GENERAL PHYSICS 1
GENERAL PHYSICS 1 PHYS 1110
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CT9l At a particular instant two asteroids in intergalactic space are a distance r 20 km apart Asteroid 2 has 10 times the mass of asteroid l The magnitudes of the accelerations of asteroids l and 2 are al and a2 respectively a1 What is the ratio 5 m2 10 m1 m A 10 1 B 110 D C l D cannot be determined CT92 A satellite is in circular orbit at an altitude of 100 miles above the surface of the Earth M Earth The satellite s prelaunch weight is its weight measured on the ground The magnitude of the force of gravity on the satellite while it is in orbit is slightly greater than its prelaunch weight the same as its prelaunch weight slightly less than its prelaunch weight much less than its prelaunch weight but not zero zero FPO CT93 Planet X has the same mass as the Earth but 12 the radius Planet X is more dense than Earth What is the acceleration of gravity on Planet X A g same as Earth B 2g C 4g D 8g E None of these CT94 A rock is released from rest at a point in space far from Earth beyond the orbit of the Moon The rock falls toward the Earth and crosses the orbit of the Moon When the rock is the same distance from the Earth as the Moon the acceleration of the rock is Ignore the gravitational force between the rock and the Moon A greater B smaller C the same as the acceleration of the Moon Moon 0 rock As the rock falls toward the Earth its acceleration is A constant B not constant CT95 At time t0 a satellite in circular orbit about the Earth is directly over Denver 300 miles above the city and traveling eastward at 16000 mph At the same time a rock is released from rest 300 miles above the city very near the satellite 0 satellite rook True or False at t0 the accelerations of the rock and the satellite are identical in magnitude m direction A True B False CT96 Two satellites A and B are in circular orbit around the earth The distance of satellite B from the center of the Earth is twice that of satellite A What is the ratio of the magnitudes of the accelerations ofAto B 21A a A 1 B22 C4 B D 12 E 14 w CT97 A planet in elliptical orbit around a star moves from the point in its orbit furthest from the star A to the closest point P The work done by the force of gravity during this movement is A zero B Positive C Negative The planet executes one complete orbit starting from point A and returning to A The work done by the force of gravity during this orbit is A zero B Positive C Negative CT98 T2 Kepler39s third law states that the ratio 13 is a constant for all the planets The period T of the Earth is 1 year An astronomical unit 1 AU is defined as the mean distance from the Earth to the Sun therefore the mean EarthSun distance is lAU Consider an asteroid in circular orbit around the Sun with radius r 2AU The period of the asteroid is A 2 years B 3 years C 232 E 283 years D 223 2 159 years E None of these CT99 Does escape velocity depend on launch angle That is if a projectile is given an initial speed V0 is it more likely to escape an airless planet if fired straight up than if fired at an angle A Yes B No CT9 l 0 A satellite is in circular orbit around a planet that has a very tenuous atmosphere extending up to the altitude of the satellite Due to atmospheric drag the speed of the satellite A increases B decreases C remains constant CT9l 1 Suppose the Earth had no atmosphere and a projectile was fired from a mountain top with sufficient speed to put it in circular orbit The magnitude of the acceleration of the projectile While in orbit would be A much less than g because it doesn39t fall to the ground B much greater than g C approximately g D Impossible to tell A CT912 Suppose a projectile is fired straight upward from the surface of an airless planet of radius R with the escape velocity vesc meaning the projectile will just barely escape the planet39s gravity it will assymtotically approach infinite distance and zero speed What is the projectile39s speed when it is a distance 4R from the planet39s center 3R from the surface A 12 vesc B 14 vesc C 19 vesc D 13 vesc E None of these is correct Ignore the gravity of the Sun and other astronomical bodies CT9l3 An object moves along the XaXis The potential energy UX vs position X is shown below UX XA When the object is at postion xA which of the following statements must be true A The velocity is negative ie along the X direction B The acceleration is negative C The total energy is negative D The kinetic energy is positive E None of these statements is always true CT9l4 A planet is in elliptical orbit around the Sun The zero of potential GMm energy U is chosen at r 00 so U0 How does the magnitude of U U compare to the KB A lUlgtKE B lUlltKE C lUlKE D depends on the position in the orbit CT9 l 4 Astronaut Dave Bowman is standing in the centrifuge of the spaceship Discovery He drops his pen and observes it fall to the floor Which statement below is most accurate A After Bowman releases the pen the net force on the pen is zero B The pen falls because the centrifugal force pulls it toward the floor C The pen falls because the artificial graVity pulls it toward the floor ll ti Set frequency to BA Q An astronaut in intergalactic space is twirling a rock on a string Suddenly the string breaks when the rock is at the cyan xx does the rock follow after the string breaks 39 0 Before the string breaks the astronaut must be exerting a force on the rock causing the centripetal acceleration D C After the string breaks no net force is acting so the rock travels at constant velocity which was horizontal to the right tangent to circle at the moment the string broke More on forces CAPA homework due at 5pm Exam 730 900pm tomorrow Room assignments on web page 15 multiple choice questions 2 sets of long answer questions Should bring 2 pencil scientific calculator and 1 side of 812 X 11 sheet of paper with notes Chapters 15 are covered httpwwwcoloradoeduphvsicsphvs11 10 Review of what is covered Ancillary stuff Vectors Chapters 13 and 3 Getting components from trigonometry Getting magnitude and direction from components Addition and subtraction Units and unit conversion Chapter 18 Review of what is covered Motion Displacement is a vector measuring change in position Velocity measures displacement per time Avera 39 quot7 r 39 39 ge veIOCIty avg At Instantaneous velocnty vdt Acceleration measures change in velocity per time v Average acceleration c7an Instantaneous acceleration c7 E Often deal with constant acceleration such as free fall Where ay 2 g Review of what is covered Using diagrams and graphs Understand and be able to draw motion diagrams Understand and be able to draw position velocity and acceleration versus time graphs Understand how to get information about one quantity by looking at another quantity For example Velocity can be found from slope of position vs time Change in velocity can be found from area under acceleration versus time graph Review of what is covered Motion with constant acceleration One dimensional equations of motion ax 0 so xf xivix Ar and vfx vix Projectile equations 2 V v At ivi At i At of motion fy 1y g yf y y 28 VtgyZV12y gyfyi yfzyiviyvfyAt ll Set frequency to BA M x Q Two stones are dropped into a bottomless pit the second stone is dropped 