Gen Physics I
Gen Physics I PHYS 2010
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PHYS2010 General Physics I Course Lecture Notes Section IX Dr Donald G Luttermoser East Tennessee State University Edition 24 Abstract These Class notes are designed for use of the instructor and students of the course PHYS 2010 General Physics I taught by Dr Donald Lutternioser at East Tennessee State University These notes make reference to the College Physics 8th Edition 2009 textbook by Serway and Vuille IX Gravitation A Newton s Law of Gravity 1 Every particle in the Universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distances between them The was rst realized by Sir lssac Newton and hence this is referred to as Newton s Law of Gravity and can be expressed mathematically as m1m2 A 7 7 Fg G 752 IX 1 a m1 and m2 E masses of objects 1 and 2 measured in kg b 7 E distance between the masses measured in m c G E constant of universal gravitation gt this is the pro portionality constant between the force and the dependent parameters 2 G 6673 x 1011 N71 IX2 kg2 d Newton s law of gravity is thus an inversesquare law 2 The gravitational force exerted by a spherical mass on a particle outside the sphere is the same as if the entire mass of the sphere were concentrated at its center 1X1 PHYS 2010 General Physics Note that r RP h radius of sphere e planet plus the height above the sphere Example 1X71 Problem 740 Page 223 from the Ser way amp Vuille textbook Two objects attract each other with a gravitational force of magnitude 100 X 10 8 N when separated by 200 cm If the total mass of the two objects is 500 kg what is the mass of each Solution Let the individual masses be represented by m1 and m2 and the total mass of the two objects be represented as M m1 m2 500 kg Then we can express the second mass as m2 M m1 The distance between the two masses is 7 200 cm 0200 In the gravitational force between them is F 100 X 10 8 N and the gravitational constant is G 667 X 10 11 N mgkgg Now make use of Newton s Law of Gravitation Gm1m2 Gm1M m1 F 2 2 r 7 Fr Gm1M m1 GMm1 Gm 0 Gm GMm1Fr2 Algebra has a well know solution to the quadratic equation of the form ax bx c 0 x i bib2 4ac 7 2a as Donald G Luttermoser ETSU IX 3 Here xm1 a G b GM andc Frg so GM i xGQMQ 4am 2G GM i G M2 4Fr2G 1 4 1 1 8 k 2 2 mom i 500kg2W0 ggtlt020031gt 2 667 x 10711 N m kg 500kg AMW gawkgimi As such m1 either equals 40 kg2 20 kg or 60 kg2 30 kg either answer is correct If we chose m1 30 kg then m2 M m1 500 kg 30 kg 20 kg Hence the solution is m1 30 kg and mg 20 kg 3 Why do objects fall independent of their mass on the Earth s surface M m a GraV1ty Fg G 139 b Motion F ma mg c Set Fg F then MEBm G mg R3 or i G M6 g i IX3 R3 IX 4 PHYS 2010 General Physics 1 Plugging in values 6673 x 1011 N m2kg259763 x 1024 kg 6378 x 106 m2 398845 x 1014 N m2kg 0 but since 1 N 1 kg ms2 and paying attention to our input signi cant digits we get k 2 g 980 S 980 ms2 kg e Since 9 can be measured from the ground 69 motion ex periments and RED can be measured astronomically 69 simple shadow length measurements at two different lat itudes Eratosthenes did this in 200 BCl we can then determine the Earth s mass with rewriting Eq IX 3 M 69 G 4 The gravitational constant G a Measuring G in the laboratory accurately is a di icult taskl b Cavendish was the rst to measure it while he was trying to determine the density of the Earth in 1798 see the textbook for details c More sophisticated experiments have been carried out since that time gt G s accuracy is only known to 5 signi cant digits G 6673041 x 1011 N 1112ng where the 41 digits are uncertain Donald G Luttermoser ETSU IX 5 gt the speed of light s accuracy is known to 9 signi cant digitsl c 2997924581 x 108 ms 1 Future space based experiments in free fall and a vac uum should increase the accuracy in the measurement of G l 5 The farther we get from the Earth s center the smaller the ac celeration due to gravity GMG g 2 6 IX 4 where r E distance from Earth s center a When 9 is measured on the Earth s surface or some other planetary surface it is called the surface gravity b When 9 is measured elsewhere 76 not on a planetary surface it is called the acceleration due to gravity and if the object is in free fall