Vector Product 35 0 The quantity AB sin Bis equal to the area of the parallelogram formed by A and B The direction of C is A perpendicular to the plane Clt 9 formed by A and B i The best way to determine this direction is to use the mm righthand rule Vector Products of Unit Vectors 25339 ixiAjgtltjlEgtltlEO ixj jxil jxlE 1Egtltji x ixl j Properties of the Vector Product o The vector product is not commutative The order in which the vectors are multiplied is important c To account for order remember AXBBXA o If A is parallel to B e O0 or 180 then A x B O 0 Therefore A x A 0 Vector Products of Unit Vectors cont 0 Signs are interchangeable in cross products AXBAXB ix i ixi Using Determinants 55quot o The cross product can be expressed as i j 12 A A A A A A A A A AgtltBAx A A y Zi x Zj x yk y Z B B B B B B Bx By BZ y Z x Z x y 0 Expanding the determinants gives AgtltB 1sz AZBy i AxBZ AZBxi AxBy AyBx12 Angular Momentum 55quot 0 Consider a particle of mass m located at the vector position r and moving with linear momentum p erFZTrxg Adding the term g X p drgtltp 27 dz Torque Vector Example 35quot 0 Given the force F 200 300 i N r 400i 500 i m zquot r x F 400i 5003N gtlt 200i 3003m 400200igtlti 400300i x3 500200jgtlti 500300igtltj 2012 N m Angular Momentum cont 55quot o The instantaneous angular momentum L of a particle z relative to the origin 0 is I defined as the cross product of the particle s instantaneous position vector r and its instantaneous linear momentum p o Lrxp 0 Torque and Angular 355 O Momentum 3 o The torque is related to the angular momentum 0 Similar to the way force is related to linear momentum drgtltp 21 dt 27 dt o This is the rotational analog of Newton s Second Law 22quot and L must be measured about the same origin o This is valid for any origin fixed in an inertial frame 33 Angular Momentum of a 33 0 Particle Example 339 o The vector L r x p is pointed out of the diagram 0 The magnitude is L mvr sin 900 mvr o sin 90 is used since vis perpendicular to r o A particle in uniform circular motion has a constant angular momentum about an axis through the center of its path More About Angular Momentum o The SI units of angular momentum are rx p kgm2 s 0 Both the magnitude and direction of L depend on the choice of origin 0 The magnitude of L mvr sin 0 bis the angle between p and r o The direction of L is perpendicular to the plane formed by r and p Angular Momentum of a Rotating Rigid Object 0 Each particle of the object z rotates in the xy plane about the z axis with an angular speed of a o The angular momentum of an individual particle is L m r2 a o L and 0 are directed along the zaxis Angular Momentum of a Rotating Rigid Object cont a To find the angular momentum of the entire object add the angular momenta of all the individual particles LZ 2L 2m2co 0 o This also gives the rotational form of Newton s Second Law 27 1d w Ia dt dt Conservation of Angular Momentum o The total angular momentum of a system is constant in both magnitude and direction if the resultant external torque acting on the system is zero o Net torque O gt means that the system is isolated o Ltot constant or L Lf o For a system of particles Ltot XL constant Angular Momentum of a Bowling Ball 0 The momentum of inertia of the ball is 25MR 2 o The angular momentum of the ball is LZ la 0 The direction of the angular momentum is in the positive 2 direction Conservation of Angular 555g Momentum cont 0 If the mass of an isolated system undergoes redistribution the moment of inertia changes 0 The conservation of angular momentum requires a compensating change in the angular velocity 0 I a If 0f 0 This holds for rotation about a fixed axis and for rotation about an axis through the center of mass of a moving system c The net torque must be zero in any case 6343 03955 36 CW Kinematics Chapter 2 o Describes motion while ignoring the forces that caused the motion 000 o For now will consider motion in one Motion In One Dimensmn dimension 0 Along a straight line xx x I 0 Will use the particle model 0 A particle is a pointlike object has mass but infinitesimal size Position PositionTime Graph o The positiontime graph shows the motion of the particle car 0 The smooth curve is a guess as to what happened between the data points 0 Defined interms of a t frame of reference 0 One dimensional so generally the X or y aXlS f o The object s position is its location with respect to the frame r I of reference V G M if r Displacement 0 Defined as the change in position during some time interval Represented as AX AX Xr39 X SI units are meters m Axcan be positive or neganve 0 Different than distance the length of a path followed by a particle Average Velocity o The average velocity is rate at which the displacement occurs Ax xf xl mege E