### Create a StudySoup account

#### Be part of our community, it's free to join!

Already have a StudySoup account? Login here

# PHYSICS I HONORS PHYS 1150

RPI

GPA 3.99

### View Full Document

## 5

## 0

## Popular in Course

## Popular in Physics 2

This 130 page Class Notes was uploaded by Diamond Kirlin MD on Monday October 19, 2015. The Class Notes belongs to PHYS 1150 at Rensselaer Polytechnic Institute taught by Staff in Fall. Since its upload, it has received 5 views. For similar materials see /class/224887/phys-1150-rensselaer-polytechnic-institute in Physics 2 at Rensselaer Polytechnic Institute.

## Similar to PHYS 1150 at RPI

## Reviews for PHYSICS I HONORS

### What is Karma?

#### Karma is the currency of StudySoup.

#### You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

Date Created: 10/19/15

Physics Honors Monday 25 August Fall 2008 First class Name and course on board read roll introduce TA7s Distribute outline lecture and lab schedules Discuss them New material for today Start thinking like a physicist Two new ideas namely 1 Solving problems with dimensional analysis and 2 F ma is a differential equation Today7s Topic 1 Dimensional Analysis Given a problem ask What is the important physics7 Then list the quantities that have something to do with that physics In many cases you can combine these things to get the answer at least to within a factor of 2 or 7139 or something like that Convenient notation distance L time T mass M temperature K Err ample What is the radius of a black hole of mass M This is a combination of gravity and relativity so use M G and c Newton7s constant G has units m3kg sec2 so G LSM lT Z Easy to see this from F GmMr2 ma Let the radius be R MmGycz where we need to nd x y and z This gives R MmLSyM yT ZyLzT Z MHEWT ZW L1 1 so z y 3y z 1 and 2y z 0 The second and third of these say that y 1 and so z 1 and z 72 Therefore we get R GMcz 2 The correct answer is GMCZ called the Schwarzchild Radius Hint Useful website for units and constants is httppdgbgov2008reviewsconsrpppdf Practice Exercise You7ve heard of String Theory7 The length of strings is supposedly the Planck Length77 since gravity is uni ed with quantum mechanics and relativity at this distance Gravity is governed by Newton7s gravitational constant G quantum mechanics by Planck7s constant h and relativity by the speed of light 0 Combine these to nd the Planck Length Note that the dimensions of h are distancegtltmomentum7 or LZMT Look up values of G h and c and compare the Planck Length to the size of a proton 10 15m Physics Honors Monday 25 August Fall 2008 Today7s Topic 2 Newton7s Second Law as an Equation of Motion The most important of Newton7s Laws is the second For motion in one dimension7 we write F ma We7ll get to F m6 later this week Physicists refer to this as the equation of motion for a particle moving through time in one dimension of space Let t be the position x of a particle at time t Then the velocity is 11t dzdt E i Have you all seen this notation before Acceleration is at dvdt 1 So Newton7s Second law can be written as z Fzt 3 where we make it clear that the force F may itself depend on position andor time Equation 3 is called a differential equation Unlike an algebraic equation7 whose solution is a value for a quantity like7 say7 x the solution of a differential equation is a function say Since t describes the motion of a particle with mass m acted upon by a force Fx1t7 we call Eq 3 the equation of motion Simple example Let Fzt F07 a constant independent of both z and t Then d1 F0 57 7 4 z 1 dt m dx F mpg lm 5 1F m E t2v0t0 6 where the last line is the solution to this equation of motion This might look more familiar to you from your high school physics course7 if we write a FOm You probably called this motion under constant acceleration 7 but of course a constant acceleration implies a constant force7 so its the same thing Each step in the integration above involves a constant of integration 1 called these con stants 00 for the rst step7 and 0 for the second step These are standard notations7 and in fact good ones The constant 110 is the velocity at time t 07 also called the initial velocity Similarly7 0 is the initial position In order to solve any equation of motion7 we will need to specify these initial conditions Practice Exercise Let yt describe the one dimensional motion of a particle with mass m that moves vertically Assume that the particle is acted on by a constant force F 7mg Isn t that peculiar a force that is epaetly proportional to the quantity in that appears in the equation of motion Assume that the particle starts out at y 0 with an initial upward velocity V Find the maximum height to which the particle climbs7 in terms of m7 9 and V Make a sketch of yt as a function of t What is the shape of the curve Mathematicians have a name for it7 and l7m sure you7ve heard it before n by simple a At In the s the sketch v is parallel everywhere rajectory ok at limiting entiate since h can change Physics Honors Thursday 26 August Fall 2008 This class will concentrate mainly on mathematics First we will review vectors and and apply them to three dimensional motion Then we ll talk about Taylor series how they can be used in approximations and then apply them to complex numbers Today s Topic 1 Vectors and Motion in Three Dimensions Introduction The vector equation F ma is a shorthand for three equations namely F33 mag and Fy may and FZ maz Notation A in book A on board Unit vectors are n in book a on board The length of a vector is A A so 1 If you add two vectors the result is still a vector C A B C 5A This means that A AA A number times a vector is still a vector The dot product 7 between two vectors is a scalar A B AB cos 9 where 9 is the angle between the two vectors This is the projection of A onto B or vice versa The cross product 7 between two vectors is a vector A X B C Watch out Some day you ll learn about polar and arial vectors The magnitude of C is AB sin 6 The direction of C is given by convention by the right hand rule This means that A x B B x A Vectors can be written in terms of coordinates in any given coordinateAsystein or reference frame For an ay z coordinate system de ne baseAvectors i j and k all orthogonal so their mutual dot products are zero and with i X j k etc Then A A B A x B A M Azk maimama Aye AZByi A213 Amen one AB12 AA OOI Ehree Dimensions A vector rt ztk gives ree dimensions The velocity vector is Vt rt and the Rhee tiectorxequation of motion is Fa tm 2 Examples Work W Ed is a gpgitigr piggct gorgue F is 0039 9mg x and e ai 0 39wh chL FRE vector along the r axis and v v aslemi the speed increases without 3 Uniform ciroular motion The motion rt traces out a circl at a constant rate second traced out eview radians cross product 11 R w27r revolutions per second df rm sin wit 7 cos wtj 3 We can show tangent to ti fleet gbmculqtinjAVSiTl Vf7 7w Sln t oswicoswtsi mt A A We couldtoritilzueto form new vectors by highgderi a i in tives of r but we shall see in Our etudeptsdxoeel tllng Wily Xi Wis aznggni to thSe circleujais Tellx ilaegps are of chief interest incidentally it is easy to Show thatNA r L 90 nt i y ote tie r i 0 A E A A V Example Ls Uniform Circular Mntjon a Iii aggwt k atmjw27nlt1 COS wt J SI I w2rt 10a 10b 1 392 s Circularmotion lay Irl trnportant role ilt dhyERl l I sim lejlt and m s impontarit caserr nffwri r circular motion which p w i Note 7 7 u r 10d 39 The acceleratiocft i directed eggamiinward and is known as the centripetal rcos mi Contlder a pa icle movujig in the w lane accor quot1 r V 17quot l gastritis resettlementh SW 39 shortw T 77771 sin 0421 where r and w are constants 1 V r I on w and the ms fj le The force that is respons1ble for to 1 Ni i l X M I TE 5052 wt 1 TE Sin a A Wort about Dimension and Units i 39 A notmcabacisdtantlteigsat r nththsnairc Using the familiar identity 53quot 9 Boshgasul39ed the dimension of the quantity For example the dimension of velocity is distancetime and the dimension of acceleration is velocitytime or distaneetimeltime 2 distance time As we shallaiscuss in Chap 2 mass distance and time are the fundamental physical units used in mechanics The trajectory is a circle To introduce a system of units we specify the standards of The particle moves counterclockv msmenienliefc l Fi issfa lgt nfcmlnand time Ordinarily we mea 1 no at 3 0 If traverses the Clr l rdi3t ilc r913 ime lh39seconds The units of velocity 39139 i w is called the angular velocity of themma t r lh us d g and the units of acceleration are meters per second2 ms39l is acc Vtl Matuit AvirD Al lrl 394 39r2cosZ at sinz ionl r constant SEC 17 FORMAL SOLUTION C radians per second Thi ratio of two lengths To avoid gross errors it sides of an equation hav example the equation F exponentials and their arg the units Li s and the righ per second 17 Formal Solution of Kii Dynamics which we shall to find the acceleration c acceleration finding the Vi integration Here is the i If the acceleration is kr can be found from the de rlvi ll T a by integration with respect given the initial velocity Vi the time interval 1 to in vat z vt Avau Al s vi an At since Ava m at At Tal iit Zita min Al The approximation becon and the sum becomes an no mi mo d Practice Exercise Find the period P of the conical pendulum 7 pictured below in terms of the hanging angle 6 the string length L and the acceleration 9 due to gravity The mass m executes uniform circular orbits with radius R If the speed is 11 then the period is just the time it takes to execute one orbit namely P 27rRv So you need to gure out 11 in terms of R or perhaps in terms of the string length L and the hanging angle 6 since R L sin 6 Since we know that F ma we just need to gure out what the force F is Clearly this force is provided by the tension T but only by part of it Physics Honors Thursday 26 August Fall 2008 Topic 2 Taylor Series Approximation Techniques and Complex Numbers It makes sense that we can approximate any function by some high order polynomial The function is expanded about the point N 0 1 Shown are the zeroth order ie 8 y a constant approximation the rst order second order and the full function The zeroth order is just the value of the func quotZeroth order tion at 0 1 and the rst order is the 1 First order 2 Second order tangent line at that point The second or Fufunction der approximation comes close to the full 00 1 2 3 function in the neighborhood of z x0 fx a0 a1z 7 x0 a2z 7 x02 a3z 7 03 11 f0 10 fio 11 f 0 2 39 12 fm0 3 39 2 39 13 1 fW JCWO f0 0 gfK ox 0 13 1 2 Examples 1 x 1 ow yam 71z 14 z 1 392 1 3 1 4 1 5 e 1xic Ex z 15 1 1 1 s1nx z 7 if 5x5 16 1 1 cosx 17 if 1x4 17 Obviously if ltlt 1 ie z is small then only a few terms maybe just one is needed in order to get an accurate approximation Complex numbers There is a number 239 such that 2392 71 Numbers formed by a real number times 239 like 2239 7239 3 or m where z is a real number are called imaginary numbers Dont let the words fool you though All ofthese are perfectly valid numbers The imaginary numbers are a group outside integers rational and irrational real numbers Numbers like 3 2239 and z z z y are called complex numbers They have a real part and an imaginary part We write for z z z y where z and y are real z Rez and y The modulus of z is 2 y2 Now here7s a neat thing m 1 1 1 1 e 1zzim23Xm34zz4 m5 12 14 r 13 15 17 z Jr x zx7 Jrgz cosx z sin This is called Euler7s Formula It is very useful in physics and engineering Practice Exercise First7 a math rule77 The derivative of an exponential is an exponential That is f e gt fs 16 Use this to nd the solution to the equation of motion for a particle of mass m subject to a force ikz where k is a positive constant Write down the equation of motion as Then show that t 06th is a solution to the equation of motion7 for any value of C so long as to has one of two possible values What are those values Next7 write the general solution77 zt as the sum of two functions 06 one for each of the two possible values of w and with two different values of 0 Call them 01 and 02 ls it obvious to you that this sum of solutions is also a solution to the equation of motion Finally7 determine the two constants 01 and Cg from the initial conditions 0 A and 0 Simplify the result as much as possible Euler7s formula will be handy What kind of motion does this describe Physics Honors Thursday 4 September Fall 2008 Let s try to stick with the 20 20 20 7 plan that is new material for 20 minutes 20 minutes to work on a practice exercise and then 20 minutes to talk about the solution Today s Topic 1 Applications of Newton s Laws Discuss Free Body Diagrams77 Statics rst Use example of one block A sitting atop another block B page 68 in KampK and take care to note the weight of each block the normal force on B from the lower surface and the force F1 F2 which block B A exerts on block A B Hence F1 MAg and N F2 MBg Third law says F1 F2 so N MA MBg which is just what you expect of course Pretty much the same thing for two masses hanging connected by a wire under tension The top wire hangs both masses Tell the story of the Hyatt Regency disaster July 17 1981 in Kansas City Missouri httpenwikipediaorgwikiHyattRegencywakwaycoapse Dynamics is very similar just don t set a 0 See practice problem and also homework Let s do the conical pendulum See Example 28 in your textbook Find the period P in terms of the angle 6 the string length L and the acceleration 9 due to gravity The speed of m is v Btu so P 27TR U To find 11 use F ma with a w2B v2B Recall Equa tions 10 Find F Tsin6 with Tcos 9 mg Putting it all together we have R2 R 2 7 2 7 2 P i 27 U2 727139 a Rm RmcosQ 2 2 2 2 7T Tsin6 7T mgsinQ L P 2W BcosQ27T cos gst g a Practice Exercise Find the acceleration a of the coupled masses on the left in 1111 terms of m1 m2 and g The mass m1 slides on a horizontal frictionless surface m2 hangs freely and the string that connects them is massless The pulley is also massless and gives no frictional resistance to motion Start by drawing free body diagrams of m1 and 7712 Identify all forces on each of the two masses and then work with the components that are in the direction of the acceleration Topic 2 The Everyday Forces of Physics Start by making a list Use vector notation o Tension the pull of a massless string T 0 Weight ie gravity near the Earth7s surface W 7771912 0 Springs ie Hooke7s Law F ikx o Frictional forces 7 Kinetic friction fk nkN opposing the direction of motion 7 Static friction f5 3 MSN as big as it needs to be to keep the object from moving Note that Ms gt Mk for the same two surfaces Demo with book on desk 0 Viscous forces FE 70v 0 lnverse square central forces Draw some pictures here Newtonian gravity Fan 7GMaMbr2ranb Electrostatic attraction Fan kQaQbr2ranb Note General central force is just Fan foams Let7s study the motion of an object under the in uence of a drag force See Example 216 Assume a mass m is moving through a viscous medium The equation of motion is do do 0 70o ma m7 or dt a W 18 Now integrate The math trick you need is dln dz 1x We have 1 t 72 dt or ln igt or 11t ooe Ctm 19 m 0 v0 i no in Note use of initial condition notation The behavior is clear The particle moves with velocity 110 at t 0 but then slows down exponentially Note Study the shell theorem in Note 21 Basically a spherical object attracts an external mass gravitationally where the equivalent object mass is all that is contained within a sphere or radius intersecting the external mass Practice Exercise Express the gravitational acceleration near the Earth7s surface 9 in terms of G and the mass ME and radius RE of the Earth Start by putting an external mass m on the surface of the Earth Then calculate the gravitational attraction of the Earth on this mass using the formula for Newtonian gravity Cavendish devised an sensitive experiment that measured G in 1798 People say he weighed the Earth What do they mean Note that Eratosthenes measured the Earth7s radius somewhere around 200 BC Physics Honors Monday 8 September Fall 2008 Today s Topic 1 Momentum and Systems of Many Particles Newton actually wrote the Second Law as 2 F dpdt where momentum p mv If mass m doesn t change with time then 2 F mdvdt ma7 so we recover the old version Now imagine an object colliding with a brick wall The collision will change the momentum of the object The force of the collision varies over the collision time Draw a figure See Example 39 in KampK We can use Newton s second law to write an expression involving the integral of the force and the change in momentum t2 Ap Ftdt E J t1 We call J the impulse The average force Favg is J divided by At t2 t1 Now imagine two objects colliding with each other Each exerts an equal and opposite momentum on each other That is J1 J27 or Apl Apg Ap1 p2 0 In other words7 the total momentum P E p1 p2 is conserved in collisions Note that this doesn t work if there is some external force acting on the masses Now look at two particles again7 but not just collisions Allow some external force Fext to act as well as internal forces Then dP2 2 i Fex dt dt dt miaim2a2 Z t The last equality follows because the internal forces cancel in the sum It is convenient to write P Mvcm where M m1 m2 and vcm drcmdt with m1r1m2r2 l mrmr 20 WW2 Mlt 11 22 H rem in which case Newton s Second Law reads 2 Fext M rem M am We call rem the center of mass ls center of momentum better For more than two particles mlrl m21 2 le N 1 N m1m2mmN M rem 711 Don t forget that this is really two or three equations for worm ycm7 and if three 20m The figures below show how this all works out Moon 4 P A 7 Qev xk 7 739775 La k ix 7 xv I s 39aquotquotr quot J 3 Aka 439 3W 12 7 C7quot gt 3937 1x 39 Earth39s Earth w39quot U Q path Path of e center of mass 1f39 i39 Am 39 H gt A Center Sun of mass 11 Practice Exercise An elastic collision is one in which kinetic energy is conserved The kinetic energy of an object with mass m and speed 1 is K mvz The total kinetic energy is the sum of the individual objects kinetic energies We will learn more about this later Let two objects one with mass m and the other with mass 2m collide elastically in one dimension They are initially heading towards each other with speed 1 Find their speeds and directions after the collision in terms of m and 1 Topic 2 Calculating the Center of Mass of Solid Objects If we are working on a solid object we could sum over all the atoms that make it up but that would be nuts Instead we use calculus and assume that the object is continuous Then the sum becomes an integral That is 1 Mdm and remMdmr We usually write the in nitesimal mass dm as a density usually called p or 0 times a volume element dV or some other measure ie dm prdV This allows the density to vary as a function of position inside the solid object The rest is in the details Here we can slice up an eggplant making the point that f dV is the sum of small pieces Here is a simple example What is the volume of a right circular cone of height h and base radius B Let 2 measure the height and slice the cone up into thin disks each of thickness dz The radius of each disk is Rh 7 zh R1 7 zh so the volume of each disk is 7TB21 7 zh2dz Therefore the volume of the cone is h h 2 22 2 h 7 2 2 7 2 7 2 dViO WR1Zhd277TRO lt17hh277TR or one third of the base area times the height Of course this is volume not mass but that7s basically the same thing for constant density If the density is not a constant you7d need to include the functional form of pr and that would be a harder problem Practice Exercise Find the position of the center of mass of a long thin rod of mass M and length L The density of the material making up the rod changes linearly from zero at one end Start by realizing that the mass of a short segment of length dx is dm Ud where 0a is the linear mass density Write a formula for 027 and determine any parameters by xing the mass of the rod to be M Then nd the position of the center of mass Physics Honors Thursday 11 September Fall 2008 Today s Topic 1 Work and Kinetic Energy in One Dimension Let s use some calculus to manipulate F mdvdt in one dimension b b b b d d d 1 1 a Fd ma 07de ma d d dt ma vdv 5mm 5771113 22 We write F but it could be a constant for example F mg yb 1 1 mg dy mg h 5771112 577102 gt v x2gh 23 ya for an object falling through a distance h ya yb starting from rest We already knew this from solving the equation of motion Eq 6 but here we ve done this a different way A harder but still easy example is a spring force F k Equation 22 becomes 1 1 1 1 k 57m 57m 24 You will play around with this on the practice exercise These are both examples of something called the WorkEnergy Theorem De ne the work in one dimension done by a force F on a mass m through a path 33a gt an as Wba De ne the kinetic energy of that object as K Then Eq 22 says that 1 2 5m for veloc1ty 1 Wm Kb K E AK 25 That is the work done on the object equals the change in kinetic energy Of course the work done for a constant force is just the product of the force times the displacenegg l mpe 43 Let s use this to calculate escape velocity77 A rocket starts at ra Re with escape velocity 11 It ends up at very large distance rb gt 00 with v 0 If the initial velocity were larger it would have a nite speed at in nite distance If smaller it would fall back to Earth 0 GMem GMem 0 GMem 1 2 2GM e Re Te0 mv 7 Re We will consider all these ideas from a purely energy 7 point of view on Monday Practice Exercise An object with mass m is acted on by a spring force F k It has zero velocity when a A Use Eq 24 to nd the velocity v V when a 0 in terms of k m and A Explain why this is consistent with t A cos cat which solves the equation of motion so long as w2 km 13 4 0 v at our result ag 43 The W In Sec 42 Newton s s 117771sz 11 which we n The qua hand side caHed the particle mo Wba K This result cisely the shortly see energy in t 1 J 1 kg The unit 01 1 a 10 ll 1 erg The unit w 1 ftlb z 1 Vertical M01 A mass m i initial speec mum altituc earth compl d1 AZV39FhESl39tYdt39a ie ivork done is just the sum of the work fort r dimenstonal case we had I A done over each of the four straight line segments Equat39c 43912 becomes A M A The force in b is called conservative and the 4 F dr gmvbz gmvaz TF5 force in a is called non conservative We will V I h b 1 t 139 IIILC M he IS called a me Inte e lscussmg a S a 0 more 010 to ev a Ii integrals in the next two 14 Physics Honors Monday 15 September Fall 2008 Today7s Topic 1 Potential Energy Conservation of Energy Consider a mass m on a spring7 ie ikz and take the time derivative of 1 1 E E77102 Ekzz 30 dE i i E mm lax 11m1 kz 11F kx 0 31 We say that E is a constant of the motion It is conserved quantity Note that dk22dz kx ln general7 let idUdz for some function So we have 71 2 dEi dUi dUdzi dU i E2mv Uzgt dt inn11 dt 77mmdzdtivFd70 32 We call Uz the potential energy and E the total energy Relating this to work7 b 17 CW 17 Wba Fxdx 7 de 7 dU 7 U17 7 Ua 33 That is7 the work done from a to b is the negative of the change in potential energy Now generalize to three dimensions This is simple7 in principle We just write 17 U1 7 U Um 7 Um emu 7 Fm dr 34 but the the integral can m principle depends on the path If it does7 then Eq 34 makes no sense We can only de ne a potential energy for forces with path independent work7 also known as conservative forces These let us write down an energy that is conserved Non conservative forces lead to a loss of mechanical energy For F FCFDC7 we have b b KbiKalVbtgtf FdriUbanIE I CdrgtE17EalVgC 35 The energy lost in going from a to b is just the work done by the non conservative forces Finally7 let7s de ne power Power is just the rate at which work is done That is dW Fdr dr E dt 34 36gt PE Practice Exercise An object with mass m falls at a constant that is7 terminal velocity 1 through the atmosphere If the drag force is f lm nd b in terms of m7 9 and 1 Then7 calculate a the change in total mechanical energy AE when the object falls through a distance h7 and b the work lV11C done by the non conservative forces Finally7 show that AE Wm 15 Topic 2 Energy Diagrams We gain insight about motion by looking more literally at energy conservation ie E 771332 U so 339 v i g E U 37 Velocity can be to the left or right but speed E is determined The square root leads to turning points to the motion which limit E the position based on the energy See the fig ure and discuss motions for different values of E Relate force plot to the energy plot Dis cuss behaviors of particles with various El and mention oscillations Draw the analogy with up and down a hill Note that in princi D O ple one can get the motion t by integrating I m 15 4 r6 3 x1 x0 12 1 WW Eq 37 We ll do this in a practice problem Neytral Siam Un table equilibrium equmbnum eqwhbnum Example The Pendulum A simple pendulum is a mass m attached to a massless string of length 6 which swings in a plane through an angle gz See page 255 in your textbook Analyzing this in terms of F ma is a little tricky but we can do it It is two dimensional but the two axes change with time One is radial in which there is no motion and the other is angular along the path a variable we ll call 3 6gb We have F3 mg singz m5 so singz 38 Nobody knows how to solve this problem analytically However if the pendulum never swings through large angles then singz gz and we recover the harmonic oscillator equation of motion with w2 96 We can also look at this in terms of energy The potential energy in terms of gz is Ugz mgh mg Ecos gz mg 1 cos gz 39 Plot this as a function of gz and oscillate about the minimum at gz 0 Also point out the equilibrium point at gz 7T This leads into a discussion of small oscillations ie 252 254 2 g 82 Ugz mg lt 2l4l 771962 m 2 0 which is a harmonic oscillator potential with k replaced by mg ie km gt gE Practice Exercise Use Eq 37 with Ua mgr and E mgh to find Assume that a h when t 0 This describes the motion of something falling from rest at a height h right Split drdt so that you get f da on the left and f dt on the right It should be obvious after integrating that the constant of integration is zero for this initial condition You ll need the following obscure calculus rule i 2xa ba ad 92 T b 41 where a and b are constants 16 the momentum and energy of each particie change due to the interaction forces Finally long after the collision c the par ticles area 39 fr d t ht l 39t Physncs Honors gal Elehaunisdgjl iglg39 terrarlher mes W39 h quotF vli p 8 Ions and velocnties xp men a we usualiy know the in la velocities v1 and v2 often one particle is initially at rest in a target Today s Topi zq hsnggrnv39gtziggie alays The e xp rimer39t rnlgni conSlStTfWTITrg th n a lms v1 and v2 With SUIt Textbook Sec g pgl igwdl lf ggalgrinciples are involved the discussion is intended to illustrate ideas we have already discussed 7 l a b Since external forces are usually negligible the totai momentum is conserved and we have Pinitial miVi t m2V2 P nal m1V1 t m2Vl2 42 Pi P Pinitial P nal For a two bodyieomsienwthis newbies ev g 44 Elastic and irmli lst39rir damsioi ismliith mt r al forces that cause the collision are c n gr vative7 then me aniiial en 5Wil be conselrve Fatihaway froin the collidin poiw7 tl ere are no forces so no litaer rParl o eiltlait lg n5 con ampaoao 389 Krile 39ns cogsereataibn of kinetic ener 9Wr lJc ll elQ th9emtB Cehh c ot39hlegioel lg 85an thWe Wi eene gy ides n additional relation between the velocities f fa 26 Q i KfZO 5 equation gK we now 3 6w If Q 0 then the collision was elastic Using the same notation as above 1 1 1 1 Einstic andimelgs c wmgpi ngmlv 577121 Q 46 I C idr i39 39r39 we tworidr f ple 4189 sgic g ol39 3 s 5 fl eb 13239s e exercigeil gut ecw mass w 51 liltel aCiCVla gv coul s rm 5 up ose hat 1 d 11 an 112 7 v onse ing rnornen urn nd ene g A Wlhitialljli agil Er 1 has speed v as shown and rider 2 is at rest After the collision 1 i at EWe idWHo Wno the right with speed 37 It is clear ttmmp mtenehalsnbaeqi enagrved and that the totaiB kinetic ene gy of the two bodies Mu 2 is the same before and aiter the collision lg gouging gmwhich the total kinetic energy Before O 3 A I 03 lt 3 tunchangecti is caligd a Elastic co lision A collision is elastic if the in ht P tiaEh Of PCes gore nconservative 1like the spring force in our After example 2112 5211 3112 537m 50 2112 6111 6v2 51 or 0 m 122 52 17 This equation has two solutions for 12 each of which gives vi by Eq 49 namely 371 gt 5 53 or 12 0 gt 7 71 vi 7 721 54 The rst of these is just the initial conditions If nothing happens when the two balls pass near each other then nothing happens If they do in fact collide however then the second ball stops and the rst one res backward at twice the velocity that it started with Collisions and the Center of Mass The velocity of the center of mass is i mlvl m2V2 v 7 55 7711 m2 Let7s call this the laboratory frame77 We can change our frame of reference to the center of mass frame77 by imagining that we are observing the collision by moving along with the center of mass In this frame the velocities of the two particles are observed to be 7712 7 V 7 7 56 V10 V1 m1 m2 V1 V2 7711 7 V 7 7 57 V20 V2 m1 m2 V2 V1 so that the momenta in the center of mass frame are given by 77117712 7 58 P10 miVic m1 m2 V1 V2 MV 77117712 7 7 59 P20 szzc m1 m2 V2 V1 MV where we have made the de nitions 77117712 V E V 7 V and E 7 60 1 2 M m1 m2 The total momentum p10 p20 0 in the center of mass frame as we expect It is easy to show that for elastic collisions in the center of mass via 010 and 120 1120 Your textbook has the details We call V the relative velocity77 and M the reduced mass77 You will come back to these quantities over and over again as you study the physics of collisions both classically and quantum mechanically Practice Exercise Suppose that the two balls in Example 418 stick together after the collision ie vi 12 Calculate the Q value for the collision Physics Honors Monday 22 September Fall 2008 First exam is this Thursday in class at 10am Today we will do some math