2 seconds after the first stone Assume no air resistance As both stones fall the difference in their speeds A increases B decreases IC stays the sameI D Impossible to tell Both stones velocities are increasing at the same rate so the difference between them 196 ms stays the same Review of what is covered Circular motion amp relative velocity Circular motion even at constant speed is an example of acceleration because the velocity vector direction changes Can divide accelera tcilor i into tangentialzand radial v v v components atan y and arad 7 Relative velocity A reference frame which moves at constant velocity can be 0 is an inertial reference frame Use vector addition of velocities to transfer from one reference frame to another Set frequency to BA Q A car traveling at constant speed rides over the top of a hill moving right At the top of the hill the acceleration is A Zero 2 m D Right E Another direction The speed is constant so there is no tangential acceleration The velocity vector direction is changing so there is radial acceleration which is perpendicular to the velocity and in this case points down Review of what is covered Forces amp Newton s Laws El r LI 397 1 2 1 9 1 gt T 1 m 1 l l r icy gt Lix a Hrf v Jil J 32 54 2 Lilli ll 11 icilll ll fki lull ill Lll ll till Elsiquotwilfullyl l ljlrl rci l gltl irci l l l l Newton s First Law A body maintains a constant velocity which can be 0 unless acted upon by a net force Newton s Second Lawzz mc7 or ZFEWQ amp may Newton s Third Law If body A exerts a force on body B then there is a force which is equal in magnitude but opposite in direction by body B on body A FA F onB BonA Set frequency to BA Q An object is being lowered on a cord at a constant speed How does the tension T in the cord compare to the force of gravity on the object A T gt FG l C T lt FG constant velocity D Impossible to tell T Constant speed in a straight line means acceleration is 0 so 2F 2 0 and therefore FG ZFsz FG20 so TFG I Exam tips Remember that displacement velocity acceleration and force are vectors and vectors have a direction and magnitude Understand the CAPA problems clicker questions and tutorials Try the example tests available on the Exam Info link using just your notes and 15 hours Take your time and check your work I Force review Force is a vector so superposition of forces is true can combine all forces into one net force using vector addition We draw a free body diagram to figure out the effect of the various forces and to calculate the net force Newton s 1st law A body will stay at a constant velocity unless acted upon by a net force Newton s 2ncl lawZ me Can split max 8 25 may A net force causes an acceleration which is inversely proportional to the mass of the object 6ij gig rm that WE k Force of gravity all feel the farce of gravity It Keeps the Earth tOgether and us from floating Off l g 1 mg E ml k g fi lm LNW m9 21 time f gifgv m L lo I Units of force From Newton s 2nol law we can deduce the units of force F 2 me has units of kgms2 also known as a newton N That is IN 1 kg ms2 Mass is an intrinsic property of an object independent of where it is measured The force of gravity depends on the gravitational field the object is in In SI units kilogram is a unit of mass and newton is a unit of force including weight In English system pound measures force including force of gravity Slug is the unit of mass lpound 1 slug fts2 A 1 kilogram mass experiences a gravitational force of about 22 lbs on Earth PhySIcs 1110 lall 2004 Lecture 34 12 November 2004 Announcements 0 Reading Assignment for Mon Nov 15 Review for Exam 3 Exam 3 Tuesday Nov 16 7309pm Same rooms as Exam 2 Covers Knight Ch 811 Monday Review in class QampA 57pm Ramaley 0250 Elliptical Orbits and Potential Energy Now we cannot use our centripetal acceleration formula since the radius is not constant Since the force will now change with distance between the sun and a planet for example this means the acceleration is not constant and we would need to solve differential equations to derive a solution However we can do an energy analysis if we use the potential energy U that comes from Newton s law for gravity By considering the work when moving an object m from r to rf while feeling the force from a planet with mass M we can derive that the correct function is GMm r U This formula can be confusing ltjust depends on the distance between the centers so it extends in all directions from a planet Also it starts from a very negative number and grows with distance Don t be confused by considering the magnitude of U At small r it is very small large magnitude and negative Also it appears to blow up to infinity when r0 But in fact this never happens the objects hit each other first and then we have additional contact forces to prevent them from going to r0 Since it is a potential energy we can always just imagine a ball rolling down the potential energy curve it will roll towards r O as it should and potential energy grows the higher above the ground greater r you go The figure shows an energy diagram for this potential energy Note that if we also plot a particle s energy E as a constant at some value there are 2 very different behaviors If E gt 0 then there is no turning L341 Physms M m FaH 2mm pumt and the pamde eseapes WE lt u n strapped andthat means mean never gm greaterth n a Ertam dws tance frumthe mass M Tm svvhat an arm s As usuakvve see thatvvhen the pamde gets muse m M 1 Wyeth Energy gruvvs and wnen n muves away u get 5 qu 5 m an eunmeax ermmne p anet speed nearlhe sun pumt a s rnuen fasterthan Wnen n 5 far away pumt b Nate thattms means that gravm dues Wurk m an EWpuca mm 7 KE 5 changmg Egt0 EscapeVelocity we Wantan emeeue eseape an urbwt vve need m make sure that n hastuta Energy greater than n We ean mug mm m nd nd the ve umty at a gwen mstanee rvvmch s necessary m emmty ve a escape ms sthe Escape v L344 PhySICS l l lU iaII 2004 lfwe start from the surface of the earth M is the mass of the earth and r is the radius ofthe earth Putting in the numbers yields an escape velocity of 11200 ms 25800 mph Note that it doesn t even matter whether you shoot the object horizontally or vertically it still escapes Energy of a Circular Orbit In the case of a circular orbit we can actually get a formula for the energy ofthe orbit as a function of radius Again we use that in a circular orbit Frmv2r ECmv2 GM r M Fr G m r GMm GMm r2 r GMm 2r Note this is just half the potential energy ofthe orbit Rotation of Rigid Bodies We now proceed to a more advanced treatment of motion where we can no longer use the particle model but must account for the fact that object have size and shape We will find that for this more complicated case we can breakthe motion into two parts the translation ofthe object through space our standard particle trajectories combined with the rotation of the object about its center of gravity Before solving this more complicated case we will focus on just the rotational motion As usual we will start with kinematics We developed angular variables for circular motion Because any piece of a rotating object is undergoing circular motion we will be able to use the same variables here L343 Physms 1MB FaH 2mm Rotational Kinem atics RecaH that when we use angu arvartab es they usuaHy must