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General Physics I

by: Alyce Abshire

General Physics I PHYS 130

Marketplace > Siena College > Physics 2 > PHYS 130 > General Physics I
Alyce Abshire
Siena College
GPA 3.6

Rose Finn

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Rose Finn
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This 46 page Class Notes was uploaded by Alyce Abshire on Tuesday October 20, 2015. The Class Notes belongs to PHYS 130 at Siena College taught by Rose Finn in Fall. Since its upload, it has received 40 views. For similar materials see /class/225360/phys-130-siena-college in Physics 2 at Siena College.


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Date Created: 10/20/15
Newton s Law of Universal Gravitation Ch 13 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 distance between them F G mlm2 g r3 G is the universal gravitational constant G 6673 X 103911 Nm2 kg2 Law of Gravitation cont This is an example of an inverse square law The magnitude of the force varies as the inverse square of the separation of the particles The law can also be expressed in vector form mlm2 2 12 r F12 2 TG Notation F12 is the force exerted by particle 1 on particle 2 The negative sign in the vector form of the equation indicates that particle 2 is attracted toward particle 1 F21 is the force exerted by particle 2 on particle 1 More About Forces F12 39F21 The forces form a Newton s Third Law actionreaction pair Gravitation is a field force that always exists between two particles regardless of the medium between them The force decreases rapidly L as distance increases A consequence ofthe inverse square law Gravitational Force Due to a g G vs 9 Distribution of Mass Always distinguish between G and g The gravitational force exerted by a finite G is the universal gravitational constant Size spherically Symmetric mass diStribUtion on a particle outside the distribution is the same as if the entire mass of the distribution were concentrated at the center For the Earth It is the same everywhere g is the acceleration due to gravity g 980 r ns2 at the surface of the Earth 9 will vary by location M Em 2 RE FgG a g Moon s Acceleration Newton looked at proportionality of accelerations between the Moon and objects on the Earth Newton s Verification Developed theory at age 23 He compared the acceleration of the Moon in its orbit with the acceleration of an object falling near the Earth s surface He calculated the centripetal acceleration of the Moon from its distance and period The high degree of agreement between the two techniques provided evidence of the inverse square nature of the law Newton s Canon Centripetal Acceleration Newton s Assumption The Moon experiences t t I Newton treated the Earth as if its mass were a Gen p a all concentrated at its center acceleration as it orbits the Earth He found this very troubling When he developed calculus he showed this assumption was a natural consequence of the quotM r r Law of Universal Gravitation M M only mass interior to your radial position is 472204 M important T2 Measuring G g Finding g from G G was first measured by The magnitude ofthe force acting on an object of Henry Cavendish in 1798 g l It mass m in freefall nearthe Earth s surface is mg The apparatus shown here i l mum This can be set equal to the force of universal allowed the attractive force my gravitation acting on the object between two spheres to i cause the rod to rotate i ix MEm quot mg I G 2 The mirror amplifies the r J L R 39 motion j 7 E It was repeated for various ri J G ME masses g T R2 9 Above the Earth s Surface If an object is some distance h above the Earth s surface rbecomes RE h GME RE 1712 This shows that 9 decreases with increasing altitude As r gt 00 the weight of the object approaches zero Variation of g with Height Altitude l1 km g ms z 1 000 733 2 Jim 5158 3 000 453 4 000 370 5 0 303 6 NM 260 7 MU 223 8 000 193 9 Hill 59 10 000 L40 50 000 013 x U Ch 13 Question 4 The gravitational force that the Sun exerts on the Moon is about twice as great as the gravitational force that the Earth exerts on the Moon Why doesn t the Sun pull the Moon away from the Earth during a total eclipse of the Sun Kepler s Laws Introduction Johannes Kepler was a German astronomer He was Tycho Brahe s assistant Brahe was the last of the naked eyequot astronomers Kepler analyzed Brahe s data and formulated three laws of planetary motion Kepler s Laws Kepler s First Law All planets move in elliptical orbits with the Sun at one focus Kepler s Second Law The radius vector drawn from the Sun to a planet sweeps out equal areas in equal time intervals Kepler s Third Law The square of the orbital period of any planet is proportional to the cube of the semimajor axis of the elliptical orbit Notes About Ellipses F and F2 are each a focus of the ellipse They are located a distance 0 from the center The longest distance through the center is the major axis a is the semimajor axis The shortest distance through the center is the minor