Note for PH 101 at UA-General Physics I (5)
Note for PH 101 at UA-General Physics I (5)
Popular in Course
Popular in Department
This 45 page Class Notes was uploaded by an elite notetaker on Friday February 6, 2015. The Class Notes belongs to a course at University of Alabama - Tuscaloosa taught by a professor in Fall. Since its upload, it has received 50 views.
Reviews for Note for PH 101 at UA-General Physics I (5)
Report this Material
What is Karma?
Karma is the currency of StudySoup.
You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!
Date Created: 02/06/15
GIANCOLI Lecture PowerPoints Chapter 5 Physics Principles with Applications 6 edition Giancoli Chapter 5 1 Circular Motion Gravitation Us A J Copyright 2005 Pearson Prentice HaH In Units of Chapter 5 Kinematics of Uniform Circular Motion Dynamics of Uniform Circular Motion Highway Curves Banked and Unbanked Nonuniform Circular Motion Centrifugation Newton s Law of Universal Gravitation Units of Chapter 5 Gravity Near the Earth s Surface Geophysical Applications Satelites and Weightlessness Keper s Laws and Newton s Synthesis Types of Forces in Nature 51 Kinematics of Uniform Circular Motion Uniform circular motion motion in a circle of constant radius at constant speed Instantaneous velocity is always tangent to circle Copyright 2005 Pearson Prentice Hall Inc 51 Kinematics of Uniform Circular Motion Looking at the change in velocity in the limit that the time interval becomes infinitesimally small we see that 51 a lim 5 Al gtO AI 51 Kinematics of Uniform Circular Motion This acceleration is called the centripetal or radial acceleration and it points towards the center of the circle Copyright 2005 Pearson Prentice Hall Inc The period T is the time taken to perform a complete revolution The frequency f is the number of revolutions per second f 1T The circumference of a circle of radius r is 27tr Hence we can also relate the speed of circular motion to the radius rand period T by 2 quotu T 52 Dynamics of Uniform Circular Motion For an object to be in uniform circular motion there must be a net force acting on it We already know the acceleration so can immediately write the force v2 IEFRmaRm r 51 Copyright 2005 Pearson Prentice HallI Incl 52 Dynamics of Uniform Circular Motion We can see that the force must be inward by thinking about a ball on a string Force on ball exerted by string Force on hand I exerted by string Copyright 2005 Pearson Prentice Hall Inc Forces in Circular Motion Note that there is no mysterious new centripetal force It s just that the resultant of all the forces acting on the body must point towards the center for circular motion to happen Likewise there is no mysterious new centrifugal force we just have the inertia of all objects Exercise Question An object is in counterclockwise circular motion when suddenly the cord is out Which trajectory best represents the motion after the cord is cut Exercise Answer Answer C 4 When no force acts on the object it continues in a straight line with constant velocity Since its velocity is always tangent to the circle it continues along trajectory C 53 Highway Curves Banked and Unbanked When a car goes around a curve there must be a net force towards the center of the circle of which the curve is an arc If the road is flat that force is supplied by friction 4 Force on car sum of friction forces acting on each tire Tendency for passenger to i go straight 1 gt Force on passenger Copyright 2005 Pearson Prentice Hall Inc 53 Highway Curves Banked and Unbanked l 39 39 a 39 try lamb Copyright 2005 Pearson renti ce Hall Inc If the frictional force is insufficient the car will tend to move more nearly in a straight line as the skid marks show 53 Highway Curves Banked and Unbanked As long as the tires do not slip the friction is static If the tires do start to slip the friction is kinetic which is bad in two ways 1 The kinetic frictional force is smaller than the static 2 The static frictional force can point towards the center of the circle but the kinetic frictional force opposes the direction of motion making it very difficult to regain control of the car and continue around the curve 53 Highway Curves Banked and Unbanked y Banking the curve can help keep