A Review of Some Calculus Rules i COS 7 COS 7 COS df du d do do d Chain rule Product rule gun i ozi 61 dz dx Partial Derivatives Gradient and Force amp Potential Energy in 3D The derivative with respect to of a function fx is just the rate at which f is changing that is AfA when the changes are very small We write f x dfdz where the right side of the equation is a much more descriptive notation Now suppose that f is a function of two or more variables x and y that is fxy We can ask how f changes with respect to x if y is held constant or vice versa We write these derivatives as dfdz and dfdy respectively and call them partial derivatives77 There are other notations but they are all bad In many cases dfdz and dfdx can be quite different but we will not get into this kind of confusing situation in this course Suppose z changes by an amount dz and y changes by an amount dy How much does ay change The answer is pretty obvious namely 51 5f d d 7d 62 f 695 96 6y i So now consider the work dW done by a force Fr over a distance dr The change in potential energy Ur Uzy a assuming that one can be de ned is dU idW 7Fr dr 63 Write the left side using Eq 62 and write out the right side in terms of components so 9U 9U 9U 7d 7d 7d iFmd 7 F d 7 Ed 64 66yy622 z yy 2 Clearly the z component of Fr is just idUdz and so forth It is nice to write this in a simple notation using vectors by de ning the gradient operator V as follows A 6 V 7 397 65 16x de 32 In this case Eq 64 is really just a statement that Fr iVUr 66 We frequently refer to Fr as a vector eld77 and Ur as a scalar eld77 19 A Useful Theorem about the Curl of a Vector Field 4 Imagine a tiny closed rectangular path in the zy plane with side dX lengths dx and dy and with the lower left corner at the point x It is very simple to calculate the work done by a force eld Fxy while going around this path since the force is approximately con dy stant along each of the straight legs We have fFdr FzydxFzdxydy X y 7 Fltzdzydygtdz7Fltzydygtdy lt67 Collecting some terms and ignoring a change in y say when moving in the z direction fF dr dzy 7 Fzy dy dy 7 dzy dy 7 dz Fdz7yF7ydy7Fd7ydyFz7y dm dx dy y 6F 6F dzdyVgtltFdedyVgtltFdA 68 We call V X F the curl77 of the eld F and have dA E dzdy and the vector dA pointed perpendicular to the plane bounded by the tiny loop Now imagine an arbitrary surface A bounded by some curve C You can tile A into a bunch of tiny rectangles and the loop integrals around any two adjacent rectangles cancels on the border In other words the tiny loop integrals 68 end up giving the loop integral around 0 Mathematically this says that foFrdrAVgtltFdA 69 This is called Stokes Theorem Obviously a force eld Fr is conservative if V X F 0 This is automatically satis ed if Fr 7VUr A Different Useful Theorem about the Divergence of a Vector Field Now consider a vector eld Er and a tiny box in three dimensional space with sides dz dy and dz The ux7 out of the box near the point x dzyz is pretty clearly Em dxy zdydz Play a game similar to what we did with Stoke7s theorem building an arbitrary volume V out of a bunch of tiny boxes For two adjacent boxes the ux out of one cancels the ux into the other so you end up measuring the ux through the entire closed surface A that contains the volume It isn7t hard to go from here to show that f Em dA v13 dV 70 A V Physicists call this Gauss Theorem mathematicians call it the Divergence Theorem The quantity V E is called the divergence of the eld We will nd some important uses of Gauss7 theorem this semester when we study uid behavior Next semester you will use Gauss7 theorem in your study of electromagnetism 20 is zero Physics Honors Monday 29 September Fall 2008 SEC 62 ANGULAR MOMENTUM OF A Hand back Exam 1 Average was 741 Questions Tell me Grades 3 60 see me Angular Momentum of a Particle i y wnere 71 is the perpendicular d line Of P This Midhi sal39ntobject moving in two dimen por onal to thesitl lsarl dfe the action of a central force AS the SketCh ggh i The gure shows that L2 rpm FyFm ya or where pL is the component of Now consider a quantity 6 de ned as Fy yF0 g E m 71 This quantity is a constant of the motion ma39yacy y a yi x may mFy ng 0 L t r If we de ne L E r X mv r X pfthlepn we see that 6 LZ We call L the angular momentum of a particle and canserved for motion in a central force eld B O x wecalii M1 the v x mv laments mamm i f Al eadnrid q alFtirlhEsNe ity is It igmhiei ftp 1Vth Soer in terms of parallel and perpen dicular gomponents of r If we write r ri r then T J I ll such that ri is perndilular 139 X p I1 r X p rLXprHgtltprLgtltp irl iPLi 39 72 lXP p Irillpl 73 L 2 r x phtri a0 Q Pa I ll ii oment arm 7 of the angular momentum since r x p 0 Parallel v 3 Second tenants Ke l i szstatandim hav s qiifelilt i of ang lar momentum conservation Following the ggrgeattrghe ght a argun dArdrrd6 so L ihdr dt UJ PJ B t shows that if tiile tlg que is zero lentiim is con ege ii tools momentum sergegiiiigi a sia i iiith 4 Method 3 m so dA hfsid dmgm h 2mple con t iietyadlthction a 938 EAgeypgemay already realize Vith linear momen um an energy conservation g u il equal Ar co struc magi ecpg 1 central force7 r M rAr 1 d3 7 r 74 2 dt 2 dt in drawin Homarmmuwed gegdg is hgrglaaice of oordinate system Show this explicitly for womentum hggoil lheeg39 X6is Eq 63 follows directly from Newton s second we talk about extended systems does angular 21 me its proper role as a new physical concept from r st at y 0 What is L for something falling am or me parallel to the y aX1s along the line a A n muc gen en in its present context considerations of angu l ad to some surprising simnlificatinnn at the mar in SEC 63 T0 For small of a triangll AA 3 ampT 4amp21 The rate at dA 2 lim Alt 0 lim ll gt0 372 E 2 c The small limit in polar angular mc Torque lf angular momentum is constant for an object acted on by a central force7 then how do we get the angular momentum to change By brute force7 we see that dL d dr dp i 7 F 75 dt dtltrgtltpgt thpert M l where we used F dpdt and V X p 0 with V drdt and p V We write 739 E r X F7 where T is the torque It is convenient to think of torque as force times a moment arm Imagine a bunch of masses hanging from a string The torque from gravity is y is up Zr gtlt F Zr gtlt imngj 771nm gtlt 95 Fem gtlt 93 76 If we hang from the center of mass7 then all moment arms are measured from here and rcm 0 So7 there is no torque and the object is balanced Sometimes we call the center of mass the center of gravity77 Practice Exercise Calculate the torque for each of the two cases in the previous practice exercise7 due to a mass in falling under the in uence of gravity7 and show that in each case7 739 dldt Angular Momentum and Fixed Axis Rotation So what is the motion resulting from a torque Consider one single particle7 inside the solid object7 located at r and acted on by F It can only move in a circle of radius r and the force in that direction is Fsin 0 If the path length it travels is called 5 then Newton7s Second Law says Fsin0 ms Multiply both sides by r7 and we have 739 mr2oz7 where 04 is called the angular acceleration The quantity mrz has replaced mass when we describe the motion in angular variables Since 04 is the same for all parts of the object7 and since 739 is either from one force at one point7 or due to a sum of such things7 Newton7s Second Law for a solid object becomes 7 Ia where I Emir 77 is called the rotational inertia or moment of inertia Notice that I is a property of the solid object and the speci ed axis Do you get the feeling that I is not really a scalar77 but not a vector either If so7 you are right It is actually something called a tensor You can prove something called the Parallel Accis Theorem which states that the rotational inertia of any body about an arbitrary adis equals the rotational inertia about a parallel adis through the center of mass plus the total mass times the squared distance between the two ares Mathematically7 we write I ICID Mhz See Example 69 in your textbook Rotational Inertia of Solid Objects Of course7 for solid objects7 we dont do the sum over all the masses in Eq 77 to get the rotational inertia lnstead7 we do the integral I frzdm This is just like calculating the center of mass7 that is f r dm7 but with another factor ofr See Example 68 in your textbook We will be doing more of these sorts of things7 include homework problem 68 22 X We choose a coordinate system whose The torque about A is from Table 61b T2ToRXFz 2 bF bF 0 The torque is zero as we expect and an is conserved The angular momentum Fall 2008 Solid objects b6th rotate andltmnslate There is a lot of physics associated with these mo tions See Chap6 Ratherlthanbalferive and review it all7 here are the rules Table 61 binc dL dzu we a a xed aX1s Rotation plus translation 0 7 means CM Pure rotation a ou L w 739 Ia K 5h 0 10a bMa L2 0w R x MVZ 0 7392 TO R X FZ with To 0a W2z 7 Ki low2 MV2 v v 10 10 Good examples to revieweiprglude the Atwood s Machine Example 6107 Kater s Pendulum 6127 and the Disk on lce 7 615 We ll do two applications in class today Objects REHHfQ39IDoW 5 I d E39 am39eefer to Example 616 A uniform drum of radius 0 and mass Consider some object Mabelinpardiustbpdn ss WH39an d IO MbZ rollin down a ramp How t f39 rtia ofthe r mabou it g Wa ec the time i ltakes for the object to get to the bottom of the does the shape ie ram 266 p i METHOD 1 Wemeesiaotmmmumm mmnbnv of friction F senalni ly e ti teiESlng just 2 F ma given by ngi ligiinatf zwgmtain W Sin 9 f Sin 9 ah dherngtiem lbout the center of m Friction makes it roll7 with torque I f co 131563 MIR2 and a ab we obtain For rolling ithout slipping we aLgo hav Mgsln i9 2 Ma f Igor a bar 2 O or Solvmg a a 39 sin 6 Mb2a METHOD 2 Choose a coordinate system whose origin 1 is on the plane about A is The torque TOODRXF7O39Z TO R X FZ bfbf W a 4m Rid w sin a RnN w W cos a z bWsin The normal force cancels W cos 9 so n since Ry 2 b and W cos 6 N The angular momentum about A is gsinQ1 6 A different approach uses torque and angular momentum This says that FMa b Mab 78 sin 6 ng sin 6 net torque v 12 4 L allow Rx MVZ Mb2w bMV 16Mb2w 1r flllibl bolf 1 Mba and get the same answer as before lb3w gives gmiewaeceiemamxm thak s clon g fztiowget d wn the ramp il energy at the start 39 M gh and there is no kinetic energy At the bottom of the ramp7 the GM has acceleration a to s T trave Tz For rolling without slipping a ha and 2 a gsm Note that the analysis woula have been even more direct if we had chosen the origin at the point of contact in this case we can calculate firectiy from O O O O s gnoxrqulowly at the bottom Fr1ction1s irreleV Since f and N act at the origin the torque is due only to W and 739 bWsin 9 Use one dimen nal Wire relationsihtoragsomethmg mwingcethmu w atquotllat l31 jaquoti 9a39is the same as Eq 78 23 110 aggynuiggg eMbQXVW ng2 51 MV2 Then i 2gb Viil lJr 79 int here Why gh a distance d with SEC 67 l The Work In Chap Kb I where IV ii We can energy ti translatii To dei motion f Mf l ME The v dRV Integrati 12w Now kinetic 6 about th To 100 10 Rotation we multi rnd rebsite KINEMATICS OF AN ULTRAELASTIC ROUGH BALL 89 W and The h elib whfbllfe o dmtelsy tem arid nomenclature 77 US131 leatheepeaeaie aea gamete and a 101quot w a atyve37attlvg0tlisneual lse ebbiealenonegme W is used for 1 brevity where R9 is the radius of the ball The moment of l Vb V0 inertia about lthel39Glilzdf tlli quotli all is IO R2 with 2 5110Garwin uses oz instead of but I switched so that there 2 is no confusion witlriagu for angular acceleration The kinetic energy of the ball itsniusti The Super Bull has velocrty V b and spin on before am 0 mp l catlon glves cing from the ma plane After the bounce the ireloci 1 2 l 2 2 l l K 2 9pin are Va and ma respectively The P m l gelo ivg l M ai ilMRaalt C 80 y is simply reversed in the collision so Kb Kw The angular momentum Gaboiitcfgh contfact lpldint see tthl gure is just subscriptions b and a refer to egrigtipong vineglmiyg glgr o V 81 39e or after a given collision and since all of the forces in the collisioncf sgaflgifolilglnlthegt contact point Lb La This im 1 COLLWa y l a simple result since Eqs 80 and 81 imply with conservation applied at the moment of inertia about tl axis i 2 2 V i z i MR2 with a for a uniform eagle Mailg T V Vb b d BltEgQ Ca T llZ Vb 82 mass and R therpw g g a g mal medsggxggiir gfhat the parallel velocity of the contact tic energy including only the spin to and the 111 portion Of 5116 133111 SEV l 0 is P1931361 mponent of the velocity V is Cb Ca l937b011I10i6r Ca V Cb Vb 83 Subtracting Eq 7 from Eq 6 we have KM Vall lzslz w3 639 the parallel velocity77 the velocity tangent to the ball s surface at the conga t oipjtu li e have 1shown that this Velocityl i s lddrisvergdd in magnitude but reverses in direction after a cbllision with the wall or the floor Subracting Eq 83 from 82 yields e CwR has been introduced to Sl lify 1 equations but in fact represegce 3 ga1g gx ll gt10bgt10 81 1 84 heral velocity of the ball s surfaceaadjalcent 1 1 1 18 wan relative to the centerl and resubstituting in Eq 7 we nd 1e angular monlimit emmllyuth hetcp inup b galfampg 9 ljglg 8931 gself iva moment to realize that I act 13 the 101 thele 1l l yenpielndicular to the oor is conserved in t e co 1s1on Why If a ball falls 8 L1wMRyer 3 ly2wdgyde then Vb 0 and 2 1Cb 47ltUb for 25 re a soli sphere That is it bougg gngiessid axfith at rlangivgrselgg itgnahat depends on how fast ltlaifsagqo i rlf ng 2 velocity after collision high 43 magnet egfeileaeietaliviligt ifii tg t r it a 45 angle so that acutecen geal sfo i Li cc39 enter Of mass 9 1 7 at andangl ius equating th i i m iimrgglafilill 1eanC Tamil b lam T tan 3 T 3 lt 39 Ty 2t an 866 f momentum bagel apla iieffcefoi veother calcrdaationswinyb ging tlgg l a CtOW 101quot bOuncmg Off the l underside of a table M39ZVa2aob2 M2Va2a0a2 Ka so that the ball on bouncing makes an angle 9atan 137 2322 with the outward normal 3 from the wall See Zllig 3a Similarly a ball 292 Practice Exercise F odsitzlee amgmiammmmenmmingfotmmemesaeai sonedpthe rdatei ii il d t 39eWQlie n ih ZZL Ehei igomtp theFoil leachcmalss r211 e itfa gx h39 bW Eh iWatosFthe j rodTEtrf is ateeewe hate 864 1 F619 tl fs39iia i s LI 02 793 Eacr2m cosm 2s BDaLcirefe Ewiadilfshe e u w 39 emeieepiedbm nagmmeeumwmtm a tl hugh meif weenie ting fgge eaaggwafers etwneeegh ttanc es along h and our non tensor formulation 0 rotationa 1ner 1a is naive RlGlD BODY MOTION and differentiating we have since 739 constant dr v dt 1 1 d0 7 cos i c030 sin6k lx2 fz ldl wr 1 co 6i 1 6 0i S COS 139 Sln 1 V2 2 l where we have used d dt w Next we shall find the velocity from v m X r Assuming that m can be resolved into components w xE fz39 we have 1quot i i 13 3 er Vi Vi Y Sin 9 r Sin 0 orne particle in the object Its poc Vi tion 1 r and its velocity is V drdt Tl e z w 1 COS m agrioijtudgnog 1s 7quot Sin d dt In oth r Vwords we wr1 in agreement with Eq 1 dr d9 As we expect there is no problemmtfea ngmlxeragy 6mm VecItor 8 ln the foliowing example e shall 5 that ap 1 ca quot greatly simplified by resolvi Vg e ietygoifgggghee gts lag6 artlc l n en axes The example also d c gl 8titat iSPlEQSInQe illeniOtatl is not necessarily parallel toa irlguWG lll ltPY t e umt VeCtOlquot 11 le 0 s n i the sk tch The robl is fi th ular 955ng maul cos a so in nd coes aoz39lg The most direct method is to calculate the angular momentum from 32ker rogOaSs origin r lcular to e m an IS p rpen Suppose a particle rotates counter clockwise in the 33y plane Let 9 be lei relative to the aXis Then7 its angular velocity would be w wllt wheree e f 39 t at equation 86 gives the correct vector velocity V wj yi er Draw a picture J h the particle at a couple of points7 and sketch r and v 25 Lsina Lsina lxample i The torque on the rod is given by 4 dLdt We can find dLdt quite easily by decomposing L as shown in the sketch We followed a similar procedure in Example 66 for the conical pendulum The com ponent Lz parallel to the 2 axis Lco a i constant H ce th re i no i malezztnmue she echeteti e knits 51 L sin a swin s withOthe rod If we choose xy axes so that L coincides with whei mth nttheiiaegv asueomentum 0f the skew r0 L caus ess this To see this let s nd dLdt and interpret th I 4h 0 lch are in ughe lk h di it sense OnlyttietEani te t l is 5P39 i ti fe ib i oft 1 t m Near all the Jy s in xamegi spin angular mom n magnitu e 1505 ag a axi 9nsts tai maw Howem a ero di m is consitb39rdeiiagaiargrw L sin a cos wt Lt L sin or cos wti L sin or sin wtj 4 L 87ralegxempmemiye L Li sin cot dL g L QWWWBB QE39Q n L sin a sin wt T E COL Sln a 1 Sln Wt J COS Wt c ggifsjgj fgrs marriy giigej Hence 7 Ear Vlam ygflmp gss Alternatively we can use a geometric approach Looking down 03 ii mghi tli39enaligisofrootmrtli en t lofohctii horizontal h plane we see tl at X gm Th dif c on 21 0 The IOAQE ELF LhAQ This is the only component of L that c 7 anges 39 IALhITLhAgut by LAIt9 tl i c z herefore l Lhm O t39Ur lh 5i91 i dt J Ln sin oz wi siewgd owg mg L Sin a w 89 I I USingsg xtezi l t tiieas vren tb f 39 uit as before The torque is zero if 0 and also if or 90 L 0 C a L n xT Helltei juel issfrt incforces which hold the vertical bearing nto the rotation axis Can you see Mg id ritic 7 W etHe soraaeo ere in both cases Imagine you re holdir Hence The Gyroscope 2A1 jil l g gOSCOp is a spinning wheel on a horizontal a g the bearing in your hand 07 is suppo geyd i rimemend only D an ru F iha c Nearl all of 1 an lar mo magnitu e E wheel a out its a le wL N Note t lso can you see i gt WWVAJ 1 lcu ed How it illus vectors 1 Rew ro not magnitude if so plane lnstantaneously apply a torqu 5 Instead giravity exerti39ssquot passes through the support 13 E L always points along the axle 39 f b We r omendturn vector L is alongnm tg lnggiifg ggs right epeeight 5N sinceithe normallaforeeeithaeqstmnmmentte otorque and tr l p ar llel to AL n tion in the ho it You may ha causesu the spe rotation is cons The torque is pl Theodmu scirm iaad yha flyvgheel is spf39 ning ragwghgaiifg like grgeeg n r momentu of thev ngpg i gmm mlg m gyroscLope s ang gy r g g ggr tum L is direcfieg39f l icallyebalance where a is themal rota ng39 whi When the gyros39ol H I 6 mi z axis it has a sm orbital angular39mr11 y we m aemrauemenceahs se 0 constant In magnitude and direction elf nnsequently we shall ignore it hJ3 The Gyro A th s We now turn y in eunderstood bj i aighitti de 739 lW and dir 13 aperpendicular to step carefully i trams thesltsissiloiheiierass e precesg on angular frequenqty Lh At the instant shown in figur tight th ton and dLdt is in the y direction In W L TIOwSHJL U Ltz e with opposing orces 90 L A simple model makes it easier to see physically why gyroscope precesses Consider two masses on an arm rotati inthe horizontal JR 01 down on the two masses so there is no net force and t e cent r of mass doesn t move If the force acts for time At then Ap j mAv FAt for each mass One mass moves up and the other moves down Thus 2i At AviFAti v 7 my 7 Agz m 26 which is the same result that we got i There must be a torque on the gyros the W Li Equation 90W 3 739 changed s L The source 1 t e torqu SO force anTat t lfwefag epivoi L is due ht of the flywheel actir The magnitude of the torque is w v is in the y direction parallel to dLdi Physics Honors Thursday 9 October we Different format today We will spend most of the period doing worksheets which you turn in These will be counted together as one homework assignment The point is to heslllomarly you get up to speed on the material we ve been covering especially the use of vectors describe motion both translation and rotation in two and three spatial dimensions Worksheet 1 Angular velocity for motion in a plane Collect the worksheets and work the problem out on the board Rigid Bodies and Conservation of Angular Momentum For a symmetric rigid body we now know that the angular momen tum vector is parallel to the angular velocity vector We write I Emir is the rotational moment of inertia about the rotation axis We say that L is conserved if there are no external torques but nobody can prove this is so starting from Newton s laws See the diagrams on r the right The torques Tij ri gtlt fij cancel only iffij works along the line r rj That is fij fji Newton s Third Law is not enough There is some higher reason for angular momentum conservation L I w where Worksheet 2 Conservation of Angular Momentum Collect the worksheets and work the problem out on the board General Angular Momentum of Rigid Bodies The Rotational Inertia Tensor Fa BODY MOTION Let cf be the torque on l j X fjl the torque on i CEO Y X fu The sum of these two tc 131 121 n X flj I j gt Since fy f1 we have 11 W 1 X To V 391 quot 39339 X fli r X fl where r is a vector from that m m 0 since cancel in pairs just as tl torque would then be 2e of an isolated system is Since neither r nor f1 flj must be parallel to r to the situation in figure angular momentum is n are equal and opposite The situation shown i central forces and we clt momentum follows tron force motion However forces to be central WI no direct bearing on whe isolated system is consen exclude the situation shc It is possible to take 1 following grounds althou forces to be central they in their simplest form ticles are idealized mass in this case the force be since the only vector deti rj from one particle to tl try to invent a force whic particle axis as shown i If all position vectors r for the particles of mass m within a rigid body are measured with respect to the center of mass then the angular momentum vector is LerpermirZmirigtltwgtltri You can reduce this expression with some tedium See KampK You ll nd for example low 2 2 LZ E m LUZ E mirizi was E miyizi coy or L 92 where I is the rotational inertia tensor a 3 X 3 matrix of numbers This is a general formu lation of the angular momentum of a rigid body Note that if w wk then LZ I LUZ where I 2771quot2 just as we had before Also as we have seen if the off diagonal elements 7 of I are nonzero then there are other terms in L How much do you guys know about matrices Can I tell you that the axes which diagonalize I are called the principle axes Worksheet 3 The Rotational Inertia Tensor for the Skew Rod Collect the worksheets and work the problem out on the board if there is time 27 Name Physics Honors Thursday 9 October Fall 2008 Worksheet 1 Angular velocity for motion in a plane Consider a particle which moves in the my plane Answer the following questions A In one sentence7 why does I39t tell us the position of the particle B The particle moves in a circle with angular velocity to It starts at Ly r7 0 Find the functions t and yt in terms of r u and t C Find the velocity vector Vt r39t in terms of r u and t and unit vectors and D The angular velocity is w 1012 Take the cross product 14 gtlt I39t and show that it is the same as Vt Refer to Sec 14 in your book for the cross product of the unit vectors Name Physics Honors Thursday 9 October Fall 2008 Worksheet 2 Conservation of Angular Momentum for Rigid Bodies A student sits in a swiveling chair that allows her to rotate about a vertical axis Her rotational inertia is IO She is holding a wheel on an axle The wheel has rotational inertia IW The axle is oriented vertically Initially7 the student is not moving7 but the wheel is spinning with an angular velocity u counter clockwise as viewed from the top A What is the magnitude and direction of the initial angular momentum vector Li B The student turns the wheel and axle over by 180 so that it again is oriented vertically7 but now the wheel rotates clockwise as seen from the top If the students angular velocity is now 9 write the nal angular momentum Lf in terms of IO7 W Lu and 9 Be careful of signs You can ignore the mass of the wheel as compared to the mass of the student C Find 9 in terms of IO7 W7 and u 312 Name Thursday 9 October Physics Honors Fall 2008 RlGID BODY MOTION Worksheet E hmheh gtoamggglom mkxensor for the Skew Rod We found the angular momentum of a rotating skew rod from first principles in Example 73 Let us now find L for the same device by y usingUseinq 92 tolcalculate the angular momentum of the skew rod A ma sles rod of ie th 2 s p ratest equal ses m Th rod A is sk rialghrne asthma last iflaeeale rflaamgglfanvelooty is w wk with angular velocity 0 Att 0 it lies in the plane The coordinates l of ihKarticleSOat any other time are 39 Us1ng p as de ned at the left7 derive LZ in terms of m p Particle 1 Particle 2 x1pcoswt 92 pcoswi y1 psinwt y2 psinwt z L 22 h whenp lcosaand h lsina The components of i can now be calculated from their definitions For instance In quot11le2 l39 212 t39 7712122 i 322 2mp39 sin2 at M I 1y i miZizi quot129222 Zmph sin wt 39 B By rearranging as yghandaiwninerEqareQQagdslmlvatrhatWe W l p2 sin2 wt 121 iiZ sin cot cos wt ph cos wt I m p2 sin wt co wt pzcos2 wt h2 ph sin wt i h s t ph s39 t pzi L33 7 77513222 w atld Ly 7 E miyizi w i The oriamon factor 2 multiplies each term Since 0 00w we have from Eq 713 L Zmphw cos wt L1 Zmphw sin wt L 2mp2w We can differentiate L to find the applied torque 139 2mphcu2 sin cot 1 Zmphw2 cos wt T 0 C COmP1 t the fOHOWHigJQRls fgraieh aiQoO ldmimnefetih ptwemasses in terms of p w and h make the Substitution ph l2 cos a sin a Particle 1 Particle 2 1 pcos cut 332 yl y2 psinwt 21 22 h D Derive L3 and Ly in terms of m p h and w Compare to our result from last Class 30 SEC 7l Princi if the the cc in Exa diagor where axes proble Rotatit The k Example 31 Physics Honors Tuesday 14 October Fall 2008 Today we nally l leave rotations of rigid bodies but we will use those topics later Accelerating Reference Frames Newton s first law of motion says something like An object at rest stays at rest unless some force acts on it Also an object in motion doesn t change its direction or speed unless some force acts on it 7 These statements are fundamental assumptions and cannot be proven In fact they aren t always true We say that an inertial reference frame 7 is one in which westerns fsl39ofaviff sometimes called the law of intertia lrflgbcldsimFORM CHEW SYSTEMS 34 law doesn t hold is an faceelenateingureferedaeoeaiframe nave exactly like a n g gravitational force the fictitious force i rm and ropor ional to sta yevani l geihe rst law The SGCOnd laWstSr yr li a rtional to the mass The fictitious forc on an extended body therefore acts at the center of mass E F mA 93 Example 82 Cylinder on an Accelerating Plank Thesoathateifablrere are no forces F then Sciiia bit a iasliab i f wit 399 m rA cheferencemfrearmie iiditiiowhiehxpl lewton s in 39formly accelerat39 system is un39 mimic iisecondilawwreal originate in the acceleration of the coordinate system not in inter action between bodies Here are two examples illustrating the use of fictitious forces Lie themgtzziazlistenaisliiinsigisi s baztg tings enthyou feel s n V tls lih h il iii dmiimi mll il mvii ver39 an w t39 it tensquot 7 o o O acce raiono in are ev 39na em ixe o e n ba i iz nfmaei fietitiousrf ercnellaFa et mA andyerything iss grie oil feifr39eis ib mip iyi uatr 39fe 39a 39 accelerating With t e car The equations of motion in the system fixed to the accelerating plan Analyze this Mays foifakrn observer standing on the street and for someone inside theear The cylinder rolls on the plank Without slipping so Outside inertiallilsgyiudInside accelerating l9 1 T E1 ct 0 MC i wi mAi m A F 39 Fm T sin 6 a39 3 3be sin 6 F ct W SEC 1247 RMLY apg gvmm etgu m2 angiwsiig wth 34 a w Tees 0 V w 0 7 cos 5 A W0 The acceleration of the cylinder iner l system is SmitheAperson outside the car says tiat TsinQ mA whiliectihelrperson 1ns1 e t e car says tl i at F ct 9 mthei litiE haefaeroa enthei mml r taeien inghyst snreallgvpretty i it trivial but more complexzsituati fls lite mores omateatetiqng g ymgghya ggggjgigg w g i3733 constan and is proportional to nch imammma iihed Bantam iifomsysiem try it n 39 n r g 39 th celerating car the ficti b t 39 h 39 I If H 39 I f x m a w J u i 29 The effectivmiian extended body therefore acts lief f39h sdemritefvdf39 39ag e era 39quotg 5 5 em a u r l u u n r n n n tational force is the vector sum of the real and fictitious forces How we a eirnie oo onasrin 39n aclrain cr O I beliefews fads ight dy morect39fnipletxg situation from Exani lfile39eg piw39you text39btaok v Consider again the car and weight on a string of Example 81 but nov Sometimes Fquot is called an inertial farce wever th ter cum to we more clearly e39rriphaslzes that Fm does not ampkyil intemder DI I an Acceleratin Plank assume that the car Is at rest With the weight hanging vertically Th The r ank accelera es to the right with acceleration A The A 8 effro agThtraS ms R mnsnwt h mtlSIlTlgin onaa p0I nSGTVGIquot quot whi h IS accelera ed al tiie rate he accelyra ion39of Re cy in ei masaaeahaia transmitsrialiedeiacssleratesntertiietr ght due in tosylsliaen dediknai rls fencehg pla riEo39s anombSettweskemhthe ipltank a e tistfaiotoreecmggeras MSA Va tE smite tester 8F nests and a a fis the frictio f e F M Jib h he ir cton sh wn anoderat 39siiisniri deatsytbisviie til it the assdsratsn ref the Fr 390 areylinder is a to the observer on the plank then the torque M f f j icagceljeyates the cylinder by 0 via fR 00 DdR Wh jeJm MR22 So first write a 23 X to lif i 39 quot quoter ysifrtlti tir39tlie i39 f ipltat relative to the accel erating frame where X locat sItlTeaplank Use this to explain why the acceleration in the inertial frame is a a A ThemWise the relationships between rolling and translational motion above to find the accezl ragonso 1 aridficcrt in terms of A w f ct M 10122 Since Io MRzZ and Fmt MA we have Galilean Relativity THE SPE Ao IgtgstgFa zlittlenmore formal about this idea of reference frames In particular we will show that Newton s laws of motion are the same in different inertial reference frames that is two frames that are moving with constant velocity with respect to each other Vlty beF emi s dis ut ibmoIkm mmniglehe notateoingami figures from Sec 114 in your textbook as correCt fo moisfedht d39cgblawgi2Wf38th