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# GENERAL PHYSICS PHYS 2001

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This 163 page Class Notes was uploaded by Elva Fahey on Tuesday October 13, 2015. The Class Notes belongs to PHYS 2001 at Louisiana State University taught by Staff in Fall. Since its upload, it has received 43 views. For similar materials see /class/223006/phys-2001-louisiana-state-university in Physics 2 at Louisiana State University.

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Date Created: 10/13/15

FREQUENCY UMBER OF WAVES PER SECOND quotH ER WAVELENGTH DISTANCE BETWEEN CORRESPONDING ADJACENT PARTS ON A WAVETRAIN PERIOD ME FOR A WAVE O PASS A POINT Infrared Visible Light RADIATIONquot Ultraviolet Xrays Gamma Rays Cosmic Rays quotELECTROMAGNETI D TPorlod tFrequoncy vSpeed I Wavelength leu direction gt wvum o lIn391 ANTINODE POINT OF MAX DISPLACEMENT POINT OF ZERO DISPLACEMENT 1 WAVELENGTH 2 COMPLETE WAVES 4CREST ANTINODE 4NODE Q TROUGH ANTINODE Important Considerations about Waves Waves transfer only energy through great distances Mechanical waves move through matter EM waves can travel through a vacuum Matter is disturbed but not transferred from one place to another The velocity of the wave is denoted by v hf Example Sound wave has a frequency of 262 Hz while traveling 330 ms How far apart are the wave crests 7 Vf330262126 m The velocity of a wave along a string or spring also depends on the properties of the material it travels in For a stretched spring or string FT m L velocity FT Tension on string in mass of string L Length of string ml Linear density EX Pb The Middle C string is under a tension of 944 N The period and wavelength are 382 X 10393 s and 126 m Find the linear density of the string f 1T Speed ofthe wave is v MT 126382 X 10393 330 ms mL Fv2 9443302 867 X 10393 kgm EX Pb A vibrator moves one end of a rope to generate a wave The tension is 58 N If the frequency of the wave is doubled to what tension should the rope have to maintain the same wavelength v 7tf 2 12 If tension is doubled then velocity is too mL v F Veloc1ty IS doubled and squared mL therefore tension is quadrupled F58N4232N EX Pb The linear density of the A string on a Violin is 78 X 10394 kgm A wave on the string has a frequency of 440 Hz and a wavelength of 65 cm What is the tension in the string F mLv2 78 X 10394 kgm286 ms2 638 n V A 39 f 65 m440 286 ms Sound Characteristics are mechanical longitudinal wave requiring a medium for transference Speed varies with characteristics of the medium in solid and liquid substances less compressible have higher elactic moduli and speed is greater Temperature helps determine speed in gases only Higher temp higher speed Speed of sound in air at sea level and room temp is 343 ms about 1000 st Pitch Sound Pitch is related to frequency and measures how high or low the sound is Freq range of average person is between 20 Hz and 20000 Hz Above this level is ultrasonic which has many uses Below this level is infrasonic though it can t be heard it does effect us physically Loudness A measure of the energy in a sound wave which is its amplitude Sound intensity is proportional to the power in the wave per area of space covered Units are wattsm2 The lowest sound intensity that can be heard by human ears is l X 103912 0 m2 and is called the threshold of hearing Another unit for sound Doppler Effect This occurs when the source of a sound wave is moving in relation to the hearer As the distance between decreases the pitch frequency rises Conversely when the distance increases the pitch falls Doppler frequency shift also occurs with ElVlR waves This becomes a valuable tool for measuring relative distance changes for objects on earth as well as in outerspace Applications are also used to determine the direction and speed of moving objects Applications of Sound A Sonar B Ultrasonic Cleaners Used to determine water depth and locate underwater objects reef submarines and school of fish Sonar unit consists of a transmitter and receiver mounted on the bottom of a ship Used by jewelers and scientists for cleaning delicate instruments and objects with hardto reach areas Consists of a metal tank filled with cleaning solution Attached to the bottom of the tank is a transmitter which produces highfrequency sound waves that couse dirt to be pulled from the tiniest spaces C Ultrasound in Medicine 1 Used to obtain a picture of the inner anatomyespecially in obstetrics to determine possible abnormalities Ultrasound can determine cancer hemorrhaging areas and can remove brain tumors once considered inoperable D Radar Devices 2 Doppler flow metermeasures the speed of blood flow using transmitting and receiving elements placed directly on the skin The transmitter emits a continuous sound which is reflected from the red blood cells and its frequency is changed in a kind of Doppler effect because the cells are moving The receiving element detects the re ected sound and an electronic counter measures its frequency To measure the speed of moving vehicles and for measuring the speed of air masses to determine wind shear the movement of thunder storms and the possibility that a tornado is forming 39aul 39suos y faM uuaram yldag 9399 EHDSH IVI Vertical position of the Slinky F35t Wavelength Vertical position of one point on the Slinky W Undisturbed position a At a Particular Time Period T g 439 5 fr 94 39 Distance rmqu i3 b At a Particular Location or Wwe g r 607 f gnu 5 39 v 25 39439 ayes Transverse Wave The velocity of the wave is denoted by v Kf Example A sound Wave has a frequency of 364 Hz while traveling 343 ms How far apart are the wave crests 9 vf V A wave has a frequency of 120 Hz and travels a distance of 90 min 3 seconds Determine if possible a itsquot speed b its period c its wavelength and 1 its amplitude Ultrasound has a speed of 1500 ms in tissue The smallest detail visible has a size about equal to one wavelength of the sound Qa Calculate the smallest detail visible with a 20 MHz ultrasound b To what depth can the sound probe effectively c How long does it take the echo to return to the probe from a depth of 010 m a The smallest detail visible has a size about equal to one wavelength of the sound The relationship between propagation speed frequency and wavelength is Vw f Solving for I gives 7 Vwf 1500 ms 20 x 106 Hz 075 x 10393m Thus details less than a millimeter should be discernible b The effective depth is 200 wavelengths Therefore 2007 200o75 x 10 393m 015 m d The time for an echo to return is the time for its round trip Since 1 vt We obtain t dv 020m1500 ms 13 x 16 Ultrasonic scanners determine distances to objects in a patient by measuring the times for echoes to return What is the difference in time for echoes from tissue layers in a patient that are 0020 and 0022 m beneath the surface respectively The speed of sound in tissue is 1500 ms dvt tdlv 00220020 2 x 1032echo comes back4 X 10 3 1500266 x 106 5 Simple Harmonic Motion SHM Continuous buck and forth repetitive motion as a spring or pendulum Stretchy materials like rubber bands and bungee cords are similar to springsthey are elastic When a force stretches or compresses a spring it is distorted Newton s 3 law requires the spring to push back with an equal force This is the restoring force of the spring which equals