At o The dimensions are length time UT 0 The SI units are ms o Is also the slope of the line in the position time grap no soon also can aw lt1 Vectors and Scalars 0 Vector quantities need both magnitude size or numerical value and direction to completely describe them Will use and signs to indicate vector directions 0 Scalar quantities are completely described by magnitude only Average Speed 0 Speed is a scalar quantity same units as velocity total distance total time o The average speed is not necessarily the magnitude of the average velocity Example 1 o A person walks first at a constant speed of 500 ms along a straight line from A to B then back along a straight line from B to A at a constant speed of 300 ms 0 Calculate the average speed for the trip it correct one H r H m variablelorlhe distance and then calculate the average speed aslhe total distancaioia time 0 Calculate the average velocity for the trip Instantaneous Velocity o The limit of the average velocity as the time interval becomes infinitesimally short or as the time interval approaches zero 0 The instantaneous velocity indicates what is happening at every point of time no soon 90 now am lt1 Example 2 o A person walks east 3 miles then north 4 miles 0 Calculate the displacement remember displacement is a vector Instantaneous Velocity A practical example You go on a long trip and you are interested to calculate your average velocity you know xand b You might also be interested in knowing your velocity at the instant you saw the police car with flashing lights behind you Instantaneous Velocity equations o The general equation for instantaneous velocity is V x 11m At dt At O o The instantaneous velocity can be positive negative or zero Instantaneous Speed o The instantaneous speed is the magnitude of the instantaneous velocity o Remember that the average speed is not the magnitude of the average velocity 2 C300 00 66 a Instantaneous Velocity graph The instantaneous velocity is the slope of 113 the line tangent to the I X vs tcune This would be the green line The blue lines show that as Atgets smaller they approach the green line Average Acceleration o Acceleration is the rate of change of the velocit y Avx fo in ax At At 0 Dimensions are LT2 o SI units are ms2 Instantaneous Acceleration graph Instantaneous Acceleration The slope of the velocity vs time graph is the acceleration The green line Av dvx dzx represents the a Ego At dt dt instantaneous acceleration The blue line is the average acceleration o The instantaneous acceleration is the limit of the average acceleration as At approaches 0 Acceleration and Velocity 1 Acceleration and Velocity 2 o When an object s velocity and acceleration are in the same direction the object is V o When an object s velocity and acceleration are in the opposite direction the object is o The car is moving with constant positive velocity shown by red arrows maintaining the same size 0 Describe the acceleration Acceleration and Velocity 3 v gt gt gt Agg gg g augv sgv a p gt gt gt gt Velocity and acceleration are in the same direction Acceleration is uniform blue arrows maintain the same length Velocity is increasing red arrows are getting longer This shows positive acceleration and positive velocity Acceleration and Velocity 4 6 V gt gt gt gt gt a 95 5 g a g quotA x gt 97 6 W t a lt lt 4 lt o Acceleration and velocity are in opposite directions 0 Acceleration is uniform blue arrows maintain the same length 0 Velocity is decreasing red arrows are getting shorter 0 Positive velocity and negative acceleration Chapter 3 Vectors Polar Coordinate System o Origin and reference line are noted 0 Point is distance rirom the origin in the direction of angle 6 ccwfrom reference line 0 Points are labeled r6 c was mmml mla Cole 9 a x y Cartesian Coordinate System 0 Also called rectangular coordinate system 0 x and y axes intersect at the origin o Points are labeled Xy Polar to Cartesian Coordinates 0 Based on forming a right triangle from rand 6 o x rcos 6 o y rsin 6 0 v 000 3C QQ 213 ltng a Q 5 5quot 4715 6 5 1 3 sin 9 COS 6 r JVj I y tan 6 Z l X quotxquot 6 L x b o am ThomsonWis Cole Cartesian to Polar Coordinates o ris the hypotenuse and 19 an angle Aquot tan 6 X x iquot rztlxz y2 5 W V Jquot 9 o must be ccw from positive x axis for these equations to be valid jarv quot 7 30 6 3938 Vectors and Scalars o A scalar quantity is completely specified by a single value with an appropriate unit and has no direction 0 A vector quantity is completely described by a number and appropriate units plus a direction Example 31 5 If I ll 39II o The Cartesian coordinates of a point in the xyplane are Xy 350 250 m as shown in the figure Find the polar coordinates of this point xi in II l iifill 391l39fl Solution From Equation 