axis b is the semiminor axis eccentricity e o la For a circle e 0 For ellipses 0 lt e lt 1 Planetary Orbits The Sun is at one focus Nothing is located at the otherfocus Aphelion is the point farthest away from the Sun The distance for aphelion is a c For an orbit around the Earth this point is called the apogee Perihelion is the point nearest the Sun The distance for perihelion is a c For an orbit around the Earth this point is called the perigee I Kepler s First Law A circular orbit is a special case of the general elliptical orbits Is a direct result of the inverse square nature of the gravitational force Elliptical and circular orbits are allowed for bound objects A bound object repeatedly orbits the center An unbound object would pass by and not return These objects could have paths that are parabolas e 1 and hyperbolas e gt 1 Orbit Examples Pluto dwarf planet has the Kepler s second Law Consequence of conservation of angular momentum 512 Grav force produces no torque so angular i Q r F l v d highest eccentricity of any planet 3 momentum is conserve 39 ePluto 0 25 MW Mathematically 1quot Halley s comet has an orbit W 39 r X P Mp r X V constant 7 With high eccentriCity b dA 12 r x v dt half area ofparallelogram u 39 el lalley s comet 097 M I r 7 W 7 J collstant zlr VIII d 2M 51 m Equal areas in equal times XX 39 i The law applies to any central force Whether inversesquare or not ll J 39 J 39 Keplers Third Law Keplers Third Law cont Can be predicted from This can be extended to an elliptical orbit the inverse s uare law 3 39 q JMSunMPlanct MPlanetv Replace r Wlth a Start by assummg 5 12 F Remember a is the semimajor axis circular orbit 2 The gravitational force V 27 T2 47139 a3 K a3 supplies a centripetal T 3 force K5 is a constant 4 2 T2 r KS Sun Sun Ks is independent of the mass of the planet and so is valid for any planet Example Geosynchronous Example Mass of the Sun Satellite 7 Using the distance between the Earth and the Ageosynchronoys satellite Sun and the period of the Earth s orbit appears t rema39 Verthe Kepler s Third Law can be used to find the same p 39 t the Earth mass of the Sun The graVItatIonal force i 2 3 supplies a centripetal force l 47239 I You can find h or v y Sun GT2 Similarly the mass of any object being a r J orbited can be found if you know information I about objects orbiting it The Center of Mass Center of Mass Coordinates There is a special point in a system or object The coordinates of the center of mass are called the center of mass that moves as if all of the mass of the system is concentrated at that point 7 7 21quot 7 Zmrz XL M yum 2w The system WI move as If an external force M M M were applied to a single particle of mass M located at the center 0f mass where M is the total mass of the system M is the total mass of the system Center of Mass position Center of Mass Example The center of mass can be located by its Both masses are on the position vector rCM X39ax39s The center of mass is V 2m on the Xaxis I ml i The center of mass Is 4quot quotm lCM M closerto the particle quotIt with the larger mass W F N Ax r is the position of the ith particle defined by 4 1 xi yj 21 Center of Mass Extended Object Think of the extended object as a system containing a large number of particles The particle distribution is small so the mass can be considered a continuous mass distribution The coordinates of the center of mass of the object are 1 1 xCM jxdm yCM ydm zCM zdm Ch 9 Problem 41 Man n u 10 20 31 A uniform piece of sheet steel is shaped as shown Compute the x and y coordinates of the center of mass of the piece quotClll Center of Mass Extended Object Position The position of the center of mass can also be found by 1 rCM Ejr dm The center of mass of any symmetrical object lies on an axis of symmetry and on any plane of symmetry Center of Mass Example An extended object can be considered a distribution of small mass elements Am The center of mass is located at position rCM Center of Mass Rod Find the center of mass of a rod of mass M and length L The location is on the Xaxis 7 or yCM ZCM 0 rim Arlv XCML2 4 Motion of a System of Particles Assume the total mass M of the system remains constant We can describe the motion of the system in terms of the velocity and acceleration of the center of mass of the system We can also describe the momentum of the system and Newton s Second Law for the system Velocity and Momentum of a System of Particles The velocity of the center of mass of a system of particles is Zmivl d A I drCM VCM The momentum can be expressed as Mvm Z lmyvy 2p pml The total linear momentum of the system equals the total mass multiplied by the velocity of the center of mass Forces In a System of Particles The acceleration can be related to a force MaCM Z If we sum over all the internal forces they cancel in pairs and the net force on the system is caused only by the external forces Newton s Second Law for a Momentum of a System of 395 System of Fjarticles Particles 7 Since the only forces are external the net