be wth ragtahs m urderfur the turmutas D be shhpte Thehgure shuws a pteee ufmrcmarmutun ch S rcw R Recachat t we use ragtahstu gesehhe the ahgte a then we eah whte x V we huwtuuh athuwtastthe ahgte changes that S WE take the We dErWaUVE utthts eguatmh we get E SR t t The ahgutary tty m ts very usetut m mtattuhat muttuh because whtte aH the WM guat pteees uf ah uhteet are muvmg wth gtttereht regu arve umues they aH haye the same ahgutarye u ty Fur t exampte 3 Wm LF39 uh a turhtahte h s uter Edge ts muvmg mueh fasteraruund than the WHEr em M t rpmmutthe g Edge FmaHy ttwe tet the ahgutaryetuety hangE WE get ah ahgutar aeeeterattuh usuaHy gehuteg wth the greek Etterot ttts dE Ed ahatuguustytu regutar acceharatmm asthe UmE dErWatWE uf the ahgutaryetuetty at VA 42 e SWEE the angu arpusmun ve umw ahg aeeeteratmh are retateg by regutarttme dEHVaUVES we eah tmmegtatety whte EIer the turmutas fur euhstaht aeeeteratmh usmg the ana ugy wtth the regutar euhstaht t r uht mutas 6 Q w AIaAI2 w w 01At w w 201A6 Eefure turmng tn mtattuhat gyhamtes WE H next gtseuss the pmnt abuutwhteh abject rutate L34 7 4 lDl Motion in one dimension 1D Chapter 2 in Young amp Freedman In this chapter we study speed velocity and acceleration for motion in onedimension One dimensional motion is motion along a straight line like the motion of a glider on an airtrack speed and velocity speed s units are ms or mph or kmhr or t1me elapsed distance traveled d t speed s and distance d are both always positive quantities by de nition velocity speed direction of motion Things that have both a magnitude and a direction are called vectors More on vectors in Ch3 For lD motion motion along a straight line like on an air track we can represent the direction of motion with a 7 sign going right gt always 7 going left vA710ms vB10 ms Objects A and B have the same speed s lvl 10 ms but they have different velocities If the velocity of an object varies over time then we must distinguish between the average velocity during a time interval and the instantaneous velocity at a particular time h 39 39t39 AX De nition average veloc1ty v E w change 1n t1me At X1 6 X2 l v X 0 initial nal V Xr Xi X2 X1 E tf ti t2 t1 At AX X na 7 Xinmal displacement can be or 7 162009 University of Colorado at Boulder Notice that A delta always means quot nal minus initialquot is the slope ofa graph ofX vs t Review Slope of a line 7 yz y1 AX X2 7X1 y Y y 0 slope slope 7 Slope X X X Suppose we travel along the XaXis in the positive direction at constant velocity v start rise A AX V n At yaXis is X XaXis is t 162009 University of Colorado at Boulder Now let us travel in the negative direction to the left at constant velocity start lt o lt o lt o lt o gtx 0 X AX slope v E lt0 At At t AXlt0 Note that v E constant 3 slope of X vs t E constant 3 graph of X vs t is a straight line But what if v at constant If an object starts out going fast but then slows down and stops x slo e E 0 sto ed Slower p pp slope gt 0 fast X The slope at a point on the X vs t curve is the instantaneous velocity at that point De nition instantaneous velocity E velocity averaged over a very very short infinitesimal time interval AX d X v E A11m E E E E slope of tangent lme In Calculus class we would say that the t E 0 velocity is the derivative of the position with respect to time The derivative of a function Xt is defined as the slope of the tangent line d X E 11m dt Atgt0 At 162009 University of Colorado at Boulder tangent line I I t X t I fast Islow V dxdt t Acceleration If the velocity is changing then there is nonzero acceleration Definition acceleration time rate of change of velocity derivative of velocity with respect to time In 1D instantaneous acceleration a E lim d V Atgt0 At dt Av average acceleration over a nonin nites1mal time interval At a E m s m units ofa a 2 s s Sometimes I Will be a bit sloppy and Just write a XV where it understood that At is either a t in nitesimal time interval in the case of instantaneous a g At is a large time interval in the case of average a 162009 University of Colorado at Boulder dv Av vf vi vz v1 dt At tf ti t2 t1 o vconstant 3 Av0 3 a0 0 V increasing becoming more positive 3 a gt 0 0 V decreasing becoming more negative 3 a lt 0 In 1D acceleration a is the slope ofthe graph of v vs t just like v slope of X vs t Examples of constant acceleration in ID on next page 162009 University of Colorado at Boulder Examples of constant acceleration in 1D Situation 1 An object starts at rest then moves to the right direction with constant acceleration going faster and faster 1 2 3 4 O H gt gt V 4 a gt 0 a constant 3 AV a constant since V vs t is straight 2 At t I1 situation 11 An object starts at rest then moves to the left 7 direction with constant acceleration going faster and faster 4 3 2 1 lt 4 4 O V 1 t 2 At A a lt 0 a constant 3 V since V vs t has constant negative slope 4 Situation Ill 1 2 0 0 o 3 5 4 V 1 alt0 aconstant since v vs t has constant negative slope 2 t 3 4 5 162009 University of Colorado at Boulder The direction of the acceleration For 1D motion the acceleration like the velocity has a sign or 7 Just as with velocity we say that positive acceleration is acceleration to the right and negative acceleration is acceleration to the left But what is it exactly that is pointing right or left when we talk about the direction of the acceleration Acceleration and velocity are both examples of vector quantities They are mathematical objects that have both a magnitude size and a direction We often represent vector quantities by putting a little arrow over the symbol like 7 or a direction of a art direction of v direction of a the direction toward which the velocity is Mg art direction of v Reconsider Situation I previous page 1 z 1 is an earlier time 2 is a later time A A gt v1 velocity at time 1 7 vim A gt v velocity at time 2 7 V nal Av quotchange vectorquot how v must be quotstretchedquot to change it into V2 VI Aquot A A gt direction of a d1rection of Av T V2 Situation II AAV 1 In both situations I and II AV is to the left so acceleration a is to the left Situation III This has been a preview of Chapter 3 a Our mantra quot Acceleration is not velocity velocity is not accelerationquot 162009 University of Colorado at Boulder Constant acceleration formulas 1D In the special case of constant acceleration a constant there are a set of formulas that relate position X velocity V and time t to acceleration a formula relates a V V0 at V t b X X0 Vot 12at2 Xt c V2 V02 2aXX0 V X 0 X0 V0 initial position initial Velocity X V position Velocity at time t Reminder all of these formulas are only Valid if a constant so these are special case formulas They are not laws Laws are always true dV Proof of formula a V Vo at Start With de nltion a E AV V 7V In the case of constant acceleratlon a a 2 1 At t2 7 t1 Since a constant there is no difference between aVerage acceleration a and instantaneous acceleration at any time 2 V1 V0 V V VO a 5 VV0at t10 t2t t See the appendiX or your teXt for proofs of the remaining formulas Example Braking car A car is m0Ving to the right with initial Velocity V0 21 ms The brakes are applied and the car slows to a stop in t 3 s with constant acceleration What is the acceleration of the car during braking F a leSi7msz At t 3s Do you understand