kx k is the spring constant which is a measure of the stiffness of the spring stiffer springs have a high k x is the amount of distortion in meters as measured from its relaxed or rest position The restoring force causes the spring to move back to its original position length However momentum carries it past its normal resting position and another restoring force is applied to return it to normal This back and forth motion is simple harmonic motion New terms for this kind of motion Cycle a complete back and forth movement one time Amplitude L where x is the maximum the farthest point away from the normal rest position in either direction unit meters Period T the amount of time s to complete 1 cycle unit seconds Frequency f the number of cycles completed per time unit Hertz Hz cycless period and frequency are reciprocals 1f T Restoring Force of a Spring ENVVVV F 1 quot u k 1 Sp Cont m m F always points where F is directly proportional to x Simple Harmonic Motion of Springs PENDULUM Period T is independent of Mass 85 Amplitude P maximum PEGmaximum Ezero if 521 KE zero m T quot PE minimum KE maximum Period T is dependent on Length T2 1V Lg Resonance Amplitude is a measure of the energy supplied to a system J determines the natural frequency of an object if left to itself If an external force causes The equation f Vibrations it may be a different frequency or the same frequency as the natural frequency of the obj ect Maximum amplitude of Vibration is achieved if the forced freq natural freq of object eX Pushing a person high in a swing by matching the swings to the push This effect is called resonance and results can be dramatic Bridges Tacoma Narrows buildings and walkways can collapse from the Vibration as in earthquakes 41 Am on a stant planet set up a simple pew ofmth 12 m With SIIM it makes 100 W vibrations in 280 s What is the acceleration of gravity on that planet 9 If the period of a simple pendulum is to be 20 5 what should be its length A wreel tg ball is hanging at the end of a long cable on a crane A student wants to estimate the length of the cable and therefore impfovises by using a simple pendulum made from a 0500 length of string and a stone The student observes that in swinging batik and forth over a small amplitude the wrecking ball makes one complete oscillation cycle in the time it takes the stone to complete ve cycles What is the length of the cable in preparation for shooting a hall in a pinball machine a tting 1 a 675 Nm is compressed by 00650 m relative to its umtrained length The hull m 00585 kg is at rest against the springnt point A When the spring is released the bail slides without rolling to point 13 which is 0300 m higher than point A How fast is the billllllttvingil lg apls I aa r7r a a rlj Mastic example 39 hmp abowstringbackforommheforereleasing tearmW39 kag l39hebowhasaspdngMof SNlm 1WhatistheelasticpoteWenergy Eliowf isthearrowtravellingasitleaves nbow allb dmhwmuchgravita omlpoten emgyv tgain I owhighw litrisein ntir Pterodactyls watch as an asteroid plunges thru Earth s atmosphere This is an artist s conception of a catastrophic event 65 Myr ago thought to have led to the extinction of the dinosaurs During the impact an asteroid 110 km across released its energy of motion as heat and light delivering the explosive equivalent of 100 million atomic bombs CHICXULUB SEISMIC EXPERIMENT 1996 BPPS MFEFHAL OfJLLEEUNVERSVTY OF 39l39EXASUNAMLPVGEOLDG CAL SURVEY CANADA 3 M5 DES 2 WITH 0 OFFSHORE GRAVITYMAGNEDKB 2 LAND UNES PASSNE RECORM Gulf of Mexico Chicxulub Chapter 6 Work and Enerqv 61 Work done by a constant force Energy What is it You all probably have an idea of what energy is It comes in many forms mechanical chemical heat light nuclear etc And energy comes in two types Kinetic Energy Energy of motion Potential Energy Energy due to position Energy is governed by certain laws which we will learn in this chapter And we will find that solving certain problems with an energy approach is often much easier than using Newton s Laws But both of these very different approaches are linked through a common concept and that is the concept of Work We all have our own ideas about what work means to us but in physics work is very explicitly defined It is defined in terms of Force and Displacement Work done on an object by a constant force W F d Note Do not confuse work with weight W Fd d lln words Work is equal to Force times distance I UnitS Forcegtltdisplacement E Nm EJ Joules Now that we know what work is we can more explicitly define Energy Energy is the ability to do work In the equation for work Fd is not necessarily the total force on the object but it is the net force or component of the force along the direction of motion ie either in the same direction as the displacement d or in the opposite direction of the displacement d Work is a scalar not a vector but it can be either positive or negative Work is positive if the force and displacement are in the same direction Work is negative if the force and displacement are in opposite directions Examples A large crate slides on a horizontal frictionless floor from left to right for a distance x A constant force F is applied to the crate in the different cases as shown Calculate the work done on the crate in each case gt gtF gt F gt H4 H4 H4 PH 1 W Fd d Fx 2W Fd d Fx x FoosQXx 3 W 2 F d Fx Negative since the force is in the opposite d direction of the displacement 4 W Fd d 2 0 Zero No work done since there is no component ofF along the displacement Example A 50kg crate is pulled across a frictionless horizontal warehouse floor by a constant force of 750 N The force is directed at 25 above the horizontal How much work is done on the crate by the pulling force if the crate is moved a distance of 152 m The crate slides in the xdirection so that is the direction of the displacement Therefore we need the total force in the xdirection to calculate the work W Fd d Fx d 117005250 d 750cos25015210330 N Example A 1200kg car is being driven up a 5 hill Africtional force directed opposite to the motion of the car has a magnitude of 524 N Aforce F is applied to the car by the road and propels the car fonvard The length of the road up the hill is 290 m What should be the magnitude of F so that the net work done by all the forces acting on the car is 150 kJ Draw the FBD for the car Choose coordinate axes Break the forces down into their x and y components The displacement of the car is along the x direction so we need the net force in that direction ZFX F f Wx Fd Work W WFddgtWF f Wxd gtFfWx W W 3 gfW51T15gjfmgsm5524120098sm50 2066 N 62 Work Enerqv Theorem How do we quantify how much energy an object has In other words how do we quantify its ability to do work on some other object If I have two identical cars one moving fast and the other slow certainly the car that is moving faster has a greater energy Thus the energy of a moving object must increase when we increase its velocity Also if I have a truck and a bicycle moving at the same speed surely the truck has much more energy than the bicycle Thus the energy of a moving object must increase when we increase its mass This energy of motion is called Kinetic Energy Any work we do on an object changes its Kinetic Energy Therefore Work and Kinetic Energy are really the same concept and This is the Work Energy