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by: Meredith Wolf

FundamentalsofPhysicsI PHYS207

Marketplace > University of Delaware > Astronomy > PHYS207 > FundamentalsofPhysicsI
Meredith Wolf
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This 36 page Class Notes was uploaded by Meredith Wolf on Saturday September 19, 2015. The Class Notes belongs to PHYS207 at University of Delaware taught by Staff in Fall. Since its upload, it has received 50 views. For similar materials see /class/207152/phys207-university-of-delaware in Astronomy at University of Delaware.


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Date Created: 09/19/15
Physics 207 Lecture 3 Reminders 0 Discussion and Lab sections start TODAY Homework Submission deadline Today BEFORE lecture begins WRITE YOUR DSC ON YOUR HOMEWORK Agenda for Today 0 1D Kinematics review cont d o Vectors review Recap o If the position X is known as a function oftime then we can find both velocity v and acceleration a as a function of time 7 t V dt V dv dzx a dt dtz Page 1 1D Motion with constant acceleration const o Highschool calculus It dt n1 1 1 a dv 0 Also recall that 7 dt 0 Since a is constant we can integrate this using the above rule to find l vjadtajdtatv0 l 0 Similarly since v 2 we can integrate againto get Xvdtjatv0dtat2v0txo Recap So for constant acceleration we find x x x0vot1312 2 V t v vo at aconst a t Page 2 Lecture 3 Act 1 Motion in One Dimension 0 When throwing a ball straight up which ofthe following is true about its velocity v and its acceleration a at the highest point in its path a Bothv0anda0 bv 0buta0 4y cv0 buta 0 Lecture 3 Act 1 Solution 0 Going up the ball has positive velocity while coming down it has negative velocity At the top the velocity is momentarily zero x 0 Since the velocity is continually changing there must t be some acceleration V In fact the acceleration is caused t by gravity 9 981 msz more on gravity in a few lectures o The answer is c v 0 but a it 0 Page 3 Useful Formula 7 2 xx vt fat 002 v v0 at 0 Solving fort o Plugging in fort Viv v v0 1 vivo 2 t 0 x x0 v0 Ea a Alternate Calculusbased Derivation W W i chain rule a dt dX dt av 3 adXVdV dX JadXaJdXJvdv aconstant Xu Xu 3 axx0 V2 V3 II Page 4 Recap o For constant acceleration gtltgtlt0votat2 vv0at a const c From which we know 2 2 v avg Zaxixo 7 Vav 500 V Example 1 the o A car is traveling with an initial velocity v0 At t 0 rate of driver puts on the brakes which slows the car at a ab 39 v0 ab X0t0 Page 5 Example 1 o A car is traveling with an initial velocity v0 At t 0 the driver puts on the brakes which slows the car at a rate of ab At what time t does the car stop and how much farther X does it travel Va ab4 X0t0 Example 1 0 Earlier we derived v v0 at o Realize that a aL1 0 Also realizing that v 0 at t tr find 0 va abt or a Page 6 Example 1 c To find stopping distance we use v2 evoz 28X 7 X0 0 Inthisoase vvf0 Xa0 andxxf a v02 2mm x 7 V02 f 2a7 Example 1 2 v v 0 So we found that tf 70 Xf 2170 ab 2 ab 0 Suppose that v0 65 mihr 29 ms 0 Suppose also that ab g 981 ms2 Findthatt 33 and X 43m Page 7 Tips 0 Read Before you start work on a problem read the problem statement thoroughly Make sure you understand what information is given what is asked for and the meaning of all the terms used in stating the problem 0 Watch your units Always check the units of your answer and carry the units along with your numbers during the calculation 0 Understand the limits Many equations we use are special cases of more general laws Understanding how they are derived will help you recognize their limitations for example constant acceleration 1D FreeFall o This is a nice example of constant acceleration gravity 0 In this case acceleration is caused by the force of gravity Usually pick yaxis upward Acceleration of gravity is down Page 8 Gravity facts 0 9 does not depend on the nature ofthe material Galileo 15641642 figured this out without fancy clocks amp rulers 0 demo feather amp penny in vacuum 0 Nominally g 981 ms2 At the equator g 978 ms2 At the North pole g 983 ms2 0 More on gravity in a few lectures Problem 0 