we llet t lil z vlrfequotUZECEl iliean Transformations but the latter is a o of his ins i1atlon but the must h ve included the following con sideratidhtor h QRMBEH not a universal constant the 39539 T598 fit5 ti il O l dti m heieamte giaietliacle It is lo t39V39ty wou39 fa Zipeflalhne titit matt Pdr iQ ledn erf them and at tfi etherO e for M 7 one at he owever t of wenv O O wail thewaeeate Maggi e qg l geg srr gf oagngln of the tions a through llb l idlj syggests isaliotmetsdeea Righttigecdi mmed System indepentttqfetfigf thqavqf the source Our inability to detect 5 absolute motion either with light or with newtonian forces implies that absolute motion hagmg rplgiwhysicsvhere R Vt Whereas most physicists regarded the absence of the ether as 94 xx Rvt a Paradefhgtn geit sgg hg getti fdir f es td liatibsitiivith a constant velocity v relative to the of the prinCIpJe of re iew w s e entia con rv tive he thEf Epgid thigh th g eeofisfglraigivftll ew i a g 1 3X13 ther w d destr Amoare tl t r tow d 39 39 as ndang Ylp pr rOsaOa tyOlyTl h gcmtifosllgggfg tr even though the veloc1tyOof the particle is as theshift istbyay tmpogseeggqgaiigaFQEgOs gs 3 1ve To the gahernghe accelerations are the some I fact Tsiieembamsddmt tE Eo ngtahialpt 39 wsteahsuvsg INew on s second law is intact In fact you dually csa39 39dtetel l wmbi Elaine me apeihinibg sbacigliargi the particle s motion Notice also that for TanaEiif gg 2 ndra ldm39 miza ot EOEESWWE that the distances and directions between a n o q I o si n c t e sameaiettaeeeassee oatgesttgeeeeer he Posu mlusfauaggpkees spyme plausibl 39talid ita i fmately wrong assumptions For example it as O sumnsgglrzigst lr at leOrOOig gOhs are measured the same by both observers but this ends up being 1ncorrect Leave t 15 ugh9m tg hheh mmttssemy teameisiaavity he laws of physics have the same form in all img gstme of gtga oa ly ilistl ir ggg s epnz J 0 frOm h I 39t f39 l 39 39t e o tO O o lohg O reafo efvgrsfg Wt Space 393 a unlverishget elmfingin fnsat39hg 3 y system Its position along the a ax1s r c x incythe egg system is The mathematical expression of the special theory of relativit O Ct is embodied in th 3 L reriOtz transformationsD t esi mMesmsbritffe equat39on foraihe wavefront along the 1 axEEaEnple 11 1 to for ral lng eVenti Indlfferent inertial wstemev hrerassumed that its velocity is C Its position along tie mystique the Omathe Omatics of relativity baggy gmglqhefeie IO y System is therefore7 fmm EqO 94O mentary algebra yiillsu O eOO The reasoning is also simple but 8 10 it has a deceptive sin piiuiy pgt rt by W g l riif OF 39dtb ea te39C CityOof the two syStems tie Galilean transformations The x Ve39OC39tY of ta p355 quotT ME 7 Qt SYS RTS lanai iii swimwear iiriiatsllqmiii ielitsieiiieiii9 Syst mv the it appears t0 We in the 33 Cfiavvt 39Nediq39eetaitenhwirtlhsapwleciftyr itmze i er oeurse this is just what you d expect However a ml n0 ore cau es I39I ura 5 39 39 c 39 O cent to exg irythgisrl a t rhi j f t i E f5 lamp99ei leigne sms rgkaoMab m eill su qaations constant c or alibolis ers 0TH the Ga ile n transform tioEs a8 say that the speed of light is gainqqigtg mer 1a rame 1 inconsis ency is w a le to the development of special relativity and the eventual recognition that Newton s laws are only an a roximation in a world w er all veloci ies re small com ared to C pp 11135 The Lorentz rans ormations p Since the Galilean transformations do not satisfy the postulate r O that the speed of light is a universal constant Einstein roposec Shay that the force of grawtyas thersaeshstweehtmehmtgqhaesnbe sewrifetaa arhether they re viewed in a primed orineinq nilisyieidreystemuecrlafesdgt ic e blim granefetma aim fen m in C vt Practice Exercise the rest system 2 y z t and the system 25 y z i which moves with velocity 0 along the positive x axis The origins coincide at t t 0 We tglga the most general transformation relating the coordinates of a given event in the two systems to be of the forrr x Ax l Bt 1126 11 39u 112l SEC 114 THE GAl 114 The Galile Let us review fo in different coc y z in which V 2 which is trar v For conveni take the axes ti if a particula rest system tem are r a r r R where R vi Since 22 is in the x x vi y y z z t t The last equat from the newto for granted the physics Equations 1 Since the laws they are unaffe ciple of relativitj by the Galilean the meaning of The Galilean Trai Consider how w bodies from obs might be to dlS orbit of one of moon and of Ju tive to an astron mi 2 Hr Fr 771ng quotroc marat lf the accelerations of m in the two systems are related by a am quotlquot A Physics Honors Thu rsday le gt lbigfmmm thequot Fall 2008 F Ftictr where Fm mA So far the argument is identical to that in O Sec 83 Our task now is to find A for a rotating system Thanks to Brian for covering class today and lab yrestarelaeyali ating A is to find the transformation connect ing the inertial and rotating coordinates and then to differentiate Remember Exam is tWO weeks frOm tOday However there isa much simpler and more general method which consists of finding a transformation rule relating the time deriva ives of an vector in in rtia and otati coor inat in order Note Section 84 in book on Equwalence Prmczpl g isomt mstgmg obu Wage gppfilng he Mam between the velocity of a particle measured in an inertial system Rotating Coordinate Systems vim and the velocity measured in a rotating system vm Recall from last class Frame accelerating with aceeleraamn sod esnita bey Newton s laws but you can x things by inventing a ctitious foi heaVFil f re ed imi xe rotation Without ans39a m39 and 50 we consider a rotating system x y 2 whose origin coincides with Familiar example For a turntable rotating at cons g ma acceleration 92R where R is the distance from Blithe rosettes 98in ilise foe t eyposrfareloeifitogie r a m S 62 X15 U ermore e e I turntable you feel a ctitious force mQQR outwaridtrwfsnaall glgffg glyEigaf ifbreegi e now that a particle has position vector rl in the m plane and r z plane Our goal today is to be more formal with this conceptt39of rotating coordinate systems 7 Z See the gure One sest or3 5oordinates rotates about the other with angularpvie d fftlyfi QR if COORD39 SYSTEM origins coincide that is we are only concerned with rotation not translation The 3 y z system is inertial Pitt ttif ttdlAthfbe median vect rt l At and from eryf of the particle in the 39 coincide Rotation is athleeg displa s e S m iS X xquot Our job is to relate the derivatives of the vector rt in the inertial unprimed and rotating primed frames Exa irriirTe c ahge 39iil this vector between times t and t At below 4 In short the rotating observer is fooled7 After the time At has elapsed he measures a change Ar based on til cs i i E i o39 actlii 39SPEHB otwirasigoogea aing position because his axes havra splfranfga CU aie39s g39 itees h eessai c ef zilatllfactsmon vecmr fa a mg e isp acement e remembers that the m1 lt95 is coordinates We know how to were 1 fo Jar kft tffheyn thine gnrneateigithabughh wewrgular velocity vector 9 Qk accordffigpt Ed atergb t ia wy from its earlier POSition and We see from the drawing at left Ar and Ar are not the saI A rtitz5lttgtltngtltrgtm W If we COmbine equatiea a tfll fs tW Gefbdi tyilig different in the two frames Since r t agdf f di en only by a pure rotation we can L t Q X r the result of SecK7 2to 97 r t rat a x r At H 33 once 9 Ar Aquot 9Xr l At At lo 86 NONINERTIElggis serbtialnlg gg Mkt At gt 0 we can relate the velocities in the two frames as Vinertial Vrotating Q X r Coriolis an lneqtrifo aleioesmtyeorn haematite mpelmielsi of r and really relates the derivative of this 98 a rotatingxteemewll imiewd itwatw d d rttamitiat vrfyrstems So we could just as well write e it skids outward as if pushed by the centrifugal force tires is not adequa ei l o kee t with a rotating system For Instance if we whirl string we instinctively say that c rd itlru I force is outward in a coordinatgilsyst a ing With h correct the rock is stationary n t eincentrifu a f balance with the tension in the string Inian inier tial there is no centrifugal force the rock is acceTera lag ad to the force exerted by the string analyzing the problem However it is esse tial a i t t th st b t in to use fictiti u forc s in i e tial 6 5y eWhene We have assugn Jghgtte ghee roggfigfn rge jttig 9 does not change With time Thus the Here a e some Examples coordinat itlthUS orce in the rotating coordinate Surface of a Rotating Liquid Eithergystem is vgid for From t standpoint of rdD erver rda n inertial frame however what has dB happened is t e s le aid orcd efteld by t eirg dlwltlh e E e ciar turning with the road There iterna yai e eeBencseeelsette ieestaia mt39 tt gie the acceleration v tor bec a me n ml the rock thils Q X Vin cIeDtis in sys m V r ifygetiex r rot rgfug ot 9 x Q X r frames system a F ct 2m 2 gtlt th m9 Q X r dB dt F0 QgtltVr0t gtltr 100 101 A bucket haSatWOStmsmsth Tweemddterrnmias Wimd hm92r for a position r in the rotation plane In a coordifa el gstem rota i g withE the bu at the r exf tendn iamfergyfl For equilibrium the total force on m water39s susecrsajesve ur old friend the centri angel static COYleSl er surface of the liquid zero ThJ X E E the 39EonEeadoSclidnghwithwtvfrimtiiomoncaradial rod tiOUS forcerbtatilrlighv tt ialrlgular velocity to Use the rotating coordinate Fo cossgysteliinFThe vector w x r is in the horizontal plane upward Fo sin irit1hetlBWer gure So Fcent mw x w x r is outward where ngcandrErcmw 2273imoeampv Oguisras mWforWFOT radialle outward with the bucket Radial E F ma so i Transverse E F ma I SO 1 15 9 e radial e ation gives 7quot t die 1 I itions T uW 2mw2Ae Jt B l gt be Ffict w n at is the irection of Q What is niece algsca edhthe coriolis force rey must be mwzr mi N 2mm 0 mwzr i L w IThe first term is new It depends on the Example 87 r pl O I Fccnt FCor Bra Wt where and B are determined from initial or the norrrial force of the rod on the bead sorne pra tice with thefdirections of oriolis and centrifugal forces on the Earth s surface V e niagnitude and direction of Q X r for a particle at the equator What if the particle is at the North Pole Do the same for Q X Q X r Now calculate in terms of Q and the radius R of the Earth the magnitude of the centrifugal force on a particle at the equator and at t he pole Next instead of being stationary let the particle fall from some height near the Earth s surface It therefore has a velocity th downward in the rotating frame similar process and determine the magnitude and direction of the coriolis force on the falling particle in terms of Um Q and R 34 Go through a SEC 85 PHYSICS Solving these wzr 4 arctan 9 Unlike solids face Hence F0 be perpendicula is therefore We can integrate We have fdz fra 7239 2 9 where we have t surface is a par The Coriolis Fort A bead slides wi speed to The bead In a coordina radial The ske Fmt is the cen39 wire is frictionles gravity In the Fcent mi N quotquot Foo a 0 Using Fem m mf mw2r 0 which has the Si 739 Ae Be where A and B Physics Honors the general features of a system of two particles interacting with a central force frF where fr is any function of the distance 739 between the particles and F is a un angular momentum and energy results to the case of planetary motion fr oz 174 and show how Motion under Centiit Wd ep39er39s empirical laws Recall our discussion ciztmbleglymtm ianglsgegtergg msgleWe found that we can two body problem as aseaep irlarabterysirithasimaepaa le ilerrarigw that o i Sfpgeei r iEhee lib bodies where M m1 7712 a di m7r1 m2r2M is the position of the centeQZlf mass other hand the two eqtitlatiranseoi mataem zararalsg laeocemlbilaedldm spabtpacting hence is called the reduced mrl39fs g wee o w Ea tsdea ft th the o eiibww hentiraill fbrceheproblem Note that if 7712 gtgt m1 TGTWSut quotXiq a e39t f pheno M themeandatiomqimosenwe ignore it iTRAL FORCE MOTION Recall that angular momentum L hr x r is conslt rved7 in both magnitude and di under a central ce r anflrssnirrhiar awhile r 2 r1 r2 l39i weld these two equations to The equations of motion are mlfl 7 Whizl quatioTs 92a and b are coupled together by r the behavior of m1 2 r f1 r2 waded pfnn on r son rut e halfshowwtaereie prop Him is easi r to handle if we replace r and r2 by r r1 r2 and M equation of motion for R is trivial sin single particle and has a straighth ters After makinMGi lala we shall show how to find a complete soluti n by using the conservation laws of mlf l sismgfg 0 it vector along the line of cen Fall 2008 Finally we shall apply these treat a 77Lng fTf 91 see what happens to the center and f mass 92a On the e there are no external forces ward solution ection7 for central forces One obvious conclusion is that tle motion must be in a plane r m1 m2 anWewgel ilrthgrah iinsiggspplarvgggrdigates T16 mag and m2 relativaimldheo ehherantgihlassraemhetsmdn shawslerQ The total We complete Summrmifcal ene fg islepends on the particular However a number of the properties of central m of f7 ce motion hoid true l we turn next to investigateZ where Ur is the potential energy with fr E Genera Propecrm g 3 equation illiquot j fi i7 is a vector equation and a iingle particle retailseatfaeieaie hrtsitatsr lsidered In trig 1 laws to find some iuce the equatigl to i Motion ls P quot 39 to a central force f6 luced mass u Hence the a nt It is easy tc show that thisimplies that thejrncgiooyo hle maria onfined to a plane pendicular to L Since the motiom mtmmsw lerality choose Mitzi the xy plane lithe F a6 0 The particles are kept apart by the 2 E i This is qualitatively similar to case 1 but on the boundary 7 r02 between two pa rti r A py the properties of the cross pr lr L is fixed in spagec eg iEgpgpwfsgtat r can only 9 PeVPequotdiCU aL a iii ilirit ii Qttm t Et ii il filt i iif ener 3 E lt 0 The motion is bounded for both large and small 239 The inEggn r af lq stgamlpssof Lt a grm29 fng T dl l l 62 1 9 VH50t 7 399 lwyr in Uerrr fr otigiiseful fo mr ealiew talleseorhel ehte rtaobeys d6 102 ldr So 62 artists U r 103 2hr anal sirngteeentrall fnreeclmetion in terms onstant dist nce from 0 another Missedurnindwheerieor spiacnnstant a hyperbola case 2 to a parabola case 3 to an ellipse a gen properties 0f the Sollltlec t iedphtential frgsrijzhiend htemial nengysofptwo gravitating h i39easi39efhe solid or Udl an equation in a Single scalalE a nova along a straight line on a collision Course since when l pWelgio centrifugal barrier to hold them apart he Capture or Comets uppose that a comegvgth E gt 0 drlQmerZsystem Frl ur Wigwam the energy diairarn ror motion under a graviiaiid in 39 approgh the sun and then swing away never 2 730mm to become a li ember or the solar s m its energy would have to be reduced to a negative value Howey 104 me tum Il of iEiSiOEmin t e Since g swx uv where v 21 aduct How move in the Weathers r07 Mali lilldut remthat 6 a the motion is restricted to r ions 5dr diitaquot ehssystemersod at the motion is Here are the vaious ossibilities nelnelepglarfenomlhatesthe equation of 35 the unbounded and bounded motion he comet39s on We 7 radial coordinate dentin change For or instance lr the come is de lected y a masslve planet like Juplr Ebbquotth are 396quots il l aitme s b e quotWeen min and Suppose that a comet is heading outward from Othe Sun toward 1 39 1quotquot with 39 if L exore mars ol rera taggecla lgit up r rl pite elocny be V For simplicity we shall assume that the orbits are I ppreciably deflected by the sun during the time of interaction In the comerJupiler center or mass system Jupiter is essentially est and the center of mass velocity of the comet is v v V own in llgure a v 39 is th Vl ill lb in the center or mass system the path of the come is deflected t he rinal speed is equal to the initial speed ll Hence the interacti erely rotates v through some angle a to anew direction v as she Fig b The nal velocity in the space iixed system is 97a cles form a bound system V1 V SEC 93 GEN EF The total ene E 11rva l 11M l where the po U0 U7 a The constant leave r unspe on the motior We can elil result is ll 2 1 E Em i This looks like dimension al further by int 1 I r L effr 2 I so that E M2 l U err is called to simply as potential U0 The formal dr 2 E dt u or Dark Matter in Galaxies We can learn a lot about one of the most important problems in physics today just by combining what we know now about circular orbits Newtonian gravity and observations of the Milky Way and some distant galaxies A circular orbit of some small mass m about some large mass M obeys Newton7s second law in the form mM 112 y GM 2 m7 wh1ch g1ves o 7 r r r for the orbital velocity Note that the period of revolution is T 27Tro so we have 4712 T2 rs GM which is called Kepler7s Law of Periods Well look at this more on Thursday Note that this is a way to determine the mass of the sun from the revolution periods of the planets One very basic result from this is that the tangential rotation speed 1 should be proportional to 177 as soon as the mass is concentrated at distances smaller than r Galaxies should look like this since there is generally a big bulge of bright stars near the center However the galactic rotation curves look very different See slides This is one of the strong pieces of evidence for dark matter in the universe For some interesting reading see Alternatives to dark matter and dark energy Philip D Mannheim UConn Published in ProgPartNuclPhys56340 4452006 You can also visit httparxivorg and get e Print astro ph0505266 Practice Exercise Use the Milky Way galactic rotation curve to estimate the mass of the galaxy contained within a distance of 16 kpc Note that 1 kpc103 pc31 gtlt 10L9 The mass of the sun is 2 gtlt 1030 kg How many sun like stars do you need to get this mass Note that the brightness of a typical spiral galaxy like the Milky Way seems to be about 1010 times that of the sun Table 21 1 in Introductory Astronomy and Astrophysics by Zeilik and Gregory The Milky Way Galaxy Visible Light Infrared Light 38 32139 300 r 275 P 250 I 225 200 S w OBIIds NOIiViOH 175 h 39 I III quot r CO THIS PAPER HI BURTON AND GORDON 1978 CO HII REGIONS BLITZ FICH AND STARK 1982 BURTON AND GORDON ROTATION CURVE ROTATION CURVE THIS PAPER Keplerian prediction for IOIO Solar M asses quot H1 I I l I l l l l l 150 O 2 6 8 1o 12 14 16 RADIUS KPC NoteThe vertical axis has a suppressed zero Nec zs9o 1c 507 Physics Honors Thursday 23 October Fall 2008 Remember Exam 2 is one weeks from today Same ground rules as before Motion under Newtonian Gravity We want to nd the motion under Newtonian gravity in three dimensions Assume an object say7 a planet with mass m orbiting a massive body M Dont worry about reduced mass M distinction now The potential energy is See Example 53 in your textbook mM 7 C 77 where C E GmM 105 Ur7G T T We use 0 because that makes it easy to generalize to other central force laws First let7s review from last class Motion is in a plane since L mr gtlt r is conserved for central forces This is easy to show using Newton7s second law dL quot 0 f 0 106 7 gtlt gtlt gtlt gtlt dt mr r mr r r r r r where the sum of forces takes the form that de nes a central force Central forces are conservative See Example 48 in your textbook So write fr 7dUdr for some Ur We know that this means that the energy 1 1 i E Em Um Em is r262 Um 107 doesn7t change with time It is a constant of the motion7 Note that we7ve written this in terms of polar coordinates r and 0 See Section 19 of your textbook for a review The magnitude Z lLl of the angular momentum is also a constant of the motion The component of velocity perpendicular to the moment arm r is 119 719 by de nition So Z mrzd 108 Once again7 for as long as there is motion7 the value of Z does not change7 even though 7 and 6 are of course functions of time They just always combine so that Z is constant Now we can nd the shape of the orbits We can combine Equations 107 and 108 to form two differential equations describing the motion as rt and 19t7 namely d6 Z dr 2 Z2 E W and a 7 Ue r where U530 E Ur 2mr2 109 However7 the shape of the orbit is some function r0 So7 divide the two to nd d6 Z 1 Z 110 E mrz E 7 Ue r r2mEr2 277107 7 Z212 The integral looks messy7 but in fact it can be done analytically You can nd it in tables Integrating both sides7 and putting the constant of integration7 7190 on the left7 we have 77107 7 Z2 rxmZCz 2mEZ2 40 0700sin 1 111 Divide through my m0 in the parentheses take the sine of both sides and solve for 7 627710 r r0 112 17 112E62m02sin0 7 00 Lets clean this up Pick 00 77T2 it7s arbitrary and de ne 2E62 r 2 0 roiimC and e 1m02 gt0 so 7 79 liecosg 113 The parameter E is called the eccentricity The curves r0 are called conic sections The forms become familiar if we work in cartesian coordinates x 7 cos 6 and y 7 sin 0 so xz2y2 76x ro gt 1762z2726r0xy2 rg 114 Case 6 0 The curve is r To a circle Setting mozr pZmr 627717 3 Crz gives 7 27710 To This is precisely what we know we should get for a circular orbit Case 6 1 ie E 0 ln cartesian form we see that the orbit is a parabola P7 115 Case 6 lt 1 ie 1 7 62 gt 0 Your book works this out in polar coordinates but lets try it in cartesian First nd the ends of the orbit by setting y 0 This leads to To 1 7 62 T0 T0 1 In n 7 d max gt 77 7 max 7 m n z 1 6 an x 1 7 6 a 2 x z Now translate to x z 7 so where so a mm roe17 62 is the halfway point between the two ends Substituting for z in Equation 114 and simplifying we nd 2 172 2 2 T0 Eb 117 lt egtz 11 1762 lt gt Clearly y ib are the top and bottom ofthe orbit Since a2 bZ17 62 we have yz E f i7 which is the familiar equation of an ellipse The parameters a and b are called the semimajor and semiminor axes The position x 0 is called the focus it is not the center 1 118 Case 6 gt 1 ie 62 7 1 gt 0 Multiply Equation 117 through by 71 but keep the sign in front of b2 so that it is still positive Equation 118 becomes 7171 119 This is the equation of a hyperbole Note that 0 is now a negative number This is enough for today You should read through your book on Kepler7s laws but we wont speci cally cover them in this course They will be important though if you should take a course on astrophysics Name Physics Honors Thursday 23 October Fall 2008 Worksheet The shows a circular orbit of a around the Sun 21 Draw on this graph a circular orbit with larger total energy b Draw on this graph an elliptical orbit with the same total energy as the original circular orbit Make the eccentricity e of the orbit equal to 45 Hint Use the de nitions of To 6 and a to show that the energy E only depends on C and a Calculations c At a point where the elliptical and original circular orbit cross7 draw the velocity vectors for each of the two orbits at that point Indicate which is which 1 Explain briefly why the magnitudes of the two velocity vectors must be the same e Of the original circular orbit and your elliptical orbit7 which has the largest angular momentum Assuming the same orbiting mass in each case Explain your answer USUIIKGLUI l5 l dlllill39dl U US HUI wraps L BHU 4 anu lrUlTl numerous problems However so far we have considered only the idealized case in which friction is absent and there are no external forces Physics Honors In this Wage E a iete biea ce the effect of etltiQOQBthe osciliator and then study the motion when the mass is subjected to a driving force which is a periodic function of time Finally we Second exam is on Thursday th qg t e f8 t 019 musuate a remarkable 39 resulty the possrbllity of predicting how a 7 mechanical system The Simple Harmonic Osmilllateep cga ian applied driving force of any given frequency First start with a review of o pvmgt i a th i y faeu39 tft91 fse f he o f39s pUt mto mono th 0 d fi gygdd d lflawlsags 2F 39 mo mi For is seneaenggeerewggg Eerosstesnstets J61E m 55 har mayg xNquot monic oscnii r who we discussed at the end of Chap 2 The if U U LJl prototype oscillator is a mass m acted on by a ring force L x Famin gke 2W whereas is thegzdishladeamntj qdi lbbum Any function t proportiona39lrt Hafdf39br giffllflgol zeggthfsBfif39ggfieral solution is 161 Bscfn wot C cos wot 12110 1 We need more information to d1etfgrfrcfilffdI airsid C Typically we use initial conditions 7 2 t B 39 t C o t 102 Everybody knows that a and megii childcny fe 333916 thing right here Solution using complex numbers We solved Eq 120 by recognition 7 A better way is to use an unsatz namely LAB 9 Substitute into Eq 120 and solve for w 10 3 w2A m 33216th gt w iwo 122 We shall use we rather than can as in previous chapters to repre Once agamv there are two poseeh Wt at r i a emseeelseiieetsohtm e are arbi traryg gnstaglgewwhicmggniube evaluated from a set of givepzi itial conditions such as the positiOn and the velocity at a particular Initial conditions Work wit39lmgolution 121 Obviously 330 C We also have 71t woB COS wot woC sin wot gt 110 wOB or B v0w0 124 Standard Form of the Solution A the Identlty eves US a dl wetaeoeweefu anun glistegsoizecreeiem form t A coiscu10tAccgg cgsvgg cos wot A sin gz sin wot 12510394 WhiCh We Compareto 121m Snt hem and d are constants To show the correspondence between Eqs 10 an v we make use of mmtrigonometric B Asingz and C Achikachtity A 11302 2 and tanng W 126 cos a 3 cos aCOS B sin asin Energy Have Ua keg2 since dUd Fspring Also K mv22 mjv22 so 1 1 1 E K U Emaing2 sin2w0t gz kA2 cos2w0t gz ikA2 127 So the total energy is the potential energy when the position a A the amplitude 7 Practice Exercise Apply the initial conditions to solution 123 Solve for A1 and A2 and then show that the complete solution is the same as 121 with B and C in terms of 330 and 110 Recall Euler s formula namely 6 9 cos9isin6 43 The Damped Harmonic Oscillator Now 2 F Fspring Fdamp 71m 7 b1 71m 7 bi The equation of motion becomes b k mi7kx7bi gt iviw 0where yz andwOEwE 128 The de nition of we is just as before Solving this by recognition is too hard so we immediately move on to our ansatz t Aem Inserting this into Eq 128 gives 7w2mww370 7 wzi 777 129 Case 1 1 lt 2140 De ne M E Aug 7 724 and the solution to Eq 128 becomes zt Ae wz cosw1t b 130 Some comments are in order First the oscillation frequency is M lt we Secondly the oscillation maximum is not a xed amplitude but decays away with a 16 time constant y2 1 2mb Finally the exponential factor in front makes the association of A and b with 0 and 00 kind of messy This isnt so interesting so we wont worry about it The red plot shows a damped oscillator with y LOO4 with the dashed line showing the exponential factor alone It is clear that the energy decreases with time In fact in the case where y lt we a situation called lightly damped it is not hard to show that 72 V E 7 7 7 1 2 74 t 7 E06 where E0 7 EkA 131 1 So 1 y E 739 is the energy decay time constant mama I xt gt39 J 4 6 Timet Case 2 1 gt 2140 In this case we write the solutions to Eq 129 as w m where 111 1 2 2 V2 7 041042 gt t 0167041 026 132 a where both 041 and 042 are positive real numbers so both terms of the solution are decaying exponential functions The blue curve in the gure is for y 31 and the same initial conditions as for the underdamped case This situation is called overdamped 77 Case 3 1 2140 This case is called critical damping Something is weird though It looks like our ansatz has broken down because we only have one solution a y2 when we know there should be two But dont worry math is good See your textbook or a book on differential equations The second solution in this case is proportional to few so t Dle m Dgte m 133 is the general solution This is the green line plotted in the gure Practice Exercise Derive Eq 131 for the decay time of a lightly damped harmonic oscillator Follow the discus sion in your textbook pages 416 and 417 Our Eq 131 is the books equation 1016 44 Physics Honors Monday 3 November Fall 2008 Hand back exam If so7 give grade statistics Some Notes 0 Wednesday is make up lab77 day need email to me or Scott by the end of today 0 Preliminary lab notebooks due in class neat Monday 0 Stay tuned Testing something new7 perhaps7 in lab on coupled oscillations Nov19 Review Simple and Damped Harmonic Oscillations The simple harmonic oscillator is based on E F ikz and has the equation of motion at aw 134 where we E xkm We solve this with the ansatz77 t Acm Substituting7 iwz flog gt w iwo 135 So7 both t Alcwot and t Agc wot are valid solutions In fact7 any linear combina tion ofthe two is a valid solution We nd A1 and A2 by applying the initial conditions The damped harmonic oscillator is based on 2F ikz 7 bi and has the equation of motion at who 7 W 136 where y E bm We again solve this with t Acm Substituting7 we nd a quadratic characteristic equation77 for w The solutions to this equation are u ig i M where M E Log 7 7 137 This is the underdamped solution7 where y lt 2mg7 which is the only case which yields harmonic oscillations So7 both t Alc Vtchlt and t Agc ltZc wlt are valid solutions Again7 any linear combination of the two is a valid solution7 and we nd A1 and A2 by applying the initial conditions For future reference7 t Alei y tZeiw1tAzei Zeiiwlt 138 is the general solution to the equation of motion for a mass m attached to a spring with force constant k and being damped out with a force proportional to velocity7 with coef cient b the parameters M and y are de ned in terms of m7 k and b by the relations given above Note that we could rewrite Eq 138 simply by using cimt coswlt i sinwlt and then de ning coef cients B1 and B2 appropriately in terms of A1 and A2 so that t Blew 2 cos w1t Bgc l 2 sin ml 139 Ac ll2 cosw1t o 140 This is exactly the same result which we got in Eq 130 from last class The Forced Harmonic Oscillator Now imagine that we are forcing 7 the damped harmonic oscillator by applying a sinusoidally varying force as a function of time That is 2 F hw bi F0 cos wt where w is now a frequency that we can control independently The equation of motion becomes F 7 w 0coswt 141 m We can st iemtahisHEeqcifa Efo39if39ifsing something of a trick Pick another variable yt so that F 2 i O To find the steadystat