Theorem and in words it says that the Work done on an object is equal to the change in its Kinetic Energy So how can we calculate the KE of an object Well from the Work Energy Theorem KEWFddmad maatz mat2mv2 KE 2 mv2 From here we can see that the KE increases linearly with mass and also increases with the square of the velocity Thus the Work Energy Theorem states the following W AKE KEf KEO 2mv mv This is work done by the total net the sum of the forces on the object acting along its displacement Now this is important If the work done on an object is positive W gt 0 then the object speeds up If the work done on an object is negative W lt 0 then the object slows down As an example consider a tennis ball that we throw straight up Example Starting from rest an extreme skier slides down a mountain that makes an angle of 25 wrt the horizontal The coefficient of kinetic friction between her skis and the snow is 020 She slides down the mountain for a distance of 104 m and then comes to the edge of a cliff Without slowing down she skis off the edge of the cliff and lands downhill at a point whose vertical distance is 350 m below the edge How fast is she going as she leaves the cliff How fast is she going when she lands Draw the FBD for the skier Choose coordinate axes Break the forces down into their components The initial velocity of the skier is 0 so KEo of the skier is also zero As she slides down the mountain the forces in the xdirection do work on her This changes her kinetic energy We can use the Work Energy theorem to calculate what her KE is at the top of the cliff 2 WAKEKEc KEO 2me ButWFddso Fd dmvc Fd is the net force in the direction of the displacement ie the xdirection 1 2 2 F d 77W 2 3WX fkdmvc 2d W 3 V lt X m m 2 2dW sin 25 ukFN V f The skier is not accelerating in the ydirection so I know from Newton s 2 d Law e uilibrium a q ZFyFN7Wy02FNWyWcos25 yfng mg 2 2dW sm 25 ukW cos 25 V c m 2dgsin 25 uk cos 25 210498sin 25 02 cos 25 492 3 v5 701 ms y To find the speed with which she hits the ground vf we use 2D kinematics Choose standard coordinate axes for this part V What do we knovxf 535m g Ty ya 35 m yf 0 v vex VCCOSZSO 635 ms v W may vcsin2 296 ms y 39g v v 4H v 4H We need to find vfy v2 vjy ZayAy V2 2 2962 2 980 35 7156 gtny 846 mS v 6352 8462 2112 v 106ms 63 Gravitational Potential Energy We ve discussed Kinetic Energy KE which is energy due to motion But an object can also have energy due solely to its position This is called Potential Energy PE In this chapter we will be focused on the PE you can get due to gravity First consider a tennis ball in a vertical freefall from some initial height ho to some final height hf 0 What would be the work done by gravity on the tennis ball ha in moving it from ho to hf quot WF d ltmagtlthf hogt ltmggtlthf hogt mglthO hfgt O h Work done by gravity WGraV mgh0 hf Notice that if the ball is moving downward then ho is greater than hf and W is positive which it needs to be But what if the ball doesn t fall straight down What if it follows a curved path V Let s flick the tennis ball off the ed e of the table with some speed v so that it follows the curved path shown hf Now what is the work done by gravity in moving the ball from ho to hf The only force that acts on the ball after it is pushed is again just gravity and gravity only acts in the vertical direction Thus to calculate the work I need the displacement that s along the direction of the force In other words the vertical displacement This again is just the difference in height ho hf WGraV mgh0 hf The work done is the same in either case Thus the work done by gravity on an object only depends on the initial and final heights and not the path taken Now let s go back to the vertical case and say the tennis ball is resting on the surface of the table I reach down and pick the tennis ball up and raise it to some height h lfl apply a force equal to the weight of the tennis ball then I can raise it at constant speed I do positive work on the ball W Fdd mgh Now just hold the tennis ball at rest at this height above the table The tennis ball is not moving therefore its KE 0 but does the ball still have energy Yes Where does the energy come from It comes from the work I put into the ball in raising it to the height h Thus the energy it now has is just equal to the work I put in This is called Gravitational Potential Energy PE mgh G Units Still force x distance gt J Joules h in the equation for PEG is relative to some arbitrary zero Huh Let s say I hold a tennis ball 2 m above the floor but a 1m high table is nearby O If I puty 0 at the ground then relative to the ground the ball s PEG mg2 m But if I sety 0 at the table top then the ball s PEG mg1 m It doesn t matter where I put the zero just as T 2 m long as I m consistent 1 m If y 0 is at the table top then the PE of the ball when it hits the ground is PEG mg1 m Example A 06kg basketball is dropped out of a window that is 61 m above the ground The ball is caught by a person whose hands are 15 m above the ground a How much work is done on the ball by its weight What is the gravitational potential energy of the ball when it is b released and c caught d How is the change in the ball s gravitational potential energy PEf PEO related to the work done by its weight a W mgho 42 06986115 b PEO mgho 069861 c PEf mghf O69815 6m d PEf PE09J36J I The change in the potential energy is negative 1 5 which means the ball lost potential energy m I Where did it go It went into the work done by its weight since PEf PEG 2 W 64 Conservative Forces NonConservative Forces and the WorkEnergy Theorem The work done on a mass by gravity is W mgh0 hf And as we showed it doesn t depend on the path taken only on the initial and final positions Thus the work done by gravity on the basketball for II either of the paths shown at the left is the same Gravity is an example of a Conservative Force 1 A force is conservative when the work it does on an object is independent of the path between the object s initial and final position 2 A force is conservative if the work it does in moving an object around a closed path is zero A closed path is one in which the initial and final positions of the object are the same As an example of a closed path let s consider a roller coaster car as it moves along the track The roller coaster track is an example of a closed path Assume that the track and the wheels are frictionless Let the roller coaster begin at the start position 2 work on it Everywhere the roller coaster goes downhill gravity does Everywhere the roller coaster goes uphill gravity does work on it And by the time you get back to where you started around the closed path the total work done bv gravity on the roller coaster is zero NonConservative Forces A force is nonconservative if the work done on an object by the force DOES depend on the path taken between the object s initial and final positions Friction is an example of a nonconservative force Air resistance on the roller coaster for example The friction does negative work on the roller coaster when it goes downhill and uphill In general both conservative and nonconservative forces can act on an object simultaneously So that the net work done by the external forces is Wext WC WNC By the Work Energy Theorem vgdzAKE gt WC WNC AKE Work done by the Work done by the