external The total linear momentum of a system of force equals the total mass of the system multiplied particles is conserved if no net external force by the acceleration of the center of mass is acting on the system 2Faxt M aCM MvCM ptot constant when EFeXt 0 The center of mass of a system of particles of combined mass M moves like an equivalent particle of mass Mwould move underthe influence of the net external force on the system Motion of the Center of Mass 395 Example g Center of Mass A projectile is fired into the air 39 DiCk FOSbury intrOduced a new high jump technique the Fosbury Flop at the 1968 Olympics Raised world record 30 cm and suddenly explodes With no explosion the projectile would follow the dotted line After the explosion the center of mass of the fragments still follows the dotted line the same parabolic pa h the projectile would have followed with no explosion The Vector Product and Torque The torque vector lies in a direction perpendicular to the plane formed by the position vector and the force vector 1 r x F The torque is the vector or cross product of the position vector and the force vector The Vector Product Defined Given two vectors A and B The vector cross product ofA and B is de ned as a third vector C o C is read as A cross B The magnitude of C is AB sin 6 0 is the angle between A and B More About the Vector Product The direction of C is perpendicular to the plane formed by A and B Direction is given by the x B f righthand rule r L fingers in direction of A J 1 curl into direction of B thumb gives direction of cross product C Be sure to use your right hand mm mm i iv 4 ltx Ch 11 Question 3 Vector A is in the negative y direction and vector B is in the negative x direction What are the directions of A x B B x A Ch 11 Question 2 Is the triple product de ned by ABxC a scalar or a vector quantity Explain why the operation ABxC has no meaning Angular Momentum The instantaneous angular momentum L of a particle relative to the origin 0 is defined as the cross product of the particle s instantaneous position vector r and its instantaneous linear momentum p Torque and Angular Momentum The torque is related to the angular momentum Similar to the way force is related to linear momentum dL 27 This is the rotational analog of Newton s Second Law 25 and L must be measured about the same origin This is valid for any origin fixed in an inertial frame More About Angular Momentum The SI units of angular momentum are kgm2 s Both the magnitude and direction of L depend on the choice of origin The magnitude of L mvr sin p q is the angle between p and r The direction of L is perpendicular to the plane formed by r and p ThevectorLrxpis The magnitude is A particle in uniform circular Angular Momentum of a Particle Angular Momentum Of a Example Rotating Rigid Object Each particle of the object rotates in the xy plane about the z axis with an angular speed of a pointed out of the diagram L mvr sin 90 mvr sin 90 is used since v is perpendiculartor The angular momentum motion has a constant 5 of an individual particle angular momentum about an y is L ml r12 a Iaa1ltthrough the center of Its n H 7 x L and m are directed along the z axis Angular Momentum of a Angular Momentum of a Rotating Rigid Object cont Bowling Ball To nd the angular momentum ofthe The momentum of entire object add the angular momenta Inertla 02fthe ball IS of all the individual particles 115M I t 2 e angu ar momen um LZLiZmiri 0 150 oftheballisLZlw i i The direction of the This also gives the rotational form of angular momentum is in Newton s Second Law the positive 2 direction sz za dt dt Conservation of Angular Momentum The total angular momentum of a system is constant in both magnitude and direction if the resultant external torque acting on the system is zero Net torque 0 gt means that the system is 39solated Ltot constant or Conservation of Angular Momentum If the mass of an isolated system undergoes redistribution the moment of inertia changes The conservation of angular momentum requires a compensating change in the angular velocity This holds forrotatlon about a fixed axis and for rotation bout ah axis through the center of mass o system The het torque must be Zero ll l any case g Ch 11 Question 12 Often when a high diver wants to turn a flip in midair she draws her legs up against her chest Why does this make her rotate faster What should she do when she wants to come out of her flip Ch 11 Question 14 Stars originate as large bodies of slowly rotating gas Because of gravitation these clumps of gas slowly decrease in size What happens to the angular speed of a star as it shrinks Explain Conservation Law Summary For an isolated system 1 Conservation of Energy 39 E Er 2 Conservation of Linear Momentum Pi pf 3 Conservation of Angular Momentum 39 Li Lf Conservation of Angular Momentum The MerryCoRound The moment of inertia of the system is the moment of inertia of the platform plus the moment of inertia of the A person Assume the person can be quot 5 treated as a particle As