y fly woy i m Sln Wt 142 t k39 A w H and de nite The differential equation that governs zt is therefore this trial solution is n01 x Bc05wtCsinwt gvzwgz Eigth 143 into liq 1023 you will f m where wehfflsf neded f vietgall that wt zt to get the motion of our oscillator F0 1 Applying lo njaW Aem yields something a little different than before Instead of nding a characte icweq ation to solve for w we instead nd an equation for the complex amplitudg 7651353 unctia 39of w That is F0 1 A somew ft mare form m cog w2 iyw present ri thelIDlElw w l g gt A The behavior of A am Note We flaming didd39in a solution like Eq 139 since it just yields zero in the di erential equationls i8 3n itquots get the constants we need to apply the initial conditions This part of thaosolutfan though dies out with time so we usually just ignore it it is called a transient T f nis easy to interpret if we write As 1 gt O Awo x Ha F0 1 144 7 39 also y gt0 7 1 l w A i Agg p y W f g limit 51 i m wg C122 C12y212 and Q5 tan W2 tag 145 when w 2 00 In other words the motion of the forced damped oscillator is given by wt Acoswt go where A and gz are strong functions of the driving frequency w These functions are shown on the left for the lightly camped case where y lt we The motion has an amplitude A that becomes very large when the driving frequency gets close to the natural frequency too This phenomenon is called resonance Note the phase The motion is in phase gz 0 for frequencies well below we and precisely out of phase gz 180 far above we Also it passes through gz 90 as w moves through the resonance The resonance has a width that depends directly on y In fact if we de ne the width Aw as the distance between the two to values for which the curve drops to half its value then Aw y The value Q E cooy is a dimensionless number which characterizes the width and can be interpreted as the amount of energy lost in one oscillation Large Q means low damping and is an important gure of merit for resonating systems 46 Physics Honors Thursday 6 November Fall 2008 Hand back exam discuss grade statistics Average on test was 76 73 Waves on a Stretched String The Wave Equation These notes follow Chapl of Physics by Resnick Halliday and Krane 5th Ed What is the equation of motion77 for a string that is stretched under some tension F y k gt The motion is up and down so analyze the forces 39F in the vertical direction on a small piece of string 91 Jquot quot92 i r at xed at We assume small amplztude waves so 655 can represent the length of the small piece See F7 the gure If F is the tension in the string then 2 F Fsin 02 Fsma1 mm 146 H 5 g 1810 Put 6m adx for linear mass density a Obviously ay jj Ty8152 since we are working at xed 55 Also for small de ections 0 ltlt 1 so sin6 6 tan6 slope 8y8x So 9y By 823 1 8y By a 823 F 6 6F 6 147 m 2 sm 1 855 2 9x 1 8152 gt 6x 9x 2 9x 1 F8252 l The left side Eq 147 is just pg8552 as 655 gt 0 This gives l 92y l 92y 148 8552 v2 9152 lt where 1112 E aF Eq 148 is called the wave equation It describes a shape of the string function y of x which changes in time hence yx t The solution is any function like b yxt l vt ie 55 E I l vt 149 y This means that the change in time is rather peculiar The shape f moves to the right or left with speed 11 This is V the de nition of a non dispersive wave Practice Exercise Which of the following solves the wave equation Eq 148 for v 1 Take derivatives or show which are of the form Eq 149 Note that the sum of two waves is also a wave ie the principle of superposition This is because Eq 148 is a linear differential equation A yxt x2 t2 then try yxt x2 t2 47 Sine Waves We almost always describe waves as if the shape at l 7175 is a sine function y 4 A gt It is handy to describe a sine wave as follows ym A f 1 x 277 7 y yx7t ym s1n 7175 150 l Timeo h v Time I We call A the wavelength7 for obvious reasons Note that for xed 1 slzfyoex 0 we have y07t ym sin 27TvtA which means that the transverse oscillations are sinusoidal with period T Av 1f where f or7 maybe7 V is the frequency The more common way to write the equation of a sine wave is yx7t ym sinkx out 151 where k 27TA7 cu 27f and q3 is called the phase constant Standing Waves Add two sine waves of the same speed and frequency7 but moving in opposite directions y1x7 t ym sinkx out and 1121 7 t ym sinkx out gt yx7t y1x7t y2x7t 2ym coswt sinkx 152 where the result follows from a trig identity This wave has zero velocity Le km kx 075 but an amplitude that varies like cos out It is called a standing wave77 Watch a quot r 1quot gt y I 1 1 S 3 s H 4 quot2T f T 1817 Something important happens when we set up standing waves on a stretched string that is xed at both ends7 like a guitar string As shown in the figure7 we can only have standing waves that t into the distance L between the two xed ends That is A 2L 7 L gt A gt 153 n 2 n n f quot2L l gt That is7 frequencies have to be integer multiples of a fun damental77 frequency vZL This is the basis of the har monic sound of nearly all string and wind instruments 48 Physics Honors Monday 10 November Fall 2008 The topic we call coupled oscillations has far reaching implications The formalism ends up being appropriate for many different applications7 some of which bear only a passing resemblance to classical oscillation phenomena This includes the mathematics of eigenvalues and eigenvectors7 for example These notes describe the elementary features of coupled mechanical oscillations We use one speci c example7 and describe the method of solution and the physical implications implied by that solution More complicated examples7 and their solutions7 are easy to come by7 for example on the web1 The following gure describes the problem we will solve X2k Two equal masses m slide horizontally on a frictionless surface Each is attached to a xed point by a spring of spring constant k They are coupled to each other by a spring with spring constant kc The positions of the two masses7 relative to their equilibrium position7 are given by 1 and 2 respectively Now realize an important point We have two masses7 each described by their own position coordinate That means we will have two equations of motion7 one in terms of i1 and the other in terms i2 Furthermore7 since the motion of one of the masses determines the extent to which the spring kc is stretched7 and therefore affects the motion of the other mass7 these two equations will be coupled as well We will have to develop some new mathematics in order to solve these coupled differential equations We get the equations of motion from F ma 7 so lets do that rst for the mass on the left7 ie7 the one whose position is speci ed by 1 It is acted on by two forces7 the spring k on its left and the spring kc on its right The force from the spring on the left is easy It is just 7km The spring on the right is a little trickier It will be proportional to 2 7 x1 since that is the extent to which the spring is stretched In other words7 if 1 2 then the length of spring kc is not changed from its equilibrium value It will also be multiplied by kc but we need to get the sign right Note that if 2 gt zl then the string is stretched7 and the force on the mass will be to the right7 ie positive On the other hand7 if 2 lt zl then the spring is compressed and the force on the mass will be negative This makes it clear that we should write the force on the mass as k0x2 7 1 1See httpmath fulIertonedumathewsn2003SpringMassModhtmli The equation of motion for the rst mass is therefore 71ml k0z2 7 x1 mil 154a We get some extra reassurance that we got the sign right on the second force because if k kc then the term proportional to 1 does not cancel out The equation of motion for the second mass is now easy to get Once again from the spring k on the right if x2 is positive then the spring pushes back so the force is 71mg The force from the coupling77 spring is the same magnitude as for the rst mass but in the opposite direction so 7kcx2 7 1 This equation of motion is therefore ik g 7 k02 7 1 mfg So now we can write these two equations together with a little bit of rearrangement k kc l 7 ch 7mi1 155a kcil ka g WiZ To use some jargon these are coupled linear differential equations To solve77 these equa tions is to nd functions z1t and z2t which simultaneous satisfy both of them We can do that pretty easily using the exponential form of sines and cosines which we discussed earlier when we did oscillations with one mass and one spring Following our discussions about single mass oscillating systems we assume the same ansatZ and write x1t alem z2t 126th 156 where the task is now to see if we can nd expressions for al a2 and in which satisfy the dif ferential equations including the initial conditions You7ll recall that taking the derivative twice of functions like this brings down a factor of u twice so a factor of 7w2 There fore plugging these functions into our coupled differential equations gives us the algebraic equations2 k k0a1 7 002 mwzal 158a 7kca1 k kca2 mwzag 158b This is good Algebraic equations are a lot easier to solve than differential equations To make things even simpler let7s divide through by m do a little more rearranging and de ne two new quantities Lug E km and a E kcm Our equations now become Lug a 7 w2a1 7 wfag 0 159a 7w02a1 tug a 7 w2a2 0 159b 2A mathematician would call collectively call these two equations an eigenvalue equation The reason becomes clearer when you write this using matrices In this case Eqsi 158 become kkc 71 a1 7 2 a1 71 kkca27mw a2 157 So what have we accomplished We think that some kind of conditions on 11 a2 and w will solve these equations Actually we can see a solution right away In mathematician7s language these are two coupled homogeneous ie 0 equations in the two unknowns a1 and 12 The solution has to be 11 a2 0 Yes that is a solution but it is a very boring one All it means as that the two masses dont ever move The way out of this dilemma is to turn these two equations into one equation In other words if the left hand side of one equation was a multiple of the other then both equations would be saying the same thing and we could solve for a relationship between 11 and 12 but not al and a2 separately Mathematically this condition is just that the ratio of the coef cients of a2 and al for one of the equations is the same as for the other namely3 7w3 w3 w 7 w2 2 2 2 2 160 w0 we 7 w iwc 4 7 2 2 2 2 we 7 W0 we 7 w 2 7 2 2 2 iwc i w0 wc 7w 2 w w w w 161 In other words these two equations are really one equation if w2 wg E wi 162 or instead if w2 w 21033 w 163 Borrowing some language from the mathematicians the physicist refers to wi and w as eigenvalues We will also refer to A and B as eigenmodes What is the physical interpretation of the eigenmodes To answer this we go back to Eq 159a or equivalently Eq 159b Remember they are the same equation now If we substitute w2 wi into Eq 159a we nd that af 1 24 164 where the superscript just marks the eigenmode that were talking about In other words the two masses move together in lock step with the same motion This happens when the frequency is w iwo i km and that is just what you expect It is as if the two masses are actually one with mass 2m The effective spring constant is 2k as you learned in your laboratory exercise where you had one cart with variable mass and a spring on each side Therefore we expect this double mass to oscillate with w2 Good This all makes sense Now consider eigemode B Substituting w2 wg 2w into Eq 159a we nd that a fag 165 In other words the two masses oscillate against each other and with a somewhat higher frequency In this case it is interesting to see the difference between we ltlt we in other words a very weak coupling spring and we gt we strong coupling spring For the weak 3A mathematician would say that we are setting the determinant equal to zero 51 coupling spring7 the frequency is just the same as we and that makes sense If the two rnasses aren7t coupled to each other very strongly7 then they act as two independent rnasses m each with their own spring constant k On the other hand7 for a strong coupling between the rnasses7 the frequency is w wc which gets arbitrarily large Sornetirnes7 this high frequency mode7 can be hard to observe A simple experimental test of this result is to set up two identical masses with three identical springs In other words7 we we In that case Lug Bwi and therefore the frequency for rnode B should be xg times larger than the frequency for mode A We should be able to make this test as a demonstration in class We have been talking about general properties of the two eigenrnodes Of course7 this doesnt tell you how the system behaves given certain starting values7 that is7 speci c initial conditions For this7 we need to understand that the two eigenrnodes can be combined ln fact7 the gen eral motion of each of the masses really needs to be written as a sum of the two eigenrnodes Each of these comes with a positive and negative frequency7 just as it did when we applied all this to the motion of a single mass and spring Recall that the sum and difference of a positive and negative frequency exponential7 are equivalent to a cosine and sine of that frequency lncorporating what we7ve learned about the relative motions of modes A and B7 we can write the motions as follows z1t Jew4t be iwAt CEMBt de wBt 166a z2t Jew4t be iwAt 7 06 7 de wBt 166b where 17 b7 0 and d are constants which are determined from the initial conditions Note that these four constants appear as they do in Eqs 166 because each frequency terrn must match up between the equation for z1t and z2t with the correct relative sign7 in order to solve the coupled differential equations An obvious set of initial conditions is starting mass 1 from rest at z 0 and with mass 2 from rest at its equilibrium position This leads to the solution z1t x02 coswAt coswBt 167 z2t x02 coswAt 7 coswBt 168 The last laboratory experiment for this course will investigate this motion Sorne number of setups will be using new electronics that allows each mass to be followed separately and simultaneously On to the plots Two cases are shown7 one with we omegao7 and the other with we LOO4 For each case7 we plot z1t with x2t and also z1t z2t with z1t 7 x2t The rst case shows clearly that the motion is somewhat complicated for either rnass7 but clearly splits into two eigenrnodes with rather different frequencies for the sum and difference The second case shows something much closer to a beat pattern7 with the energy shifting from one mass to the other and then back again 5 1O 15 20 25 30 031oa1 X1 X2 15 X X2 1 W E n n 39 HM W 1 U l l w Physwcs Honors Thursday 13 November F2 2008 quotRydsy we Sign ahew emean mm msde Tmuwe xwm mueuuae u ya by discusxmg the M9131 subsem 0 52mm Meme Ih 15 away 0 sumbhmg canle syseus by aveagng me their Prwmuesa ah ewmah thl should wek 1 they exe aembhaam eweu The meal Gas beus umsene abe lled wuh avey lsxga humbe dpexucles Thepemcles ere veysmau aha den mlmecl wnh eaeh mhex Ihee ee an many baxueles aha they exe mmng se smele yvuf 1 brassuuaemhewans qu ae s as39eady See the gure Put N gas mexe m aubmb dadslmg39eh 1 M uxehaszveleeuye h bemae 0 the w aha muus memeuum change m The ume bewee beuheas 5 2m se mum 2m ThemaeuuaemhewanamLus39hesumenhexees v de lhemnlaculesamvldedbylheexeadlhewe Ihaus 1 v N whee we xemynze ha he vnlumen hecuhe e v 13 New the uuahuw m beeeheee gsbaeax about the z uueaum aha mbe N1 veyxaxee se we expect my e5 e3 e3 aw whee we an e the Lbheah Square veeem she an mdecules have he same messa K mm 1 he awean bheue eee ex he mdecules see we aah hew wme ev NEW 170 sane 2m 2 huh 111 ch 1 mega m the es Fa Wean sesame he mmpmelure T e maas ed 111 dm mml uhus aahaa Km K The cmvelsm constant mm mpmelure m eeey e u be mm as mm mm and the value 1c 1 38 x 10mm We asey hanaum e 141 m aaeh e he three uuaeueus m whmh the uaxuee aah move h ehe wards K 3141 9V NIcT 171 2 as the Meet 5m m ye may have aheady see this In a husm aeese aeuh m amneeu way The water to mum moLexwhere amexe is Awdm s number Nu 6 02 x 1033 dbaxudes Thsl 15 N nNu whee ms lhenumhm afde Chemm s elxv wme R E NAka that is w mr mass 32 emu at m o e one 10 pa whee abasaax 1 one he at be suuee mee Islam humbe N exeeee mebaeuxes m the hug the ayeuaee 512291 1e WA am the aveaee humbe d cdllsm mm the be wans In one saaehu Pucmcal Aspects af Tmpmthhhe Thekmd extempehethhhewewehhhet de ned he teehea absolute 2mmtm Semethhhee ht he teehea the Kehmtemhhehethhhe eeehe As he ee phvshee he meeheh ht he the only hhha a tehhpehethhhe thet mettehe Thehe ehe mheh eehhhhehh mpehethhhe eeehee hheea hh etehydet Me ehthehhheh hehtheh he hheeh hhhheh by phyehehete The twe adhexyeeehee ehe pehhehheht TF eha oehehhhe Tc h e Tc 315 eha Th Tezz 172 Thhe he he hhemeeh hhthehhhethm te tempehethhhe xhhtheny people Wemed te meeehhhe hew oehehhhe eha body tehhpehethhhe eha freezing hmhht a thhhe h the tee ex thehheht Them e eeheee the hue phvsheex meehhhee tehhhehethhhe based eh hhtehheh eheheye eha the eheehhte tehhhhehethhhe etehe wee hem ht wee chm m hehe the eehhe ehhhe ex megeev ee ceehhhehhht hhht the monithesmh et the hxeeewheethehe wehhha he hehhehheh khhethe ehehey Sehhethhhee we teen thet pehht ehsnlum hehew Thmnal Expehehhh A Meteheh pmhmty that Used ta Deane Tmpmthhhe Meet ehhae expend whe they eet hm The heeeehh he ehhtethe max temp meehe hem h hheheehhheh hem Sdlds ehe hhe etehe ed by h 339 e i ttehhehhehhew ehhhe ehhh Mh eweheg heeh hhte h 2 heheh eepeh e the f ehey he did e The emeet he hetheh hheex end het Iexg bx tymteeh tempehethhhee R eh39hae we whte A a1 AT m whehe ch he teened the mm o thew WWW29nd expheeeee the hheetheheh ahsnga hh laugh AL1 degee 0f tehhpehethhhe ehehge end wehee mm ehhd te sdld ahee the hheethmehahexhgeheemehweweetm aAATmzwehemdemeeeexeeehd AV aaV AT bx wehhhhe Rehhehhheh m new expehehme Nae thet mud ahhe wetehh ehe mae eemphheeted theh mild phaeme Beeches 1 Determhhe the tempehethhhe et whheh the Fehhehhheht end oehehhhe eteehee eehhehae that he Eve the eehhe wehhe hr the mpehethhhe Heweyou eveh heeh ehhhtehdeehh easy eh mt wheh thhe wee the ehh tehhpehethhhev 2 miehh why ehhh demhhtheh K aJeT hr the tempehethhhe 0f eh haeeh gee wehhhi het hhehe eehee Ifwe whete ht hh tehhhe exTF eh Te hheteea 0f eheehhte tempehethhhe T 56 Physwcs Honors Monday 17 November F2 2008 amen kat and Hm We ere 3mm sum telhng shout SYSth mg the one Pm M m saumhnum state 1 e ch s1 mes Thegeshesmmmmx e mythstwewmm l Razenxdeelgs mama2 them ea 1 me e st thememwakmhedmmmegsbmemm This gure PMstsmevmsusvnlume mthegs m the I mm It s m ed a w dlsgem w m quenutxes hedmgdm dm memt wen mg eddmgm infmate gt r mg molecules a at the mess M 5mm m hm a me thet walk w s dune m m yaxhy the pm We have aw Fdh 7mm m 1 s thefaoe echngmthewt m Lhedn ecnm whmeAxstheexeem39thepxstm end man m that Wm Increases h Nowmxmdatehngthegsfmm we 5 WWW v vgmmn 99ehd v v4 elmgPeLh 1 The Pressure 1 We emaent my mLh mess the walk m m the 95 s pAV 175 9104 K In mntxest elmg Path 2 the vnlume nes ndt change so the wnzk dame elmgthm peLh mmmgm cmmems mm am mm mm News n39wegnfmm 5 th msteednielmgpelh Lhu l elh dm lnmA LhmedSSlccend thmhmkup k7 mmmmmmws WM mo Jam 7 v9 79m WM an We Sq the walk dme 1 gang mm one stale to ehdthew W1 depend on the path telen m the W mam We m m U 1x V and also we Wham T as Ytele New begusetheymlvdepemdmthevamxesdtheysetenymetxme kaWxshdte stete vexlshle 1t depends on the Path taken to at hetwem Mama states Next let s thmk shout whet hsppemed m the 95 m guinng B th elmgmlh 1 Fa SlmthlY assume 1t 1 en ee 95 Mm assume thet the 95 supP y Valve 1 turned aft 37 57 no gas molecules are added or taken away Then since pV NhT7 the temperature must have increased7 since p pf has not changed but the volume has increased from V1 to Vf Since the temperature has gone up7 so has the internal energy U The work done on the gas7 Eq 1757 is negative7 so that would have decreased the internal energy Where did the edtrd internal energy come from It came from the thermal reservoir7 by transferring energy in the form of heat Heat Q is a new concept It is a different way of transferring energy7 other than work W7 in a thermodynamic system Energy is conserved Always ln thermodynamic systems as well as mechanical ones Our expression of energy conservation for the gas in the cylinder can be written as Q W AU 177 This is called the First Law of Thermodynamics Note the signs lf heat is added to a system7 then Q is positive If the gas expands against the piston7 then W is negative Speci c Heat Capacity We pause for a moment and talk about heat in a general sense This will be useful later Add heat and the internal energy goes up7 so the temperature increases For solids and liquids7 the volume doesn7t change7 so work is not an issue For all materials7 the amount by which the temperature increases depends on the mass and the speci c properties of the material This is all collected into the de nition of the speci c heat capacity 0 namely 1 Q 77 178 C in AT where m is the mass of the object7 Q is the heat added or lost7 and AT is the change in tem perature The speci c heat capacity is tabulated in many places for lots of substances We write 0 me for heat capacity77 Clearly7 0 depends on how much matter is present Speci c heat capacity depends on temperature7 but it is not a strong function of temperature for most substances for common temperatures Under these conditions Tr Q mT ch mcTf 7 179 For an ideal gas under constant volume7 Q AU 32NhAT Therefore 0 32Nh 32NNANAk 32nR We write CV 3R2 for the molar heat capacity77 at constant volume Real7 monatomic gases show molar speci c heat capacities very close to this value Diatomic or polyatomic gases show higher values7 re ecting the additional degrees of freedom in those molecules For an ideal gas that expands under constant pressure7 W 7pAV 7NhAT Therefore Q AU 7 W 52NhAT and the molar heat capacity at constant pressure is Op 5R2 Again7 real monatomic gases agree with this very well In addition7 one expects that Op 7 CV R 831 Jmol K We commonly write Op 5 E 7 7 180 7 CV 3 for a monatomic gas A table is included which demonstrates these concepts 58 Cychc pmesses Ah thductmh The world mehe em duse a soeeued wemteeee m them dynsmms These ee Pimesses thet epeete m eeyde enmgup et the me pleee mm which they stet We39ll do me wmh thee hext dsss mt 1AM Stan wmh esxmple ee end see whet we eeh xeexh hem ht The gureshnws ecyehepheeee for eh haeex Es 1t gas fusn mmA msbymuemgthepzesshheetmstehtmh t the hut heet Qt end Inwmng the tehpehethhe A I h the the val eet steht meme V k m whheh eee heel Q2 1 added lorelse the temp me I 2 he dame eh t W h f A by miuamg the vduhhe eha hheeesmg the Pressure hh V such away thet the tehpeetuhe hehehs mteht h w h she the mm s cyehe the ehextge m heme eexgy exumd the eyde must he Tee Iheehee by the rst yew 0f Lhamndynemwi 2 w o whee Q end w epheeht the het weh exound the Me 5 he m the gure W2 lt 0 Fee a 175 oh the mm hende W gt 0 me the vdume aeeeeses Rhythemeewa gt we smee the eeeuhde the curve de ned by path 3 e hege theh thet m peth 2 The weh dme elmgPeLh 1 e Thameelhene twmkxswW2Wagt0 endsc hem theelQlt0 I39hslxsenemwmk he done eh ther eha heet xeeves ht We teen Such ethmgewnyemzoh mhsteex the Me epeetea amuse the heet e edded end the 95 he weh eh the 12mm We ten the eh W These conclusions wm always be the tee a cydm Pimessese Xegxdless 0f the curves thet hound them Iheemsteht valumEPelh 3 e eehexeh not72m It allows the ewe e IcTaVe whee euthteeeheehmw heme weth eegtmee VG MemeVA 7 M47 gt0 m 3 She the Path s eh mhem edeheuc 1t he eheeetened by Q o 0 m eetuhe eh ncw The haeel Es lew gves my gar Dmdmg these gves us dz 7 t g e V V m hteeethgthe ehhethm e We Fa hhhts celled A em 5 t the say the my 7 V5 7 VA 7 y e 7 e we e he e w e emteht m Th e the equetm thet aeenhes eh edeheuc path eh epv dmgem 59 Physics I Honors Fall 2008 Useful Tables for the Study of Thermodynamics Taken from Physics7 by Halliday7 Resnick7 and Krane7 Fifth Edition Heat Capacities of Some Substances COE iCIents Of Lmear EXpanSlon Substance Speci c Heat Capacity Jkg K 76 W Lead 129 Tungsten 135 Lead 29 Silver 236 Aluminum 23 Copper 387 Brass 19 CO 17 Carbon 502 pper Aluminum 900 Steel 11 0 d 1 9 Brass 380 Pr mar E 335 3 2 Granite 790 yreX g ass Glass 840 lnvar alloy 07 Fused uartz 0 5 Mercury 139 q Water 4190 Molar Heat Capacities of Ideal and Real Gases Op V p 7 V Gas Jmol K Jmol K Jmol K y Monatomic ldeal 208 125 83 167 Helium 208 125 83 167 Argon 208 125 83 167 Diatomic ldeal 291 208 83 140 Hz 288 204 84 141 N2 291 208 83 140 02 294 211 83 140 Polyatomic ldeal 333 249 83 133 002 370 285 85 130 NHg 368 278 90 131 Physics Honors Fall 2008 Homework Assignment Due Thursday 20 November 1 The following diagram shows the problem in coupled oscillators that we solved in class gtX1 k m Using the parameters wo E km and u E kcm nd the motions z1t and z2t for the following sets of initial conditions a x10 x20 A and 210 i20 0 b x10 x20 0 and i10 0 V c x10 x20 i20 0 and i10 V 7 iiz VA 2 As a result of a temperature rise of 32 Cquot a bar with a crack at its center buckles upward as shown L0 in the gure on the right The distance L0 377 m between the bar supports is xed and the coef cient A of linear expansion of the bar is 25 gtlt 10 6K Find m 7 A the distance to which the center rises 4 LD gt 3 In an experiment 135 moles of oxygen 02 a diatomic molecule are heated at con stant pressure starting at 11 C How much heat must be added to the gas to double its volume 4 Gas within a chamber undergoes the process shown in the pV diagram below 40 30 I 4 T a f 0 g 20 x m k 1 O 0 0 1 2 3 4 5 V L Find the net heat in Joules added to the system during one complete cycle Physics Honors Thursday 20 November Fall 2008 Review Statistical Mechanics and Thermodynamics Concentrate for now on gases For a monatomic ideal gas7 the equation of state77 is 2 pV KNkTnRT 184 which relates the state variables p7 V7 and T to the number of molecules N through a fundamental constant k The internal energy U N32kT NltKgt where K is the average molecular kinetic energy The 3 counts the number of degrees of freedom7ie three because all a monatomic gas molecule can do is move in three directions lf7 for example7 the molecule is diatomic and can rotate7 there are ve degrees of freedom If the gas container has movable walls7 then work W can be done on the gas If it is in contact with some kind of thermal bath7 then heat Q can be transferred into the gas These change the state of the gas by an amount the depends on the path taken between the states The rst law of thermodynamics says that AUWQ 185 where by convention7 W is the work done on the gas In effect7 this equation de nes Q Entropy A New State Variable Now we de ne a new state variable S called entropy Actually7 rather than de ning entropy itself7 we7ll de ne the change in entropy AS between two states We have As 186 T reversible The notation implies that the path for the integral is reversible that is7 connected by a series of equilibrium states For example7 AS AQ T for a reversible isothermal path stopcock closed Vacuum If System Here7s a trickier example7 a class problem of free expan sion77 The walls are perfectly insulated7 so no heat comes in to the gas Opening the stopcock lets gas molecules move into the evacuated chamber7 until there is equal density everywhere What is the change in entropy 39 S 39aquot quot 39 a Initial state i You are tempted to say AS 0 since AQ 0 However7 the 39 Irreversible process shown is not reversible We need to compute AS process f dQ T along some reversible path Since no work is done and no heat is transferred7 AU 0 So7 choose an isothermal expansion from Vi to Vf Therefore dQ dW pdV and stopcock open Vf dV Vde V 3 A 19 f ASi T Nk V NklnltVigt gt0 187 Vi V b Final statef 2405 62 The Second Law of Thermodynamics Note that free expansion of a gas Eq 187 always yields an increase in entropy This is a system completely isolated from its surroundings and the expansion process is irreversible ln a reversible process like isothermal expansion or compression heat is transferred through a common thermal reservoir so we don7t expect any change in net entropy if we include both the gas and the thermal bath with which it is in contact So now we state the Second Law of Thermodynamics namely When changes occur within a closed system its entropy either increases for irreversible processes or remains constant for reversible processes lt never decreases ln other words AS 2 0 Adiabatic Processes in an Ideal Gas A process carried out in thermal isolation is called adiabatic lt is characterized by Q 0 so W AU 32NkAT 32nRAT nCVAT lf the temperature change is in nitesimal then we can write dW 7pdV nCVdT The ideal gas law gives dpV pdV Vdp anT or Vdp nCVdT anT ndeT Dividing these gives us Vd C dT C d dV pdV nCVdT CV 13 V lntegrating this equation is simple For limits called A and B it just says that PB VB VA 7 lo 7 7 lo 7 lo 7 or V7 constant 189 gltpA v gVA gVB p gt This is the equation that describes an adiabatic path on a 12V diagram 188 The Carnot Engine The gure on the right shows a cyclic process which is bounded by two isotherms at two temperatures TL lt TH and two adiabats which connect those two temperatures This is called a Carnot cycle The cycle proceeds clockwise so negative work is done on the gas from A to B and heat QH enters the gas During the isothermal contraction C to D the work is positive and so heat QL leaves the gas Remember that AU 0 along an isotherm and Q 0 along an adiabat Since AU 0 around the cycle we must have V WDAWHQHWBCWLQLO 190 0 VA VD vB VC 2408 Now WH lt 0 and WL gt 0 Futhermore since paths H and L are isotherms WHQH 0 WL QL Therefore WDA W30 0 and also QH 7WH gt 0 heat enters along H and QL 7WL lt 0 heat exits along L Since by areas lWLl lt lWHl we have lQLl lt lQHl so net heat enters the gas Furthermore the work done around the cycle is WWDAWHWBCWLWHWLlt0 191 and the gas does work on the piston Heat in work out This is an engine 63 How ef cient is this engine In