it also is often called the freefall acceleration Example 1X72 Mt Everest is at a height of29003 ft 8840 m above sea level The greatest depth in the sea is 34219 ft 10430 m Compare the Earth s surface gravity at these two points Solution Let he be the height of Mount Everest and hS be the greatest depth of the sea Using Eq IX 4 and the fact that the radius of the Earth at sea level at the Earth s equator is RED 6378077 X 106 m we get the distances for the highest and lowest points of the surface from the center of the Earth as rh REB he 6378077 x 106 m 8840 x 103 m IX 6 PHYS 2010 General Physics 6386917 x 106 m 7 1 REB hs 6378077 x 106 m 10430 x 104 m 6367647 x 106 m Which gives the surface gravities of GMEB i 66730 x 1011 N mZkg259763 x 1024 kg 9h 7 6386917 x 106 m2 GMEB 66730 x 1011 N mQkg259763 x 1024 kg 91 7a 6367647 x 106 m2 and x 100 x 100 0605 the surface gravity of the lowest point on the Earth s surface is a little more than half of a percent larger than at the highest point B Kepler s Laws of Planetary Motion 1 Johannes Kepler 1571 1630 was a German mathematician and astronomer who used Tycho Brahe s observations of Mars to derive the 3 laws of planetary motion The data showed that the Copernican model of heliocentric Sun centered solar system was correct except that the planets move in elliptical and not circular paths around the Sun as Copernicus had assumed 2 The laws a Law 1 The orbit of a planet about the Sun is an ellipse with the Sun at one focus The so called elliptical or bit The equation for the ellipse in Cartesian coordinates Donald G Luttermoser ETSU IX 7 when the foci are on the x axis is 571 IX5 a semimajor axis b semiminor axis focus i Semimajor axis a Half of the longest axis of an ellipse ii Semiminor axis b Half of the shortest axis of an ellipse iii 1 Astronomical Unit AU is the length of the Earth s semimajor axis 1 AU 14960 x 1011 m iv The relative atness of an ellipse is measured by the eccentricity e 1 a2 b2 6 a IX6 IX 8 PHYS 2010 General Physics v The distance from the center to either focus is given by x a2 b2 vi The ellipse is just one type of conic section If a b Eq IX 6 gives 6 0 and we have a circular orbit gt a second type of conic section 216 a circle vii If we let a get bigger and bigger such that a gt b then Eq IX 6 gives 6 m Wa aa 1 as a gt 00 When this happens we have a parabolic orbit Such an orbit is said to be open both cir cular and elliptical orbits are closed and never return A parabolic orbit is achieved when the ve locity of a satellite just equals the escape velocity Uesc 39 viii There also are orbits that are more open than parabolic orbits gt the so called hyperbolic or bits These orbits have 6 gt 1 and can be achieved if v gt vase b Law 2 A line joining a planet and the Sun sweeps out equal areas in equal amounts of time law of equal areas i This means that planets move faster when closer to the Sun in its orbit than when it is farther away ii Objects in very elliptical orbits don t stay near the Sun for a very long time gt comets iii Perihelion Point on an orbit when a planet is closest to the Sun 7 perihelion distance Donald G Luttermoser E TSU iv Aphelion Point on an orbit when a planet is farthest from the Sun Ta aphelion distance v The perihelion and aphelion of a solar orbit can be determined from the sernimajor axis and the eo oentrioity with 7 a1 e IX 7 Ta a1 e IX 8 also note that 7 Ta 2a IX 9 ifA1 A2 area then t1 t2 time c Law 3 The square of the orbital period T of any planet is proportional to the cube of the sernirnajor axis a of a planet s orbit about the Sun harmonic law T2 olt a3 IX10 IX 10 3 PHYS 2010 General Physics Kepler s laws were empirically deduced Nearly 100 years after these planetary laws were developed Newton came along and showed theoretically why they are valid a Gravity for a planet in orbit about the Sun E 9 G M9 M10 Fg rg i M1002 7 r b Centripetal force F6 c Set these two forces equal GMQ Mp Mp1 7 2 T 7 7 2 GMo U 6 1 Note however that the orbital velocity of a planet can be expressed as the length of the circumference C of the orbit divided by its orbital path For demonstration purposes we will assume that the orbit is circular but the same result is obtained for elliptical orbits then C 27W 