why we haVe set V 0 in this problem NegatiVe acceleration means that the acceleration is to the left 162009 UniVersity of Colorado at Boulder lD9 Let39s stare at the formula X X0 vo t l 2at2 until it start to make sense You should always stare at new formulas turning them over in your mind until they start to make a little sense 2 X XO 2 Vet l2at V V V how far you travel how far you would travel if how much more agt 0 or less a lt0 v constant a 0 ou travel compared to how far you would have gotten if a 0 Gravitational acceleration Experimental fact In freefall near the surface of the earth all objects have a constant downward acceleration with magnitude g 98 ms2 g gt 0 by definition The term free fall means that the m force acting on the object is gravity 7 no other forces are acting no air resistance just gravity A falling object is in freefall only if air resistance is small enough to ignore Later when we study gravity we will find out why g constant 98 ms2 for all objects regardless of mass For now we simply accept this as an experimental fact Things to notice 0 The acceleration during freefall is always straight down even though the velocity might be upward Repeat after me quotAcceleration is not velocity velocity is not accelerationquot All objects regardless of mass have the samesize acceleration during freefall Heavy objects and light objects all fall with the same acceleration so long as air resistance is negligible Example Object dropped from rest What is the position velocity and acceleration at l s intervals as the object falls Choose downward as the direction so that a g If we instead chose upward as the positive direction then the acceleration would be in the negative direction a 7g Remember the symbol g is defined as the magnitude of the acceleration of gravity g gt 0 always by definition Often we call the vertical aXis the yaXis but lets call it the XaXis here 0 X0 0 v0 0 3 X 12 a t2 v a t from constant acceleration formulas gEIOms2 X5I2 v10t 162009 University of Colorado at Boulder lD 10 3 45 30 10 Notice that you can compute the acceleration a by taking any pair of t V values and computing a g Vf iv You always get a 10 msz At tf iti Example Projectile Motion A projectile is red straight up from the ground with an initial speed of lvol 10 ms ODescribe the velocity vs time Assume negligible air resistance Choose upward as the direction and set the ground at y 0 3700 V010ms ymax a 7g 798ms2 v v0at voigt 0 A Gr h f t ap 0 VVS V red slopeconstant7g v0 v 0 at apex i Vo L hits ground OWhat is time to reach maximum height ymax v lOms AtthemaXimumhei tv03 0 v7 t t l0s gh 0 g g lOms2 OWhatisymaX MethodI yt y0 v0tat2 vot gt2 v v4 0 3 Att1s y ym vot gt2 101 051012 10 5 5m 162009 University of Colorado at Boulder lD 11 Method 11 Use v2 v02 2ayyo At apeX v 0 a 7g and y 7 yo ymax 0 ymax SO we haVe Z 102 0 V02 7 2gym ym V 2g 210 Comments about projectile motion 9 The acceleration is constant straight down magnitude g only if we can ignore air resistance Real projectiles like cannonballs moving through air are strongly affected by air resistance Play with a simulation of projectile motion with and without air resistance Go to httpphetcoloradoedu and nd the simulation called Projectile Motion OThe formula y y0 v0 t gt2 is a quadratic equation so there are two solutions that is two values oft for a given value of y These two times correspond to on the way up and on the way down ORecall that the direction of 5 direction of A not the direction of 7 How does this square with vertical projectile motion rstart apex V 1 2 l l T j l Ej 1 going down Projectile s V1 N t a going up Notice that quotdeltavquot is always downward regardless of the direction of the velocity Qualitative comments about acceleration OYou can E acceleration but you cannot feel constant velocity If you are in an airplane traveling at constant velocity heading NW at 600 mph say and the ride is smooth then you can eat dinner juggle fall asleep exactly as if you were at rest In fact if the ride is perfectly smooth there is no way to tell that you are moving relative to the ground except by looking out the window Prof Einstein says that it makes just as much sense to say the you and the airplane are at rest and the ground is moving backwards All that can be said is that the airplane and the ground are in relative motion Which object is at rest depends on your frame of reference 162009 University of Colorado at Boulder 1D 12 But you can tell right away if you are accelerating If you are accelerating forward you feel yourself being pushed back into the seat Actually the seat is pushing you forward it just feels like you are being pushed back 7 more on that later 91f you are in a car there are two different ways that you can be accelerating forward 1 Start at rest and then oor it 2 Move in reverse at high constant velocity and then apply brakes In both these cases you are accelerating in the forward direction In both these cases you feel exactly the same thing you feel yourself being pressed back into the seat Puzzle for later when you are in a chair that is accelerating forward why does it feel like there is a force pushing you backwards OAcceleration and velocity are completely different things Acceleration is the time rate of change of velocity Be aware that the rate of change of something at the something a rateofchangeofv i V Example The radio weatherman says quotThe temperature is 480 and the temperature is falling at 100 per hourquot iwwmim m4 T7w 92 d If all you know is ll f then you know nothing at all about T And if all you know is T you know nothing at all about They are completely different knowledge of one tells you nothing about the other 162009 University of Colorado at Boulder lD 13 Important Math Appendices Proving the constant acceleration formulas We can derive the constant acceleration formulas on page 8 without using calculus Let s derive X Xo vot l2at2 velocity When a constant a plot ofv vs tis a straight line since a is A the slope of v vs I In this case the average velocity is halfway v v between the velocities at the start and nish v V We can also write the average velocity using the de nition V AX X X0 At t 0 39 V0 Some rearranging and substituting gives 0 t time X X0Vt X0 V0Vt Now substitute v V0 at proven on p8 X X0 V0V0att X0 Vot at2 Done To prove v2 v02 2aXX0 combine the formulas v Vo at and X Xo v0 t l2at2 eliminating time t I39ll let you do that Just enough about derivatives You don t have to know a lot about derivatives in this course But what you do have to know you have to know eXtremely well So please study this appendiX carefully The derivative of a function fX is another function f X df dX de ned by lim W The derivative of fX is the slope of the tangent dX AXgt0 AX AXgt0 line to the curve fX vs X In this chapter we have been considering functions of time X Xt and v vt Position X and d velocity v are related by V d Velocity v and acceleration a are related by a d There are 4 important theorems about derivatives that we will need again and again in this course 162009 University of Colorado at Boulder 1D 14 Theorem 1 The derivative of a constant is zero I Proof 1 Function fX A where A is a constant The plot fX vs X is a straight line with zero slope x Slope 0 Remember that the derivative