34 r x2 y2 J 350 m2 250 m2 430 m and from Equation 33 250 m 350 m tan6l 0714 X 6216 Vector Notation o When handwritten use an arrow K o When printed will be in bold print A 0 When dealing with just the magnitude of a vector in print an italic letter will be used A or A o The magnitude of the vector has physical units o The magnitude of a vector is always a positive number 3 j j 6 Vector Example 0 A particle travels from A to B along the path shown by the dotted red line 0 This is the distance traveled and is a scalar o The displacement is the solid line from A to B o The displacement is independent of the path taken between the two quot5 M points 0 Displacement is a vector Adding Vectors 0 When adding vectors their directions must be taken into account 0 Units must be the same 0 Graphical Methods 0 Use scale drawings 0 Algebraic Methods 0 More convenient Equality of Two Vectors Two vectors are equal if they have the same magnitude and the same direction A B ifA Bandthey point along parallel lines A of the vectors shown are equal Adding Vectors Graphically g J 00 CC QQ 0 Adding Vectors Graphically final When you have many vectors just keep repeating the process until all are included 0 The resultant is still drawn from the origin of the first vector to the end of the last vector CDC at Negative of a Vector o The negative of a vector is defined as the vector that when added to the original vector gives a resultant of zero Represented as A A A o o The negative of the vector will have the same magnitude but point in the opposite direction Adding Vectors Rules final 0 When adding vectors all of the vectors must have the same units 0 All of the vectors must be of the same type of quantity 0 For example you cannot add a displacement to a velocity Subtracting Vectors 0 Special case Of vector Vector Subtraction addition 0 If A B then use A XE B it 0 Continue with standard vector addition procedure it B CA B x e 200 Thomsonl mols Cole Multiplying or Dividing a Vector by a Scalar The result of the multiplication ordivision is a vector u The magnitude of the vector is multiplied or divided bythe scalar If the scalar is positive the direction of the result is the same as of the original vector u If the scalar is negative the direction of the result is opposite that of the original vector Chapter 5 The Laws of Motion Applications of Newton s Law 0 Assumptions 0 Objects can be modeled as particles 0 Masses of strings or ropes are negligible 0 When a rope attached to an object is pulling it the magnitude of that force T is the tension in the rope 0 Interested only in the external forces acting on the object 0 can neglect reaction forces 0 Initially dealing with frictionless surfaces Objects in Equilibrium o Ifthe acceleration of an object that can be modeled as a particle is zero the object is said to be in equilibrium 0 Mathematically the net force acting on the object is zero 2110 21g Oand ZFy 0 053 86 0 CU Equilibrium Example 1 o A lamp is suspended from a chain of negligible mass o The forces acting on the lamp are c the force of gravity F9 0 the tension in the chain T o Equilibrium gives ZFy0 T7Fg0 TF g Equilibrium Example 2 o Conceptualize the traffic light 0 Categorize as an equilibrium problem 0 No movement so acceleration is zero 0 Analyze Webassign Problem 0 A bag of cement of weight 425 N hangs from three wires as suggested in the figure below Two of the wires make angles 71 550 and 72 260 with the horizontal Assuming the system is in equilibrium find the tensions in the wires Newton s Second Law Example 1a o Forces acting on the crate 39 ii o A tension the magnitude of forceT o The gravitational force lt8 5454 g o The normal force n exerted by the floor 09 CO 0 8G Newton s Second Law Example 1b 0 Apply Newton s Second Law in component form Newton s Second Law Example 9 9 9 03 0 Suppose the person pulls with a force that is equal to gt 100 N and the friction force 4quot between the crate and the floor is 20 N solve for the i acceleration if the mass of the crate is 10 kg llllll l n Note About the Normal Force The normal force is not F always equal to the gravitational force of the object For example in this case ZFyningF0 andn FgF o n may also be less than Fg Inclined Planes Forces acting on the object The normal force n acts perpendicularto the plane The gravitationalforce Fg acts straight down Choose the coordinate system with X along the incline and y perpendicular to the incline Replace the force of gravity with its components What is the acceleration if the car is sliding along the icy incline Take the incline angle to be 10 Multiple Objects 0 When two or more objects are connected or in contact Newton s laws may be applied to the system as a whole andor to each individual object o Whichever