The pilot of a hovering helicopter drops a lead brick from a height of 1000 m How long does it take to reach the ground and how fast is it moving when it gets there neglect air resistance 100E Page 9 Problem 0 First choose coordinate system Origin and y direotion 0 Next write down position equation 1 2 y7y0v0yt Egt 1000177 0 Realize that vay 0 it 1 2 fit y V0 29 gt lt 1 y yo 3th Problem 0 Solve for time twhen y 0 given that yo 1000 m 2 t y0 2X1000 143s g 981ms 0 Recall 2 2 vy v0y 2ay y0 Solve for vy vy iJZgyo 7140rns Page 10 Lecture 3 Act 2 1D free fall Alice and Bill are standing at the top of a cliff of height H Both throw a ball with initial speed v0 Alice straight down and Bill straight up The speed of the balls when they hit the ground are VA and vB respectively Which of the following is rue a VA lt VB 3 VA Vs c VA gt VB Alice vquot Bill Vu Lecture 3 Act 2 1D Free fall 0 Since the motion up and back down is symmetric intuition should tell you that v v We can prove that your intuition is correct Equation v2 7 v5 2egH e H 0 This looks just like Bill threw 339quot the ball down with speed Vquot so vquot V vquot the speed at the bottom should quotquotquotquotquotquotquotquotquotquotquotquotquotquotquotquotquotquotquot quot be the same as Alice s ball H yo Page 11 Lecture 3 Act 2 1D Free fall 0 We can also just use the equation directly Alice v2 avg 2eg0 7H Bill v2 7v 2eg07H same Vu Vectors review 0 mi dimension we could specify direction with a or sign For example in the previous problem ay g etc o In 2 or 3 dimensions we need more than a sign to specify the direction of something 0 To illustrate this consider the position vectorr in 2 dimensions Example Where is Philadelphia Choose origin at Newark Choose coordinates of distance miles and direction NSEW In this case r is a vector that points 40 miles northeast Philadelphia Page 12 Vectors c There are two common ways of indicating that something is a vector quantity A A Boldface notation A Arrow notation A Vectors o The components of rare its Xyz coordinates r rxryrz Xyz o Considerthis in 2D since it s easier to draw rX X r cos 9 whererr ry y rsm e y Ky earctanyX e x Page 13 Vectors o The magnitude length of ris found using the Pythagorean theorem y rr1lx2y2 x o The length of a vector clearly does not depend on its direction Unit Vectors o A Unit Vector is a vector having length 1 and no units It is used to specify a direction U Unit vector u points in the direction of U Often denoted with a hat u I 0 Useful examples are the Cartesian unit vectors i j k point in the direction ofthe X y and z axes Page 14 Vector addition 0 Consider the vectors A and E Find A B c We can arrange the vectors as we want as long as we maintain their length and direction Vector addition using components 0 Consider C A B a c AXH M Bxi Byj AX B Ay 8 b c Cxi Cy 0 Comparing components of a and b CXAXBX cyAyBy Page 15 Lecture 3 Act 3 Vectors 0 Vector A 021 0 Vector B 302 0 Vector C 142 What is the resultant vector D from adding ABC a 354 b 425 0 524 Lecture 2 Act 3 Solution D Axi ij Azk Bxi BY 8210 cxi w czk AX BX Cxi AY BY CYj AZ BZ CZk 031i204j122k 425 Page 16 Recap of Lecture 3 0 Recap of 1D motion with constant acceleration Text 26 0 1D FreeFall Text 26 example 0 Review of Vectors Text 31 32 amp 34 0 For Wed look at textbook sections Chapter 41 thru 44 Page 17 Physics 207 Lecture 28 Announcements Final hwk assigned this week final quiz next week Review session on Thursday May 19 230 400pm Here Today s Agenda 0 Recap Angular Momentum o Rotation about a fixed axis L 103 Example Two disks Student on rotating stool o Angular momentum of a freely moving particle Bullet hitting stick Student throwing ball 0 Comment about 10 not true if I is changing 0 Vector considerations of angular momentum Bike Aneel and rotating stool Angular momentum L dL 0 m where and TEXT dL 0 In the absence of external torques TEXT E 0 Total angular momentum is conserved Page 1 Angular momentum of a rigid body about a fixed axis 0 Consider a rigid distribution of point particles rotating in the xy plane around the z axis as shown below The total angular