other words7 how much work do you get out7 for heat that you need to put in Let us de ne an engine ef ciency just that way7 namely W W W 7 6El ll H LllQHl lQLl17lQLl 192 lQHl lQHl lQHl lQHl We can go further by realizing that entropy change is zero around the cycle That is QH QL lQHl lQLl AS7 77770 193 TH TL TH TL Therefore lQLllQHl TLTH and the ef ciency of a Carnot engine is TL i i E 1 7 T7 Carnot engine ef c1ency 194 H Of course7 operated counter clockwise around the cycle7 a Carnot engine becomes a Carnot refrigerator Instead of e ciency we speak of coe cient of performance This is the ratio of what you want ie the heat QL taken out of the low temperature reservoir7 and what you pay for7 namely the work W you have to do on the gas We write g lQLl TL lWl lQHl7lQLl THeTL A very large coef cient of performance is needed for a refrigerator that operates between two temperatures that are close to each other K Carnot refrig coeff of performance 195 There is no such thing as a perfect engine7 ie one with E 1 This requires TL 0 for a cold reservoir7 that is one with no internal energy There is also no such thing as a perfect refrigerator7 ie one with an in nite coef cient of performance This would mean that lQHl lQLl E Q and AS QTH 7 QTL lt 07 which Violates the second law These statements can be put together to show that No real engine can have an ef ciency greater than that of a Carnot engine working between the same two temperatures77 That is7 Eq 194 is the most ef cient an engine can ever be In other words7 by Eq 1927 any real engine must always discharge some heat QL to its surroundings Don7t Forget Short homework assignment due in class on Monday Physics Honors Fall 2008 Homework Assignment Due Monday 24 November 1 Two identical blocks7 each of mass M and made from a material with speci c heat capacity 07 are kept thermally isolated from each other and from their surroundings One block is at temperature T1 and the other is at temperature T2 gt Tli The blocks are then put in thermal contact with each other7 but still thermally isolated from their surroundings 8n 5 Use the relationship between heat transfer and speci c heat capacity to prove that the nal equilibrium temperature of the two blocks together is T T1 T22i Find the change in entropy for the system consisting of the two blocks You can imagine a reversible process by which one block is taken slowly to its nal temperature7 and then the other block is taken to this same temperature7 but contact with a slowly changing thermal bathi 2 The gure on the right shows a cyclic 43900 process7 plotted as a TS diagram7 rather 39 than a pV diagrami 300 7 2 77777 77 ai Explain why this is a Carnot cyclei V b Calculate the heat that enters H 200 ci Calculate the work done on the gas 7 l 00 O l l 0 02 04 06 s JK 241 9 Ludwig Boltzmann is one of the few people who understood en tropy He came up with a way to understand it in terms of disor der which he wrote as W His result in the form of the equation S k log W is etched on his grave stone LVDWIG BOLTZMANN 18514 16 66 Physics Honors Monday 24 November Fall 2008 Remember that homework and nal lab notebooks are due on Monday 1 Dec How to Build a Star Hydrostatics of really big ball of gas Start with the balance of pressure against gravitation Then use ideal gas law to relate to temperature Will assume that stars like the sun have uniform density Finish with the behavior of white dwarf stars aka degenerate Fermi gas More reading in order of increasing sophistication 0 Introductory Astronomy and Astrophysics 4th Edition Zeilik and Gregory esp Chap10 o Order of rnagnitude theory of stellar structure George Greenstein Am J Phys 55 804 1987 Erratum Am J Phys 56 94 1988 0 Stars and statistical physics A teaching emperience Roger Balian and Jean Paul Blaizot Am J Phys 67 1189 1999 Begin First set up the notation Build our star with shells of radius r Total mass of star is M and radius is R Density is p M47TR33 assumed to be independent of r Write Mr be the mass enclosed at radius r ie MR p47rr33 MrR3 Now balance pressure pr against gravity Take a small piece of area A thickness dr and mass dm pAdr at radius r Outward force from difference in pressure must equal the gravitational force from mass inside radius r Recall shell theorems Write it out pAilt10dpgtA GMrdm GMr2pAdr r r 7 i MUMU dr 7 G T2 196 This equation expresses hydrostatic equilibrium for a large spherical self gravitating object like a star or a planet or a moon It is written here in a way that holds for any density function pr but we will be taking p to be a constant Let7s use this to determine the pressure p0 at the center of a star ie r 0 Make the replacements for constant density in Eq 196 to get R dpi r3 M 17 3GM2 g 7 3GM22 dri lt gt 47TR33T27 47TR 3T pr 7pc 87TR6T At the surface of the star the pressure is zero by de nition ie pR 0 Therefore 3GM2 87TR4 00 197 Next let us try to estimate the temperature at the center of a star This requires us to know the equation of state which relates the gas pressure and density to the temperature for the matter that makes up the star Understanding the equation of state is a big deal and 67 an area of active research interest For now we will use an equation of state based on an ideal gas that is Eq 184 namely pV NkT If we assume that the star is made of N identical particles with mass m then p NmV and p kam 347T MmkTR3 In fact stars are hot so hot that matter is a plasma of free nuclei and electrons The nuclei are from mostly hydrogen about 25 helium and a few percent of heavier elements For now though it is close enough to assume the star is 100 hydrogen atoms Therefore the temperature at the center of a star is T 7147139R3m i1GmM C k 3 Mpc k 2R 198 Can you see why we would not be able to get this answer from dimensional analysis The White Dwarf Star We found some properties of stars assuming they consisted of an ideal gas but the as sumptions used to get Eq 184 don7t always hold Instead we need another equation of state In deriving Eq 184 we wrote Fm AgoAt 2pm2Lvm and that is still ne For N particles this gives a pressure P NFmLZ NpUV We temporarily switch to P to avoid confusion with momentum Also we are making a rough estimate so put pm and 111 to p and 1 Our concern will be with electrons so write an electron density 715 NV in which case P nepu At very high densities the ideal gas law assumptions are violated The electrons are so close together they start to violate the Pauli Exclusion Principle The distance between them is d 1n5 3 and the Heisenberg Uncertainty Principle says that pd 71 so that p Q I m 3 Putting 1 pme and switching back to p for pressure the equation of state becomes 712 2 12 712 p 53 10 He Mi3 hiyame 7771 N E715 mfg E 199 where n is the number density of what would have been atoms and we make the somewhat incorrect assumption that the star is still made up of hydrogen atoms with mass m This is an odd sort of equation of state Combined with Eq 197 it leads to a star whose radius decreases with mass like 1M13 As the mass increases and the radius decreases the electrons are more and more con ned until they are moving as fast as they can ie 1 c The equation of state becomes 43 p n5 full3 c he ngZl N ha 7143 he 200 m Combined this with Eq 197 gives an equation where the radius drops out One solves for a mass that is the highest possible mass of a white dwarf star Doing a more careful job leads to a value of 144 solar masses This is the Chandrasekhar limit77 after the physicist who rst derived it What do you suppose would happen for a star of mass larger than this Physics Honors Fall 2008 Homework Assignment Due Monday 1 December 1 This problem asks you to carry through some numerical calculations that lead to an understanding of what it is like in the center of the Sun The following web sites contains useful data httppdgbgov2008reviewsconsrpppdf httppdgblgov2008reviewsastrorpppdf a Use Eq 197 to estimate the central pressure of the Sun Express it in units of atrno spheres7 where one atmosphere is the air pressure at the surface of the Earth F7 Use Eq 198 to estimate the central ternperature Express it in Kelvin 0 Recall that temperature is related to the average kinetic energy of the gas particles What is the average kinetic energy of the hydrogen atoms in the center of the Sun Express the answer in electron volts eV What does this tell you about the kind of matter in the center of the Sun A typical binding energy77 of an electron in an atom is around 10 eV 2 Show that Eq 199 leads to a star77 with radius R and mass M where R X 1M13 Physics Honors Monday 1 December Fall 2008 Last class I will miss all of you7 but dont be a stranger Hand in lab books Third midterm exam is this Thursday Final exam is Wednesday7 3 Dec 3 6pm7 this room Conservation Laws and the Continuity Equation Our last class will concentrate on how to write down a general conservation principle Then7 we7ll apply this to uid dynamics The basic idea is simple lmagine some substance that is conserved In other words7 it cannot appear or disappear unless it is somehow supplied If we talk about the amount Q of this substance in some region R of space7 then the only way to change this amount is to let some ow into or out of the region Make this statement mathematical The rate of change of Q is dQdt De ne some quantity J Jrt to be the ux density77 of the substance That is7 it is the amount of the substance owing perpendicularly through some small area per unit time So7 through some small surface dA dA 7 an amount J dA JdA cos ows per unit time Here7 fl is a unit vector perpendicular to the surface7 and b is the angle between f1 and J If we let 8 denote the surface of R7 we then write our conservation condition as in EiijiJdA 201 The negative sign is there because of the convention that elements of area dA point outward on S Therefore7 if the integral is positive7 there is a net ux out of R so Q decreases Now let p pr7 t be the substance density Then Q is the integral of p over R7 and dQ 7 d 7 6p RigRMn dViRadl 202 The derivative becomes a partial derivative inside the integral7 because all we care about is the explicit dependence on time That7s all that would be left after we carry out the integral over space For the right side of Eq 201 we will make use of a result we rst mentioned in Eq 70 Imagine that R is small and rectangular7 with one corner at x y z and with sides of length dz dy7 and dz In this case fJ dA Jmz dyzdydz 7 Jmxyzdydz s Jyxy dyzdxdz 7 Jyyzdzdz Mtg2 d2d96dy Jz7y72ddy Jz d 7 7 Jm 7 7 wride dx 61m My 611 7 7 d d d 203 6x6y62zyz 70 A J dAdt In other words Jr t pv is the amount of mass owing perpendicularly through some small area per unit time The gure shows ow from P to Q in a closed region Specialize to p pr independent of time Then Eq 201 says that the integrated ux over the surface is zero Assume the areas A1 and A2 on the right are small so that the velocities V1 and V2 do not vary much over them and are perpendicular to them Then pllel pzvaz 208 If the uid is incompressible liquids that is the density prt constant everywhere then the continuity equation becomes UlAl 2142 209 1603 le water ows faster in constricted pipes 71 Physics Honors Fall 2008 Practice Exercises on the Continuity Equation 1 A pipe of diameter 345 cm carries water moving at 262 msec How long will it take to discharge 1600 m3 of water Answer 1h 49 min 2 The figure below shows the flow of an incompressible liquid through a straight pipe with a constriction midway between the input and output r 4 quota x 5 1 7quot Q gt I i iij l 604 Let 2 measure the distance along the axis of the pipe with 56 and y perpendicular to 2 What are Jag and Jy at values of 2 near points P and Q For a value of 2 near the midpoint between P and Q ie in the middle of the constriction make a sketch of Jag as a function of 56 Use your sketch and Eq 207 to explain why v2 gt v1 3 The figure shows the con uence of two streams to form a river One stream has a width of 82 m depth of 34 m and current speed of 23 msec The other stream is 68 m wide 32 m deep and flows at 26 msec The width of the river is 107 m and the current speed is 29 msec What is its depth Answer 39 m 1627 4 The figure shows a familiar effect namely the necking down77 of a stream of water when it falls under gravity Water adheres to itself so that no pipe is necessary to keep the mass continuous Use the continuity equation and conservation of mechanical energy a to explain this effect If the cross sectional area A1 is 12 cm2 and that of A2 is 035 cm2 and their separation h 45 mm at what rate B does water flow from the tap Answer R 34 em3see l l A 72 Honors Physics Monday 27 August Fall 2007 First class Name and course on board read roll introduce TA7s Distribute outline lecture and lab schedules Discuss them New material for today Start thinking like a physicist Two new ideas namely 1 Solving problems with dimensional analysis and 2 F ma is a differential equation Dimensional Analysis Given a problem ask What is the important physics7 Then list the quantities that have something to do with that physics In many cases you can combine these things to get the answer at least to within a factor of 2 or 7139 or something like that Convenient notation distance L time T mass M temperature K Err ample What is the radius of a black hole This is a combination of gravity and relativity so use M G and 0 Let the radius be R Mwacz where we need to nd m y and z Newton7s constant G has units m3kg sec2 so G LSM IT Z This gives R ML3yMin72yLzTiz MmiyL3yzT72yiz L1 so z y 3y z 1 and 2y 2 0 The second and third of these say that y 1 and so z 1 and z 72 Therefore we get R GMcz The correct answer is GMCZ called the Schwarzchild Radius Hint Useful website for units and constants is httppdgb gov2007reviewscontentssportshtmlconstantsetc Newton7s Second Law as an Equation of Motion The most important of Newton7s Laws is the second For motion in one dimension we write F ma well get to F m6 later this week Physicists refer to this as the equation of motion for a particle moving through time in one dimension of space Let t be the position x of a particle at time t Then the velocity is 11t dzdt E 1 Have you all seen this notation before Acceleration is at dvdt i So Newton7s Second law can be written as z Fzt 1 where we make it clear that the force F may itself depend on position andor time Equation 1 is called a differential equation Unlike an algebraic equation whose solution is a value for a quantity like say x a the solution of a differential equation is a function say Since t describes the motion of a particle with mass m acted upon by a force Ft we call Eq 1 the equation of motion Simple example Let Fzt F0 a constant independent of both z and t Then i i 7 d1 7 F0 7 7 dt 7 m dx F it E gota o0 1F pt EEO not no where the last line is the solution to this equation of motion This might look more familiar to you from your high school physics course if we write a FOm You probably called this motion under constant acceleration but of course a constant acceleration implies a constant force so its the same thing Each step in the integration we did involves a constant of integration 1 called these constants 110 for the rst step and x0 for the second step These are standard notations and in fact good ones The constant 110 is the velocity at time t 0 also called the initial velocity Similarly 0 is the initial position In order to solve any equation of motion we will need to specify these initial conditions Practice Exercises 1 Astronomers observe that distant galaxies recede from us at a nearly uniform rate That is a galaxy at a distance d recedes at a speed i Hd where H 75 kmsecMpc is called the Hubble Constant The mega parsec Mpc is a standard cosmological measure of length equal to 31 gtlt 1022m What is the age of the universe You are welcome to make estimates instead of looking up exact values but that is up to you Did you know that the number of seconds in a year is very close to 7139 gtlt 1077 2 Let yt describe the one dimensional motion of a particle with mass m that moves vertically Assume that the particle is acted on by a constant force F 7mg Isn t that peculiar a force that is epaetly proportional to the quantity in that appears in the equation of motion Assume that the particle starts out at y 0 with an initial upward velocity V Find the maximum height to which the particle climbs in terms of m g and V Make a sketch of yt as a function of t What is the shape of the curve Mathematicians have a name for it and 17m sure you7ve heard it before Honors Physics I Thursday 30 August Fall 2007 Today we will work with Newton7s Second Law in terms of vectors7 ie 13 m6 First7 however7 lets go over some preliminaries Calculus Tools Rules we will use I will try to review the calculus rules before we use them Your math class will show you where these rules come from Last class used t ntn l which means the same as ftndt We will worry about the case n 71 later We also used a t 1 where a is a constant This class lm0 3437 Chain Rule and w v Jrqu Product Rule Vectors in 2D for now See Textbook Appendix H Just scratching the surface of an important concept in physics7 but right now you can think of vectors as shorthand for working in two or three directions at the same time Notation is a unit vector in the z direction for z So7 we write a vector 6 am ayj The two values am and ray determine the vector 55 Draw a picture ofa stick vector with components labeled on my axes like Fig2 2 Examples Fziy F17vmivyj iier The length of a vector 6 is l l a gal2 That7s enough for now well develop more concepts with vectors as we need them later Newton7s Second Law again f is really two separate differential equations Fm mi and Fy my Solve for the motion by considering the equations separately Hopefully that7s easy to do7 but it depends on F Sometimes Fm can depend on y and vice versa That can make things hard7 but it can also make them interesting Maybe different coordinate systems will help The standard example is projectile motion7 namely Fm 0 and Fy 7mg Section 4 3 The solutions of the two differential equations follow very simply from last class t 0 mot 1 and yt yo vyot Egt2 So7 we have solved the equation of motion For any time t we now know where the particle will be located It is at Ft tz39 ytj Of course7 to answer any question numerically7 we need to specify the initial conditions F0 moi ij and 170 010239 11 So what path does the particle follow Simplify by setting 0 yo 0 That is7 de ne the origin to be the place from where the projectile is launched Then it zvmo and 12110 1 g 2 7 777 2 y M096 21 The path is a parabola7 which passes through the point my 00 The range of the projectile is the other value of z for which y 0 Something new Lets talk now about a special class of forces These are called central forces because they always point back to some central point y F y Diagram of a central force in two dimen sions The geometry of the rectangles makes it clear that FyFm ys7 or equiv F alently Fy 7 yFm 0 X X Consider a new quantity which I will call Z7 which is de ned as follows 6 2 m 961 7 yet lt3 Since s t and y yt are functions of time7 we also expect Z Zt However7 dZ i i i i Emyxyiy7yz mzy7y szinm0 The time derivative of Z is zero In other words7 Z does not change with time This is your rst example of a constant of the motion We say that Z is conserved Our derivation shows that this is true for my central force Picture of stuff to come Z is called the z component of angular momentum Indeed7 it is the z component of the three dimensional vector L E F gtlt We7ll explain cross products and angular momentum in more detail later Practice Exercises 1 Find the range R of the projectile whose arc is given by Eq 2 Also nd the maximum height h which the projectile achieves Try to express your answers in terms of the initial speed 00 E l ol and the launch angle b tan 1vyOvmo 2 Consider a one dimensional force F ikz Show that E E miz ksz is a constant of the motion Honors Physics Thursday 6 September Fall 2007 Hand in homework Pep talk Don t be scaredll Recommend Schaurn s Math Handbook TooLay Kinetic potential and total mechanical energy First Any questions on the labs what we re looking for etc Introduction Practice Exercise 2 from last Thursday 30 August For F k show that E E 771332 k is a constant of the rnotion 7 Go slowly a mm km ma kJJ ma F 0 So E is conserved This E just works for this k but it is obvious how we generalize this De ne U so that dUda Now E E 771332 Then mdsi ma39 dE dU dUdx i dt dt d9 dt dU F mdgt ma 0 We call 771332 E K the kinetic energy 7 U is the potential energy and E is the total mechanical energy Only E is conserved The Complete Solution to motion in one dimension Emphasize that there is an alternative to solving the equation of rnotion 7 if one discusses motion in terms of energy 1 E E m2 U 2 E U 4 1vj m Two ways to look at this a Velocity can be to the left or right but speed 7 E is determined The square root leads to turning points to the motion which limit the position based on the energy 14 53 139 E2 Energy Textbook Figure 12 8 Label K abd U at some position 3 Relate force plot to the energy 10 plot Discuss behaviors of particles with var ious E and mention oscillations Draw the analogy with up and down a hill 7 1 a u o a Neutral Stable Unstable equilibrium equilibrium equilibrium b In principle one can get the motion t by integrating Eq 4 We ll do this in a practice problern Work and Energy in one and two dimensions Line integrals Traditionally we start talking about work before talking about energy because this has some practical use in engineering applications and because it helps make the distinction between conservative forces for which a potential can be de ned and non conservative forces like friction for which it cannot Math concept Dot Product Textbook Appendix H 4 A B AmBm AyByAsz AHB cos b where b is the angle between A and B Note that 1 and 0 so this all works ne if we write A Ax Ag and B B1 By and just multiply Work is force applied through a distance 5 Successive ways to write work W Constant force in one dimension W F5 Variable force in 1D over small distance dr dW Fdz Variable force in 1D between points a and b W dW Constant force at an angle b W F5 cos b 139 3 Variable force over a small distance d dW d Fmdx Fydy Variable force between points a and b W d Variable force for a closed loop path W d This is a line integral T Line integral over a closed path The second relation makes it clear that U 7Wx but there is an important subtlety lf potential energy is to make sense it cannot depend on the path one takes to get from point a to point b In other words we can only de ne a potential energy Uz for a force a with the property f 1 d 0 that is the work done around a closed path must be zero We call such forces conservative because we can come up with a Uz for them and therefore talk about conservation of mechanical energy Friction or drag or things like that is the prototypical non conservative force The work done by a constant frictional force f is W ifs regardless ofthe direction of travel Clearly there will be nonzero in fact negative work done over a closed loop Practice Exercises 1 Use Eq 4 with Uz mgx and E mgh to nd Assume that z h when t 0 This describes the motion of something falling from rest at a height h right Split dzdt so that you get fdx on the left and fdt on the right It should be obvious after integrating that the constant of integration is zero for this initial condition You7ll need the following obscure calculus rule dx 2xa bx xa bx T b where a and b are constants 2 Show that the force is non conservative by integrating the work around a closed loop and showing that the result is non zero Pick a simple loop for example a rectangle parallel to the z and y axes and with one corner at 00 and the other corner at 11 Try the same thing for and show that it is conservative Honors Physics Thursday 6 September Fall 2007 Solution to Practice Problem 2 Problem Show that the force y is non conservative by integrating the work around a closed loop and showing that the result is non zero Pick a simple loop for example a rectangle parallel to the z and y axes and with one corner at 0 0 and the other corner at a b Try the same thing for Fzy and show that it is conservative Solution Divide the rectangular loop into four sections I II III and IV as shown below YA 0 4 We can therefore write the work around the closed loop as W f d d Fd II I II F d Pd 5 III IV and do the integrals one by one This is a good idea because d is parallel to either z or y in each of these four sections H 00 an X First consider the force yi that is Fm y and Fy 0 Along paths II and IV we have d idy respectively Therefore P d iFydy 0 along each of these two segments Also along path I d d so P d Fmdx ydz However y 0 all along path I Therefore the integrals along lines I II and IV in Eq5 are all zero V On the other hand along path III F d iFmdx iydm ibdx Hence Fug aibdz fab 7r 0 III 0 Therefore W 31 0 around the closed path and is non conservative Now do Once again the integrals along lines II and IV are zero however 13 d azdx and P d a7d I 0 III 0 You can do the integrals if you want they are iaZQ but it is already clear that they are the negative of each other and will cancel in Eq5 Therefore for this force the work done around the closed path is zero At some point in the future you will learn about a mathematical concept called curl77 and something called Stoke7s Theorem These will be more formal ways of evaluating these sorts of integrals around closed loops Honors Physics Monday 10 September Fall 2007 Some Recap Energy Work and Line Integrals Ask for questions re solution of Practice Problem 2 from Thursday Uniform Circular Motion First let s talk about how to describe the motion ie kinematics Then we ll talk about the forces that can bring about this motion ie dynamics By uniform circular motion we mean motion in a circle with constant speed 11 The velocity however is constantly changing in direction Since velocity is changing the acceleration is nonzero even though the speed is constant Demonstrate with the pink string and small mass Where s the force Does it do work This is a good example of how to analyze a derivative in physical terms as opposed to mathematical ones Refer to the gure below The acceleration vector is d1 A17 172 771 Z 39 a dt Alimo At 0136 297nm Take it slowly I Y m 9 P P2 v2 Textbook Figure 4 16 showing the geome p v1 9 try used to physically nd the derivative ff 2 V2 d1 dt The speed 11 of the particle moving 2 in the circle is the magnitude of any velocity vector along the path The circle radius is r and the path length subtended by any angle gz in radians is just rgz The inset shows that A17 1 points towards the center of the circle and 2 has mag nitude 2 x 1 sin 6 E The figure shows that 5 points towards the center of the circle That s important The geometry also allows us to write 2 sine i v2 1 2vsin9 v a a 1m 1m i 0 gt0 29rv r 0 gt0 9 7quot Later when we study Taylor Series and their applications we ll see that sin 96 gt 1 as 9 gt 0 but it s easy enough for you to check Use your calculator to nd sine for small values of 9 in radiansl You ll see that as 9 gets smaller and smaller sine gets closer and closer to 9 itself Now that we know the kinematics of uniform circular motion we can describe the physical processes behind anything moving in such a circle Applications include motion on race tracks tension in spinning mechanical systems binary star behavior and dark matter and black holes in galaxies to name a very few Today we ll apply it to the conical pendulum One Application The Conical Pendulum Demonstrate it with the pink string and large mass Emphasize how the behavior is very different from that of a swinging pendulum Talk a little about oscillations Here s an interesting case where the oscillations have nothing to do with exchanging potential energy for kinetic energy So maybe oscillation is a bad name for this phenomenon You say tomato Our goal Find h eriod P of the pendulum in terms of any parameters that matter Textbook Figure 5 18 The mass m exe cutes uniform circular orbits with radius R If the speed is 11 then the period is just the time it takes to execute one orbit namely P 27TRU So we have to gure out 11 in terms of R or perhaps in terms of the string length L and the hanging angle 6 since R LsinQ We gure this out from the acceleration a v2R directed towards the center of I the circle Since we know that F ma we just need to gure out what the force F is Clearly this force is provided by the a m a tension T v The component of the tension that is directed towards the center of the circle ie in the direction of i is just F Tsin 6 But the upward component of the tension must balance gravity mg so my T cos 6 This gives 2 F mg tan6 771 and v2 Rgtane 2 IL and so P 7TR 27 R 27 C089 11 gtanQ 9 When we learn about the swinging pendulum you ll see that the period is to a good ap proximation 27nL g It is interesting to note that the conical pendulum period approaches this value as the cone angle gets very small Practice Exercises 1 Ignore our derivation of centripetal acceleration That s the name given to the ac celeration towards the center in circular motion Instead assume that something moves with a speed 11 in a circle of radius r and use dimensional analysis to make a guess at the acceleration How does it compare to the right answer 2 See Exercise 5 43 An object sitting on a horizontal surface will experience a force of static friction f Mmg that resists horizontal motion A small object sits on a turntable 13 cm from the center When the turntable runs at 33 1 3 RPM ie revolutions per minute the object stays put However at 45 RPM the object ies off Derive limits for the quantity u called the coefficient of static friction You may have seen turntables which run at these speeds in museums or perhaps in your parents closets Honors Physics Thursday 13 September Fall 2007 Hand in homework