the person moves toward the center of the rotating quot1 f 7 platform the angular speed V will increase To keep L constant Ch 11 Question 13 In some motorcycle races the riders drive over small hills and the motorcycle becomes airborne for a short time lfthe motorcycle racer keeps the throttle open while leaving the hill and going into the air the motocycle tends to nose upward Why does this happen g Ch 11 Question 15 lfglobal warming occurs over the next century it is likely that some polar ice will melt and the water will be distributed closerto the Equator How would this change the moment of inertia of the Earth Would the length ofthe day increase or decrease Motion of a Top The only external forces acting on the top are the normal force n and the gravitational force M g The direction of the angular momentum L is along the axis of symmetry The righthand rule indicates thatrerergisin the xy plane Causes precession slow rotation of top about 2 axis Gyroscope in a Spacecraft The angular momentum of the spacecraft about its center of mass is zero A gyroscope is set into rotation giving it a nonzero angular momentum The spacecraft rotates in the direction opposite to that of the gyroscope so that the total momentum of the system remains zero Can also use gyroscopes to keep a spacecraft pointed in a set direction Hubble Space Telescope Angular Momentum as a Fundamental Quantity The concept of angular momentum is also valid on a submicroscopic scale Angular momentum has been used in the development of modern theories of atomic molecular and nuclear physics In these systems the angular momentum has been found to be a fundamental quantity Fundamental here means that it is an intrinsic property of these objects It is a part of their nature g Fundamental Angular Momentum Angular momentum has discrete values quantized These discrete values are multiples of a fundamental unit of angular momentum The fundamental unit of angular momentum is hbar kgm2 hi1054x10 27 Where h is called Planck s constant Classical Ideas in Subatomic g Systems l Niels Bohr Certain classical concepts and models Niels BOhrW39dlS a DaniSh PhySiCiSt are useful in describing some features He adopted the then radical idea of of atomic and molecular systems discrete angular momentum values in Proper modifications must be made developing his theory ofthe hydrogen A wide variety of subatomic phenomena atom can be expained by assuming discrete Classical models were unsuccessful in values of the angular momentum describing many aspects of the atom associated with a particular type of motion Bohr s Hydrogen Atom The electron could occupy only those circular orbits for which the orbital angular momentum was equal to n where n is an integer This means that orbital angular momentum is quantized Uniform Circular Motion Ch 6 g Uniform Circular Motion cont A force Fr is directed toward quot A force causing a the center of the circle 39 centripetal acceleration acts This force is associated with tewerd the center Of the an acceleration aC C39rCie Newton s Second causes a Change in the Law along the radial direction Ciiieetieh 0f the VeiOCity gives vector If the force vanishes the object would move in a straightline path tangent to the circle 7 1 v EFzmaczm r Centripetal Force g Conical Pendulum HW Prob 9 The force causing the centripetal acceleration The object iS in equilibrium in is sometimes called the centripetal force the vertical direction end undergoes uniform crrcular ThIS IS not a new force It IS a new role for a motion in the horizontal force direction It is a force acting in the role of a force that Newton s 2nd Law says causes a circular motion 2F 0 I ii I 2 mg I Motion in a Horizontal Circle The speed at which the object moves depends on the mass of the object and the tension in the cord The centripetal force is supplied by the tension The force of static friction supplies the centripetal force The maximum speed at which the car can negotiate the curve is given by 1 l v a SEE m F U j 2 N my 7 1 f 4 HI I 1 IN III Solve for v Does it depend on the mass of the car Horizontal Flat Curve Banked Curve quotx NJ NsinO These are designed with friction equaling zero There is a component of the normal force that supplies the centripetal force 7 4 S 39 ZFI I I39lf Fl I Ni my 390 NI m LooptheLoop This is an example of a vertical circle At the bottom of the loop b the upward force experienced by the object is greater than its weight 2 2F 2 ml 39 7 g 1 N mg m I e 2314 vmmmm Dali I LooptheLoop Part 2 NonUniform Circular Motion At the top 0f the CirCte The acceleration and force C the fCree exerted 0 have tangential components e Omth IS less than F produces the centripetal 39 S we399 acceleration a 2 ZFU m t F1f produces the tangential K 397 r acceleration N my 111 2F 2F th g 2004 mumBrooks Cola Vertical Circle with NonUniform Speed Top and Bottom of Circle The gravitational force exerts The tension at the bottom is a maximum a tangentlal force on the The tension at the top is a minimum mg object T l 1 If 