non conservative forces conservative forces Now let s assume that the only conservative force we have is gravity Thus WC is just the work done by gravity but we already know how to calculate that WC 2 WGraV mgh0 hf mgho mghfz PEG PEf APE Thus WC 2 APE This allows me to write the work done by the nonconservative forces as chzAKEAPE This is the more general form for the Work Enerqv Theorem 65 Conservation of Mechanical Energy Energy can come in many forms and an object can posses two types Kinetic and Potential Thus the total energy of the object referred to as the Total Mechanical Energy can be written as Em KE PE gmvz mghl We can rearrange the Work Energy Theorem WNC AKE APE KEf KEG PEf PEG 2 KEf PEf KEG PE Ef E0 W E E In words this says that the work done by the non NC f 0 conservative forces is equal to the difference between the final total mechanical energy of the system and the initial total mechanical energy of the system Now here s the cool part If the work done by the nonconservative forces is zero ie WNC 0 then Ef 2 E0 Ef E So as long as the work done by the nonconservative forces is zero 0 then total mechanical energy of an object remains constant at all points along an object s path 1 2 1 2 We can write the above equation out as 3mvf mghf Emvo mgho This shows us that kinetic energy mv2 can be converted into potential energy mgh and vice versa but the sum is always a constant of the motion This is a statement of the Principle of Conservation of Energy Energy can not be created or destroyed It can only be converted from one form to another KE PE EKEPE vonms OJ 600mm 600 OOOJ quotages As the bobsled slides down the frictionless track notice how the 7 gravitational PE is converted into KE 200 OOOJ 400 OOOJ 600 OOOJ fi the top is aquot At the bottom it is all KE l 400 000 J 200 000 J 600 000 J I I But the sum of the PE and KE is the same everywhere 600 OOOJ OJ 600 OOOJ O Example Let s drop our physics book from rest at a height of 2 m above the floor With what speed does the book hit the floor We of course know how to do this problem 2 2 with kinematics 2 2 E0 2W0 mgho vyf vyo ZayAy But let s solve it by using conservation of energy 2 m The only force that acts on the book is its weight gravity and gravity is a conservative force Thus W 0 NC Ef 2mv mghf This tells me that the total mechanical energy of the system is conserved or Ef 2 E0 ln words the total energy of the book at the top when I drop it must be equal to the total energy of the book when it hits the floor 1 2 1 2 3M20mgh0 3mvfm g lf 2 As is often the case some of these terms are equal to zero Mgho vf So the initial PE of the book gets completely converted into KE right before it hits the floor and the mass drops out as we know it should gho v gt vf l2982 So where does this energy gol Example A block slides on a frictionless looptheloop track as shown in the figure a What is the minimum release height h required for the block to maintain contact with the track at all times and make it completely around the loop Give your answer in terms of the radius of the loop r b For this release height what is the speed of the block when it exits the loop So now I m just left with W Wow This problem is very hard if we try to use Newton s laws and mechanics lt s trivial if we use conservation of energy Draw the FBD on the block when it s at the top of the loop From here I can identify the centripetal force 2 P 2 mV r The minimum speed the block needs to stay in contact with the track occurs when FN O 2 3 g zvz I So this is the minimum speed the block needs to have to not fall off the track at the top of the loop L Ea 2 mgho Since the track is frictionless the work 4 done by the nonconservative forces is zero and Ef E0 The total mechanical energy of the block has to be the same everywhere along the track 1 2 1 2 Emvo mgh0 3mvf mghf Of the variables in the above equation which ones do we know v0 O hozh g zfpfgf2 fgrgth3r2r vf J b To find the speed at which it exits the loop at the bottom we hf 2 use conservation of energy again Etop Ebot top 0 mvt20p mgth 2 mv 0t mghbot gt I gG 7 31751 h OP r vbot J5 vbot h 0 Notice that neither a nor b depends on the mass of the block Example A 55kg skateboarder starts out with a speed of 180 ms He does 800 J of work on himself by pushing with his feet against the ground In addition friction does 265 J of work on him In both cases the forces doing the work are nonconservative The final speed of the skateboarder is 600 ms a Calculate the change in the gravitational potential energy APE PEf PEO b How much has the vertical height of the skateboarder changed C Is he above or below the starting point a WNC Ef E0 KEf PEf KEO PEO 2mv mv PEf PEO APEWNC mv vj 8026555l82 62 APE b APE mghf h0 1086 01 410 1086 z z mg 5598 0 Since hf h0 lt O ho gt hf so he is below where he started 67 Power We ve discussed what work is but what about the time it takes to do the work Let s say that both of us have 50 identical boxes to lift up on to a 1m high table We each have the same amount of work to do But what ifl lift my 50 boxes in 10 minutes and you lift yours in 20 minutes I ve done the same amount of work as you but I ve done it in half the time There must be a way of quantifying this Units There is It s called m W Work J Average Power F I POWCI Watts t Tune 3 TABLE 7 3 Typical Values of Power Approximate Source power W Hoover Dam 134 itquot Car moving at 40 mph 7 7 Hquot Home slow 12 739 illquot Sunlight killing on one square mctrr 1380 RC frige rn for n l 7 Telex isum IUD Person walking up stairs 150 HLuuun 17mm 20 1 horsepower hp 7457 W We can also write power in a couple of other ways From the WorkEnergy Theorem we know that W AE Thus I can write the Average Power as I3 w Also from the definition of work P Fd d Fd v gt t l Example A person is making homemade ice cream She exerts a force of magnitude 22 N on the free end of the crank handle and this end moves along a circular path of radius 028 m The force is always applied parallel to the motion of the handle If the handle is turned once every 13 5 what is the average power being expended 13 2 E Fd d F27zr 22217z028 t t t Example In 20 minutes a ski lift raises four skiers at constant speed to a height of 140 m The average mass of each skier is 65 kg What is the average power provided by the tension in the cable pulling the lift W Fd mtotgd 4msgd 7 t t t 46598l40 120 17 3000W3kW Example A 19kg block slides down a frictionless ramp as shown in the figure below The top of the ramp is 15 m above the ground and the bottom of the ramp is 025 m above the ground The block leaves the ramp moving horizontally and lands a horizontal diStance d away39 Find 0quot Erop Conservation of energy allows me to find the speed the block has when it leaves the ramp E E 15 m 1 2 top 1bot 2 I i Mia m ghtop 3 m vbot mghbot vbot 2ghtop hbot l d v 29815 025 495 ms From kinematics d vboz liS just the time the projectile is in the air in other words the time it takes the box to fall a height of 025 m 2A Ay ayt2gtt ygtt M20235 ay 98 dvm r 495gtlt023 Ch 9 Rotational Dvnamics 91 Torgue Forces cause linear accelerations F ma But what causes angular accelerations a In other words what is the rotational analog of force Units It s Torgue Force x distance A A A 139 rgtltF rFs1n6 N39m r is the distance from the axis of rotation to