1 7 7 T T e Plugging this equation into the force equation above gives U2 E T 7 4727 2 7 GMQ T2 T 7 47973 T2 GMQ Since we have assumed a circular orbit here 7 a the semimajor axis and 4W2 T2 GAIQ a3Ka3 IX 11 Donald G Luttermoser ETSU IX 11 4W2 G MG i KG 297 x 1049 s2rn3 ii KG is independent of the planet s rnass gt K only depends upon the larger central ob ject s rnass iii T2 olt a3 is valid whether the orbit is circular where a r the orbital radius or elliptical where a is used intact though the elliptical orbit solu tion has a slightly different form for the KG con stant The proof for elliptical orbits requires ad vanced calculus hence we will not show it here 4 We can base other planets in solar system with respect to the Earth T2 K 3 3 I 9 13 i IX12 TEB KG 169 169 Since TEE 1 yr and a6 l AU Eq IX 12 becomes 2 3 T10 GP 1 yr2 1 AU3 01quot Example 1X73 The Voyager 1 spacecraft hasjust passed the 100 AU mark in its distance from the Sun At this time the Death Star from an evil galactic empire intercepts Voyager and figures out which planet sent it based on the gold record that was included on the spacecraft The Death Star alters its course and starts to orbit the Sun from that point and sets its heading such that it will reach perihelion at the Earth s location Calculate the following about its orbit a the semimajor axis in AU b the eccentricity and c IX 12 PHYS 2010 General Physics the length of time in years it will take to get to Earth Solution a We are given that 7 100 AU and since it starts its free fall orbit from the position of Voyager 1 that position marks it aphelion position 7 100 AU note that we do not have to convert to SI units for this problem Using Eq IX 9 we can easily calculate the semimajor axis of the Death Star s orbit a 1 AU 1 AU aW 505AU Solution b Using Eq IX 7 and solving for e we get 13900 e i a i 1 505 AtU i 1 00198 7 0980 nearly a parabolic orbitl Solution c First we need to calculate the period of a complete orbit from Kepler s 3rd law then the amount of time it will take to go from aphelion to perihelion is one half that period As such T 2 a 3 505 AU 3 7 7 553129 15 T6 lag 100 AU 0 l X 0 T 7 129gtlt105 100 yr T 359 yr Since this is the full period of the Death Star s orbit the time to get to Earth will be t T 180 yr Donald G Luttermoser E TSU C Conservation of Energy in a Gravitational Field 1 Up until this point we handled conservation of mechanical energy for gravity for a constant surface gravity e a constant accel eration due to gravity Here we will relax that requirement and let 9 vary with 7 as shown in Eq IX 4 a b For a constant acceleration due to gravity we have seen from Eq VI ll that PE mgh IX l4 where m is the mass of the object 9 is the surface gravity e acceleration and h is the height above the ground However this is not the most general form of the potential energy of a gravitational eld it is an approximation to the general form In higher level physics potential energy is related to a force eld by the equation dPE 7 F vPE r IX15 where 13 139 as described by Eq IX l for a gravita tional eld and the del symbol 6 is the spatial deriva for Eq IX l5 to be valid we only use the component of v in the f direction e tive in three dimensions ddT since 139 only points in the r direction i Since this is an algebra based course we won t solve that differential equation with calculus How ever by realizing that the derivative symbol d just means n n tes mally small delta A E change of and we can approximate Eq IX l5 with APE Gm A e n 916 G M m where the 0 terms are the initial values PE PEG 72 IX 13 IX 14 PHYS 2010 General Physics ii Now if we de ne the initial PEO as 0 at the center of the gravitating body so 7 0 0 Eq IX 16 becomes PE ltGmgt 7 7 GMm T iii The calculus solution to the differential equation above Eq lX l5 would have given exactly the same answer iv As such the most general form of the potential energy of a mass in a gravitating eld is i G M m PE i T IX17 c With this general form of the gravitational potential en ergy we can see how the potential energy equation near the Earth s surface Eq lX 14 arises i Let s say we have a projectile that we launch from the ground While on the ground 7 RED the radius of the Earth which gives a