739tOID 0 then Look at the components of Fg J The tension at any point can 0 2F mt be found 1 7 39 y 313 i m 7 mg m V l quot g T my 39050 HIT mg you can then solve for minimum a speed at top I mmmmmmmmmmm is T my 050 l m b HW Prob 17 i 3 I Motion in Accelerated Frames g Centrifugal Force A fictitious force results from an accelerated From the frame Ofthe j frame of reference passengcter b game d A fictitious force appears to act on an object in the hpepgggsr O pus er owar 7 same way as a real force but you cannot identify From the frame of the Earth V 5 a second object for the fictitious force 39 the car applies a leftward fo ce on the passenger The outward force is often called a centrifugal force 3 39 V It is a fictitious force due to the acceleration associated with the iquot 39 car s change in direction WWWW Coriolis Force Fictitious Forces examples Although fictitious forces are not real forces This is an apparent force they can have real effects caused by changing the I r Examples radial posmon of an object In 7 a rotating coordinate system The result of the rotation is the curved path of the ball Objects in the car do slide You feel pushed to the outside of a rotating platform The Coriolis force is responsible for the rotation of weather systems and ocean currents Fictitious Forces in Linear Systems gtu The inertial observer a sees Tsinl9ma ZFy Tcosl9 mg 0 The noninertial observer b sees V F39Tsin6 F ma Z a ctztzous l a T H v lH I mg i l 77v 7 i 7 7 7 7 1 a 7L 3 7771 gs33 395f no VJquot xt wj 39 39 13 l quotl39 A ullllltllll u l ZF39V T0056 mg 0 1 m Fictitious Forces in a Rotating System n Noninm liul observer i 39 x I r T T T l T quot4 E l nroumn l According to the inertial observer a the tension is the centripetal 2 force mv T IA The noninertial observer b sees 71112 T 0 r T F ctitiom g Motion with Resistive Forces NOTE this will not be on exam Motion can be through a medium Either a liquid or a gas The medium exerts a resistive force R on an object moving through the medium The magnitude of R depends on the medium The direction of R is opposite the direction of motion of the object relative to the medium R nearly always increases with increasing speed Air Resistance R Proportional To v2 For objects moving at high speeds through air the resistive force is approximately equal to the square of the speed R 12 DpAV2 D is a dimensionless empirical quantity that called the drag coefficient p is the density of air A is the crosssectional area of the object v is the speed of the object R Proportional To v2 example Analysis of an object R falling through air accounting for air vl resistance 2F 2 mg D0Av2 2 ma ag1ZAV2 mg Terminal Speed The terminal speed will occur when the acceleration goes to zero Solving the equation gives 2mg VT D pA l 7 mg Table 6 Terminal Speed for Various Objects Falling Through Air Some Terminal Speeds Object Mass kg CrossSectional Area m2 1111115 Sky liver 75 070 60 Basthall radius 37 cm 014quot 42 X 10 3 43 Goll lmll radius 2 cm 0046 l1 X 10 3 ll Hailsme radius 050 cm 48 X 10 1 79 X 10 5 14 Raindrop radius 020 on 23 X 10quotquot 1 X 10quot1 90 2004 prismBrows Cole ChapTer 1 ScienTific MeThod SignificanT Figures ScienTific NoTaTion UniT Conversions EsTimaTing ScienTific MeThod ObservaTions amp ExperimenTs Model 39 Theory musT make predicTions musT be TesTable TesTing TesTing TesTing v Simple is beTTer Occum s razor Thi nkPErShare Conversion 60 mileshour mihr is how many meTerssecond ms 1 mile 1609 m The speed of sound is 340 mS WhaT is The speed of sound in mihr If ou counT 2 seconds beTween ligthing an Thunder how far away is The ligthing sTrike 7 9L SignificanT Figures 0 Leading zeros donH counT n ZerosafTerdecunalpmnTcou T All numbers To The lefT of decimal counT Coun r Significan r Figures Propaga ring Significan r Figures How many sig figs do The following MulfiplicafionDivision have resul39l39 is limi39l39ed by number wi39l39h asf 000956 number of significan l39 figures 200 Addi39rionSub rrac rion 800 V l resul39l39 has same number of decimal 02155 places as leasf accura l39e number 39 ThinkPairShare Pair Acfion v Propagafing Significan r Figures Measure your or your par rners heigh39r 125 X 15 in uni rs of your hand Give heighf and 125 15 uncerfainfy 12515 f What is error 2 error uncer39l39ain39l39yheigh l39 x 100 Pair AcTion Measure your hand in inches ConverT your heighT and error o inches 7 Does The error range encompass your acTuaI heighT v i g 7 V Our adopTed uniTs lenchh meTer m isTance Traveled by lighT in a vacuum in 1299 792 458 second a mass kilogram k mass of a specific plaTinum iridium alloy cylinder kepT aT The InTernaTional Bureau of WeighTs and Measures Sevres France v Time seconds 5 39 9 192 631 770 Times The period of vibraTion of radiaTion from a cesium aTom Express as a number beTween 1 and 10 Times a power of 10 107 x 1033 999 x 106 Provides a convenienT way To express VERY large and VERY small numbers Wri re in Scien rific No ra rion Examples Try heSe 15000 15 x 104 1250 425425x 101 00046 00032 3 2 