the point of contact of the force Axis of rotation F l 0is the angle between r and F Torque is positive for ccw rotations and negative for cw U rotations Example You are installing a new spark plug in your car and the manual specifies that it be tightened to a torque that has a magnitude of 32 N m Using the data in the drawing determine the magnitude F of the force that you must exert on the wrench Tszsin 317 rsinQ 32 028sin1300 Example What is the net torque produced by the forces F1 and F2 about the rotational axis shown in the drawing The forces are acting on a rigid rod and the axis of rotation is perpendicular to the page lnclude both magnitude and direction L 115m L 270m I r 7 Axis F1100N 9190o 92117quot F1 produces a negative cw torque 2391 F2 produces a positive ccw torque 72 Tim 2 72 71 2391 13E sin 91 From the figure r1 115 m and r2 270 m 12 r2F2 sin 62 61 90 62 117 rm 2 r2172 sin 92 1qu sin 91 27040sin1170 11510sin 90 Since the net torque is positive the rod rotates ccw 92 Rigid Bodies in Eguilibrium A rigid body is a fixed collection of masses ie it s not a point particle Examples of rigid bodies I l Every object is in principle a rigid body In general a rigid body could be moving linearly translational motion and it could also be spinning rotational motion Thus for a rigid body to be in dynamic eguilibrium it must not be accelerating translationally or rotationally F T t39 E 39l39b 39 x I rans a Iona qqu rIum 2F 0 Therefore the followmg must be true y Rotational Equilibrium The second part just says that the sum of the external torques must be equal to zero Thus for rigid objects in equilibrium there are no linear accelerations or angular 93 Center of Gravity The weight of a rigid body can cause a torque on an object without any other external forces acting on the object X39smm39m Take a hammer for example Let s drill a hole in the end of the handle and hang it from a peg This becomes the axis of rotation for the hammer If I release the hammer it will swing or rotate about this axis There must be a torque on the hammer causing this rotation Where does it come from It s due to the weight of the hammer But where does the force act It acts at the CM of the hammer Now you can see that there is a positive torque on the hammer due to its weight causing the hammer to rotate ccw T rF sm 6 rW Define Center of Gravity cg The Center of Gravity of a rigid body is the point at which the weight of the body can be considered to produce a torque due to its weight If gravity is the only external force acting on a rigid body then the rigid body will be in dynamic equilibrium if the axis of rotation is glaced at the center of gravity of the body Pivot fulcrum 94 Newton s 2quotd Law for Rotational Motion Torque is the rotational analog of force F 2 ma l 1 1 ms the rotatIonal analog of acceleratIon 777 Z a So what Is the rotatIonal analog of mass First let s consider a point mass that is undergoing circular motion and accelerating By Newton s 2nd Law there is a tangential force T FT 2 maT And there is a torque produced by this force T VFT Sil ltg VFT 2 239 rmaT But CIT ra So 239 rmra mr2a gt I is called the moment of Inertia It is the rotational analog of mass F 2 ma Translational Motion Newton s 2nd Law 1 l l T I 0 Rotational Motion I For a point mass I mrz A tipping point is reached when the center of gravity of an object is on the wrong side of a potential pivot point The weight at the cg produces a torque that is not balanced by another torque in the opposite direction See figure 912 in the text page 257 The moment of inertia I not only depends on the mass but also on how the mass is distributed relative to the axis of rotation 2 Unrts kgm What ifl have more than one point mass How do I calculate the moment of inertia Axis of rotation 2 I Z mirl i J gt quot 1 2 Imr2mr2 11 22 r1 J r2 739 So what about a rigid body which is just a continuous group of point masses To calculate Ifor the rigid body we have to add the contributions from each infinitesimal point mass That s complicated I have to integrate over the mass distribution 2 1 j pr dV For many objects with geometrical symmetry the moment of inertia has already been calculated Moments of Inertia for Some Common Geometricall S mmetrical Ob39ects Rod about Solid cylinder or Hoop about 30W cemer dlSC symmetry aXlS symmetry axis Sphere I lle I we2 2 1 I I UL E l I IR E I 1 MR 2 I I a 4 ME I MR3 I ML 12 2 Solid cylinder Hoop about Thin SDHBHCBJ Rod about central diameter diameter shell m is unique for a rigid body but depends on where the axis of rotation is located ln linear motion for a given force the acceleration is greater for a smaller mass For a given torque it is easier to rotate an object with a smaller I moment of inertia Strategies for solving equ brium and nonequ brium problems 1 Draw the FBD for each object in the system 2 Choose a convenient and appropriate set of x and yaxes for each object in the system 3 Choose an appropriate axis of rotation which often times will be obvious 4 Apply Newton s 2nd Law to the x and y components of the forces acting on the objects 2F o ZFy0 5 If the object is in translational equilibrium then 6 If the object is accelerating translationally then 2 max 2F ma y y 7 Calculate the torques on the object caused by each external force about the chosen axis of rotation 8 If the rigid object is in rotational equilibrium no angular acceleration 21m 0 9 If there is an angular acceleration then 2 10 10 Solve algebraically the equations obtained in steps 5 6 8 and 9 for unknown quan es Example A mass of 1 kg m is attached to the zeroend of a meter stick A second mass of 02 kg m is attached to the meter stick at the other end 100cm position If the mass of the meter stick is 01 kg where is the cg of this system located ie at what location could a fulcrum be located so that the system is in equilibrium y Drawthe FBD sz f r1 W quotW W2 cm The value we want to determine is r1 Since the meter stick is in dynamic equilibrium the following must be true 2F 0 Since we only have yforces only the bottom two equations matter 250 2FnewempWH FnemempWH 27H Let s keep breaking down the torque equation 4W sing 7r W sin m frsz 5111192 0 m m gt iWi rme rsz0 Dnmginmmmgirzmg FerWsz0 We don t know what the normal force is We could find it but it doesn t help us find the cg of the system For that we need to analyze the torque equation W1 Wm w W2 There are 3 unknowns in the torque equation so we need other relationships between the r values 0 5 5 057 From the figure the followmg must be true r rm rm r rlrz10 gtyz107yl Plug these into the torque equation rlml 7057y1mm 7 1074quotZ 0 r1m1705mm rlmm 710mz rimZ 0 gt r1ml mZ mm 05mm 10mz 05 10 05 01k 10 02k 4 mm quot392 srl g g0192m mlmzmm 1kg02kg01kg 95 Rotational Work and Energy Work Force x Displacement W 2 F d Rotational Work Rotational Force x Rotational Displacement mot T 9 This is the work done by a constant torque The units of rotational work are Joules J so 0must be in radians Translational Kinetic Energy KETMH mv Rotational Kinetic Energy KE 2 So for an object with both translational motion and rotational motion the total KE is given by 2 KB KER mvz 1a2 Tran In principle now we have another term which can contribute to the total mechanical energy of an object KEW KER PEGW 2 WV2 1af mgh As before if WNC 0 