potential energy of GMm PE Rea IX 18 ii Now when the projectile reaches its highest point above the ground h it is a distance of 7 RED h from the center of the Earth At this point it has a potential energy of G Mg m PE Rh IX 19 Donald G Luttermoser ETSU IX 15 iii The change in potential energy between these two points is G MED m G MED m RED h RED 1 1 GM 7 69 m Re R6 h Ree h Ree G M W Baa Baa h R69 R69 M Ree h Re Re Re h GM APE GMmlt iv lf RED is much greater than h which it will be for experiments near the Earth s surface h lt R69 As such Rh RED and the equation above becomes h h APE GM m GM m 7 EB ltRRgt EB R39 7 GMmh R v Now remembering our de ning equation for sur face gravity 69 Eq lX B G MED g 7 R3 we use this in the potential equation we just wrote G M APE m EBhmgh R2 69 hence we have proven Eq IX 14 from rst prin ciples 2 We can now use this general potential energy equation to gure out high trajectory orbital and space trajectory problems First IX 16 PHYS 2010 General Physics let s develop a relationship between the initial velocity no of a projectile and the maximum height h it will reach a If we were limiting ourselves to vertically directed trajec tories near the Earth s surface we would have KEiPEi KEfPEf 1 2 1 2 Emvomgyo Emu mgy yo 0 since it represents the ground and the projectile will reach its maximum height y h when the velocity goes to zero 1 0 From this we can solve for the initial velocity and get vo 1Zgh b Using the more general form of the potential our initial position will be on the surface of the Earth 7 R69 and we will reach a distance from the center of the Earth of 7 RED h which we will simply write as 7 The conservation of mechanical energy then gives KER PER KEr PEr 1 2 GMEBm i 1 2 GMEBm 2 mUO RED 7 2 m U 75 c Once again 1 0 at the top of the trajectory and as such the initial velocity is 1 2 GMEBm i 0 GMEBm 2 0 RED 7 71 1 2 GMEBm GMEBm 7 mU 2 0 RED 71 1 2 1 1 E I n UO i GMm 2 2GMm lt1 1 RED 71 m Donald G Luttermoser ETSU IX 17 l l u 2GMEB lt7 7 IX20 ED 1 Now if we once again de ne g to be the acceleration due to gravity at the Earth s surface e surface gravity we can rewrite Eq IX 3 to read G M6 9 R39 e Plugging this into Eq IX 20 gives 1 1 U0 and replacing 7 by RED h we get l l O 2R2 7 7 U 9 Gall26 Ream R h R 2R2 lg 69 ltRRh RRhgt Rh R 2R2 9 ltR Rhgt o lX 22 v A REE h f Finally one can immediately see that if h lt R69 then RED h RED and Eq IX 22 becomes 2 R vow Zgh a IX 18 PHYS 2010 General Physics As can be seen this equation as written above is just an approximation to a more general equation e Eq lX 22 Example 1X74 Problem 745 Page 223 from the Serway amp Vuille textbook A satellite of mass 200 kg is launched from a site on the Equator into an orbit at 200 km above the Earth s surface a If the orbit is circular what is the orbital period of this satellite b What is the satellite s speed in orbit c What is the minimum energy necessary to place this satellite in orbit assuming no air friction Solution a The radius of the satellite s orbit is rRh638gtlt 106 n1 200gtlt 103m 658 x 106m m 200 kg is the satellite s mass and Me 598 X 1024 kg is the Earth s mass The orbital velocity will just be the tangen tial velocity of the circular orbit Since the gravitational force provides the centripetal acceleration we have F Fg vim i GMEBm m 7 7 2 7 7 GM Uirb i 7 EB GM verb ED 5 667 x 10711 N mZkg2598 x 1024 kg 658 x 106 m 7 779 x 103 ms Donald G Luttermoser ETSU IX 19 For a circular orbit the orbital period is T i 27W 27T658 X 106 m verb 779 x 103 ms 531 x 103 s 148 hr Solution b The orbital period was computed in part a verb 779 x 103 ms Solution c The minimum energy to reach orbit can be determined by calcu lating the difference between the total mechanical energy of the satellite in orbit and total mechanical energy of the satellite prior to launch The satellite itself prior to launch is not moving but the surface of the Earth is moving due to the Earth s rotation we will ignore the Earth s orbital velocity here since we are not leav ing the Earth s gravitational in uence Hence minimum launch energy Emin KE PE0rb KE PEEL The Earth s rotational period is the de nition of a day so Trot 1000 day 86400 s From this we get the Earth s rotation velocity of 277136 