Today Newton s philosophy gravity circular orbits and dark matter Mass What a concept Is it just F a inertia or does it have a deeper meaning gt Gravity Newton was brilliant Recommend Gleick s book Newton s Law of Universal Gravitation Two point masses m1 and 7712 separated by a distance r attract each other along the line between them The magnitude of this force is m m a g or all together we write F G7 F G where G 667 X 10 11Nm2kg2 is called the gravitational constant and was first measured by Henry Cavendish in 1798 with a torsion balance See textbook Fig14 5 The earth is nearly a sphere with radius RE For an object with mass m on the earth s surface use Shell Theorem see below to get force from gravity That is mME an R3 mg 9 R33 Easy to see how this scales with planets of different size and density since M p gnRg Note the old adage Cavendish weighted the Earth The Shell Theorems Shell Theorem 1 A uniformly dense spherical shell attracts an external particle as if all the mass of the shell were concentrated at its center Shell Theorem 2 A uniformly dense spherical shell exerts no gravitational force on a particle located anywhere inside it See your textbook for the proofs Good example of intellectual usefulness of calculus Gravitational Potential Energy Textbook Figure 14 11 Gravity is a con servative force The work W f F d done by gravity is independent of the path No work is ever done on a circular arc and the same work is done when chang ing the radius r regardless of the angle Since I can make up any path by crawling along arcs and changing radii my assertion is true To find the potential energy due to gravity just realize that F dUdr where mM r Ur G for a mass m attracted to a mass M at a distance r Classic application is escape velocity Imagine launching something straight up from the surface of something with mass M and radius R How fast do I need to send it so that it barely escapes ie makes it in nitely far away with zero velocity Total mechanical energy is conserved ln nitely far away the gravitational potential energy is zero and it is not moving so the kinetic energy is zero too So the total energy at the beginning is also zero 1 Mm 2GM EmoziG0 o T 7 For earth this is 112 kmsec See Table 14 2 for other objects especially neutron star7 Circular Orbits and Dark Matter Just use our result from last week for centripetal acceleration to get mM 112 GM G 2 mi o 7 r r r for the orbital velocity Note that the period of revolution is T 27Tro so we have 4712 2 7 3 T 7 GM which is called Kepler7s Law of Periods Well look at this more next week Note that this is a way to determine the mass of the sun from the revolution periods of the planets One very basic result from this is that the tangential rotation speed 1 should be proportional to 177 as soon as the mass is concentrated at distances smaller than r Galaxies should look like this since there is generally a big bulge of bright stars near the center However the galactic rotation curves look very different See slides This is one of the strong pieces of evidence for dark matter in the universe For some interesting reading see Alternatives to dark matter and dark energy Philip D Mannheim UConn Published in ProgPartNuclPhys56340 4452006 You can also visit httparxivorg and get e Print astro ph0505266 Practice Exercises 1 Imagine an object with a mass M and radius R which is so dense that the escape velocity equals c the speed of light Write an equation for R in terms of M and other necessary quantities Compare this to dimensional analysis problem we did in lecture on the rst day of class 2 Use the Milky Way galactic rotation curve to estimate the mass of the galaxy contained within a distance of 16 kpc Note that 1 kpc103 pc31 gtlt 10L9 The mass of the sun is 2 gtlt 1030 kg How many sun like stars do you need to get this mass Note that the brightness of a typical spiral galaxy like the Milky Way seems to be about 1010 times that of the sun Table 21 1 in Introductory Astronomy and Astrophysics by Zeilik and Gregory The Milky Way Galaxy Visible Light uxg Infrared Light c m a E m w 1 VS WM GEEdS NOUVLOH a m RAmus KPC 5 NGC 2590 39 IC 507 Honors Physics Monday 17 September Fall 2007 Before we begin anybody want to talk about the dark matter plots some more Today How Kepler s Three Laws ca 1605 follow from physics ala Newton ca 1700 Discuss a little history including Tycho Brahe 1546 1601 References Textbook Sec14 7 Just states Kepler s Laws does not prove them Elementary Derivation of Kepler s Laws Erich Vogt AmJPhys 641996392 See also The special joy of teaching first year physics Erich Vogt AmJPhys 752007581 Available to you from the web site httpscitationaiporgajp Today I will break my own rule and hand out class notes before the new stu Kepler s Three Laws 1 The Law of Orbits All planets move in elliptical orbits having the Sun at one focus Draw a reminder of what an ellipse looks like Textbook Fig14 14 Label the semi major axis a and the distance from the focus to the center as ea Discuss eccentricity e a little 2 The Law of Areas A line joining any planet to the Sun sweeps out equal areas in equal times Mention that this is a consequence of conservation of angular momentum which we discussed in class a couple of weeks ago 3 The Law of Periods The square of the period of any planet about the Sun is proportional to the cube of the planet s mean distance from the Sun We saw this was true for circular orbits so you can guess it will hold for ellipses if we define mean distance appropriately Erich Vogt s Proofs The rst law is the hardest to prove so we start there Begin with energy E lt 0 1 M EKU mu2 G m r and divide through by m and then by the positive quantity Em to find v22 Em 1 6 We will show that Eq 6 is the equation of an ellipse Now remember the constant of the motion we derived for central forces Check Appendix H on cross products 6 my39 IFX mur singb muh 7 where gb is the angle between 17 and 77 so h has a geometric meaning Figure 1 from Vogt s paper The geome where h is the perpendicular distance from the focus to the tangent at point p The ellipse shown has b a2 Tonge try for the bound elliptical orbit of a planet P509 at point p around the Sun at the focus The ellipse parameters ab and c are b r 6 shown as well as three alternative pairs of l h x coordinates a and y r and 9 r and h 0 C 1 focus Note that h S r everywhere along the path Replacing 11 with h in Eq 6 gives b2 2a 7 7 7 1 8 hg T where a GM27Em and b Zm72Em12 Even though it looks screwy7 Eq 8 is the same as the familiar equation for an ellipse namely 531 9 Vogt demonstrates this equivalence in his paper lt7s mostly algebra with some calculus This is a cool result It gives the semimajor a and semiminor b axes ofthe elliptical orbit in terms of the physical constants E and Z for the motion The second law is simple to prove The area dA of the skinny triangle swept out time dt is just h1 dt2 so that dA 1 Z R 7 E which is constant Clearly7 Kepler7s second law is a statement that angular momentum Z is conserved for central forces 7pm The third law is pretty easy to prove7 too Since the area swept out per unit time dAdt is a constant7 and the area of an ellipse is 7Tab7 so the period must be T abdAdt Now realize that we can write b ZmaGM12 Putting this all together7 we get i 7TaZm aGM12 27Ta32 i 4W2 3 T 7 x T2 i 7 Z2m GMWZ GMa So7 it would seem that Kepler interpreted the semimajor axis a and the mean distance77 of the planet to the Sun It would be interesting to look up what he actually wrote and how it best translates to modern English Practice Exercises Just one for today In terms of the standard equation for an ellipse7 Eq 97 the eccentricity e is given by a2 7 b212 ea See Vogt7s paper for details Derive an expression for the eccentricity in terms of the constants of the motion E and Z7 and the physical constants G M7 and m Then write E and Z for a circular orbit of radius R7 and nally demonstrate that e 0 using your expressions You can either plug your expressions for E and Z into your expression for e or step back a bit and just show that they give you a b Honors Physics Thursday 20 September Fall 2007 First exam one week from today in this classroom Some relevant items 0 Homework due on Monday but not Thursday 0 See past exams posted on the course web page 0 Lab period next week devoted to review Bring questions 0 Don t memorize Bring whatever materials you d like to the exam The Simple Pendulum Work with the diagram in your textbook Figure 17 10 Figure 1710 The simple pendulum The forces acting ont he pendulum are the 7 tension T and the gravitational force mg L of X which is resolved into its radial and tan gential components We choose the 3 axis m to be in the tangential direction and the y axis to be in the radial direction at this x particular time jgigcase l7 10mg First talk about the motion in terms of energy The potential energy is due to gravity only The tension T does no work on the pendulum bob since it is always perpendicular to the direction of motion Let the zero of the potential energy be when the pendulum is hanging still Then the potential energy due to gravity is U mg x height mg X L Lcos 6 mgL1 cos 6 The potential energy is plotted on the 2 right Two total energy values are indi cated Note the turning points and em 5 E3mgL2 phasize that the bob will oscillate between them This is just what you know a pendu lum does We could solve for the motion using the energy but we won t Instead 05 EmgL2 we ll go back to the equation of motion ie Newton s second law and refer back 0 to Figure 17 10 U9mgL 0 67 To solve this with the equation of motion rst use a to measure the distance along the arc as in Fig17 10 The only force that acts in this direction is the tangential component of gravity namely mg sin 6 So F ma becomes mg sin6 mi or sin6 10 after we cancel m on both sides of the equation and realize that a L6 This is the differential equation we must solve in order to nd 675 ie the motion of the pendulum in terms of the angle as a function of time However this equation is impossible to solve exactly and instead we resort to an approxi mation lf 6 which we always measure in radians is small then sin6 6 In this case i Mo 11 where w2 E gL This equation is easy to solve In general we have 6t Acoswt B sinwt 12 where A and B are arbitrary constants set by the initial conditions If the pendulum starts from rest at an angle 6 60 then 6t 60 cos wt 13 Mention concepts of frequency f w27T and period T 1f 27Tw and L T 27TgtltCOFFSCUOHS 2 Lgtlt 11260 7r 7 isini g 4 2 Application to Einsteinls Equivalence Principle The observation that the period of a pendulum does not depend on the mass of the bob is a direct consequence of the equivalence of inertial and gravitational mass In other words no mass appears in Eq 10 because we canceled it on the left and right However if we write the gravitational force as may and Newton7s second law as F mIa then we would have T 2 lt2 mo 9 If anyone ever discovers that the period of a pendulum is affected by the mass of the bob or by its composition then it would imply that gravitational and inertial mass are not exactly the same However very sensitive tests have been done and nothing has been found To Einstein this was trivial See your Textbook Sec14 9 Einstein did not view gravity as a force7 Instead it is the shape77 of space and time That is mass or energy since E mcz makes space change in such a way that an object follows a natural path which looks as if it is being moved by a force called gravity7 Another way of saying it is that a projectile doesn7t follow a curved path it follows a straight line but in a curved space In this view there is just one kind of mass not two Practice Exercises 1 Prove that Eq 12 satis es Eq 11 for all values of A and B and that Eq 13 satis es the initial conditions 2 Write down equation for conservation of energy for a pendulum that starts from rest at 6 60 Your expression should contain only L g 6 6 and 60 For small 6 cos6 17622 Use this to eliminate cosines from your equation and rearrange to get an equation for 6 d6dt Now integrate this equation to get Eq 13 Honors Physics Thursday 20 September Fall 2007 Solution to Practice Problem 2 Problem Write down equation for conservation of energy for a pendulum that starts from rest at 6 60 Your expression should contain only L g 6 6 and 60 For small 6 cos6 17 622 Use this to eliminate cosines from your equation and rearrange to get an equation for 6 d6dt Now integrate this equation to get Eq 13 Solution The total energy E whatever it is will not change with time and will always equal 1 E inn2 mgL17 cos 6 1 i mL262 mgL1 7 cos 6 where we know that the velocity at any point on the arc is 1 i L6 Now since the pendulum starts from rest at 6 6077 we know that 1 0 when 6 60 Therefore the equation for conservation of energy is 1 i mgL1 7 cos 60 EmL262 mgL1 7 cos 6 1fthe pendulum starts out at some angle 60 then it will never be at an angle 6 that is greater than 60 So if we use the small angle approximation for 60 then we can use it for 6 too This turns the energy equation into 1 1 x 1 EmgLeg E771L2192 5mm which is easily rearranged to give which as you know by now can also be rearranged to give d6 7 dt lt63 7 62gt 7 Why did we choose to take the 7 sign when we formed the square root Well we are starting out at 6 60 and the angle will decrease from there so we expect that lt 0 Now we want to integrate this last equation What are the limits of integration Well again we have 6 60 when t 0 We also have some value 6 at some time t that is 6t So we can write 9 d6 t 12 7 wdt 90 lt63 7 62 0 where we put primes on the dummy variables77 of integration What7s in a name The integral on the right is just 7wt The integral on the left is trickier We can make a change of variables though namely 6 60 sinu d6 60 cosudu which lets us carry out the integral on the left very simply namely 6 sin 166 sin 166 d6 12 o 60 cosudu o du SinilwQO i Z 90 93 7 92 7r2 60 cosu 7r2 2 Finally then we put the left and right integrals equal to each other and get sin 1660 7 iwt sin 1660 gim 660 sin 7 wt coswt or nally 6 60 coswt which is indeed Eq 13 There are a lot of little tricks that we used to come up with this solution Perhaps a simpler way to approach it would be for me to ask you for the period T and you might realize that this is just four times the time for the pendulum to swing from 6 60 to 6 0 Right I would not have expected you to work this all the way through but it was worth it for you to try You will see more and more of these sorts of mathematical manipulations as you move upward in your physics education Honors Physics Monday 24 September Fall 2007 First exam on Thursday See comments from last week Notes posted on web site Today Taylor Series and their applications including 6 cosm z39sinx and then on to simple harmonic motion and the solution in terms of imaginary exponentials Math Topic Taylor Series This is a way to approximate a function y f by an expansion about a point 0 in terms of an in nite polynomial in m 7 x0 and the derivatives dydx f dZydxz f and so forth evaluated at 0 id fz0 f z0 f z0 etc The formula is 1 fW 960 fWOWC 950 gfWOWC 9502 39 quot Check that everyone knows what the factorial means derivative notation Graphic depiction of a Taylor expansion 10 The function is expanded about the point p 0 1 Shown are the zeroth order ie 8 w a constant approximation the rst order 6 second order and the full function The 4 i i i i i i i i i i i i i i i i i i i i zeroth order is just the value of the func quotZer0th order tion at 0 1 and the rst order is the quotquot FII St order 2 l Second order tangent line at that point The second or Fu function der approximation comes reasonably close 00 1 2 3 4 to the full function in the neighborhood of 0 lmportant examples expanded around 0 0 unless otherwise stated 1 z 1 i1ax72xozozi1x2 1392 13 14 15 1 E i 1 1 s1n z7 x3 z5 6m 1 1 cosx 17Ez21z4 Obviously if ltlt 1 ie z is small then only a few terms maybe just one is needed in order to get an accurate approximation Math Topic Imaginary and Complex Numbers All you need to know for now is that there is a number 239 such that 2392 71 Here7s some more information though that well use at some later time Numbers formed by a real number times 239 like 2239 fixg or m where z is a real number are called imaginary numbers Dont let the words fool you though All of these are perfectly valid numbers The imaginary numbers are a group of numbers outside integers rational and irrational real numbers Numbers like 3 2239 and z z W are called complex numbers They have a real part and an imaginary part We write for z z z39y where z and y are real z Rez and y The modulus of z is xxz 12 Now7 here7s a neat thing m 1 1 1 1 y e 1zz zz2 m3 1zz4 5m5 7 1 2 1 4 r 1 3 1 5 7 17iz Jrgz 1zx7 cosx 239 sinz This is called Euler7s Formula It is very useful in physics and engineering Watch Simple Harmonic Motion Probably the most important equation of motion77 in all of physics is that given by the force F 71 This is called Hooke7s Law and describes the force from a simple spring The potential energy for a spring is Uz kxz27 since F 7dUdz The potential energy has a minimum at z 07 so an object with nonzero total energy will oscillate about the minimum Note also that my potential energy that has a minimum at some z 0 can be written as the Taylor series 1d2U dU UW W950 a 95 750 E W Fm mmo m Uz0 7 02 702 where k U x07 since the rst derivative is zero for a minimum That is7 it looks like a spring force7 if we dont get very far away from the minimum The term Uz0 doesnt matter since we can always add a constant to U and the force doesn7t change Also7 we can just translate to x z 7 x0 if we want to do without the ugly x0 everywhere Now7 here7s the neat part The equation of motion is just 71 mi i i 77 m Try a solution of the form t Aem Ansatz Obviously imam z wzt and 7w26m 7w2zt In other words7 our ansatz works ifw i We write t Ac1M Be M 14 where A and B are typically determined by the initial conditions 0 and You will see this approach over and over again in your studies of physics Practice Exercises 1 We7ve seen that the potential energy for a pendulum is U0 mgL1 7 cos 0 Use a Taylor expansion to show that if 6 ltlt 17 then Ut9 looks like the the potential energy for simple harmonic motion What is the effective value of k What is the period T 2 Beginning with Eq 147 show that a simple harmonic oscillator that starts from rest at position 0 0 moves like zt 0 coswt Honors Physics Monday 1 October Fall 2007 Hand back rst exam Average over 757 so no curve Under 607 please make an appointment to see me Mention line integral problem something similar will be on the nal Perhaps spend a few minutes on questions regarding the exam Can also answer questions individually after the break Today we will just scratch the surface of the subject called oscillations You7ll see more of this in our course7 then next semester7 and then much more in your advanced courses Review Simple Harmonic Motion Also Phase Review how we got to Eq 14 Start with F 71 which leads to i Solve this with our ansatz t 6 Plug this in to the differential equation of motion to nd w iwo where we E lkm Warning Slight notation changes So7 we write zt Be uot 067Wi with 0 0 B C and 110 lw0B 7 0 CL which tells us that B zo 110z39w02 an O 0 7 volw02 Combine this to get Mt 0 COSWOI sinw0t W0 Write this in a different better way Put B A25i and O fl6445 Then Mt geiwot g67lwot ACOSltw0t b 15 We call A the amplitude max value of zt and b the phase of the oscillation Drag Forces Damped Harmonic Motion Not all forces are conservative although you will come to appreciate that energy is always conserved We just need to be careful to include all kinds of energy An important example of a non conservative force is drag This is the force on an object passing through some uid like water or air We write FDrag 7lm ibz 16 where b is a positive constant that depends on the shape of the object and the particular uid Note that drag is always acting against the direction of motion7 so the work it does will never be zero around a closed loop ln fact7 if 2F FOODS FDrag 7dUdz 7 bi then 1 dE Cl dU Emi2 x mi 7 Eamp dx and the total mechanical energy always decreases What a drag for designers 92 2F 7 FOODS 7sz Now lets apply drag to our oscillator7 which is no longer simple 7 but still harmonic The equation of motion is ZFma 7kx7bimi which we rewrite as i2vzw z0 17 where again Lug E km and also 7 E b2m This looks tough to solve7 but we can put our ansatz t em in and see what happens We nd from Eq 17 that w272i ywiwg0 which is easy to solve for w Using the quadratic equation7 we get 22 i 7424w2 b k b 2 wmi wgiv2iii 77 gt 18 2 2m m If the oscillator is not too strongly damped7 that is y lt we then the solution looks like t 677i Be ubt 06 where M 7 drug 772 is the frequency of the oscillator Dont forget about the 27f The amplitude of this oscillator decreases with time Of course it does lts total mechan ical energy is decreasing We call this a damped harmonic oscillator The damping time constant is 739 E 177 and is the characteristic time scale with which the amplitude decreases A useful measure of the damping rate is Q for quality factor de ned as the energy of the oscillator at any given time7 divided by the amount of energy lost over one oscillation cycle7 ie Q E EAE Your third practice problem lets you explore this quantity a bit more There are no oscillations if y gt we over damping or if y we critical damping For each of these cases7 t is made up of purely real exponential functions Why We will leave the study of these cases to an advanced course in mechanical oscillations Forced Oscillation and Resonance This is a fascinating subject7 worthy of several classes7 but we wont be able to do much with it in our course Imagine a driving force FDrive Foam The equation of motion is 2F ma ikx ibx Foam mi or i 27 ng Fem em 19 Try the ansatz t AH and see how the amplitude A depends on w the driving frequency Remember that A can be complex7 so consider both the magnitude and phase What happens when w 1 This is a familiar phenomenon called resonance Practice Exercises 1 For simple harmonic motion7 determine the amplitude A and phase o in terms of the initial position x0 and initial velocity o0 Sketch the two cases o0 0 and ii x0 07 and indicate the phase on your sketch Does the answer agree with your formula 2 What are the dimensions of b in the drag force de nition7 Eq 16 How can you combine b7 m7 and g to get something with the dimensions of velocity What do you think this velocity is physically Prove it by solving the equation of motion for an object in free fall but subject to drag7 ie 2F 7mg 7 by Why are the signs the way they are This is not a simple differential equation to solve7 though7 so ask for help if you need it 3 Find an expression for Q in terms of y and we that is7 in terms of b7 m7 and k Honors Physics Thursday 4 October Fall 2007 New stuff today Momentum then applied to systems of particles then on to rigid bodies Will introduce center of mass aka center of momentum in terms of three dimensional integrals But don t let the fancy math language fool you Momentum Impulse and Conservation of Momentum Newton actually wrote the Second Law in terms of momentum 17 m1 as a d d1 4 ZFiEimEima where we can take the mass m out of the derivative if it is a constant F Now imagine an object colliding with a brick wall The collision will change the momentum of the object The force of the collision varies over the collision time See the gure on the right Figure 6 7 from the textbook We can use Newton7s second law to write an ex pression involving the integral of the force and the change in momentum t2 4 4 A17 Fath J t1 We call fthe impulse The average force Favg is J divided by At t2 7 t1 0607 Now imagine two objects colliding with each other Each exerts an equal and opposite momentum on each other That is J1 7J2 or A171 A172 AltI71 172 0 In other words the total momentum p E 171 172 is conserved in collisions Note that this doesnt work if there is some external force acting on the masses Spend a little time talking about momentum conservation as invariance under space trans lation as opposed to energy conservation as invariance under time translation Many Particle Systems and Rigid Bodies Look at two particles again but not just collisions Allow some external force Ext to act as well as internal forces Then c113 1171 61172 a s a EEEmlalm202ZFext The last equality follows because the internal forces cancel in the sum It is convenient to write P Mum where M m1 m2 and 17m dfcmdt with a a 1 Fem m mm 771272 20 m1m2 in which case Newton7s Second Law reads 2 EM Mam Mdcm We call 77cm the center of mam7 although a better name is probably center of momentum This all carries through if there are more than two particles in the system We write a m1F1m2772 mN77N 1 N Tcm 7 mm 21 m1m2mN M Don7t forget that this is really two or three equations for SUCH gem and if three Zcm See textbook7 Eq7 12 See gures 7 8 and 7 97 which show how this all works out j Moon L l 4 7 i 1 15 kg Earth39s 7 Earth 1 path Path of 739 center Of quot1355 f3939 39393 52 Center Sun of mass 0708 0709 Calculating the Center of Mass of Solid Objects If we are working on a solid object7 we could sum over all the atoms that make it up7 but that would be nuts lnstead7 we use calculus7 and assume that the object is continuous Then the sum becomes an integral That is 1 Mdm and 77cmMdm77 We usually write the in nitesimal mass dm as a density usually called p or 0 times a volume element dV or some other measure7 ie dm pFdV This allows the density to vary as a function of position inside the solid object The rest is in the details Here is a simple example What is the volume of a right circular cone of height h and base radius B Let z measure the height7 and slice the cone up into thin disks7 each of thickness dz The radius of each disk is Rh zh R1 Zh7 so the volume of each disk is 77R21 zh2dz Therefore7 the volume of the cone is h h 2 dV 7rR21 zh2dz7rR2 1 2Z2gt7r122 0 0 or one third of the base area times the height Of course this is volume7 not mass7 but that7s basically the same thing for constant density If the density is not a constant7 you7d need to include the functional form of pF and that would be a harder problem Practice Exercises 1 An elastic collision is one in which kinetic energy is conserved Let two objects7 one with mass m and the other with mass 2m collide elastically in one dimension They are initially heading towards each other with speed 71 Find their speeds and directions after the collision7 in terms of m and v 2 Find the position of the center of mass of a long thin rod of mass M and length L The density of the material making up the rod changes linearly from zero at one end Start by realizing that the mass of a short segment of length dsc is dm 0xdx where 01 is the linear mass density Write a formula for 0567 and determine any parameters by xing the mass of the rod to be M Then7 nd the position of the center of mass Honors Physics Tuesday 9 October Fall 2007 Today Describing the motion of solid objects also known as rigid bodies Note It is convenient to describe rotational motion about the center of mass Then let the object move according to external forces by following the CM and deal with rotations about the CM separately However we don t have to do it this wayl Kinematics Describing the Motion of Solid Objects Specify an axis of rotation Then motion of all y parts of the body are described by some angle of rotation ng 57 about that axis even though 5 and 7 will be different for different pieces of the object See Fig 8 3 P 1 So gt takes the place of displacement x for one fquot dimensional translational motion Change ingz39S 1 l is Agt and angular velocity is w E dgtdt A xquot i and angular acceleration is a E dwdt d x Everything else follows For example rotation under constant acceleration is Eq 8 7 1 92505 5M2 wot 250 Be careful of unitsl The correct units are radians which are truly dimensionless Some times people use degrees 180 7r radians or revolutions 27f radians so watch out ls this the same to we used in simple harmonic motion Yes If the angular velocity is constant then the time T it takes for one complete revolution is given by wT 27r Vector Representations of Rotational Kinematics We can write LB or 07 but we can t write g2 This is because finite rotations do not commute Do the trick with the textbook However tiny rotations do commute so 6 is a vector and so is Q dgtdt Since J is a vector 07 Ltdcit is too So what direction is Q It has to be along the axis since that s the only direction that isn t changing By convention we use the right hand rule to choose the positive direction We won t stress it but it is good to see the relationships between linear and angular variables especially in vector form See Figs 8 10 and 8 11 in your textbook y Z Z Rsin 9 Hsine 2 hi 0810 0811 R where must be specified at some point within the object located For example 17 X Rc3gtlt 17 Do you see what happens if Q is constant See practice by R Also a a Q X Dynamics Torque and Rotational Inertia Forces cause translational motion but torques cause rotational motion The torque T from a force F applied at a point speci ed by F is 7 rF sin6 rFL 71 where sin6 is the angle between Fand We call ri rsin6 the mo ment arm77 of the force Clearly 6 0 means no torque no matter what the force ln vector notation 7 FX F 0902 lmagine a bunch of masses hanging from a string The torque due to gravity is El n X Fn X X 7Fem X lf we hang from the center of mass Fem 0 So there is no torque and the object is balanced Sometimes we call the center of mass the center of gravity77 So what is the motion resulting from a torque Consider one single particle inside the solid object located at Fand acted on by lt can only move in a circle of radius r and the force in that direction is Fsin6 lf the path length it travels is called 5 then Newton7s Second Law says Fsin6 sz39 