is the slope of the tangent line to the curve The slope is zero so the derivative is A zero Proof 2 Start with the de nition of derivative gt X alim mw hm 1im0 dX Axgt0 AX Axgt0 AX Axgt0 AX Axgt0 AX Theorem 2 If fX AXn where A and n are constants then f An Xquot391 X Let s prove this for the special case n 2 fX AX2 3 f 2AX X Again we start with the de nition of derivative In any proof you have to start with something you know to be true De nitions are always true by de nition i E hm g lim fXAX fX A dX Axgt0 AX Axgt0 AX fXAX fX AXZ fXAX AXAX2 Af 2 2 fX fXAX fX AXAX AX I Ax Ax LA I X XAX AX2 2XAXAX2 AX2 AX 2AX AAX Taking the limit AX gt 0 we have f 2AX Done X 162009 University of Colorado at Boulder lD 15 Theorem 3 The derivative of a constant times a function is the constant times the derivative of the function d Af gXAfX where Aconstant 3 d g w Aamp bi bi dX Proof d Af A f AX f Hm AfXAX Afx lim x x dX Axgt0 AX Axgt0 AX A hm fXAX fX 2 Ag AXgt0 AX dX Theorem 4 The derivative of a sum is the sum of the derivatives w m fX gXhX 3 dX dX dX Proof lim 1111 Moo z dgxhx fxAx fx z MgxAXhxAXgxhx bi bi AX mu AX dxg 1imgXAX gx hxAx hx dg dh AX AX Exercise Starting with the constant acceleration formula Xt Xo vo t l2at2 use your knowledge of calculus to prove that v v0 at Hint take the derivative dX dt and use the theorems above Example A rocket in space has position as a function of time given by X X0 At3 where A is a constant What is the velocity and acceleration of the rocket Hey You should try to work this out yourself before looking at the solution below Solution Xt X0 At3 Take the first derivative to get velocity 162009 University of Colorado at Boulder 1D 16 dX dx0 At3 dAt3 dt dt dt dt Notice how we needed the theorems V 03At2 Now take another derivative the second derivative to get the acceleration 2 d3At2 adVd 32At6At dt dt dt Notice that this is a case of nonconstant acceleration so none of our constant acceleration formulas applies here Comment about notation The acceleration is the second derivative of position wrt time dv d dx Z a We usually write the 2quotd derivative like this a d X dt dt dt dt2 3 The third derivative1 1 d X is written d X dt dt dt dt3 162009 University of Colorado at Boulder l Vectors A and B have the same magnitude If vector A is in the X direction and vector B is in the 7y direction what is the direction of vector D AiB red green 1 Purple None of these yellow blue Answer upper right green 2 You know the acceleration of a particle at all times What do you know about the direction of the velocity of the particle Green You know nothing about the direction of the velocity Yellow You know that the direction of the velocity is either parallel to or antiparallel to the direction of the acceleration Answer You know nothing about the direction of the velocity Knowing the direction of the rate of change of velocity the acceleration tells you nothing at all about the direction of the velocity Velocity is not acceleration acceleration is not velocity 3 A simple pendulum is swinging back and forth What is the direction of the acceleration of the pendulum mass at the moment when it is at maximum displacement to the right Pink zero Blue Yellow Purple None ofthese Green Answer Yellow Lower left The acceleration is in the same direction as Av v v1 where v1 is the velocity just before the mass stops and v is the velocity just after it stops V V2 Av A freebody diagram of the forces on the ball when it is at the extreme right position is shown Suppose now the ball swings back and forth between horizontal positions When the ball is in the horizontal position the tension in the string is Pink zero Yellow not zero Answer The tension is zero By the same Av v v1 argument as in the question above one can show that the acceleration is straight down when the mass on the extreme right with the string horizontal Since the acceleration is straight down the net force must be straight down so there can be no horizontal component to the net force so the tension must be zero Alternatively one can say that the radial part of the acceleration is ar vzr and since v 0 instantaneously ar must be zero so there must be no force along the radial direction 4 A computer monitor is thrown upward It rises to a height h and falls back down to Earth During the ight the work done by the force of gravity is Pink mgh Green 2mgh Blue zero Yellow mgh Purple 2mgh Answer Work done by gravity is zero Work is negative on the way up positive on the way down During the ight the work done by the drag force of air resistance is Pink zero Yellow positive Green negative Answer Negative The drag force is always in the direction opposite the displacement so the work done is always negative both on the way up and on the way down 5 A Cadillac and a Volkswagen have a headon collision and stick to together in a mangled pile of metal and bloody limbs The police determine that the wreckage is in the exact same spot where the two cars collided Detective O39Newton who got an A in Physics 1 110 writes in his report that just prior to the collision the two cars had the same Pink magnitude of momentum Yellow kinetic energy Blue mass Green speed Answer Momentum Which car had the larger speed Pink Cadillac Yellow Volkswagen Answer Volkswagen 6 Schwinn decides to redesign its model X2000 28speed quotRoad Warriorquot bike 12995 The rims of the wheels are made lighter but more weight is added to the frame so the total weight of the bike remains unchanged Compared to the old bike when the new bike is moving at a given speed its total kinetic energy is Pink less Yellow greater Blue the same Answer less same translation KE but smaller rotational KE in the wheels Compared to the old bike when the new bike is accelerated from rest up to a given speed does the rider have to work less more or the same Pink less Yellow greater Blue the same Answer less Wm KEf iKEi 7 A distraught student standing at the edge of a cliff throws one caculator straight up and another calculator straight down at the same initial speed Neglecting air resistance the one which hits the ground with the greatest speed is the one thrown BLUE upward YELLOW downward PINK neitherithey both hit at the same speed Answer They both hit at the same speed Same question but now do not neglect air resistance The calculator which hits with the greatest speed is the one thrown BLUE upward YELLOW downward PINK neitherithey both hit at the same speed Answer downward 8 A aming physics text is dropped from an airplane ying at height h at constant horizontal velocity velocity and speed v0 Neglecting air resistance the text will BLUE quickly lag behind the plane YELLOW remain vertically under the plane PURPLE move ahead of the plane GREEN it depends how fast the plane is ying Answer text remains under the plane Vx V0 constant What is the speed with which the text hits the ground Blue V0 Zgh Yellow V0 zgh Purple Neitherdon39t know Hint For lD motion along X V2 Vo2 2aXX0 2 2 l 2 Answer Neither Correct answer is Speed Vx Vy V0 2g h 9 Consider an exhausted physics student standing in an elevator that is moving upward with constant velocity The upward normal force N exerted by the elevator oor on the student is BLUE larger than YELLOW identical to PINK smaller than the weight mg ofthe person Answer