then Etot is a constant of the motion Rolling Motion Release a solid sphere from rest at the top of an incline and let it roll without slipping to the bottom The total mechanical energy at the top of the ramp must be equal to the total mechanical Et energy at the bottom of the ramp Etop Ebot Ebot Rottop PEtop 113Transb0t IltElRotb0t Ebot PEtop KETmsbot KERotbot 1 2 1 2 Emvbol 3101702 Notice for the sphere which is released from rest and rolls without slipping the initial gravitational PE goes into both the translational KE and the rotational KE at the bottom of the ramp Example We claim that a solid disk of radius r and mass m will roll faster down an incline than a ring of radius r and mass m since the moment of inertia of the disk is smaller than that of the ring Use conservation of energy to calculate the translational speed of the ring and disk at the bottom of the ramp and show that it is faster for the disk Since gravity is the only conservative force that does work on the objects by conservation of energy the total mechanical energy at the top of the ramp for each object must equal the total mechanical energy at the bottom of the ramp m E Ebot top ls E bot Igzgransmp ngottop PE top IltF Transljot 11 3R0t1jot PE bot PEtop 2 KB KERotbot Transljot 1 2 1 2 Emvbot Elwbot 2 Vb t 2 v mm 70 Plug this in above mgh mvb0t I bat SOlve this for Vbot r v 2mgh This is the speed that each object has at the bottom of the bot m L ramp Notice that it depends on the moment of inertia I 2 r Ring Disk 2 I mr2 I mr 2mgh 2mgh v v bOtrin 2 bOt is 2 g m quot42 d k m i 4 vbOtring vbOtdisk g The disk has a greater translational speed at the bottom of the ramp and thus arrives there first 96 Anqular Momentum To this point we have looked at all the following concepts in terms of both translational and angular rotational motion Translational Rotm Displacement X 6 Velocity V 0 Acceleration a 0 Force F 139 Mass m I Work F d 39 T 6 Energy 71112 IC2 Momentum mv Ia la is the rotational or Angular Momentum L Z 0 L Ia Units Moment of lnertiax Angular velocity kgm2 x rads kgm2s Remember when we studied linear momentum we found that when the sum of the external forces was equal to zero then the total momentum of the system was a constant of the motion e It was conserved 2 Fe 2 O gt Pf P0 xt For rotational motion we find a similar conservation law If the sum of the external torques is equal to zero then the total angular momentum of the system is a constant of the motion ie it is conserved 27820 gt Lf 2L0 This is the Principle of Conservation of Anqular Momentum l If the moment of inertia of a I system increases then its Lf Lo 3 Ifwf Iowa gt wf 0 wo angular velocity must f decrease to conserve angular momentum and vice versa Example An artificial satellite is placed into an elliptical orbit around the Earth The satellite s closest approach to the Earth called the perigee is rP 837 x 106 m from the center of the Earth lts greatest distance called the apogee is rA 251 x 107 m from the center of the Earth The speed of the satellite at its perigee is vP 8450 ms What is the speed of the satellite at its apogee Gravity is the only force that acts on the satellite and it always points in toward the center of the Earth The angle between r and FG is zero so the Apogee gravitational force exerts no torgue on the satellite vA Since the sum of the external torques is zero I know that the total angular momentum is a constant of the motion LApogee LPerigee AcoA P6013 cca hceol tl ri hi gfae llt etr t 3t lits2 8h llil Wsrcg lleo 3 tall i igxl v sen stall a wlieeaiiamgmggteat39iterg is san taangenuai speeds a v r 837 106 8450 mrjxt A Wm P 2 vA 2820 ms A P Fa gtlt Ch 10 Simple Harmonic Motion 101 The Spring Force Probably all of you have some familiarity with a spring If you push on a spring it pushes back If you pull on a spring it pulls back And the harder you push or pull on the spring the harder the spring pushes or pulls back If you re careful not to pull too hard then you will find the following to be true FApplied 0C x In words the force you apply to the spring is directly proportional to the spring s displacement We can make this proportionality an equality by multiplying the right side by a constant FA lied kx PP The proportionality constant k is called the spring constant Units ForceDisplacement Nm The spring constant is a measure of the stiffness of the spring The larger the spring constant k the stiffer the spring Forces which are proportional to the displacement like the spring force are called elastic forces F Elastic region lnelastig region lfl pull or push on a spring with force F then Newton s 3rd Law tells me that the spring pushes or pulls back with an equal and opposite force This force is called the Restoring Force For an ideal spring the restoring force is given by Unstralned length of the spring lfl pull on the ball with force F the spring pulls back on the ball with restoring force F n MinnMin q I F kx The minus sign in Hooke s Law indicates that the restoring force F always points in a direction opposite to the displacement of the spring When the displacement is proportional to the restoring force but in the opposite direction like Hooke s Law F kx the motion that results is called Simple Harmonic Motion SHM So what does SHM look like Let s consider a mass hanging vertically connected to a spring L L When I attach the mass to the unstretched spring the weight of the mass pulls it down The mass now hangs in a new equilibrium position Notice how the forces cancel If I now pull down on the mass with some force r quot F the spring pulls back with an equal but Wirrfg t39Lw opposite restoring force And ifl release the mass then it executes SHM about its equilibrium position The maximum distance the mass moves both above and below the equilibrium position is A called the Amplitude of the Motion So the mass is undergoing SHM Let s make a plot of the yposition of the mass as a function of time yl A A V The position of the mass is periodic or harmonic in time ie it repeats Notice it oscillates between A and A the amplitude and the motion repeats continuously in the absence of friction Let s define a few quantities that we will use to characterize the SHM l A A A T V Period The time it takes to complete 1 cycle T Units s Freguency of complete cycles per unit time f Units cycless Hz Hertz Notice one complete cycle is completed each period thus T21 Example What is the amplitude period and frequency of the following motion y 025 m 025 In Amplitude A 025 m Pe od We have 4 cycles in 12 s so the period is 12 s 4 cycles 1 f7 3s 033 Hz 102 Displacement Velocitv and Acceleration in SHM The Reference Circle Let a point Q move on a circle of radius A at constant angular speed a if Look at the projection of point Q on the xaxis Let s call it point P A As point Q moves around the circle at constant 6 speed what kind of motion does point P have y I x x x From the figure 0059 Z 2 gtx A0056 But 6 wt xt A cos wt This is the displacement of point P as a function of time A6 At If 6 27 then point Q has gone once around the circle Thus the time it takes At to go once around is just the period T 022 And since T1f 022 Remember a is the angular velocity speed in rads a A6 a 39 At Angular velocuty 79 Angular frequency Now let s look at the velocity of point P as a function of