mrzb Multiply both sides by r and we have T mr2a The quantity mr2 has replaced mass when we went from translational to rotational motion Since oz is the same for all parts of the object and since 7 is either from one force at one point or due to a sum of such things Newton7s Second Law for a solid object becomes 2 Io where I Zmnri 22 is called the rotational inertia or moment of inertia Notice that I is a property of the solid object and the speci ed axis Do you get the feeling that I is not really a scalar but not a vector either lf so you are right lt is actually something called a tensor7 You can prove something called the Parallel Amis Theorem which states that the rotational inertia of any body about an arbitrary accis equals the rotational inertia about a parallel accis through the center of mass plus the total mass times the squared distance between the two acces Mathematically we write I Icm MhZ Rotational Inertia of Solid Objects Of course for solid objects we don7t do the sum over all the masses in Eq 22 to get the rotational inertia lnstead we do the integral I ferm This is just like calculating the center of mass that is fr dm but with another factor of r See Fig 9 15 Practice Exercises 1 Follow Figure 8 11 and show that for an object spinning at some constant angular speed w then the acceleration d at some point has the correct direction and magnitude for the centripetal acceleration 2 Find the rotational inertia of a long thin homogeneous rod of mass M and length L about an axis through one end and perpendicular to its length Now use the parallel axis theorem to get the rotational inertia for an axis through the center Honors Physwcs Thursday 110cwber F2 2007 Hem m humewak m a ka mean yesmdsy s w heme hmu39fyvu wem ume mmeke meeewemems xwm key m m leh when yedmgymr lsh heme meme 5 eeammehemwe expemem Angular Mammbum af a ngle Fannie Rash a a e 2 Emzyr an mm on mhea mnstem eme mmquot upeme em on w ammloms Mmunn acne em cf mammal symmeew end eemex fumes New we knew me 2 me Laummnml eme vector 1 eweweuwnmezewxw seewemameew mu w w e the enge hewee Feud 5 mew gure When we an Vog s dmvelm a Kepleva lawseweusedl wmwhee hvm Sequ 7 Ifmsteeiwewmeumsleedm fy c 2 wee m Thus angular mememumw 5 me mm cf mnmmwmquot wee ee mzquequot 1 me mumml cf mew NextwemxmanourselveswnehlheveamVx hzshndmsssm ehellspenm a ya body mung about e xed ems See gure me In your mum w e WWW 1M when L wr 1 the meumel mans 1 hr swim about the Lexis end w 1 the memmae ex me en veeew m We whde new body me new meme n e em ewe p m W meme memes instead a teem em 0 muse we wean like we we eemeumgme i In W he would he be me e lees m m We smgje mes w Canada however e Symmmcnyd body See gure 10 m m hank Rat may mess m Lhme 1 one symmeumllv plewi mthe Wane suds unhe ems In ths mse ewmhmg eeem e 2 mmeme wu en whm summing me me ex new me me we mm emee meme we gazezeneue mememum z e m in mm nyd beam The symmeey m be mm me em ex mama Newton s Second Law for Angular Variables What changes the angular momentum Z for a single particle Let7s see d d d d i i Eamprxm x5rx axmmererf where we used Newton7s Second Law ie F d dt Well this is nice Newton7s Second Law for a single particle looks the same in angular variables as it does for translational variables This is what I meant when I said the tension in the horizontal rod supplies the torque which changes I in the discussion above For a system of particles we can do the same thing for the total angular momentum i i i i i di i L12 24 gt EZT7L However in the sum over torques all of the internal torques will cancel since they will all be paired up at the same point but with equal and opposite forces So this last equation can actually be written as the sum over external torques alone that is dL Tex i 23 27 n d lt gt This is analogous to Newton7s Second Law for a system of particles in terms of the center of mass and external forces There is actually a big surprise though from this equation Consider the behavior of the bicyi cle Wheel in Figure 1075 One end of the axle is on a pedestal If you hold the other end nothing odd hapi pens But if you let go of the end the wheel doesn7t fall It precesses around the pedestal u This is just what you expect from Eq 23 For a spinning wheel E is directed along the axle When you let go the only external torque around the pedestal pivot point is from gravity The direction of the torque Fgtlt is sideways This is the direction in which the vector E changes This is the same principle as a spinning top See Section 1075 Of course the classic application of Eq 23 is the conservation of angular momentum If there are no external torques then dEdt 0 and E is a constant Skaters speed up their spinning if they pull in their arms for example See also Figures 1012 10713 and 10714 for more examples Practice Exercises 1 Consider a simple pendulum with a mass m hanging from a string of length L Write expressions for the total torque 7 on the mass and its angular momentum l in terms of the pendulum angle 9 Show that ET d1 dt yields the same equation of motion for the pendulum as in Eq 10 ie 7gL sin9 See Sample Problem 102 in your book 2 Spinning neutron stars are sometimes observed as pulsars intense radio astrophysical objects with periods of fractions of a second The sun is almost a sphere which rotates once every 25 days A neutron star is made up of tightly packed neutrons If the Sun collapsed into a neutron star what would be its rotation period neutron occupies a cube with side length about 2 X 10 15m See Appendices B and C for Whatever else you need Honors Physics Monday 15 October Fall 2007 New stu today Motion of uids Some gases mainly incompressible liquids Thursday we will look more closely at the mathematics of vector elds Note This is an introducetion to continuum mechanics aha classical eld theory How to Look at a Fluid Fluids are continuous but not rigid Instead of force and mass we talk about pressure and density to formulate kinematics and dynamics We will refer to regions of a uid that are enclosed by a surface which doesn7t have to be real Force on a small area element of a surface is AF Size of area is AA De ne uector area ele ment AA by giving it a direction that points outward from the enclosed volume Then AF pM where p is the pressure The SI unit is a pascal Pa Atmospheric pressure is just about 10 x 105 Pa at the ocean bottom m 10 x 108 Pa See Table 1571 for more examples A small volume element AV will have a mass Am The density p is just the mass per unit volume so p AVAm If p is a constant throughout a uid of total volume V and mass m then p mV Volume changes in response to pressure If something of volume V changes by an amount AV in response to a pressure change Ap then the value of AP W gt 0 is pretty much the same for all volumes of any given material It is called the Bulk Modulus For liquids B m 109 Nm a number so large that we generally speak of incompressible liquids For gases B N 105 Nm much smaller Consider a static uid at rest but under gravity The weight of the uid on top increases the pressure underneath Consider the forces on a horizontal slug of uid Fig1572 Up is positive Mass is dm pdy so gravity is ipgAdy Pressure up mumi on bottom face is pA Pressure down on top face is 7pdpA The dy39i sum of yeforces is zero so PA P dPM 7 pgAdu 0 dp 1quot 7 7 24 Heoerencelevmyn dy pg a In a I5 02 For a constant density uid like water this means that Ap ipgAy or p p0 pgh where h is the depth below the surface of the water pool For an ideal gas remember from chemistry or wait a few weeks p is proportional to p ie p hp Then dpdy ihgp so 10 1006 where 100 is the pressure at sea level Also p0 hpo is the density at sea level so My 1006 where a 1hg pOpog m 9 km for the Earth7s atmosphere Pascal s Principle and Archimede s Principle In ut Pascal noticed that pressure transmitted itself equally pFLAi A07 through the volume of a uid This is obvious for an 5 do A v19 fa incompressible uid but true in general Typical applica 7d T Mg tion is the hydraulic lever pictures to the right Figure l on F 15 8 See Figure 15 9 for an example of a realistic car jack with a uid reservoir and alternating values 0W 1508 Archimedes Eurekal realized that a body will sink until it dis places enough uid to displace its own weight This is easy to see p Ib Imagine a blob of water inside the tank surrounded by identical 39 Ix V water Figure 15 10 It doesn t move so forces all balance Now 15 owmer replace the water blob with another material so the blob now has LgquotVquot mass m It sinks if the density is greater or oats if the density is I mg less The Continuity Equation and Bernoulli s Equation Some mass 6m ows through a screen of area A in a time 6t The volume of water swept through is just A X u at where u is the speed owing perpendicular to the plane of A Obviously 6m pAu 6t We call dinat pAu the mass ux So what The point is that mass is conserved so the mass ux has to be the same at any point in a ow This means that pAu or just Au if the uid is incompressible must be the same everywhere This is called the equation of continuity and is very useful practically The heart of uid dynamics is something called Bernoulli s Equation It combines the con cepts of uid ow with conservation of energy See Figure 16 6 in your textbook A blob of water with mass 677i is pushed through the pipe in some time 6t The kinetic energy change is AK 6mu The work done by gravity is v Outlet 6mgy2 yl and the work done by the di erence lizngf in pressure is p1A161 1921426562 p1 p26mp 5 So the work energy theorem W AK implies that Inlet 2 m p gm m constant 25 This is Bernoulli s equation Practice Exercises 1 As water falls from a faucet it necks down Use the continuity equation and F conservation of energy to ex 5T plain why From the ratio g gt A Oi 141142 in the gure on 5 E j V1 1 the right derive an expres l l 3 i A2 sion for the speed 111 in terms V of of Oi and 11 1605 2 Consider a 1mgtlt1m window in a tall building A 30 mph wind blows past the window tangential to its surface Use Eq 25 to calculate the outward force push ing on the window pane You can assume atmospheric pressure on either side of the window when no wind is blowing You may want to look up the John Hancock T 07116 in WikiPedia Honors Physwcs Thursday 18 chber F2 2007 Samad Mzdtem am 1 0712 ww m may memeue numhm wLh mhmk guresfmm mm Chap m Bald Vacbax new and Flux A mm is any meme function depeee end maybe Mme m Simple exsmple e dmmy p My pm A vecmx mm is just some veem mummy mm n has memueny emcem megmude m reman e any mm m speee end Mme We wl seam egmem vem we end 91min me vdmly eld exenma mum vem m dsy us 3 we s eeh dsw a amen Jngmn dime See hmemeem exemhe megawae age 13 show even eld m2 cms39sm Heme shows a Eoumequot Wespeekememmevwexevee edmmgeeemeexeea m nw meue lotheeread M1iA theplsnexselmeexggeiemmd MdAasi ewes meme end wme a a a see new hdnw we me e In Honor Phyncs m a www39vvvv HHva perm m y 4 u should be obvious w you w now me eu we need to do to nd the ux Lhmugw some 1313 eeeA 15m mcezzezeevee meeeee Whammy Ma 5 111 but thede wl always depend on me spee e exeeem vem mm mvnlved The Dwmgmcg Thgmwn aka Gauss Themm wheexeemeewmehweexemcegezm Ba edsurbaee ehm wewnm a See There 1 en lmpa39sm worm which Wm pme end use a 41 my 26 39 where the inlays m the 11th 1 over the vnlume emckwi by the exeet end at at at 7 e e E a E eeehehthehweeheev etetetett Weheehtethehtheeeweeethtehehtheheet Our bxmf 1 ample Imagine e very me enclosed surface ehux thh m pem el m the ewe htehee wth ehe eehheh et teeth ehh the ethe et Matthew new t We ehee the hex e eheh we tehteee the hteeex wth em 27 a 11 ewe ye mam ez they mam fame z 4 Z what which htewee the them he the hm hm hwe build ehtewehhhe eht e tth We theh the ux m one sxde of ehux 1 the nQQIWe of the ux out 139 the We of 1t nwhme m eteythhe but the eweeh ehheee eeheeh The cthhhmty Equaman the heat ths hme New hehehhheh the mess ux mJ Me Fox etth eheea we eeh whte ths e a wheet ewe ee the meee hewhgthmteh ehehhthetyetheeeheh hhht hmehe a M 1 mtheh c1 no deslIGYEde so the net mess nwmg throng eclmed surf Maybe It would he the em a elmg skmny hme The men wtn hemmemme hmhmeht he Physics 11 The mdnsed mess he pet the huge ddmxty ewe vexuhhe 5e a a w umgt Dwehgame Theorem FA 26 The J m 1 because pmtwe uxxs mess nwmg mm which heeeeeee the mess Btmyng the the hemetwe hhme makes ht ehettheh dehwetewe Thee heehehg the e the hex mtwehhee ves he i lt e at m t ea 28 Thus he the oehthhhty Equetm whtteh h e we funnel wet Fexlye we test wete thet pAw wee eeeheteht which 5 wet 1 d1 o m ehthe eteeeeht the h ehe Meme 1 The male funnel statement says ths ame mature e tube thh elmeer 39speh m the zdn ectmn For mteht dmmtye scheh lt 0 end sew6y lt o matea gt o tokeep 7 Thet the Speei heeeeee pmhte heethee See hextheeth 161713 A hm e eeheh meteteehehv n the eteeea hhe hteezeh M ds xs tee eheeha eehhe Peth Chase exectengulex ckmi Pethe end show that M d 0 hr the velmty dd h Emma ehewetehd thet ta d 0 he the veheetty ed eh the ngqt Ihet use the dd m the 11th 1 mteumel M w e w v mm v v mm v Honors Physics Monday 22 October Fall 2007 Second Midterm Ewam is This Thursday The topic we call coupled oscillations has far reaching implications The formalism ends up being appropriate for many different applications7 some of which bear only a passing resemblance to classical oscillation phenomena This includes the mathematics of eigenvalues and eigenvectors7 for example These notes describe the elementary features of coupled mechanical oscillations We use one speci c example7 and describe the method of solution and the physical implications implied by that solution More complicated examples7 and their solutions7 are easy to come by7 for example on the web1 The following gure describes the problem we will solve k m Two equal masses m slide horizontally on a frictionless surface Each is attached to a xed point by a spring of spring constant k They are coupled to each other by a spring with spring constant kc The positions of the two masses7 relative to their equilibrium position are given by 1 and 2 respectively Now realize an important point We have two masses7 each described by their own position coordinate That means we will have two equations of motion7 one in terms of i1 and the other in terms i2 Furthermore7 since the motion of one of the masses determines the extent to which the spring kc is stretched7 and therefore affects the motion of the other mass7 these two equations will be coupled as well We will have to develop some new mathematics in order to solve these coupled differential equations We get the equations of motion from F ma so lets do that rst for the mass on the left ie7 the one whose position is speci ed by 1 It is acted on by two forces the spring k on its left and the spring kc on its right The force from the spring on the left is easy It is just 71ml The spring on the right is a little trickier It will be proportional to 2 7 x1 since that is the extent to which the spring is stretched In other words7 if 1 2 then the length of spring kc is not changed from its equilibrium value It will also be multiplied by kc but we need to get the sign right Note that if 2 gt 951 then the string is stretched7 and the force on the mass will be to the right7 ie positive On the other hand7 if 2 lt zl then the spring is compressed and the force on the mass will be negative This makes it clear that we should write the force on the mass as kcx2 7 951 The equation of motion for the rst mass is therefore 71ml k0z2 7 x1 mil lSee httpmath fulIertonedumathewsn2003SpringMassModhtmli We get some extra reassurance that we got the sign right on the second force7 because if k kc then the term proportional to 1 does not cancel out The equation of motion for the second mass is now easy to get Once again7 from the spring k on the right7 if x2 is positive7 then the spring pushes back so the force is 71mg The force from the coupling77 spring is the same magnitude as for the rst mass7 but in the opposite direction7 so 7kcx2 7 1 This equation of motion is therefore ik g k02 7 1 mfg So now7 we can write these two equations together7 with a little bit of rearrangement k WW1 kc2 imii kcil ka g WiZ To use some jargon7 these are coupled7 linear7 differential equations To solve77 these equa tions is to nd functions z1t and z2t which simultaneous satisfy both of them We can do that pretty easily using the exponential form of sines and cosines7 which we discussed earlier when we did oscillations with one mass and one spring Following our noses7 we write x1t 116th z2t 126th where the task is now to see if we can nd expressions for 11 a2 and in which satisfy the differential equations You7ll recall that taking the derivative twice of functions like this7 brings down a factor of m twice7 so a factor of 7w2 Therefore7 plugging these functions into our coupled differential equations gives us the algebraic equations k k0a1 7 002 mwzal 7kca1 k kca2 mwzag This is good Algebraic equations are a lot easier to solve than differential equations To make things even simpler7 let7s divide through by m7 do a little more rearranging7 and de ne two new quantities Lug E km and a E kcm Our equations now become 1ng a 7 w2a1 7 wfag 0 29 7w02a1 Lug w02 7 w2a2 0 30 So7 what have we accomplished We think that some kind of conditions on 11 a2 and to will solve these equations Actually7 we can see a solution right away In mathematician7s language7 these are two coupled homogeneous ie 0 equations in the two unknowns a1 and 12 The solution has to be 11 a2 0 Yes7 that is a solution7 but it is a very boring one All it means as that the two masses dont ever move The way out of this dilemma is to turn these two equations into one equation In other words7 if the left hand side of one equation was a multiple of the other7 then both equations would be saying the same thing7 and we could solve for a relationship between 11 and a2 2A mathematician would call collectively call these two equations an eigenvalue equation The reason becomes clearer when you write this using matricesi but not al and a2 separately Mathematically this condition is just that the ratio of the coef cients of a2 and al for one of the equations is the same as for the other namely 2 2 2 2 iwc i w0wciw 2 2 2 7 2 w0wciw iwc 4 7 2 2 22 we 7 WO lTwciw 2 7 2 2 2 iwc i w0wciw 2 w may In other words these two equations are really one equation if mgd or instead if 2 2 2 7 2 w w0 2wc wB Borrowing some language from the mathematicians the physicist refers to wi and w as eigenvalues7 We will also refer to A and B as eigenmodes 7 What is the physical interpretation of the eigenmodes To answer this we go back to Eq 29 or equivalently Eq 30 Remember they are the same equation now If we substitute w2 wi into Eq 29 we nd that 11 12 where the superscript just marks the eigenmode that were talking about In other words the two masses move together in lock step with the same motion This happens when the frequency is w iwo idlem and that is just what you expect It is as if the two masses are actually one with mass 2m The effective spring constant is 2k as you learned in your laboratory exercise Therefore we expect this double mass to oscillate with w2 Good This all makes sense Now consider eigemode B Substituting w2 wg 2w into Eq 29 we nd that i In other words the two masses oscillate against77 each other and with a somewhat higher frequency In this case it is interesting to see the difference between we ltlt we in other words a very weak coupling spring and we gt we strong coupling spring For the weak coupling spring the frequency is just the same as we and that makes sense If the two masses aren7t coupled to each other very strongly then they act as two independent masses m each with their own spring constant k On the other hand for a strong coupling between the masses the frequency is w we2 which gets arbitrarily large Sometimes this high frequency mode77 can be hard to observe A simple experimental test of this result is to set up two identical masses with three identical springs In other words we wo In that case wg Bwi and therefore the frequency for mode B should be xg times larger than the frequency for mode A We should be able to make this test as a demonstration in class We have been talking about general properties of the two eigenmodes Of course this doesnt tell you how the system behaves given certain starting values that is speci c initial conditions 3A mathematician would say that we are setting the determinant equal to zero For this7 we need to understand that the two eigenmodes can be combined ln fact7 the gen eral motion of each of the masses really needs to be written as a sum of the two eigenmodes Each of these comes with a positive and negative frequency7 just as it did when we applied all this to the motion of a single mass and spring Recall that the sum and difference of a positive and negative frequency exponential7 are equivalent to a cosine and sine of that frequency lncorporating what we7ve learned about the relative motions of modes A and B7 we can write the motions as follows 10 aeiwAt b67iwAt Ceiw3t d67iw3t 20 aeiwAt b67iwAt 7 66111131 7 d67iw3t where 17 b7 0 and d are constants which are determined from the initial conditions An obvious set of initial conditions is starting mass 1 from rest at z 0 and with mass 2 from rest at its equilibrium position This leads to the solution z1t x02 coswAt coswBt 31 z2t x02 coswAt 7 coswBt 32 Practice Exercises 1 Show that these equations satisfy the equations of motion and the initial conditions 2 Make a plot of z1t and x2t Choose some value for 0 and um Also choose a value for Log that represents a weak coupling77 system Explain the motion of each of the two masses in terms of conservation of mechanical energy It would be helpful if you rewrite Eqs 31 and 32 using the trigonometric identities 7 1 cosu cos1 2cos u 1 cos cosu 7 cos1 2st Sin u g 1 Honors Physics Monday 29 October Fall 2007 This week waves Next Monday a glimpse of the future with Persans Also preliminary lab books due on Monday but I need to gure out grading time given traveling The Wave Concept Waves transfer energy through local motion The motion can be transverse or longitudi Typicalslringelemem nal and the energy transfer can be in one two or three dimensions Example of a 1D trans verse wave is the stretched string 2D transverse a quot A iTypical spring eiemeni wave water surface ripples 3D longitudinal wave a ltigt is sound in air Seismic waves are 3D both trans quot if verse and longitudinal Light is 3D transverse A Typical string element wave may or may not be periodic or pulsed 1301 Mathematical Description General and Sinusoidal Illustrate with transverse waves in one dimension ie the stretched string The transverse dimension of the string is 3 position along the string is 95 so the description of the wave is the function 305 t See textbook gures 18 4 and 18 5 X J xPivt 4 vii 4 13 77gt P Time I fKv39 gt xpm b 4 vi gt 7 The shape of the wave is at but this shape moves down the string at some speed 11 Write yx 0 so this is the shape at t 0 Some time later the wave has the same shape but where I has moved to 55 I vt so yx t vt A wave can also move in the negative direction ie yx t vt That is Wm 15 fW fI i 7115 33 where v is some positive number called the speed of the wave So what is actually moving Any point on the string moves up and down ie transversely as the wave moves by Figure 18 5 shows the motion of some point on the string as the pulse represented by yxt vt moves past it Note Fig18 5 plots 3 as a function 0ft for ared at The point rises from zero and then falls back as the pulse passes Extremely important but advanced point We almost always describe waves as if the shape at I vt is a sine function This is because any general function f can be built out of a possibly in nitely long linear combination of sine functions The proof of this is part of something called Fourier Analysis which we wont cover But it is true A sine wave is obviously periodic It is handy to describe a sine wave as follows y r717 2n y x yx t ym sin I vt 34 Lr V Time 0 V1 Time 2 1806 We call A the wavelength for obvious reasons Note that for xed 55 say 55 0 we have y0t ym sin ZWUtA which means that the transverse oscillations are sinusoidal with period T Av 1 f where f is the frequency We also have Af 11 Sometimes we write V instead of Geeky physics joke Question What s new Answer cM The more common way to write the equation of a sine wave is yx t ym sinkx wt 35 where k 27rA w 27rf and ng is called the phase constant See textbook on phase and phase constant Compare two waves with same k and w but different Waves on a Stretched String The Wave Equation What is the equation of motion for a wave Lets get it from F ma on a string The motion is up and down so analyze the forces y I 1lt F in the vertical direction on a small piece of string 91 1 39 7T 7777777777 quot 7762 at xed at We assume small amplitude waves so gt l 655 can represent the length of the small piece See F 39 Figure 18 10 If F is the tension in the string then ZFy Fsin62 Fsin61 6may 7 5x l 1810 Put 6m a x for linear mass density a Obviously ay y39 82y 8152 since we are working at xed 55 Also for small de ections 0 ltlt 1 so sin0 0 tan6 slope 8y8x So 8y 9y 82y 1 8y By a 82y 7 7 5 7 7 7 7 77 905 8x1 I8t2gt6x 905 8551 F8252 The left side of the last equation is just 82y8x2 as 655 gt 0 So de ning 1112 E aF 82y i l 82y 8x2 7 v2 9152 which is called the wave equation because its solution is in fact a wave That is a function of the form Eq 33 solves this differential equation for any function f This is simple to prove All you need is the chain rule Fsin02 sin61 F lt 2 2 36 Realize that Newtonls Second Law has given us a formula for the speed 11 of the wave in terms of physical parameters in this case the tension F and density a of the string Practice Exercises 1 Write the wave form yx t ym 4ka e ww wl in the standard way shown in Eq 35 and determine the phase constant qt 2 See multiple choice question 7 in Chapter 18 Find which of the following wave forms solves the wave equation Eq 36 for v 1 Show this by taking the derivatives and by showing which can be written in the form Eq 33 You might take a look at the Principle of Superposition Secl8 7 which we will cover on Thursday A mt x2 t2 Honors PHYS CS Thursday 1 November F2 2007 Homewmk met nhheh aheehheehm 0f aimeyequot hhexudmg E sndmgweyesquot end emhhdw Renew For ohenhhhehehehheh nehevese hhemm exeeteetehed smug F mquot heehae te 1121 e z m whehe e ehha F he the smug tehehehh eha h he the hhse heeh hhhht hehgth The heehtheh ahaehehtheh shhethm he sued the wehe ehhhethehhw ehha hte edhhthehh he e wehev movmgm the hhght e eh heht whth eheeede Ahthehhgh we ashes ht a the ethetehs etehhg the we ehhhethehh thhhhe ehht te he ev 39ywhme hh hethhhe phhmhpxe at mpmaehm The ehhhh n wn waves he eheh ewewe This he she the uphhhehhexe exsupehheshthehhw eha he sey m Pme ht he tehhe hsehhee the we ehhhhethehh he e whamehehthex ehhhethm let j z m eha 122 xvi bah he wees he they seh mlve the weve ehhhethehh We dehhh thet 1121 1 h he eheeewewe Salem s hehhght hh eha eee 1 Vs ht walks K 1h end I hmh sdve the wee equeume theh ee aees y A 1e We set thet the twe wss kahtehvrehev whth esh mheh man he eh emMe thet hheahths eemph thhde exthe eweheu weve You heve prehe My heem heete hm enhha wsee who ah t smmme Waves New euppse we hhtehrehe twe ehhe wss a the eehhe speed equmx eha Human weveehgvhhhht ehhe he mmngte the hhght eha the mheh he hhemngm the Mt Ihet he whehe the heet equsmy muewe hm etng haehthty Ihhe he av y pemuhex wave 1t hee hem veeehty h e Icz Icz e 02 but eh emphthhae thet vehhs hh thhhe whth engulex equmcy M It he sued e Elendmg weve w This becomes a little clearer when we watch it happen See Fig18 17 7 7 f ad a I I a u b 39 l C If i I o i o V o o n o 1 I 3 U 1 41quot 1 21quot 41quot t T 1817 Something important happens when we set up standing 4 L pl waves on a stretched string that is fixed at both ends7 like r r39 a 1 a guitar string As shown in Fig18 207 we can only have quot standing waves that fit into the distance L between the frT two xed ends That is b rl 39 l 1 n2 A 2L 7 1 7 r a That is7 frequencies have to be integer multiples of a fun iT f T l damental frequency v 2L This is the basis of the har d V rl monic sound of nearly all string and wind instruments 1820 Sound Waves Sound waves are pressure waves A disturbance compresses some continuous H medium7 and the compression propagates We expect to come up with a differential equation for the behavior of pressure 191 775 in a medium7 starting with F ma7 and that this differential equation would in fact be the wave equation This is of course true7 but we wont do it here This wave equation would involve the medium properties7 like density p0 and bulk modulus B Note that the density would also fluctuate for a compressional7 ie longitudinal7 wave7 so we talk about the mean density p0 of the medium All of the phenomena associated with waves7 like interference7 beats7 and harmonic frequencies7 are readily observed with sound It is interesting to consider the real world values for sound waves7 in terms of their speed and intensity See Tables 19 1 and 19 27 and Figure 19 5 in your textbook Practice Exercises 1 Consider two sine waves of the same velocity7 but slightly different angular frequencies cu and w Aw Find the beat frequency which modi es the amplitude of the resultant wave See the trig identities we used for the practice problems on coupled harmonic oscillators 2 Use dimensional analysis to come up with an expression for the speed of sound7 in terms of the bulk modulus B and density p0 of the medium Recall from How to look at a fluid that B E ApAVV Compare your answer to Eq19 14 in your textbook Honors Physwcs Thursday 8 November F2 2007 Hmnewmk Ihm m preummexy leh beds on Mandsw Theme to pee for eovemg ales m Mmdsw The mm for Lndsy 1 Laemz nemeromeuomen wme ere the equeums me relsle me seemeus of speee end ume hewee Wm ameeu xeeeme frames