identical The net force is zero since the velocity is constant 10 A student attempting to make her 93911 pot of coffee is so shaky she drops the coffee lter A coffee filter oats gently downward at some constant quotterminalquot velocity v i vco nsta nt The net force on the coffee filter is BLUE Upwards YELLOW Downwards GREEN Zero Answer Zero Since the velocity is constant 11 A relieved student driving away from CU on Dec 21 wishes to cause her car to accelerate How many controls in an ordinary car are speci cally designed and intended to accelerate the vehicle BLUE one PURPLE two YELLOW more than three GREEN three Answer I can think of three The accelerator pedal the brake and the steering wheel Possibly also the clutch Some people use the clutch to slow the car 12 At some instant in time two asteroids in deep space are a distance r20 km apart Asteroid 2 has 10 times the mass of asteroid 1 What is the ratio of their resulting acceleration due to the gravitational attraction between them a1az Blue 10 Yellow llO Green 1 Pink None of thesedon39t know Answer 10 This is a problem with constant acceleration Green True Pink False False As they get closer the force of gravity increases and so does the acceleration 13 An unhappy student works out his aggression by attempting to knock down a large 69 In1 wooden bowling pin by throwing balls at it The student has two balls of equal size and mass one made of rubber and the other of putty The rubber ball bounces back while the ball of putty sticks to the pin Which ball is most likely to topple the bowling pin BLUE the rubber ball YELLOW the ball of putty PINK makes no difference GREEN need more information Answer the rubber ball F ApAt For the rubber ball Ap2mv while for the putty Apmv 14 Suppose a PingPong ball and a bowling ball are rolling toward you Both have the same momentum and you exert the same force to stop each How do the time intervals to stop them compare BLUE It takes less time to stop the PingPong ball YELLOW Both take the same time PURPLE It takes more time to stop the PingPong ball Answer F ApAt Same F same Ap for each ball so At must be the same Which ball travels further while slowing down BLUE the PingPong ball YELLOW Both travel the same distance PURPLE The bowling ball Answer The PingPong ball The momenta of the two ball were identical before slowing and p mv so the PingPong ball must have been going really fast to have the same momentum as the bowling ball Since the two balls slowed to a stop in the same time the distance traveled is greater for the faster ball 15 Consider a light rotating rod with 4 point masses attached to it at distances 2 and 4 as shown The axis of rotation is thru the center of the rod perpendicular to its 0 0 0 0 length The 4 masses are moved to new positions so that all masses are now 3 units from the axis Did the moment of inertia I Pink increase 4 4 4 i Green decrease PW Yellow remain the same l Answer Decreases 2242 20 lt 232 18 16 Two wheels initially at rest roll the same distance without slipping down identical inclined planes starting from rest Wheel B has twice the radius but the same mass as wheel A All the mass is concentrated in their rims so that the rotational inertia s are I mR2 Which has more translational kinetic energy when it gets to the bottom BLUE Wheel A PINK Wheel B PURPLE The kinetic energies are the same YELLOW need more information Answer The translational KE s are the same For rolling motion KE KE KEmt mV2 Im2 for a hoop tot trans Z V mv2 mR2 mv2 mV2 For a rolling hoop the total KE is always half translational half rotational The two hoops big R and small R have the same mass m and the same total KE mgh The bigger I of the larger hoop is compensated for by a smaller n 17 A physics text ofmass m sits at rest on a wooden board m inclined at an angle 9 above a aming hibachi The coefficient of static friction between the book and the board is us How does the magnitude of the force of friction W between the book and the board compare to the weight mg I of the box Pink mg gt Fmc Green mg lt Fmc Purple mg Fmc Yellow mg can be either greater than less than or equal to Ffric depending on the size of us Answer mg gt FmC Draw the freebody diagram to see this Also realize that FmC does NOT equal usN Fmc is less than or equal usN It is only equal is the book is about to slip CT120 How many degrees in 1 radian A 1 rad 27 degrees B 1 rad 1800 C 1 rad 100 D 1 rad 5730 E Radian is not a measure of angle so the question makes no sense CTlZOb A student sees the following question on an exam A ywheel with mass 120 kg and radius 06 m starting at rest has an angular acceleration of 01 rads2 How many revolutions has the wheel undergone after 10 s Which formula should the student use to answer the question A W m az B C q f 1 2az 2 C M M22 2667 15 CT121 BIG BEN and a little alarm clock both keep perfect time Which minute hand has the bigger angular velocity co A Big Ben B little alarm clock C Both have the same 0 G CT122 A small Wheel and a large Wheel are connected by a belt The small Wheel is turned at a constant angular velocity 0 5 How does the magnitude of the angular velocity of the large Wheel 0L compare to that of the small Wheel S AzcosoaL BzoasgtoaL CzcosltoaL DL There is a bug S on the rim of the small Wheel and another bug L on the rim of the large Wheel How do their speeds compare ASL BSgtL CSltL CT123 A ladybug is clinging to the rim of a spinning wheel which is spinning CCW very fast and is slowing down At the moment shown what is the approximate direction of the ladybug39s acceleration A Blt 39 C 4 D E None of these CT124 A student in Physics 1110 sees the following question 1 on CAPA set 10 Due Friday Nov1 An engine ywheel turns with constant angular speed of 100 revmin When the engine is shut off friction slows the wheel to rest in 2 hours What is the magnitude of the constant angular acceleration of the wheel Give the answer in units of revminz The student writes 0 030Ott 0 0 an 275100reym T 120 min 39 Does the answer come out correctly with the desired units 0 0 SO IOLI 2T A Yes B No CT125 Three forces labeled A B C are applied to a rod which pivots on an axis thru its center cos450 SiII45 1 11414 c L A 4 2F L2 lt gt o l I 45y B 4 A F L4 F Which force causes the largest magnitude torque A A B B C C D two or more forces tie for largest size torque CT126 A mass is hanging from the end of a horizontal bar which pivots about an axis through it center but it being held stationary The bar is released and begins to rotate As the bar rotates from horizontal to vertical the magnitude of the torque on the bar A increases B decreases C remains constant As the bar rotates from horizontal to vertical the magnitude of the angular acceleration 0c of the bar A increases B decreases C remains constant CT127 A mass m hangs from string wrapped around a pulley of radius R The pulley has a moment of inertial and its pivot is frictionless Because of gravity the mass falls and the pulley rotates The magnitude of the torque on the pulley is 9 A greater than ng B less than ng C equal to ng Hint Is the tension in the string mg CT128 Two wheels with fixed axles each have the same mass M but wheel 2 has twice the radius of wheel 1 Each is accelerated from rest with a force applied as shown Assume that all