time y VT Point Q has some tangential speed vT Since P is the projection of Q onto the xaxis the velocity of point P is the xcomponent of the velocity of point Q From the figure we see that v s1n6 x gtvx st1n 9 V T We can write vT as VT 2 rd 2 Ad And 9 2 wt Notice the velocity of point P is a periodic function Thus vt Aasinat t In Ime V Ez 21Acoswt Aasinat At df df Acceltiation of point P as a function of time Since the velocity of point P is not constant it must be accelerating yl k The acceleration of point P is just the x component of the acceleration of point Q The acceleration of point is Q is the centripetal acceleration ac From the gure we see that a x cos6 x gtax accos6 a C 2 ac M A02 Thus ax 2 AQ2 cos6 at 2 AQ2 cos mt This is the acceleration of point P as a function of time a d At df df AaSinaf M2 coswt Displacement velocitv and gcceleration of point P as a function of time xt A cos wt vt Aa sin wt at Aa2 coswt F 2 ma 2 mAa2 cosmt ma2A cosmt mm t F ma2x kx Since F is proportional to x and in the opposite direction the motion of point P is SHM F b 2 k romaove k ma 3a m Smaller masses and stiffer springs larger k result in higher frequencies of vibration Concept Question A mass of is attached to one end of a spring and is free to slide on a frictionless horizontal surface The mass is pulled back from its equilibrium position and released What is the speed of the mass at I T2 1 Maximum 2 Zero 3 Need more info 39i m x0 xx Concept Question A mass of is attached to one end of a spring and is free to slide on a frictionless horizontal surface The mass is pulled back from its equilibrium position and released When is the speed of the mass a maximum T T2 3T4 T4 PWNT 103 Potential Enerqv of a Spring Elastic Potential Energy The elastic restoring force like gravity is a conservative force Therefore we can define a potential energy for the elastic restoring force Let a spring force move a mass m from x0 to xf lt lt l l xf x0 What is the work done by the spring in moving the mass from x0 to xf We know the work done by a constant force is W Fd d Fd xf x0 But the spring force is not constant since it depends on x FxoFxf f Instead we use the average force to compute the work F Result W kx12 This is analogous to the result we found for the work done by gravity W mgho mghf In the gravitational case the gravitational PE is For the spring force PEE 2ka PEG 2 mgh This is the Elastic Potential Energy The elastic PE like the gravitational PE is energy of position It depends on x which is relative to some arbitrary zero usually the unstretched length of the spring Now we have still another term that can contribute to the total mechanical energy EKE Trans KERot PEG PEE E l 2 2 1 2 1 2 mv 31a mgh3kx Example A mass of 15 kg is attached to one end of a spring k 325 Nm and is free to slide on a frictionless horizontal surface The mass is pulled back 075 m from its equilibrium position and released Find the speed of the mass as it passes back thru the equilibrium position We Will work this problem two ways First we will use the equations for SHM that we earlier derived and secondly we will use conservation of energy l x0 x075m a E m 1 vt 2 Am sin at We know that the time it takes the block to get to the x 0 position is 14T 1 tz 1 Ti2 l vt Aasina Aasin Aa a 2d 325Nm gtvt O75m 35ms 0 lt gt M E0 Conservation of mechanical enerqy fPEEf KE PEE KE Trans Trans 0 1 2 1 21 2 1 2 Emvf3f EV O3Cxo 1 21 2 Emvf kx0 k 2 2 gtVf X m 0 15kg 2 325 Nm 075 m2 Concept Question A slinky is held vertically by one end What happens to the bottom of the slinky when the top is released 1 It moves down 2 It moves up 3 It remains stationary 104 The Simple Pendulum A pendulum is simply a mass attached to a stringrope or rigid rod that once set into motion will swing back and forth under the influence of gravity Clearly the motion of the pendulum is I periodic ie it repeats in time But is the motion of the pendulum simple harmonic motion SHM Let s draw the FBD for the pendulum Assume it is swinging on the way up ccw Break the forces down into their x and y components Notice the force acting along the direction of motion is the xcomponent of the weight and it acts in the xdirection For SHM we need the force Wx to be Wx proportional to 6 but in the opposite 5mg 2 W Z Wx mg Sln 6 direction The force is not proportional to 0 but to sine Thus the pendulum motion is strictly not SHM But for small angles sinaz 6 so that Wx mg9 So in a smallangle approximation pendulum motion is SHM F mg6 for the pendulum in the smallangle approximation The displacement of the pendulum mass which is the distance it travels along its arc is given by S L9 mgs m F g S C0nstant gtlt D1splacement L L kx mg k Thus for the pendulum k Since a m we find the following result for the frequency of pendulum motion Wow The frequency of the pendulum motion only a Z i depends on the length and not the mass L This is the period of the pendulum 27 T 27239 Again this is only true for small Since T g angles Concegt Question A bowling ball and a pingpong ball are each tied to a string and hung from the ceiling The distance from the ceiling to the CM of each object is the same Which object would have a longer period of motion if they were set swinging Neglect air resistance and frictional effects 1 The bowling ball 2 The pingpong ball 3 They each have the same penod Example What does the length of a pendulum have to be so that its period on the surface of the earth is 1 second L T2 T2 12 T27r gtL4 7 z72025m 106 Resonance Resonance When a driving force acts on a system at its natural frequency of vibration large increases in the vibrational amplitude will result Ch 5 Uniform Circular Motion An object traveling at constant speed on a circular path is undergoing Uniform Circular Motion UCM Example Spinning a ball on a string ulu gt I 0 o o Since the speed is constant in UCM it s sometimes useful to talk about the period of the motionI which is the time it takes to complete 1 revolution The distance of one revolution is just the circumference of the circle 2m distance 2m 2m ThIs tIme Is the period T veIOCIty v Notice units are s V Example How long does it take a plane traveling at a constant speed of 110 ms to fly once around a circle with a radius of 2850 m Once around is just the period T 2 m W 163 s v m s Look at UCM from above again The speed is constant but is the velocity constant NO The velocity changes direction as the ball moves around in a circle Since the velocity is changing direction the ball must be accelerating What is the direction of the acceleration D It points inward toward the center of the motion This acceleration is called the Centripetal Acceleration ac How do we calculate the centripetal acceleration We calculate it the same way we would for any acceleration We need to find the change in the velocity of the ball over time Let s see how the velocity of the ball changes as it moves from one point to another along the circle What is the change in the velocity Av vf v0 vf v Notice that the direction of Av is in toward the Vf center of the circular motion Now look at the section of arc completed by the ball as it moves for a time 4 tr to But in the limit that At 0 s s So I end up with two similar triangles r s S r6 r r v v amp vt s39 r6 W v r 2 Rearrange this and we get AV V Tr C But this is just