27T638 X 106 m 464 Ur Tm 86400 x 104 s 18 Hence the minimum launch energy is lt1 2 2 Emin imv 7 v 2 orb 7 2 rot RED l2 Gm l2 Gm 2Uorb 7 2Urot RED 2 2 vorb Urot 1 2 l 69 Re 7 m IX 20 PHYS 2010 General Physics 779 1 3 2 464 2 N 2 667 x 10111 598 x 1024 kg x g 1 1 ll 638 x 106 m 658 x 106 m 200 kg 302 x 107 m2s2 190 x 106 mZs2gt 643 x 109 J 3 The absolute magnitude of the potential energy due to a large gravitating body is often times larger than the kinetic energy of the object in motion a We can de ne the total mechanical energy Etot of body in motion as the sum of the kinetic and potential energies Etot KE PE IX 23 b Since PE is negative in a gravitational eld Etot lt 0 216 negative for projectile e bound trajectories 4 For an object to just overcome any gravitating body s like the Earth s potential eld an object has to be launched with w total energy that is PE to escape the primary body s 69 Earth s gravitational eld gt the escape velocity a Hence to calculate the escape velocity from the surface of a large body of mass M and radius R we just have to set the initial kinetic and potential energy sum to zero and solve for the velocity 1 2 EtOt imvesc GMm i R IX24 0 Donald G Luttermoser E TSU b C d 5 Astrophysics has shown that stars evolve gaseous nebulae and collapse down until thermonuclear reactions start in their cores A star then lives for billions and billions of years except for the most massive ones until their nuclear fuel This gives the equation for the escape velocity or es cape speed 2GM R IX25 UESC As can be seen from Eq IX 25 the escape velocity does Lot depend upon the mass of the rocket Using the values for Earth in Eq IX 25 we get 2GM Re 7 vase IX 26 or vase 112 kms to leave the Earth s gravitational eld For an object to escape the gravitational eld of a primary body it must achieve a velocity greater than or equal to the escape velocity v 2 vase runs out 6 Eventually every star dies leaving one of three types of corpses 3 b White dwarfs Low mass stars become these types of objects Here the weight of the star is balanced by de generate electron pressure Neutron stars High mass stars supernovae te blow themselves apart leaving a core of solid neutrons Here the weight of the star is balanced by degenerate neutron pressure They start out as IX 21 IX 22 C PHYS 2010 General Physics Black holes zero volume objects known as singularities As the stel lar core collapses the escape velocity from the star gets higher and higher until a point is reached where vase 2 c the speed of light i Once this point is reach the light that the burnt out core emits due to its high temperature cannot escape the surface of the star gt this burnt out core is call a black hole as a result of this Note that the hole part comes from general relativity which we don t have time to discuss ii The region around a black hole singularity where U556 c is called the event horizon iii The event horizon lies at a distance from the sin gularity which astronomers call the Schwarzschild radius Rs after the astronomer who rst calcu lated it 2GM S 62 IX 27 iv Using solar values the Schwarzschild radius of the Sun if it were shrunk to the size of a black hole would be about 3 km Example 1X75 Solution a The radius is found from Eq IX 27 2GM RS 2 C Ultra high mass stars collapse down to a What is the Schwarzschild radius of a body with the mass of the Earth b What would be the average density of that body if its radius were the same as the Schwarzschild radius Donald G Luttermoser ETSU IX 23 2667 x 1011 N mZkg2598 x 1024 kg 300 X 108 ms2 886x10 3m Solution b The density is determined from M 7 Me i 3M6 V i 7TR i 47TR i 3598 x 1024 kg 47T886 x 10 3 m3 205 x 1030 kgm3 p PHYS2010 General Physics I Course Lecture Notes Section III Dr Donald G Luttermoser East Tennessee State University Edition 24 Abstract These Class notes are designed for use of the instructor and students of the course PHYS 2010 General Physics I taught by Dr Donald Lutternioser at East Tennessee State University These notes make reference to the College Physics 8th Edition 2009 textbook by Serway and Vuille III Motion in One Dimension A Displacement 1 The displacement of an object is de ned as the change in its position 2 It is given by the difference