x 10393 12 15000 15 x 104 2 wg Problem Solving Techniques v Clearly lisf problem number Given Draw diagram Show your work r Check your UNITSIIIIIIIlllllllllllllllllllllll Prefixes you should know oUsing Table 14 on p7 wri l39e power of 10 39l39ha39l39 corresponds 0 The following prefi s micro 10 6 milli 10393 cen39l39i 10392 kilo 103 mega 106 giga Assume The function f depends on he measured quan39l39i39l39ies x y and z x y and 2 have errors associa39l39ed wi39l39h hem OX Oy 02 Th Theer Inf39 39ve by The Golden RaTio The Golden RaTio 1 618 is a raTio seen frequenle in naTure in The sTrucTure of many planTs Si animals Measure The golden raTio This is supposedly manifesT in several ways in The human bo y MosT famous is The raTio of your heighT To The eighT of your navel CalculaTe golden raTio for you and your parTner EsTimaTe your measuremenT uncerTainTy amp propagaTe your error Turn in your resulTs I Static Equi brium Equilibrium implies the object is at rest static or its center of mass moves with a constant velocity dynamic Static equilibrium is a common situation in engineering Principles involved are of particular interest to civil engineers architects and mechanical Elasticity eng39mers Chapter 12 Static Equilibrium and Equilibrium Summary 39 Static vs Dynamic Equilibrium There are two necessary conditions for equilibrium In this chapter We Will concentrate on static The resultant external force must equal zero 2 Tnis l5 a statement ortranslational equilibrium The acceleration oitne center of mass ofthe oblect zero when v eWed from an inertial frame ofrefererlc The resultant external torque about any axis must must be e 239 0 This is a statement of rotational equilibrium The angular acceleration must equal zero equilibrium The object will not be moving Dynamic equilibrium is also possible The object would be rotating with a constant angular velocity In either case the E1 0 EquilibriumiEquations We will restrict the applications to situations in which all the forces lie in the xy plane These are called coplanarforces since they lie in the same plane There are three resulting equations 2FX0 2Fy0 2110 I Ch 12 Question 3 Can an object be in equilibrium when only one force acts upon it Ch 12 Question 4 a Give an example in which the net force acting on an object is zero and yet the net torque is nonzero b Give an example in which the net torque acting on an object is zero and yet the net force is nonzero g39 Ch 12 Question 6 If you measure the net force and the net torque on a system to be zero a could the system still be rotating with respect to you b Could it be translating with respect to you Axis of Rotation for Torque Equa on The net torque is about an axis through any point in the xy plane The choice of an axis is arbitrary If an object is in translational equilibrium and the net torque is zero about one axis then the net torque must be zero about any other axis l Center of Gravity All the various gravitational forces acting on all the various mass elements are equivalent to a single gravitational force acting through a single point called the center of gravity CG Center of Gravity cont The torque clue to the gravitational force on an object of mass M is the force Mg acting at the center of gravity of the object If g is uniform over the object then the center of gravity of the object coincides with its center of mass If the object is homogeneous and symmetrical the center of gravity coincides with its geometric center Weighted Hand Example Model the forearm as a rigid bar The weight ofthe forearm is ignored There are no forces in the Xdirection Apply the first condition for equilibrium 2Fy O mg 500 N tl 300 In 39 330011 Weighted Hand Example cont Apply the second condition for equilibrium using the joint 0 as the axis of rotation 21 0 Generate the equilibrium conditions from the freebody 0i 7 i 39 Ladder Example The ladder is uniform r 80 the weight ofthe ladder acts through its geometric center its center of gravity There is static friction between the ladder and the ground diagram H mg i Solve for the unknown R I forces F and R m m 7 Ladder Example 2 Ladder Example Extended P Draw a freebody diagram for the ladder The frictional force is f m Let 0 be the axis of rotation Apply the equations for the two conditions of equilibrium Solve the equations m Add a person of mass M at a distance d from the base ofthe ladder The higher the person climbs the larger the angle at the base needs to be Eventually the ladder may slip Ch 12 Question 12 A ladder stands on the ground leaning against a wall Would you feel safer climbing up the ladder if you were told that the ground is frictionless but the wall is rough or that the wall is frictionless but the ground is rough Justify your answer g Ch 12 Problem3 For what value ofx will the beam be balanced at P such that the normal force at O is zero 5 47 I O 0 39 cc L as gt g Units of Force Table 51 System of Units Mass Acceleration Force 81 kg ms z N kgms z KS uslmnnrl slug I39lsz 1h Slllg l lSi quot IN 225 III 