momentum around the origin is the sum of the angular momenta of each particle Lzrxp zmrxv ZmrvR since rand v are 39 39 perpendicular V1 We see that L is in the z direction Using v w r we get Lzmr2w R Analogue ofp mvl Angular momentum of a rigid body about a fixed axis 0 In general for an object rotating about a fixed 2 axis we can write L2 10 o The direction of L2 is given by the right hand rule same as w c We will omit the Zsubscript for simplicity and write L 10 Page 2 Example Two Disks 0 A disk of mass M and radius R rotates around the z axis with angularvelocity 03 A second identical disk initially not rotating is dropped on top ofthe first There is friction between the disks and eventually they rotate together with angular velocity 03 Z 032 Example Two Disks 0 First realize that there are no external torques acting on the twodisk system Angular momentum will be conserved 0 Initially the total angular momentum is due only to the disk on the bottom iMsz Lil1 1 2 l Page 3 Example Two Disks 0 First realize that there are no external torques acting on the twodisk system Angular momentum will be conserved 0 Finally the total angular momentum is due to both disks spinning 2 Lf1w12w2MR of 1 Example Two Disks 1 Since L L gt EMRza MRZm An inelastic collision since E is not 1 gt conserved friction Z Z Li lwr Jr Lr Page 4 Example Rotating Table 0 A student sits on a rotating stool with his arms extended and a weight in each hand The total moment of inertia is 1 and he is rotating with angular speed 03 He then pulls his hands in toward his body so that the moment of inertia reduces to If What is his final angular speed oar Wi 9 Example Rotating Table 0 Again there are no external torques acting on the student stool system so angular momentum will be conserved Initially L 1le Finay L Ifwf Li 10 Page 5 Lecture 28 Act 1 Angular Momentum o A student sits on a freely turning stool and rotates with constant angular velocity 1 She pulls her arms in and due to angular momentum conservation her angular velocity increases to 0J2 In doing this her kinetic energy a increases b decreases c stays the same ltgt 0 11 L 4 E 11 Lecture 28 Act 1 Solution 2 0 Klw2 usingLagt o L is conserved 2 lt 1 Igt K2 gt K K increases ltgt 0 11 L 4 E 12 Page 6 Lecture 28 Act 1 Solution 0 Since the student has to force her arms to move toward her body she must be doing positive work 0 The workkinetic energy theorem states that this will increase the kinetic energy of the system ltgtw1 Angular Momentum of a Freely Moving Particle c We have defined the angular momentum of a particle about the origin as L r x p o This does not demand that the particle is moving in a circle We will show that this particle has a constant angular momentum Page 7 Angular Momentum of a Freely Moving Particle 0 Consider a particle of mass rn moving with speed v along the line y d What is its angular momentum as measured from the origin 00 Angular Momentum of a Freely Moving Particle c We need to figure out L r X P o The magnitude ofthe angular momentum is L rx p rpsine prsin e pd p x distance of closest approach 0 Since rand p are both in the Xy plane L will be in the z direction right hand rule Lz pd y Page 8 Angular Momentum of a Freely Moving Particle 0 So we see that the direction of L is along the z axis and its magnitude is given by L2 pd mvd o L is clearly conserved since d is constant the distance of closest approach of the particle to the origin and p is constant momentum conservation Example Bullet hitting stick 0 A uniform stick of mass Mand length D is pivoted at the center A bullet of mass rn is shot through the stick at a point halfway between the pivot and the end The initial speed of the bullet is v1 and the final speed is v2 What is the angular speed 03 of the stick after the collision Ignore gravity M rn D I D4 D initial final Page 9 Example Bullet hitting stick 0 Conserve angular momentum around the pivot z axis 0 The total angular momentum before the collision is due only to the bullet since the stick is not rotating yet LI p x distance of closest approach my M rn D I D4 D V1 initial 19 Example Bullet