I cars from skidding In fact for I every banked curve there is one 1 speed where the entire centripetal l force is supplied by the 1 a horizontal component of 39 7 3R the normal force and no friction is required This occurs when 2 71 FN 8111 0 m r Copyright 2005 Pearson Prentice Hall Inc 56 Newton s Law of Universal Gravitation If the force of gravity is being exerted on objects on Earth what is the origin of that force Newton s realization was that the force must come from the Earth He further realized that this force must be what keeps the Moon in its orbit Copyright 2005 Pearson Prentice Hall Inc 56 Newton s Law of Universal Gravitation The gravitational force on you is onehalf of a Third Law pair the Earth exerts a downward force on you and you exert an upward force on the Eanh When there is such a disparity in masses the reaction force is undetectable but for bodies more equal in mass it can be significant Moon Gravitational force exerted on Moon by Earth Gravitational force Earth amp exerted on Earth by the Moon Copyliulll 2005 Pealsun Frenl ue Hall Inn 56 Newton s Law of Universal Gravitation Therefore the gravitational force must be proportional to both masses By observing planetary orbits Newton also concluded that the gravitational force must decrease as the inverse of the square of the distance between the masses In its final form the Law of Universal Gravitation reads quot r 54 where G 667 x 10 11Nm2kg2 56 Newton s Law of Universal Gravitation Fiber The magnitude of the gravitational constant G can be measured in the Mirror 5 A laboratory Scale 3 This is the Cavendish experiment Light K SOUICC Copyright 2005 Pearson Prentice Hall Inc 57 Gravity Near the Earth s Surface Geophysical Applications Now we can relate the gravitational constant to the local acceleration of gravity We know that on the surface of the Earth mmE mg G 2 r E Solving for g gives 8 mE 55 2 i E Now knowing g and the radius of the Earth the mass of the Earth can be calculated m gr 980 ms2638 gtlt 1061102 598 X 1024 E G 667 gtlt 10 Nm 2kg2 39 57 Gravity Near the Earth s Surface Geophysical Applications TABLE 51 Acceleration Due to Gravity at Various Locations on Earth The acceleration clue to Elevation g gravity varies over the Location m In52 Earth s surface due to NeWYork 0 9803 altitude local geology San Francisco 0 9800 and the shape of the Denver 1650 9796 Pikes Peak 4300 9789 EaLth WhICh Is 01 qu39te Sydney 0 9798 Sp er39cal39 Australia Equator 0 9780 North Pole 0 9832 calculated Copyright 2005 Pearson Prentice Hall Inc The period T is the time taken to perform a complete revolution The frequency f is the number of revolutions per second f 1T The circumference of a circle of radius r is 27tr Hence we can also relate the speed of circular motion to the radius rand period T by 2 quotu T Satellites For an orbiting satellite assume circular orbit the centripetal force is provided by the gravitational attraction Him 39112 i T2 7 Since it s orbit is circular motion we can also relate period T radius r and orbital velocity v 2m T quotU Satellites Combining these we can find the orbital radius and velocity of any satellite Warning Note that this is radius of orbit ie distance from center If we consider a satellite above the Earth then we must add the radius of the Earth rE 6380 km to the height of satellite above the Earth to get the full radius of orbit r rE height 58 Satellites and Weightlessness Satellites are routinely put into orbit around the Earth The tangential speed must be high enough so that the satellite does not return to Earth but not so high that it escapes Earth s gravity altogether 27000101111 309mm circular elliptical Copyright 2005 Pearson Prentice Hall Inc 58 Satellites and Weightlessness The satellite is kept in orbit by its speed it is continually falling but the Earth curves from underneath it Without owl Wih raVity Copyright 2005 Pearson Prentice Hall Inc 58 Satellites and Weightlessness Objects in orbit are said to experience weightlessness They do have a gravitational force acting on them though The satellite and all its contents are in free fall so there is no normal force This is what leads to the experience of weightlessness w ill w w ill l J 3931 JIJL M I quot