3 m4 Tmnwml mohi Coi g Gravitational Force The gravitational force F9 is the force that the earth exerts on an object This force is directed toward the center of the eanh Its magnitude is called the weight of the object Weight Fg mg Applications of Newton s Law Assumptions Objects can be modeled as particles Masses of strings or ropes are negligible When a rope attached to an object is pulling it the magnitude of that force T is the tension in the rope Interested only in the external forces acting on the object can neglect reaction forces Initially dealing with frictionless surfaces Objects in Equilibrium If the acceleration of an object is zero the object is said to be in equilibrium Mathematically the net force acting on the object is zero Eon ZFX 0a11d 0 a Equilibrium Example 1a Equilibrium Example 2a A lamp is suspended from a Example 54 I chain of negligible mass T Qonceptuallze the traffic The forces acting on the lamp I39ght are Categorize as an equilibrium problem No movement so acceleration is zero the force of gravity Fg the tension in the chain T Equilibrium gives ZFNOAT7Fg0 TFg g Equilibrium Example 2b Objects Experiencing a Net Force Analyze H Need two freebody 1 If an object experiences an acceleration there diagrams must be a nonzero net force acting on it Apply equilibrium equation to the lightand find T3 Apply equilibrium equations to the knot and find T1 and T2 Draw a freebody diagram Apply Newton s Second Law in component form Newton s second Law Ex 1 Forces acting on the crate A tension the magnitude of force T The gravitational force F The normal force n exerted by the oor Apply Newton s Second Law in component form ZR 2 T 2 ma 2F n Fg 0 gtn 2Fg Solve forthe unknowns If T is constant then a is constant and the kinematic equations can be used to more fully describe the motion of the crate 91 Note Aboutithe Normal Force The normal force is not always equal to the gravitational force of the object For example in this case ZamiFger andnFgF n may also be less than Fg 1 Inclined Planes Forces acting on the object The normal force n acts perpendicularto the plane The gravitational force Fg acts straight down Choose the coordinate system with X along the incline and y perpendicular to the incline Replace the force of gravity with its components Components of Gravity on an 4 Inclined Plane i Multiple Objects When two or more objects are connected or in contact Newton s laws may be applied to the system as a whole andor to each individual object Whichever you use to solve the problem the other approach can be used as a check Multiple Objects Example 1 First treat the system as a whole 2 I mma Apply Newton s Laws to i F 11 P the individual blocks ax Solve for unknowns Check P21 P12 Multiple Objects Example 2 Forces acting on the objects Tension same for both objects one string Gravitational force I Each object has the same 9 acceleration since they are I connected quot Draw the freebody diagrams T Apply Newton s Laws I Solve for the unknowns g Multiple Objects Example 3 i r Drawthe freebody diagram for each object One cord so tension is the same for both objects Connected so acceleration is the same for both objects Apply Newton s Laws Solve for the unknowns ProblemSolving Hints Atwood Machine in Detail Conceptualize the problem draw a diagram Find the tension T in the string Categorize the problem and the acceleration a Equilibrium 2F 0 or accelerating 2F m a I Analyze Draw freebody diagrams for each object Include only forces acting on the object I Establish coordinate system Be sure units are consistent Apply the appropriate equations in component form T Solve forthe unknown s Finalize Check your results for consistency with yourfree body diagram Does your answer make sense L Inclined Atwood Machine Forces of Friction Find the tension T in the string and the acceleration a on a microscopic scale surfaces are rough There will be a resistance to r the motion This is due to the interactions quotquot3 between the object and its environment This resistance is called the force of friction Friction can be modeled in a simple way Forces of Friction cont Force of Friction Friction is proportional to the 391th 3 normal force The diredion Of the 750 frictional force is m m An opposite the direction of k39 et39 f t39 40 motion and parallel to m n n g the surfaces in contact 1quotquot 2K static friction E 30 Static I in tic The coefficients of r 54 U E 0 mm mmquot friction are nearly quotm U The force of static friction is 51A I39llZwZEFEOF 50 6390 7 0 independent of the area i 3 generally greaterthan the force PP L39 A utFN f contaCt 11 of kinetic friction Fimgl i oniFisliding mm ii mm m mil w 7 iilmiiil iniiiiii IIiIiimih til illlllvi depends on the surfaces in quot quotW contact Ch 5 Problem 37 Friction Example 1 A car is traveling 500 mih on a horizontal Th bl k d d i e oc issi ing own highway a If the coeffICIent of static friction the plane so friction acts up n 4 between the road and tires on a rany clay is the plane r X 0100 what is the