the mass of the wheels is concentrated in the rims so that the moment of inertia of each is of the form I M R2 hoop formula In order to impart identical angular accelerations to both wheels how much larger is F2 than F1 Recall that E let F F1 2 C9 Wheel 1 radius R mass M Wheel 2 radius 2R mass M AIFZZFl BIFZZZFl C F2 2 4F1 D F2 2 8F1 E None of these CT 129 Consider a rod of uniform density with an aXis of rotation through its center and an identical rod with the aXis of rotation through one end Which has the larger moment of inertia C E axis AIIcgtIE BIIcltIE CIICZIE Consider two masses each of size 2m at the ends of a light rod of length L with the aXis of rotation through the center of the rod The rod is doubled in length and the masses are halved What happens to I A U2 L 2 2m 2m AIIAgtIB BIIAltIB CIIAZIB CCCCC 10 1s perp d lar to the A sm mom e11 icu all mass 111 is placed on the rim of the nt of inertia of this syste A MmR2 w B less h C gre er th In a t at diagram The disk has mass M and radius R 39 disk What is the CTlZl l A sphere a hoop and a cylinder all with the same mass M and same radius R are rolling along all with the same speed V Spherelbostk Which has the most kinetic energy A Sphere B Hoop C Disk D All have the same KE CT1212 Two light massless rods labeled A and B each are connected to the ceiling by a frictionless pivot Rod A has length L and has a mass m at the end of the rod Rod B has length L2 and has a mass 2m at its end Both rods are released from rest in a horizontal position 2m m Which one eXperiences the larger torque A A B B C Both have the same size 1 Which one falls to the vertical position fastest A A B B C Both fall at the same rate T Hint 0 f Answer The torque decreases T 2 FL I F r Sine F L decreases as the mass falls When the bar is vertical the torque is zero Answer The magnitude of the torque is than ng The tension in the string is less than mg because the mass is accelerating down FT Fnetzmg Fsza gt Fszg ma a a L Torque TZFTRZIOLZI mg R Two equations FT mgma FT I aR2 in two unknowns FT and a gt can solve CT1213 A mass m is attached to a long massless rod The mass is close to one end of the rod Is it easier to balance the rod on end with the mass near the top or near the bottom 5 Hint Small oc means sluggish behavior and X f A easier with mass near top B easier with mass near bottom C No difference mg Physics 1110 Fall 2004 Lecture 23 18 October 2004 Announcements 0 Reading Assignment forWed Oct 20 Knight Part Summary 91 0 Exam 2 tomorrow evening Room changes See web QampA this afternoon in G1BZO 57pm Review for Exam 2 General Comments 0 Do problems from easy to hard order Check answers carefully for calculatoralgebra errors Update formula sheet don t forget Ch 13 material Do not assume that MC answer existence implies correctness Stay calm Chapter 4 Force and Motion 0 What are forces What kinds are there Contact vs Long range Net force is the vector sum of all forces acting on one object Identifying forces System and environment Newton s 2nd Law mass and acceleration FBDs Newton s 1st Law inertiaa0 motion Chapter 5 Dynamics Motion Along a Line 0 Analyze analyze analyze o Pictorial rep variables known and unknown ll 0 FBD 0 Apply Newton s 2nd law 0 Solve equations Equilibrium staticdynamic gt Net force 0 a0 1D acceleration problems Mass weight and apparent weight Friction kinetic static rolling Opposing direction Static friction formula is maximum limit Words in problems give hint L231 Physics 1110 Fall 2004 Drag cross section and terminal velocity Ramp problems Chapter 6 Dynamics ll Motion in a Plane Analyze o More variables need component subscripts o FBD 0 Apply Newton s 2nd Law 0 Solve equations Trajectory X vs y X vs t and y vs t plots Constant acceleration problems Two 1D problems Projectile motion only gravitational force variable definition Relative velocity problems Check formulas intuitively Try to recognize the similarity in all of these problems Chapter 7 Dynamics Motion in a Circle Uniform Circular Motion what is it Radius radians arc length period Angular velocity omega vs t plots rtz coordinate system velocity in rtz components for uniform circular motion acceleration rtz components for uniform circular motion Dynamics Analyze as before 0 Snapshot rtz variables 0 FBD careful with rtz 0 Apply Newton s 2nd Law with ucm acceleration values 0 Solve equations Horizontal circular motion turntable road cunes ball on string Vertical circular motion topbottom are easiest be careful with normal forces Nonuniform motion R is constant tangential speed and angular velocity changes 1D Motion on circular path centripetal acceleration now changing in time Constant acceleration formulas Questions L232 2Dl Motion in 2D Velocity and acceleration are vectors They can have m direction When we are considering motion in the xy plane these vectors can point anywhere in the plane A common example of motion in 2D is Projectile motion Consider a projectile red from a cannon with an initial velocity v0 with a direction of 9 above the horizontal y v0X v0 cosO VY V0 9 v0y v0 sme 4 X Acceleration is a vector and can have any direction But in the special case of acceleration M solelyto gravity the acceleration is always straight down Review of 1D motion d v d X a v dt dt From these two equations we can derive for the special case a constant a v vo at b X Xo vot l2at2 2 c v v02 2aXX0 v v d v o 2 X0 v0 initial position initial velocity X v position velocity at time t 1182009 University of Colorado at Boulder 2D2 AX Suppose that a 0 In th1s case V constant and V V E v0 constant 2 v x Xe Vet Ifa9 0then X X0 vot at2 position if a 0 how much more agt 0 or less alt 0 distance you go ifa z 0 End of 1D motion review Now 2D Motion dv dv dv lt3 a dt X dt y dt mi gt4 5 and vd r ltgtVxdXV dt dt Special case a stant ay constant This is exactly like the 1D motion case except now we have separate equations for Xmotion and ymotion We can treat the Xmotion and ymotion separately These are the X and ycomponents of the vector 2 XX0 v0xt axt X vX voxaxt equations rfovot at2 yy0voytaytzY vvoat vy v0y ayt Example Horizontal Ri e A ri e bullet is red horizontally with v0 100 ms from an initial height of yo 20 m Assume no air resistance How long is the bullet in ight How far does the bullet go before it hits the ground 2m N N x I x Key idea in all projectile motion problems treat X and ymotions separately 1182009 University of Colorado at Boulder 2D3 The motion along the ydirection vertical motion is completely independent of the motion along the Xdirection horizontal motion X X0 0 Y y0 2m vex 100ms v0y ax0 a g 98ms2 The time to hit the ground is entirely controlled by the ymotion yoVoytgt2 5 0yo7gt yozgt a a 0 2y0 gtz t2 2y t 2y 22 064s g g 98 Now we look at the Xequations to see how far along the Xdirection the bullet traveled in 064 s y La 0 aX 0 3 vX constant v0X 100 ms X X0 voxt ax t2 voxt 100064 64m T T Why vx constant The force of gravity is straight down There is no sideways force to change vx assuming no air resistance 0 Another question What is the speed of the bullet as it falls vx constant v0x vo vy v0y ayt igt v v 0 is As the bullet travels its vx remains constant while lvy l grows larger and larger yl Vox Vo Vx VyI i v Vx 1182009 University of Colorado at Boulder 2D4 speed