Iquotl I I L l 39y l I 1 l l a a 0 b a gup n Prentice Hal Copyright 2005 Pearso my ll 6 la gdown Linc 5 Sa tgg and ch mmgggmggw Mm mmr y EMS fm g a m apparent wc g gh glggmggg bmamge We grav at cma Wm g i x g gn Hi Earn be xp m 311 Earth ags chl a Emit rm y br gf y a 59 Kepler s Laws and Newton39s Synthesis Kepler s laws describe planetary motion 1 The orbit of each planet is an ellipse with the Sun at one focus Planet Copyright 2005 Pearson Frenli ccccccccc cl 59 Kepler s Laws and Newton39s Synthesis 2 An imaginary line drawn from each planet to the Sun sweeps out equal areas in equal times Copyright 2005 Pearson Prentice Hall Ich 59 Kepler s Laws and Newton39s Synthesis The ratio of the square of a planet s orbital period is proportional to the cube of its mean distance from the Sun TABLE 52 Planetary Data Applied to Kepler s Third Law Mean Distance from Sun s Period T s3T2 Planet 106 km Earth years 1024 km3y2 Mercury 579 0241 334 Venus 1082 0615 335 Earth 1496 10 335 Mars 2279 188 335 Jupiter 7783 1186 335 Saturn 1427 295 334 Uranus 2870 840 335 Neptune 4497 165 334 Pluto 5900 248 334 Copyright 2005 Pearson Prentice Hall Inc Kepler s 3rd Law from Newton s Law of Gravity We can derive Kepler s 3rol Law from Newton s law of gravity We assume a circular orbit Again the centripetal force needed for circular motion is provided by gravitational force 9 G fm 3921 9 m I l pr u quot4 And usmg again 0 fill Kepler s 3rd Law 2 We can find that T2 Z 47 Jr3 GM Hence if we consider the ratio of periods and orbital radii of two planets 32 7 13 Kepler s 3lrcl Law T2 T2 The first expression for Kepler s 3rol Law also allows us to determine the Mass of the Sun from the orbital data of the planets 59 Kepler s Laws and Newton39s Synthesis Kepler s laws can be derived from Newton s laws lrregularities in planetary motion led to the discovery of Neptune and irregularities in stellar motion have led to the discovery of many planets outside our Solar System 5 OA 750 Jupiter a Sun o a 3 0 MJ 47 Planet 0 Ursae e Majoris 3 MJ U 391 A B C ps1 on C Andromedae g Copyright 2005 Pearson Prentice Hall lnc 510 Types of Forces in Nature Modern physics now recognizes four fundamental forces 1 Gravity 2 Electromagnetism 3 Weak nuclear force responsible for some types of radioactive decay 4 Strong nuclear force binds protons and neutrons together in the nucleus 510 Types of Forces in Nature So what about friction the normal force tension and so on Except for gravity the forces we experience every day are due to electromagnetic forces acting at the atomic level Summary of Chapter 5 An object moving in a circle at constant speed is in uniform circular motion It has a centripetal acceleration dR There is a centripetal force given by 222 EFR maR m r The centripetal force may be provided by friction gravity tension the normal force or others Summary of Chapter 5 Newton s law of universal gravitation m1 m2 2 FG Satelites are able to stay in Earth orbit because of their large tangential speed Backup 52 Dynamics of Uniform Circular Motion There is no centrifugal force pointing outward what happens is that the natural tendency of the object to move in a straight line must be overcome If the centripetal force vanishes the object flies off tangent to the circle o 7 V 1 l DOES N T HAPPEN S I HAPPEN 5quot l w I flyquot 5 IV J 1 quotCIquot 5 3 y 1 E b i 54 Nonuniform Circular Motion i tan F I 39 R l l I I I I I I I I I If an object Is movmg In a Circular path but at varying speeds it I must have a tangential component to its acceleration as well as the radial one 5 8 I aR I l I l I b Copyright 2005 Pearson Prentice Hall Inc 54 Nonuniform Circular Motion This concept can be used for an object moving along any curved path as a small segment of the path will be approximately circular Copyright 2005 Pearson Prentice Hall Inc 55 Centrifugation A centrifuge works by spinning very fast This FOF EC 62 th means there must be a by 11qu1d very large centripetal force The object at A would go in a straight k line but for this force as it is it winds up at B Copyright 2005 Pearson Prentice Hall Inc
Are you sure you want to buy this material for
You're already Subscribed!
Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'