between its nal and initial coordi nates Aye E 06f xi displacement III l 3 Displacement is a vector quantity gt has direction and magni tude A56 E 02 02 Ill 2 Example III71 A car starts at a position 30 m from a stop sign and continues past the stop sign until it comes to rest 120 m past the sign What is its displacement X gt I I Xi 0 Xf xi30m xf120m Axxf xi120m 30m120m30m150m Note that since 150 is positive the car is moving in the positive 06 direction Ill 1 Ill 2 PHYS 2010 General Physics 4 Unit vectors a b c d Actually in Example lll l displacement should have been represented as a vector A0 instead of a scalar A06 note that using vector notation in one dimension is actu ally not needed since we always know that we will alway be along the x or some other axis A unit vector has a magnitude of unity e l and a direction gt it conveys the directional information of a vector quantity i Cartesian x direction Zor 56 ii Cartesian y direction or 7 iii Cartesian Z direction E or 2 iv Polar r direction 7 v Polar 6 direction In this class we will use the hat notation e 92 for the unit vectors for Cartesian coordinates instead of the ijk notation in order to make it consistent with what we will use for polar coordinates Note that your textbook does not use the unit vector notation As such we can write a vector in component form a AAziAygA22 1113 whereas your textbook writes a vector as the sum of its Cartesian vector components the vector arrow is Lot writ ten in your textbook but it is implied a a XAIXyAZ Donald G Luttermoser ETSU Ill 3 Once again we shall be using the rst version of vector components in this course In Eq Ill 3 AI Ag and A2 are all scalars gt they only contain the vector s magni tude of the given vector component e In Example Ill l we could have solved the problem us ing vector notation Afi Jc 120 m02 30 m02 150 m02 B Average Velocity 1 The average velocity U is de ned as the displacement divided by the time interval during which the displacement occurred or in vector notation 111 4 2 The average velocity of an object during the time interval It to tf is equal to the slope of the straight line joining the initial and nal points on a graph of the position of the object plotted versus time Example 11172 In Example III 1 assume it takes 5 seconds to move Am What is the car s average velocity Again xii 120m 30m il50m iKtitf ti 5s 0s 5s 30 ms in the positive X direction Ill 4 PHYS 2010 General Physics 3 When plotting the value of a variable as a function of some other variable the path on that plot that the object takes is called a trajectory Example 11173 From the graph below calculate the average velocity of the trajectory plotted x slope of connecting line is average velocity xf xiiBm 21ni1n1 U tf tii 3s 1s 77s 05 ms in the positive X direction C Instantaneous Velocity 1 The instantaneous velocity 17 is de ned as the limit of the average velocity as the time interval At becomes in nitesimally small Donald G Luttermoser E TSU a In equation form 17 E Alitrnoiif E 17 Ill 5 where the dfdt notation is from calculus and is the time derivative of vector 06 Note that we will not be using calculus in this course I just introduce it here to give you a avor of physics at the next level b The notation Alirn means that the ratio AgeAt is to be evaluated as the time interval At get shorter and shorter approaching zero c Graphically the slope of the tangent line on the position time curve is de ned to be the instantaneous velocity at that time slope of this line I is average velocity Ill 5 Ill 6 PHYS 2010 General Physics quot slope of this line is average velocity slope of dashed line is instantaneous velocity lim At gt 0 2 The instantaneous speed of an object which is a scalar quan tity is de ned as the absolute magnitude of the instantaneous velocity Hence from this de nition speed can never be nega tive 3 From this point forward the word velocity will mean instan taneous velocity and the word speed will mean instantaneous speed Donald G Luttermoser ETSU Ill 7 D Acceleration 1 The average acceleration 6 during a given time interval is de ned as the change in velocity divided by that time interval during which the change occurs III6 Example 11174 A parked car is hit head on by another