hitting stick 0 Conserve angular momentum around the pivot z axis 0 The total angular momentum after the collision has contributions from both the bullet and the stick L 7 mv 2 Ia where is the moment of inertia f 7 2 4 F of the stick about the pivot OJF 041 D V2 final 20 Page 10 Example Bullet hitting stick 1j MD2 0 Set L L using 12 mv1mv2MD2mF Egt initial final Example Throwing ball from stool o A student sits on a stool which is free to rotate The moment of inertia ofthe student plus the stool is 1 She throws a heavy ball of mass Mwith speed v such that its velocity vector passes a distance dfrom the axis of rotation What is the angular speed 03 of the studentstool system after she throws the ball M top view initial final Page 11 Example Throwing ball from stool o Conserve angular momentum since there are no external torques acting on the studentstool system L 0 Mvd L0IanMvd Igt 03 I M top view initial final 23 Review 0 A freely moving particle has a definite angular momentum about any axis o If no torques are acting on the particle its angular momentum will be conserved o In the example below the direction of L is along the z axis and its magnitude is given by L2 pd mvd y 24 Page 12 Lecture 28 Act 2 Angular momentum 0 Two different spinning disks have the same angular momentum but disk 1 has more kinetic energy than disk 2 Whioh one has the biggest moment of inertia a disk l b disk2 c hotehough info 25 Lecture 28 Act 2 Solution 1 2 7 2 2 1 2 K Ioo I L 39 2 2 OJ 2 usith lw Ifthey have the same L the one with the biggest will have the smallest kinetic energy SLINW f L12m2 I lt I2 disk 1 disk 2 26 Page 13 When does I 10c not work dL 0 Last time we showed that TEXT E o This is the fundamental equation for understanding rotation o If we write L 1m then Ioad I dt da d1 17 mi dt dt d rEXTIo 60 We can t assume I Ioc when the moment of inertia is changing gid m dt dt 27 When does I 10c not work Now suppose TEXT 0 I or cod I 0 or a 7 gd I dt I dt So in this case we can have an ocwithout an external torque 28 Page 14 Example 0 A puck in uniform circular motion will experience rotational acceleration if its moment of inertia is changed 0 Changing the radius changes the moment of inertia but roduces no torque since the force of the string is along the radial direction since r X F 0 gt 12 OJZgtOJ1 l The puck accelerates without external torquell Lecture 28 Act 3 Rotations o A puck slides in a circular path on a horizontal frictionless table It is held at a constant radius by a string threaded through a frictionless hole at the center ofthe table If you pull on the string such that the radius decreases by a factor of 2 by what factor does the angular velocity of the puck increase Page 15 Lecture 28 Act 3 Solution 0 Since the string is pulled through a hole at the center of rotation there is no torque Angular momentum is conserved 2 L1 11m mR2m1 L2 12m2 r17an mR2m1 177sz Review Angular Momentum and TEXT EXT 0 where dL 0 In the absence of external torques TEXT E Total angular momentum is conserved o This is a vector equation 0 Valid for individual components Page 16 Review 0 In general for an object rotating about a fixed 2 axis we can write L2 loJ o The direction of L2 is given by the right hand rule same as w Angular momentum is a vector Demo Turning the bike wheel 0 A student sits on the rotatable stool holding a bicycle wheel that is spinning in the horizontal plane She flips the rotation axis of the wheel 180 and finds that she herself starts to rotate What s going on Page 17 Turning the bike wheel 0 Since there are no external torques acting on the student stool system angular momentum is conserved lnitially L Lwy Finally1LFW LWF LS Lecture 28 Act 4 Rotations o A student is initially at rest on a rotatable chair holding a wheel spinning as shown in 1 He turns it over and starts to rotate 2 If he keeps twisting turning the wheel over again 3 his rotation will a stop b double 0 staythe same lt73 igtltgt 9 3 36 Page 18 Lecture 28 Act 4 Solution 1 LIVET t LIVET 1 LIVET LW 1 LW 1 393 LW 1 turrlgitng Q Page 19


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