you use to solve the problem the other approach can be used as a check Multiple Objects Example 1 0 First treat the system as a whole F 2 m quotM F m a 4 1 2 x system x Apply Newton s Laws to the individual blocks Solve for unknowns l 5 In Check P21 P12 1 39 g 1 Multiple Objects Example 2 o Forces acting on the objects V Tension same for both I objects one string Gravitational force Each object has the same 1 acceleration since they are connected Draw the freebody diagrams 39l It Apply Newton s Laws o Solve for the unknowns l 3921 0 Draw the freebody diagram for each object One cord so tension is the same for both objects Connected so acceleration is the same for both objects 0 Apply Newton s Laws 0 Solve for the unknowns Problem Soving Hints Newton s Laws o Conceptualize the problem draw a diagram 0 Categorize the problem 0 Equilibrium 2F 0 or Newton s Second Law 2F m a o Analyze 0 Draw freebody diagrams for each object 0 Include only forces acting on the object ProblemSolving Hints Newton s Laws cont Analyze cont 0 Establish coordinate system 0 Be sure units are consistent 0 Apply the appropriate equations in component form 0 Solve for the unknowns Finalize 0 Check your results for consistency with your free body diagram 0 Check extreme values Forces of Friction 0 When an object is in motion on a surface or through a viscous medium there will be a resistance to the motion 0 This is due to the interactions between the object and its environment 0 This resistance is called the force of friction Forces of Friction cont 0 Friction is proportional to the normal force fs usn and fk W7 0 These equations relate the magnitudes of the forces they are not vector equations 0 The force of static friction is generally greater than the force of kinetic friction o The coefficient of friction p depends on the surfaces in contact Forces of Friction final 0 The direction ofthe frictional force is opposite the direction of motion and parallel to the surfaces in contact 0 The coefficients of friction are nearly independent of the area of contact Static Friction 0 Static friction acts to keep the object from moving o lfF increases so does 1395 x o lfF decreases so does j s Kinetic Friction o The force of kinetic friction acts when the object is in motion 0 Although pk can vary with speed we shall neglect any such variations 0 fkpkn ail 5 3969 O O O I 3 0 Some Coefficients of Friction Table 52 Coef cients of Friction39 quot 1 n rlnn xcl H74 0 7 Aluminum un Merl on n 47 umm u Hrrl n 1 in RUM u 39UHL rrlt l l h mluni and Mining 01 mm lhl M nu a l Wuxvrl wnml un H l umw IL LI 2M4 wnntl l In muw um nm m Innml lubrimu I ma 04m Il39K39 Ull Il r H I UH Trllun nu 1141quot um am 5 IIm39iquuinh in human um wm mnw t hm ilu39 will II in Friction in Newton s Laws Problems 0 Friction is a force so it simply is included in the 2F in Newton s Laws 0 The rules of friction allow you to determine the direction and magnitude of the force of friction Friction Example 1 The block is sliding down the This setup can be used to Friction Example 2 plane so friction acts up the I 4 experimentally determine the Ings in 6 coefficient of friction Wm g 39 6 I 54 l 7 mg 0 Draw the freebody diagram including th 1 Motion force of kinetic frictio o Opposes the motion o Is parallel to the surfaces in contact 0 Continue with the solution as with any 1 Newton s Law problem Chapter 5 The Laws of Motion The astronaut orbiting the Earth in the Figure is E553 preparing to dock with a Westar VI satellite Th satellite is in a circular orbit 700 km above the Earth39s surface where the freefall acceleration is 821 msz Take the radius of the Earth as 6400 km Determine the speed of the satellite Determine the time interval required to complete one orbit around the Earth answer V 7630 ms time 974 min Force o Forces are what cause any change in the velocity of an object o A force is that which causes an acceleration o The net force is the vector sum of all the forces acting on an object 0 Also called total force resultant force or unbalanced force Zero Net Force 0 When the net force is equal to zero 0 The acceleration is equal to zero 0 The velocity is constant 0 Equilibrium occurs when the net force is equal to zero 0 The object if at rest will remain at rest 0 If the object is moving it will continue to move at a constant velocity DO mil Classes of Forces 3 a Contact forces involve physical contact between two objects 7 b Field forces act through empty space N N if c No physical contact is it 39 required gt2 0 000 0636 Fundamental Forces 3 o Gravitational force 0 Between two objects 0 Electromagnetic forces 0 Between two charges 0 Nuclear force 0 Between subatomic particles 0 Weak forces 0 Arise in certain radioactive decay processes Inertial Frames 0 Any