car Three seconds after the collision the car that was parked is now traveling a 8 kmhr What is the average acceleration of this car in SI units Vzi 4234a Xli0 ti0 v1i0 BEFORE X X1553 th s vlrB kmhr gt V2i 0 I 39 AFTER 397 5 i TM 1711 i i i tf ti i 3s 0s 73hrs km 1 1 hr km 2667 77 A 7407 x 104 7 A s hr 3600 s x s2 x k 103 7407 x 104 in 711192 7407 x 101 992 32 1 km 32 07407 ms2 92 07 ms2 i Remember round the number such that it is consistent with the signi cant digits of the input numbers Ill 8 PHYS 2010 General Physics 2 The instantaneous acceleration of an object at a certain time equals the slope of the velocity time graph at that instant in time 239e tangent line of vt at that point v A instantaneous X acceleration vi aver e acceleration vi Iti Itf V t A17 17 d did 1292 52 limi 277lt7gt7 Ill 7 AHO At dt dt dt dt2 where the 1256 dt2 is the double derivative of the displacement with respect to time from calculus E OneDimensional 1D Motion with Constant Acceleration 1 Equations of 1 D motion a Let initial conditions be described with the subscript o gt 060 U0 amp to 0 from this point forward with a few exceptions we will set the initial time to zero b Let the nal conditions be arbitrary gt no subscript is written xvamp t Av c For constant accelerat10n gt 7 26 the slope 1s con stant 239e it does not change with time average acceleration instantaneous acceleration Donald G Luttermoser ETSU Ill 9 i Velocity 16 instantaneous velocity from acceler ation de nition vf viiv voiv vo a 7 tf ti 15 0 t or atv vo and nally 1 U0 at for constant a Ill 8 ii Average velocity U for constant a Ill 9 iii Displacement 06 x0 U7 letxo0 t thenU or 06 Ut and 1 0 v t 111 10 Substituting Eq Ill 8 for 1 gives 1 x E voatvot l E 2 at t lt 1tgtt vo 7a 2 l 06 vat Eat2 for constant a Ill ll Ill 10 PHYS 2010 General Physics iv Velocity e instantaneous velocity yet another way Take Equation Ill 8 and solve for t v u at U v vo at v vogtt7 Plug this value oft into Equation Ill 10 x 1 vvotvvo E x i vvov Uo i Ug U Uoi UO U Ug 7 2a 7 2a 112 v2 x 2a 2am 1 v3 or v2 U 2am for constant a Ill 12 1 Summary Equation Information Given by Equation U U0 at Velocity as a function of time U2 U3 2am Velocity as a function of displacement 06 vat at2 Displacement as a function of time 06 U Uot Displacement as a function of time amp velocity 4 77 Note In the above equations a is constant and the motion is in a straight line starting at the origin x0 0 Donald G Luttermoser ETSU Ill 11 at to 0 If the initial position and position are not zero the equations of motion take the following form Equation Function 1 v0 at to vhf 2 i 2 v i vo 2a06 060 Mac x x0 11005 150 at to2 0615 xxovvot to xvt 2 Problem Solving Strategy for Accelerated Motion in 1 D a Make sure all of the units of the parameters in the problem are consistent b Choose a coordinate system and draw a picture c Make a list of all of the quantities given in the problem and a separate list of those to be determined 1 Select those equations that will allow you to determine the unknown parameters e variables e Make sure your answer is consistent with the diagram drawn in part Example 11175 A car starts from rest at a stop sign and accel erates at a constant rate to the posted speed limit of 50 kmhr The car reaches the speed limit after traveling 107 m What is the average velocity and acceleration of the car How long does it take you to reach the speed limit a Units consistent NOl Ill 12 PHYS 2010 General Physics 06 107 In amp v 50 krnhr You want 81 units so change krn into In k v 50f1 103 50 x 104 mhr Change hours to seconds In 1 hr 5 1 47 1389 14 v 0 X 0 hr 3600 S ms n1s b Choose a coordinate system and draw picture vo0 v50 kmhr gt i i x xo0 x107m to0 c Quantities given 1500 0600 vo0 v14n1s 06107n1 Unknown quantities If U a 1 Choose equations U for U U2 U 2am for a v vo at for t e Calculate results and check with gure vov7014rnsigm 2 2 2 S U Donald G Luttermoser ETSU Ill 13 v2v32ax gt 2axv2 v3 v2 v2 a 206 14 ms2 02 i 142 n12s 2 a 2107 m 214 m 916 x 101 m s 2 a 092 ms2 v vo vvoat gt atv vo gt 157 71 t14n1s 201522n1s2 092 ms In s 3 Freely Falling Bodies a Galileo Galilei is the father of experimental physics gt carried out a variety of experiments in mechanics the study of motion