minimum distance in which This setup can be used to the car will stop b What is the stopping exper39m n ally detrm39ne k rigqua the coeffICIent of friction 139 q distance when the surface is dry and ii mgran is 0 600 S i tan 9 9 e For us use the angle where ng H the blockjust slips For pk use the angle where the block slides down at a F F constant speed a ThinkPair Share What is the normal force for a block resting on an incline g Friction Example 2 Drawthe freebody diagram including the 1 Mminn force of kinetic friction Opposes the motion Is parallel to the surfaces in contact Continue with the solution as with any Newton s Law problem quotlg Friction Example 3 Friction acts only on the object in contact with another surface Draw the freebody diagrams Apply Newton s Laws as in any other multiple object system problem ThinkPair Share What is the normal force on m Rotational Kinetic Energy An object rotating about some axis with an angular speed to has rotational kinetic energy even though it may not have any translational kinetic energy Each particle has a kinetic energy of K 2 mv2 Since the tangential velocity depends on the distance r from the axis of rotation we can substitute V w r RotationalKinetic Energy cont The total rotational kinetic energy of the rigid object is the sum of the energies of all its particles KR Z mllfwz 1 KR 3 71 2 2 Where I is called the moment of inertia Rotational Kinetic Energy final There is an analogy between the kinetic energies associated with linear motion K 12 mv2 and the kinetic energy associated with rotational motion KR 12 Iwz Rotational kinetic energy is not a new type of energy the form is different because it is applied to a rotating object The units of rotational kinetic energy are Joules J Moment of Inertia The definition of moment of inertia is I anmi 139 The dimensions of moment of inertia are ML2 and its SI units are kgm2 We can calculate the moment of inertia of an object more easily by assuming it is divided into many small volume elements each of mass Am Moment of Inertia cont We can rewrite the expression for l in terms of Am V th the small volume segment assumption I l prde lfp is constant the integral can be evaluated with known geometry othenNise its variation with position must be known Notes on Various Densities Volumetric Mass Density gt mass per unit volume p m V Face Mass Density gt mass per unit thickness of a sheet of uniform thickness t 0pt Linear Mass Density gt mass per unit length of a rod of uniform crosssectional area it m L pA Moment of Inertia of a Uniform S Thin Hoop Since this is a thin hoop all mass elements are the same distance from the center I Irzdin ledn1 IMR2 Moment of Inertia of a Uniform Rigid Rod The shaded area 39 i has a mass dm 2L dx 1 Then the moment of a k inertia is x J 7 2 7 DZ 2 Ii39frdmijiLzx L a x lt I i 7 lt gt Moments of Inertia g ParallelAxis Theorem In the previous examples the axis of rotation coincided with the axis of symmetry of the object For an arbitrary axis the parallelaxis theorem often simplifies calculations The theorem states I lCM MD 2 l is about any axis parallel to the axis through the center of mass of the object lCM is about the axis through the center of mass D is the distance from the center of mass axis to the arbitrary axis ParallelAxis Theorem Example The axis of rotation quot m goes through 0 mu The axis through the center of mass is shown The moment of inertia about the axis through 0 would be 0 CM MD 2 Moment of Inertia for a Rod g Rotating Around One End The moment of inertia of the rod about its center ET Iquot 1 is quot I illL2 7U A D is 12 L L Therefore LA I rm IID1 I im M lm 12 2 3 Rolling Object The red curve shows the path moved by a point on the rim of the object This path is called a cycloid The green line shows the path of the center of mass of the object Pure Rolling Motion In pure rolling motion an object rolls without slipping In such a case there is a simple relationship between its rotational and translational motions Rolling Object Center of Mass The velocity of the center of mass is A vbM R Ra dt dt The acceleration of the center of mass is vaM R la a 7 Rot CM dr dr g Rolling Motion Cont Rolling motion can be modeled as a combination of pure translational motion and pure rotational motion x H m1 wu 7M rrll Total Kinetic Energy of a Rolling Object The total kinetic energy of a rolling object is the sum of the translational energy of its center of mass and the rotational kinetic energy about its center of mass K 12ICM a 12 MVCM2 Total Kinetic Energy Example Accelerated rolling motion is possible only if friction is present between the sphere and the incline The friction produces the net torque required for rotation Total Kinetic Energy Example cont Despite the friction no loss of mechanical energy occurs because the contact point is at rest relative to the surface at any instant Let U 0 at the bottom of the plane 39 Kf UfKiUi K 12 ICM IR 2 VCM2 12 MVCM2 U Mgh U K 0


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