They exe me meme E mtmn Newmn ee hmulsled esummeme me e umvese so eke Wm oxeewee mnez mmngelspesiuxdeuvemlhemha z nee e m we speed e dzdt m me 2 me me n moves wnh suede dz it e e u m me 2 frame You hm Mus u you ee m moymeee end we ehe thrown o me yumd m me uueeuom you ee movme 011m eepees co move much moe slova to you we m me peeom wuoumew n Mexwe ueweeu eee o euueuoe me descnhed eemeuy end meeeem nee euue um predm the stmoe haw because they can be 1 to denve e weve saueum whee me wee speed e WW 2 e we Epesi e new Msxwe mum lmow uom Newmn we gm mee euueuoe hes emhmg m dc wee meeeuee frames Eveyome should measure the me speei of kg 2 xw ess nuke Meme speed so Mexwe end Newmn exemmmmu end omeo hmhe hem mue be wrong meeu xeehzed me me women we me Newmn assumed me ume we umveee maeeu me ume a me 2 chsmme en n L ue me eeme e me ume 2 a me 2 chsmex They really an 1199 1 Le spams Mew y Whez exe the ohseveme emsequemees o e heme me seme for en chsavm oozeuae me euewe bystudwng inmrvelf oxspeeeee endumeAt See Me Zortend my helm A e T y e L A 0 A V v V M e On me Lem em heem homee up end aowm hewee two mumoe m e eme eued s n nevelsedls39sme ZaneumeA 2 1 ZLuc Thu ehndm clmk Lee the frame 9 move m the new we espeed meeme m e eme 5 thl us me ume mlmvel A men ohseve m m The dock moves ewe new en emoummtmlhls ume interval and the light travels a distance 2L where L2 Lg uAt22 That is uAt2 2L2 7 2L02 cAt2 7 cAt02 Divide through by 02 and solve for At to nd At yAtO where 37 1 y E 4 1 7 uZCZ Obviously y gt 1 Therefore the time interval is longer for the observer on the ground We say the clock ticks more slowly for the observer who is moving This phenomenon is known as time dilation There is a related phenomenon called length contraction See your textbook for the discussion The Lorentz Transformation The equations that give you x and t in terms of z and t are called the Lorentz Transfor mation They are not hard to derive but we wont do it here Just imagine two reference frames one moving past the other and let them coincide origins at their respective zeros of time That is z t 00 when zt 00 Then let a pulse of light move out from the origin and require that both reference frames see it move at a speed c If the relative speed of the reference frames is u then you nd that x yx7ut 39 t yt 7uz02 40 You can of course solve these for x and t in terms of x and 1 but you know that the answer has to be the same as switching u to 7u right That is z yz ut 41 t ytuzCZ 42 It is easy to prove that for the quantity As called the invariant interval A52 E At2 7 Ax2 At2 7 Ax2 That is As is the same for all reference frames Natural Units Why should we bother to carry the 0 around all the time Think of it as a conversion factor between space and time That is measure time in seconds but measure distance in light seconds which is the distance light travels in a second This in fact puts the Lorentz Transformation into an interesting form We have x yz 7 yut t yt7 yuz But 72 7 yu2 71 so we can write 7 cosha for some value of 04 which gives sinha yu In this case the Lorentz Transformation is written as x cosh ozz sinh Ozt t sinhozx cosh ozt This looks rather like a rotation but using hyperbolic cosines and sines instead of circular ie normal sines and cosines Honors Physics Thursday 8 November Fall 2007 Exercise Name This exercise will be graded as a homework problem Please hand it in The Lorentz Transformation given by Equations 39 40 41 and 42 change reference frames in special relativity In this activity you will go through the mechanics of a Lorentz trans formation graphically You are encouraged to check your answers using the equations but by working through this on a graph you will get a better physical feel for what these trans formations are really about Imagine that your friend David is on a new space shuttle heading for a Centauri Your reference frame is designated by x t and David7s by 25 Answer the following questions using the accompanying graph which gives you a z t coordinate axis Get numerical values as best you can The graph includes a dashed line 0 showing z ct Also shown are the x and t axes that is the lines along which t 0 and x 0 respectively The black dots show the points f 00 01 10 071 and 710 H What is David7s velocity in your reference frame You will nd it useful to draw the world line77 for a particle with a xed value of x 2 What is your velocity in David7s reference frame 3 1n David7s reference frame he measures the time between two events to be 1 sec What do you say the time is You can let the events be at the same place in David7s frame 4 In your reference frame you measure the time between two events to be 1 sec What does David say the time is To do this draw a line parallel to David7s x axis 5 In your reference frame you measure the distance between simultaneous events to be 1 light sec What does David say the distance is 6 ln David7s reference frame he measures the distance between simultaneous events to be 1 light sec What do you say the distance is To do this you need to draw a line parallel to David7s t axis 7 Consider an object moving between two points The rst point is at z 0 and t 0 The second point is at z 2 and t 12 How fast is this object moving in your reference frame 8 When and where are these points in David7s reference frame 9 What does this tell you about an object going faster than the speed of light Honors Physics Thursday 8 November Fall 2007 Exercise Solution Imagine that your friend David is on a new space shuttle heading for a Centauri Your reference frame is designated by x t and David7s by 1 Answer the following questions using the accompanying graph which gives you a z t coordinate axis Get numerical values as best you can The graph includes a dashed line 0 showing z ct Also shown are the x and t axes that is the lines along which 1 0 and x 0 respectively The black dots show the points t 00 01 10 071 and 710 H What is David7s velocity in your reference frame You will nd it useful to draw the world line for a particle with a xed value of x Pick a xed point in David7s frame say x 0 For At 1 we nd Ax 12 so u AmAt12 This gives 7 1 1712 2 115 10866 3 What is your velocity in David7s reference frame Now pick z 0 and follow it along for At 1 One gets back to m 0 for Am So u Ax At 712 712 9 1n David7s reference frame he measures the time between two events to be 1 sec What do you say the time is You can let the events be at the same place in David7s frame Pick the two events at x 0 and with t 0 which is also It 0 and t 1 Read it off the 1 axis to get t m 12 115 r In your reference frame you measure the time between two events to be 1 sec What does David say the time is To do this draw a line parallel to David7s x axis Now stick at z 0 and look at t 0 which is also 1 0 and t 1 Draw a line parallel to David7s x axis and see that the t 1 point intersects David7s 1 axis at a little more than t 1 say at t w 12 115 9 In your reference frame you measure the distance between simultaneous events to be 1 light sec What does David say the distance is Pick the two points at t 0 simultaneous and x 0 and m 1 Using the same approach as above it is easy to see that the Lt 10 has a value of x a little large than 1 that is z m 12 115 Dont be confused by the length contraction effect You are the person measuring a length here not David because you7re the one making the measurement using simultaneous events Your length measurement is contracted relative to David7s This is easy to see mathematically using the invariant relativistic interval A52 Alt2 7 A952 At2 7 Ax2 03 5 00 p which must have the same value in all reference frames Therefore if At 0 Ax2 Ax At2 and so As has to be larger than Am ln David7s reference frame he measures the distance between simultaneous events to be 1 light sec What do you say the distance is To do this you need to draw a line parallel to David7s t axis Look at the points on the plot at at t 00 and t 10 These are simultaneous in David7s frame Just read the z value off the graph and get z 115 Consider an object moving between two points The rst point is at z 0 and t 0 The second point is at x 2 and t 12 How fast is this object moving in your reference frame The speed of this object is AzAt 4 that is four times the speed of light When and where are these points in David7s reference frame The later time point is close to t 2 705 What does this tell you about an object going faster than the speed of light The object is moving backwards in time in David7s frame Honors Physwcs Monday 12 November F2 2007 506101 une lest nenn mpm enne wee nemexy mmndvnsmms end Stemsth quotRydsy we Meahsmcs w Sane d 011 may he mm mm yam Chanle class Tennpeupuee A Myscmaus Cancept You en unnk you Im weenpedemeav 15 pm he eenem u e mmpheeled eepeeeuv m une concept nennex 0N whet wnen we went mlhmk about u eenennneeny he dennmen nenee e w edndnpnnnn w I39haexssmnethmgcelled un 1979mm oZhgmdymmm which stems 1 swam A m 5 am eachmzhyvml edeman un t e thud We a wed A m 5 am rat37ml equbenzm we end omen New nnnen e exew ennee we nexenm denned Wme ednnpnnnnyv pm at least n lee us nee eenneunng eened e lhennmnemx m nneeenne eennennge lmp mme nee w be male mag mmlhe meenmgdlmpemuxewhen we study eeeened smmm mechanics adamghulk memes el une nnexeenxen eeede ene ene common mnpeneme eeedee need In Everydayhheehhm31 nennen 5 used nnnen by phvsmsts w e even eunnd ehsnluld lmpexelure eeede Tne Wm 11de eeeJee exe pennednew Tye ned py une 199mgme e pnne end pedy mmpmelure end ceenne Tce denned py Lhe eezmgend pedmg Pan39s afpure wew Tne Kelvm T eeede e peeed en ehsnlum zmd end une eeeeued inpls 17mmquot afpure water We neve Tc T e m 15 a TF ch 32 Tne ndeenx ebsdule zmd wm pe eeenen eaen we lAexn some ung about gees By exmepexemg dewn nm Ymm lmpmeluxe n eeenne that pxeuy nnnen en was would have movdumen wemhdlhan mT 0 Thmnal Expansan Mm ends expend whm uneygm hm Tneneeeen eenpue m m end 2142 mam zennp nneene mm eneeeee Sq sends expend fgd on n Tne Jim 1 penned lmeex end na Iexg mnin pennpedemnee 15p xuimi we wme A e1 AT u wnede a Is eened une mm o Lmsvz39r ewm expresses une Miami chsnga m xengun AL1 degee dmmpmelure enenee end venee from mild to exdnd See male m Sme qe ectmnsl dnenee I smell we expend AA ZaA AT bx expennenexeuneee exee endAV my AT for vnlume Rehehhe me new exmmms use the uids hhe welg ee mee eehptheted theh mild The meal Gas A ehehhst w1hten youthet mest gee et norms memes ehdtehpeetuhesreuew 11V nRZ u whee 11 1s the meme T 1s the ehsdute ehpeetuse V 1s the vnlume e the eehtem end B 1s e eesteht use w 1s the humhe e1 males e1 gas Wheth e mexev Let us try te uhdesteha th1s es phvsmsts In the Plume we wm gm e hette reehng a1 whet empeetmew eetueuy meehs See Hem put N gas molecules eeeh afmess m 1h e exeeuse hes z veheety e 1t beam 0 the we end mus ehe1t ch The ma hetwee hmhees 1s me se mum 2m 7 Ihepzeeuseehthewenete L1sthesum dchefacs en the m s dwlded 1 we Nm 1 FEE PET TltFE gt whee we heme the the wehheethe ehe s y L2 w the 1heht1ty 1h peethese 1s the eveeee value em wnteh vi Thee 1s nmehmgspem t the 2 1111mm end the humhe N 1s veyxeee se we expect thet e5 e3 e3 gm whee we teen er he eh ghee eeetyw ahee en mdecules hee the see mess K gem 1s the eveheg h1het1e e1eey e1 the mdecules So we eh hew wnte 11v wax Beck mwhsl ahemms m1 e mde Thu 1 the number dmnlmulgs dmded by Avanm s huhhe N1 e 02 x 10m Thsl t NNA quotHumeng we heme E11 11 wth pVTtRZ where K rkT end B m1 Th1 shew thet mpeettue 1e the me 95 meesuhes the eymeg kmm e1eey e1 the mdAcuhs Bqlenmmle n we hha h 1 38 x lojlKe wh1eh we en Boltzmann s eehe steht w Physmstsamt usueuy usede e11 N1 e11 1hstee11 wnte the eelyslew es 11v NIcT A5 wm tehe th1s meleehe ewmah mlhmnndynsmms tethe e Thesde pmme Eeeeses 1 Deemhe the tempeetme et wh1eh the pehhehet end cesms sexes eehehae the 1st me the see value b the mpeettue Heeyeu ee heeh ou39sldem ede e heht whe the we the eh tehpeheteev 2 A lexg el slab emetet hes ehexe 1h themhhneent Thesxeh 1s wemed bymcleeml lg 1ts tempeetme Dees the dismem e the hexe heheese aeeeese mm the see e depend m theme exp s1 1e1t e the p t etew a qulsm why he deemtm K 31cT m the tempeetme e1 eh 111ee1 9s wmld he male se1se 11we whee 1t 1h tems e1TF e Tc 1hstee11 e1 ehsehte lmp mme T Honors Physics Thursday 15 November Fall 2007 Today How to build a star I was going to do Maxwell Boltzmann distribution for molecular speeds but this will be more fun Note I will be traveling before and after Thanksgiving so Monday Nov19 Heat and the First Law will be covered by John Cummings then Monday Nov26 Entropy will be covered by Angel Garcia and then Exam3 on Thursday Nov29 on material through Monday Nov19 How to Build a Star Hydrostatics of really big ball of gas Start with the balance of pressure against gravitation Then use ideal gas law to relate to temperature Will assume that stars like the sun have uniform density Finish with the behavior of white dwarf stars aka degenerate Fermi gas More reading in order of increasing sophistication 0 Introductory Astronomy and Astrophysics 4th Edition Zeilik and Gregory esp Chap10 o Order of rnagnitude theory of stellar structure George Greenstein Am J Phys 55 804 1987 Erratum Am J Phys 56 94 1988 0 Stars and statistical physics A teaching emperience Roger Balian and Jean Paul Blaizot Am J Phys 67 1189 1999 Begin First set up the notation Build our star with shells of radius r Total mass of star is M and radius is R Density is p M47rR33 assumed to be independent of r Write Mr be the mass enclosed at radius r ie MR p47rr33 MrR3 Now balance pressure pr against gravity Take a small piece of area A thickness dr and mass dm pAdr at radius r Outward force from difference in pressure must equal the gravitational force from mass inside radius r Recall shell theorems Write it out pdA p dpA GiMOEdW G MT Adr or r r dp 7 MUM a 7 7G7 46 This equation expresses hydrostatic equilibrium for a large spherical self gravitating object like a star or a planet or a moon It is written here in a way that holds for any density function pr but we will be taking p to be a constant Let7s use this to determine the pressure p0 at the center of a star ie r 0 Make the replacements for constant density in Eq 46 to get dp i ltrgt3 M 1 3GM2 7 3GM2 2 dr 7 pr 7pc 87TR6T MW iTMWT R At the surface of the star the pressure is zero by de nition ie pR 0 Therefore igow pc 7 87TR4 47 Next let us try to estimate the temperature at the center of a star This requires us to know the equation of state which relates the gas pressure and density to the temperature for the matter that makes up the star Understanding the equation of state is a big deal and an area of active research interest For now we will use an equation of state based on an ideal gas that is Eq 45 namely pV NkT If we assume that the star is made of N identical particles with mass m then p NmV and p kam 347T MmkTR3 In fact stars are hot so hot that matter is a plasma of free nuclei and electrons The nuclei are from mostly hydrogen about 25 helium and a few percent of heavier elements For now though it is close enough to assume the star is 100 hydrogen atoms Therefore the temperature at the center of a star is i 1 47TR3 m 7 1 GmM C 7 E 3 Mp0 7 2 2R Can you see why we would not be able to get this answer from dimensional analysis The White Dwarf Star 48 We found some properties of stars assuming they consisted of an ideal gas but the assump tions used to get Eq 45 don7t always hold Instead we need another equation of state In deriving Eq 45 we wrote Fm pmvwL and that is ne For N particles this again gives a pressure P NFmL2 NpUV We temporarily switch to P to avoid confusion with momentum Also we are making rough estimate so put pm and 111 to p and 1 Our concern will be with electrons so write an electron density 715 NV in which case P nepu At very high densities the ideal gas law assumptions are violated The electrons are so close together they start to violate the Pauli Exclusion Principle The distance between them is d Uni3 and the Heisenberg Uncertainty Principle says that pd 71 so that p ling3 Putting i pme and switching back to p for pressure the equation of state becomes 712 2 12 232 53 13 13 7 53N7 537 p 715 ne 7mg me mane meN m8 49 where we make the somewhat incorrect assumption that the star is still made up of hydro gen atoms with mass m This is an odd sort of equation of state Combined with Eq 47 it leads to a star whose radius decreases with mass like 1M13 See the practice eccercise As the mass increases and the radius decreases the electrons are more and more con ned until they are moving as fast as they can ie i c The equation of state becomes 43 p 715 7171 c he iii3 he N l3 hc 50 Combined this with Eq 47 gives an equation where the radius drops out One solves for a mass that is the highest possible mass of a white dwarf star Doing a more careful job leads to a value of 144 solar masses This is the Chandrasekhar limit77 after the physicist who rst derived it What do you suppose would happen for a star of mass larger than this Practice Exercises 1 Use Eq 47 to estimate the central pressure ofthe Sun Express it in units of atmospheres Use values which you can look up in Appendix C of your textbook Then use Eq 48 to estimate the central temperature 2 Recall that temperature is related to the average kinetic energy of the gas particles What is the average kinetic energy of the hydrogen atoms in the center of the Sun Does this give you a hint as to the source of the Suns energy 3 Show that Eq 49 leads to a star77 with radius R and mass M where R x 1M13 Honors Physics Monday 19 November Fall 2007 These notes prepared by John Cummings Thermodynamics is the only physical theory of universal content which within the framework of the applicability of its basic concepts 1 am convinced will never be overthrown 7 Albert Einstein First law of thermodynamics We all as physicist have great faith in the law of conserva tion of energy Yet we seem to see it violated every day constantly If I slide a book across the table to a friend the books slows and stops hopefully close enough to my friend for him to reach it Has energy been lost Where did it go JP Joule the English physicist carried out careful measurements around 1850 to show that mechanical energy was converted into heat Using falling weights to drive mechanical agitators in tanks of water and carefully measuring the temperature of the water tanks he showed that the mechanical energy was showing up as heat in the tanks Once we understand that heat is another form of energy this allows us to write the law of energy conservation for a thermodynamic system as Q W AEW 51 where Q is the heat that ows into the system W is the work done on the system and Em is the internal energy of the system Internal energy What is this internal energy It may seem like a cheat at rst to save conservation of energy but lets take advantage of what we know about ideal gasses to understand it better You know that the average kinetic energy of an ideal gas molecule is 3 ltKgt 5km lt52 and so for a sample of N molecules we see 3 Em N kT 53 So the internal energy is determined by temperature at least for an ideal gas and is the total kinetic energy of the gas molecules It turns out as derived by JC Maxwell that the theorem of equipartition of energy says that the energy of a molecule is shared equally among all of its independent degrees of freedom 7 kT for each degree of freedom If a molecule has more degrees of freedom such as a diatomic molecule that has 2 rotational degrees of freedom the internal energy will be given by N kT 54 at low temperatures like room temperature for most molecules At higher temperatures vibrational modes become available and the internal energy becomes Nng 55 Heat capacity The heat capacity of a body is de ned to be i o 7 AT 56 where Q ls the heat energy transferred to the body and AT ls the resultlng temperature change Thls ls often normallzed to a unlt mass to obtaln a property of the materlalgt rather than a partlcular ohlect glvlng C Q 7 57 c m m where c ls called the cpecc c heat of the materlal k A large area speci c heat means a glven amount ofwater heat wlll ralse the level temperature very llttle whlle a small area speci c heat means the level temperature a lot W rk n an ideal gas The work W done on a thermodynamlc system can be studied by conslderlng a cyllnder and plston wlth cross sectlonal area A contalnlng an ldeal gas at a pressurep The work done on the gas as we move the plston ls WFsdatrpA dc 52 W e p dv 59 Thls suggests a very useful way to vlsuallze work done on a thermodynamlc system the area under a pV cuwe A t m moves from one state to another lt traces a path on the W dlagram and the work done on the system ls the area under the curve 5 E w Vt V 2313 There are many ways to get from one polnt to another on a pv dlagram and not all wlll use or produce h l ls thls fact hl h L to the system ln a way that changes heat lnput lnto work 7 a heat englne Practice Exercises 1 What ls the shape of an lsotherm a llne of constant temperature on a pV plot 2 When I open my dlshwasher after the dry cycle glassware and sllverware are dry but 7 plastlc contalners are stlll We Why a Whlch wlll bum your mouth more plzza stralght from the oven or bread stralght from the same even Why 4 I want to have hot colleegt hutl have no electrlclty so I hook up a paddlewheel to a pulley and 1 kg welg t M cup holds 0 5 and ls at 20 C The welght qulckly reaches termlnal Velocityl How fax must it fall to boil the water Honors Physics Monday 3 December Fall 2007 Last real class Where did the time go Hand back Exam3 if available Final exam Thurs 13 Dec 3 6pm Ricketts 203 see sisrpiedu No lab on Wednesday We7ll be available for help on lab books7 and in general Thursday Homework due7 Lab books due7 class devoted to course review Thanks to John Cummings for class on First Law of Thermodynamics Thanks to Angel Garcia for class on the Second Law No notes7 but there will be one problem on the nal exam based on homework from Chapter 24 Need to leave time at the end of this class for course evaluation forms The Principle of Least Action What is Physics I think of it as asking Why and not being satis ed with the answer you get Physicists are always trying to understand why nature is the way it is7 and looking for the simplest answer to that question The best answer we have today for Why is Because it minimizes the action This quantity action is built from various symmetry principles that are realized differently in different physical systems A physical system will behave in the way that minimizes the action from the start to the nish of some process As you learn more physics7 you will see how the action is constructed for mechanical sys terns7 electromagnetic systems7 and the role that action plays in the formulation of quantum mechanics Try looking up something called Fermat7s principle and its applications to op tics Today7 though7 well do one simple example7 which connects directly onto the physics with which we started this course Example Motion under Constant Acceleration Consider some object with mass m moving with a constant acceleration 1 Obviously the force is F ma and the potential energy function is Uz imaz for motion in one dimension z Let7s form a quantity L that is a function of z and 2 by 1 1 Lz7 57mg 7 Uz Emz z maz 60 Even though Lzz 7 called the Lagrangian 7 looks kind of like the total energy E except for the 7 sign7 it is a very different beast so dont confuse the two Before we de ne the action7 we need to introduce the concept of a path For one dimensional mechanical motion7 a path through time is some way of getting from position zl starting at time t17 and ending up at position 2 at time t2 There are an in nite number of paths connecting the points zhtl and 27t2 Here s the Why question Why does the object follow the path that it does We know from our physics studies so far that an object undergoing constant acceleration follows the path zt iatz Dot zo but we got there from Newton7s Laws Newton was a smart guy but he was only talking about motion and not some vast underlying principle Can we get a better answer to our Why7 question The better answer is this The object follows the path zt which gives the smallest value for the action which is the integral of Lz over the path 5 t La 92 dt 61 The notation St indicates that you get a different value of S for a different path xt but of course there is no dependence any more on x since it disappears after the integral is carried out Let7s work out the integral for the path we believe is correct ie t at2 votx0 You will carry this out for other paths in the practice exercises and you7ll nd that your answer is larger than what we get here assuming the Principle of Least Action is correct Measure m in meters and t in seconds and start the object out at zhtl 00 and go through to zgt2 11 The righ 77 path that connects these two points is quadratic in time so t t2 ie a 2 Equation 60 becomes Lz 2mt22mt2 4mt2 So 5 01 4th dt gm There you7ve just calculated the action for the path zt t2 The units of action are Energygtlttime MLZT so we really have S 47713 mZsec in this example This value itself isn7t so important What7s important is that this should be the smallest value you can get for S for any path t that connects 1111 00 and zgt2 11 In later physics courses you will use something called Calculus of variations77 to show that minimizing the action is equivalent to writing mi idUdm F that is Newton7s Second Law Even Maxwell7s Equations can be derived by minimizing the action The Principle of Least Action has a deeper meaning than Newton or Maxwell Practice Exercise Try some other path that connects zhtl 00 and zgt2 11 and show that the action Sxt is something larger than 47713 Any path t It will do if n 2 1 Whats wrong with zt You might want to show that n2 2 S t 7 K H m 227171 n1 and plot S for different values of 71 Another path that you can try is t sin7Tt2 The integral is only a little tricky but you can nd it in your Schaum7s outline The answer works out to something involving terms with powers of 7139 but again the result is greater than 47713 Can you think of other paths to try Honors Physics Thursday 6 December Fall 2007 Final Review The nal exam will be on Thursday 13 December from 3 6pm in Ricketts 203 see sisrpiedu The format is 20 multiple choice problems worth two points each no partial credit and six short problems worth ten points each Here7s a quick summary of what we learned this year 0 General Topics 7 Dimensional analysis 7 Uniform circular motion Einstein7s equivalence principle Coupled oscillations and the eigenvalue problem 7 Introduction to special relativity and Lorentz Transformations The Principle of Least Action as a foundation for physical law 0 Mathemtics topics 7 Vector algebra Dot products and cross products 7 Taylor expansions Complex numbers and imaginary exponents Line integrals and surface integrals Vector elds and the divergence theorem 0 F ma as the equation of motion 7 Physics in terms of differential equations 7 Motion in two and three dimensions vectors Constants of the motion eg Z and E Frictional forces Drag forces proportional to velocity 7 The simple pendulum 0 Work and potential energy 7 Work as W d Potential energy as F 72 7 Path Independence and Conservative Forces 7 Various typical potential energy functions 0 Gravitation and orbits Newton7s law of gravity Gravity near the surface of the Earth 7 The shell theorems Gravitational potential energy 7 Circular orbits The Dark Matter problem Elliptical orbits and Kepler7s Laws o Oscillations in one dimension Harmonic approximation for potential wells77 7 Simple harmonic motion Amplitude and phase Damped harmonic motion Forced oscillations 0 Systems of particles and solid objects Momentum and impulse Center of mass Conservation of momentum Calculating the center of mass for solid objects Rotational kinematics Angular velocity and angular acceleration as vectors 7 Torque rotational inertia and 739 oz Calculating the rotational inertia of solid objects 7 The parallel axis theorem Angular momentum of a particle and a solid object 7 Conservation of angular momentum o Fluids Hydrostatics Pascal7s Principle and Archimede7s Principle 7 The Continuity Equation and Bernoulli7s Equation 7 The Continuity Equation in terms of vector elds 0 Waves 7 Properties and mathematical description of waves Waves on a stretched string The wave equation Superposition and standing waves 7 Sound waves 0 Thermodynamics Temperature heat and the First Law 7 Work and heat in closed cycles Entropy and the Second Law 7 Kinetic theory and the Ideal Gas Law 7 Thermal expansion and speci c heat 7 Static properties of stars White dwarf stars

### BOOM! Enjoy Your Free Notes!

We've added these Notes to your profile, click here to view them now.

### You're already Subscribed!

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

## Why people love StudySoup

#### "I was shooting for a perfect 4.0 GPA this semester. Having StudySoup as a study aid was critical to helping me achieve my goal...and I nailed it!"

#### "When you're taking detailed notes and trying to help everyone else out in the class, it really helps you learn and understand the material...plus I made $280 on my first study guide!"

#### "Knowing I can count on the Elite Notetaker in my class allows me to focus on what the professor is saying instead of just scribbling notes the whole time and falling behind."

#### "It's a great way for students to improve their educational experience and it seemed like a product that everybody wants, so all the people participating are winning."

### Refund Policy

#### STUDYSOUP CANCELLATION POLICY

All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email support@studysoup.com

#### STUDYSOUP REFUND POLICY

StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here: support@studysoup.com

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to support@studysoup.com

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.