i Near the surface of the Earth bodies fall at the same rate of acceleration independent of the body s mass ii Air resistance can affect the rate at which a body falls Ill 14 PHYS 2010 General Physics b Bodies in free fall i A freely falling body is an object moving only un der the in uence of gravity regardless of its initial motion ii Objects thrown upward or downward and those released from rest are all falling freely once they are released even if they are initially going up iii Once a body is in free fall all objects have an ac celeration downward gt this free fall acceleration is called the surface gravity g7 iv A gravitating body s surface gravity depends upon the total mass and size of the gravitating body v The Earth s surface gravity at sea level is g 980 ms2 980 cms2 32 ftsg HI13 vi Actually bodies dropped in the Earth s atmo sphere will experience a frictional force from air re sistance in addition to the gravitational force In this course we will ignore air resistance but it is important in reality Example IIIAG Problem 253 Page 52 from the Serway amp Vuille textbook A model rocket is launched straight upward with an initial speed of 500 ms It accelerates with a constant up ward acceleration of 200 ms2 until its engines stop at an altitude of 150 m a What can you say about the motion of the rocket Donald G Luttermoser ETSU Ill 15 after its engines stop b What is the maximum height reached by the rocket C How long after lift off does the rocket reach its maximum height d How long is the rocket in the air a What can be said about the motion of the rocket after the engines stop The rocket will continue upward but start to decelerate due to the Earth s gravitational eld until the upward velocity reaches zero The rocket then begins to fall back to the ground with an acceleration equal to the Earth s surface gravity e 980 rnsg b What is the max height y y1150m t1 V1 4 I L 39 Ymax 32 g 980 ms2 Step 2 a1 200 ms2 Engines shut off y1 vD 500 ms l if Rocket while engines are on a1 200 ms2 Step 1 Step 1 First we will need to calculate the velocity of the rocket when the engine is turned off 01 gt we are given 11 200 rns2 yl 150 mamp 00 500 ms so use 0 113 201y1 500 ms2 2200 ms2l50 m 250 x 103 n12s2 600 x 102 mZs2 Ill 16 PHYS 2010 General Physics 01quot 250 x102 1112 600 x102 mZs2 0 BLQO gtlt102 rn2s2 310 x102 mZs2 310 x103 mZs2 010310 X 103 rnZs2 557 X 101ms 557 rns Now we need to calculate yum E yg Step 2 We will need yl a2 and 01 also 02 0 rocket comes to rest Our initial velocity is now 01 and our nal velocity is 02 so we can write 2 7 2 7 2 02 7 01 ZagAy 7 01 212 yg 01 and note that ymax yg and that the acceleration is now the downward acceleration due to gravity a2 g 980 n1s2 From 0 0 212 yg yl we solve for ygz 11 vi v3 v 1112 1111 212 Zg 2 2 U1 U2 y ymax yl 557 ms2 02 2980 rnsg 310 x 103 1112 196 x 101 ms2 ymax 150 rn 150 n1 72 150m158gtlt102rn 150m158m yum308m 2 72 150 m 158 x 1034K m S Donald G Luttermoser ETSU Ill 17 c How long does it take to reach yum Break the time into 3 parts t to t1 t2 to 0 which was our initial time t1 is the time when the rocket reaches yl and t2 is the time the rocket reaches ymax as measured from yl so use Uf U1 at t1 U1 UO a1 7 557 ms 500 ms 200 ms2 57 m s 1 s2 285729 200 m s2 s S Q Ug U1 02 0 557 ms 980 ms2 557 s2 568 980 s S ttot1t2029s56 s858s d How long is the rocket in the air We just gured out how long it took to get to yum we now have to nd out how long it takes to fall then add the two numbers together III 18 PHYS 2010 General Physics Step 3 Use y vat at2 and solve for t Note that no 0 the velocity at the top of the trajectory v at ymax yg from part b hence 1 1 015 7 t2 7 t2 y 2 a 2 a or 2 2 t2 7y gt 7y a a y n YoYmax9voo9 t00 gt Set up a new problem o9 Want to solve for t 3 a g what we are given is shown in the gure to the left 0 v t 0 Y Here yygroundymax0 308m 308 m a g 980 ms2 So ti 2 308 m i 616 s2 7 980 ms 7 980 t 629 s2 793 s 156 where 155 the time it takes to fall from the maximum height to the ground from part c Donald G Luttermoser ETSU III 19 The total time in the air is then just the sum of the time found in part b and the time found in part C them 751 756 or ttotal S i S
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