reference frame that moves with constant velocity relative to an inertial frame is itself an inertial frame 0 A reference frame that moves with constant velocity relative to the distant stars is the best approximation of an inertial frame 0 We can considerthe Earth to be such an inertial frame although it has a small centripetal acceleration associated with its motion Newton s First Law Alternative Statement o In the absence of external forces when viewed from an inertial reference frame an object at rest remains at rest and an object in motion continues in motion with a constant velocity 0 Newton s First Law describes what happens in the absence ofa force 0 Also tells us that when no force acts on an object the acceleration ofthe ob39ect is zero Inertia and Mass 0 The tendency of an object to resist any attempt to change its velocity is called inertia 0 Mass is that property of an object that specifies how much resistance an object exhibits to changes in its velocity More About Mass 0 Mass is an inherent property of an object 0 Mass is a scalar quantity 0 The SI unit of mass is kg Mass vs Weight 0 Mass and weight are two different quantities 0 Weight is equal to the magnitude of the gravitational force exerted on the object 0 Weight will vary with location 0 Mass does not varry with location Weig ht Example 0 If a man weighs 820 N on Earth what would he weigh on Jupiter where the acceleration due to gravity is 259 ms2 Newton s Second Law 0 When viewed from an inertial frame the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass 0 Force is the cause of change in motion as measured by the acceleration o Algebraically 2F m a Force Examples 1 A 700 kg object undergoes an acceleration given by a 200 i 700 j msZ Find the resultant force acting on it and the magnitude of the resultant force answer 14i 49j N F51 N 2 An electron of mass 911x10 31 kg has an initial speed of 400x105 ms It travels in a straight line and its speed increases to 600x105 ms in a distance of 400 cm Assuming its acceleration is constant determine the force exerted on the electron answer 228gtlt10 18 N in the direction of motion More About Newton s Second Law 0 EF is the net force 0 This is the vector sum of all the forces acting on the object o Newton s Second Law can be expressed in terms of components 0 ZFX m ax o ZFy m ay 0 ZFZ m az Gravitational Force 0 The gravitational force Fg is the force that the earth exerts on an object o This force is directed toward the center of the ea h 0 Its magnitude is called the weight of the object 0 Weight Fg mg More About Weight 0 Because it is dependent on g the weight varies with location 0 g and therefore the weight is less at higher altitudes 0 Weight is not an inherent property of the object Gravitational Mass vs Inertial Mass In Newton s Laws the mass is the inertial mass and measures the resistance to a change in the object s motion In the gravitational force the mass is determining the gravitational attraction between the object and the Earth Experiments show that gravitational mass and inertial mass have the same value Newton s Third Law o Iftwo objects interact the force F12 exerted by object 1 on object 2 is equal in magnitude and opposite in direction to the force F21 exerted by object 2 on object 1 39 F12 39 F21 0 Note on notation FAB is the force exerted by A on Newton s Third Law Alternative Statements o Forces always occur in pairs 0 A single isolated force cannot exist 0 The action force is equal in magnitude to the reaction force and opposite in direction 0 One of the forces is the action force the other is the reaction force o It doesn t matter which is considered the action and which the reaction 0 The action and reaction forces must act on different objects and be of the same type Q 8 05 CO ActionReaction Example 0 The force F12 exerted by object 1 on object 2 is equal in magnitude and opposite in direction to F21 exerted by object 2 on object 1 F12 39 F21 Free Body Diagram 3 o In a free body diagram you want the forces acting on a particular object o The normal force and the force of gravity are the forces that act on the monitor g Applications of Newton s Law 0 Assumptions 0 Objects can be modeled as particles 0 Masses of strings or ropes are negligible 0 When a rope attached to an object is pulling it the magnitude of that force T is the tension in the rope 0 Interested only in the external forces acting on the object 0 can neglect reaction forces 0 Initially dealing with frictionless surfaces Objects in Equilibrium o If the acceleration of an object that can be modeled as a particle is zero the object is said to be in equilibrium 0 Mathematically the net force acting on the object is zero ZF0 ZFX Oand ZFy 0

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