magnitude ofvelocity V vex2 gt 2 The speed is a minimum at t 0 when vy 0 the moment when the bullet leaves the gun The speed is maximum when vy is maximum just before the bullet reaches the ground Don t forget that we are assuming no air resistance For a real ri e red in real air the bullet s speed is usually maximum when leaving the barrel and then air resistance slows the bullet down as it travels Example A projectile is red on an airless world with initial speed v0 at an angle 9 above the horizontal What is the minimum speed of the projectile Answer v0x v0 cos 9 X v v cos a Proof a g gt vyv0y gt y A Vy0 Vox Here the speed is minimum at the top of the trajectory where vy 0 Review of acceleration 1Dad V 2Dad Vz vlVI dt dt At At Av vz vl means v2 2 Av 3 V1 Av Av is the vector you add to 2 to get v2 V2 The direction of a is the same as the direction of Av since a Av gtlt positive numberl At V1 quotquotquotquotquot quot He V1 v21AV X The direction of the acceleration of gravity is the direction of AV straight down 1182009 University of Colorado at Boulder 2D5 quotShoot the Monkeyquot Experiment A hunter aims a ri e at a monkey hanging in a tree The ri e res at the same instant that the monkey lets go and drops Does the bullet hit the poor monkey Answer Yes First consider the situation w E gravity If no gravity the bullet goes in a straight V line and the monkey does not fall So the monkey is hit The height of the bullet with ay 0 is y vOsin t HP 0 With gravity on the bullet falls below the straightline path by a distance l2g t2 which is exactly the same distance that the monkey falls So the monkey falls into the path of the bullet Poor monkey Circular Motion and Acceleration Circular motion consider an object moving in a circle of radius r with constant speed v T period time for 1 complete revolution 1 cycle 1182009 University of Colorado at Boulder 2D6 A distance 2717f speedv v m T An object moving in a circle is accelerating because its velocity is changing changing direction Recall the de nition of acceleration d 9 Av 7 e a z t V V2 V1 velocity V can change is two ways 1 t At At 39 A Magnitude can change g V1 AV V d1rection can change gt gt gt V v 2 V1 For circular motion with constant speed we will show that 2 a v 1 the magmtude ofthe acceleration 1s a a E r 2 the direction of the acceleration is always towards the center of motion This is centripetal acceleration quotcentripetalquot quottoward centerquot Notice that the direction of acceleration vector is always changing V therefore this is not a case of constant acceleration so we cannot use the quotconstant acceleration formulasquot Is claim 1 sensible I I I I N I 73 2 Check units r Think to get a big a we must have a rapidly changing velocity Here we need to rapidly change the direction of vector V gt need to get around circle quickly 2 need either large speed v or a small radius r gt a vzr makes sense Proofis given below Is claim 2 sensible Observe that vector AV is toward center of circle vz V1 Direction of a direction toward which velocity is changing Av V2 1182009 University of Colorado at Boulder 2D 7 Example acceleration on a merry go round Radius r 5 m period T 3 s n V 105ms V V2 2 1 a H 23 g s r 8ms2 A human can withstand an acceleration of about 5 g39s for a few minutes or 10 g s for a few seconds without losing consciousness Proof of a V2 r for circular motion with constant speed The proof involves geometry similar triangles It is mathematically simple but subtle Consider the motion of a particle on a circle of radius r with constant speed V And consider the position of the particle at two times separately by a short time interval At In the end we will VI take the limit as At gt 0 We can draw a vector diagrams V1 representing r1 A T and 2 Av v Av Notice that these are similar triangles same angles same length ratios Also note that r and vl vz v AV Ar V At Because the triangles are similar we can write A little algebra gives v r r Av V V v2 E F1nally we take the 11m1t At a 0 and get acceleratlon a 2 r r 1 182009 University of Colorado at Boulder Physics 1110 Fall 2004 Lecture 27 27 October 2004 Announcements 0 Reading Assignment for Fri Oct 29 Knight 106107111 Energy Conservation with Gravitational Forces Recall from last time that we have defined two forms of mechanical energy so far 1 Kinetic energy Kmv2 and 2 Gravitational Potential Energy Ug mgh The total mechanical energy E is then just the sum E K Ug Ifthere are no friction or drag forces when can use energy conservation to analyze problems These are so simple we will do them in the context of concept tests after a first example Example Problem A ball of mass m is launched with an initial velocity v0 at an angle 6 above the horizontal from a height of y What is the speed ofthe ball just before it hits the ground y O lfwe used a regular force analysis we would find that the weight was the only force so we would have a constant acceleration of g downward and no acceleration in the horizontal force We could then get the equation to find the vertical velocity componentjust as the ball hits the ground then square and add to the square ofthe horizontal component then take the square root In an energy analysis we calculate the initial total energy before and equate it to the final total energy after shown below Note that the value we take for the y origin doesn t matter because only the difference between initial and final matters So different people who might choose different y origins can find different values for the potential energy of an object itjust won t affect the final value of any regular kinematic quantity L271 Physics 1110 Fall 2004 E mv mgyl 1 2 Ef 5mvf mgyf EiEf 1 2 1 2 imvo mgyi 3me 771gi V V 2gyiyf Vf V 2gy yf As shown in the concept tests we can also apply energy conservation in problems where the motion of the object is constrained to some path by forces perpendicularto the path such as normal forces or rodsropes used to keep an object in a circular path The classic example case is a roller coaster Here the actual path is very complicated and a force analysis would be incredibly long and tedious whereas an energy analysis takes two lines of algebra usually Spring Forces We have discussed springs for quite awhile and now it is time to get quantitative As we can measure in class the force from a spring is proportional to the amount it is stretched from its equilibrium length that is the length with nothing pulling on the spring The magnitude of the force is different for different springs so each spring has a constant of proportionality called the spring constant k Finally if you stretch a spring so that its length is larger than the equilibrium length it pulls back towards the equilibrium position so the formula forthe force has a minus sign to express this direction Putting all this together gives Hooke s law for springs F ks se kAs L272 Physms M m FaH 2mm Nme mum s a uwerentuse ufthe new mutater smce newuena meansthe uwerenee between the anua spnng wean andthe Equmbrmm Ength nutthe na mmus mma A su nme thattms s a eempenem Equatmn se scumd be X y Dr 1 Make sure yuu undersiand the gure be uvv Whmh gwe EXamp Es elf huvv the furmu a Wurks Next Ecture WE H an same cuncept 551 premems usmg spnngs L213
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