the acceleration NIX ac is the Centripetal Acceleration Since there is an acceleration Newton s 2nd Law tells us there must be a force What direction would this force point It would point in the same direction as the acceleration thus the force pulls in toward the center of the circular motion This is called the CentriQetal Force FC mac gt F Consider the ball spinning on a string again There is no force pulling the ball to the outside The only forces acting on the ball are its weight and the tension force If the tension is made large enough then the weight becomes negligible The sensation you have of being thrown to the outside when your body undergoes circular motion is a centrifugal force effect They are not real forces but fictitious When you move in a circle you are accelerating and thus you are not in an inertial reference frame Newton s laws are only valid in inertial reference frames Identifying the centripetal force in problems The centripetal force is not a new force It is the vector sum of the radial forces acting on an object undergoing circular motion Therefore the centripetal force should NEVER appear in a free body diagram Let s view an object from above that is undergoing circular motion Let s say several forces act on the object This is the complete FBD Since the object moves in a circle there must be a centripetal force The centripetal force is the vector sum of the radial forces The radial forces are 1 2 and 3 Forces 1 and 2 point in toward the center and force 3 points away from the center 2 Thus the centripetal force would be F F1 F2 F3 m C WHEN THE PELLET FIRED INTO THE SPIRAL TUBE EMERGES WHICH PATH WILL IT FOLLOW NEGLECT GRAVITY WHEN THE BALL AT THE END OF THE STRING SWINGS TO ITS LOWEST POINT HEAVY THE STRING IS CUT BY A SHARP RAZOR WHICH PATH WILL THE BALL THEN FOLLOW STthiG BALL dKlEDH I LSOW 38 G39l OM NMOHS SHin SH lO HDIHM 3903 SH 310 an ava HHL DNO39IV NOILISOd amsm NV wozu aans OL mmo 3H1 MOTIV ONV MEIIA dOL 3m NI NMOHS SV x310 3H1 NO avg v NEILSVj 00A BSOddnS 3903 3H1 1le aonsmm u 23wi 02503321 v axn gt510 9NLVLOZI v NO 0mm SI parao wwws v n Example A car moving with a constant speed enters an unbanked curve with a radius of curvature of 50 m If the coefficient of static friction between the car s tires and the road is 09 how fast can the car safely take the curve Draw the FBD that acts on the car FN Since the car is moving along a curved path circle there must be a centripetal force From the FBD the centripetal force is the vector sum of the radial forces The only radial force we have is the frictional force and it points in toward the center of the motion mv Thus the centripetalforce is FE f 7 r This sets the condition we want The maximum static frictional force equals the centripetal force If the centripetal force exceeds5 then the carwill slide off the curve v2 M M rluxg 500998441 Notice the result doesn t depend on the mass of the car Example A ball on the end of a string is twirled in a circle of radius 15 m at a constant speed of 27 ms What angle does the string make with the horizontal Draw the FBD Choose a coordinate system Break the vectors down into their components From the FBD we can identify the centripetal force as 2 mv F T I Since the ball is not accelerating in the ydirection 2 Fy 0 Ty W 0 T 174 r 9815 tan6 y gg 20 6640 Tx 17M r v2 272 gt 55 Satellites in circular orbits There are many different satellites and other objects currently orbiting the earth How did they get there Let s start shooting projectiles horizontally each time with increasing speed But if I shoot the projectiles fast enough so that they cover a large distance the ground beneath does not stay flat since the earth is curved Let s say the elephant s trunk is 1 m above the ground The last projectile he fired is moving very fast so it covers a lot of ground as it falls due to gravity By the time it falls a distance of 1 m the earth s surface has curved away by 1 m so the quot projectile is still 1 m above the surface The projectile always stays 1 m above the surface Thus the projectile is in a 1m high orbit When a satellite is launched a rocket takes it above the atmosphere and then we give it enough horizontal speed such that it falls toward the earth at the same rate as which the earth curves away beneath it Thus satellite motion is circular motion and we should be able to identify a centripetal force What forces act on the satellite when it s in orbit The only force we have is the gravitational force FG is also a radial force and the only one thus I FE 2 G mm my 7 Now I can solve for r the radius of the satellite s circular orbit Notice the radius of the orbit does not depend on the mass of the satellite but only on its speed Thus for each orbital speed v there is one and only one unique radius of orbit Note r is the distance from the center of the earth to the satellite Example How high above the surface of the earth do geosynchronous satellites orbit A geosynchronous satellite is one in which its orbital period matches the rotation rate of the earth such that the satellite is always over the same spot on the earth s surface m lts orbital radius is given by I G e 2 We need to find the satellite s speed v V 2 Its orbital period must be 24 hrs T 24 hrs 863400 3 E V 2717quot me V Plug this into our equation for the orbit I 6 2 2mm 2 24 2 Rearrange r3 G G 598x10 2864OO 755X1022 4n 4 r 423 x107 m This is the distance from the center of the earth To get the distance above the surface of the earth we have to subtract the earth s radius 423x107 638x106 359x107 m gt 22300 miles 56 Apparent Weightlessness and Artificial Gravity Let s say you are an astronaut hanging out in the International Space Station which is just a big satellite in orbit above the earth at an altitude of 340 km The space station is falling toward the earth due to the gravitational force and so are you It s analogous to being in an elevator when the cable is cut both you and the elevator accelerate downward at the same rate In both cases your apparent weight would be zero Have you ever felt the V x gravitational force No l 1quot 7 l d quot130quot l J x of i if When gravity is the only force Ll Orbit quot 339 that acts on you you are in free quot fall and therefore you 39 experience no forces at all Helm E The only force you have ever felt is the electrostatic force a Freefa II b So astronauts living in space appear to be weightless mama Of course there is a force acting on them F Gr Being weightless for extended periods of time can have significant physiological effects on the human body not all of them positive We evolved under the influence of gravity and thus thrive in that environment The ultimate fate of the human race will be determined by our ability to live and work in space It would be nice to have a way to make artificial gravity in space so that we feel comfortable living there Remember Einstein s Principle of Equivalence A uniformly accelerated reference frame is exactly equivalent to a gravitational field in the opposite direction This is great Because look what we can do if we have a circularlyshaped space station We can spin it Let s look at the FBD for the guy in the station There s only 1 force Since he is moving in a circle we can identify a centripetal force From the FBD it is clear that FNFC The normal force FN is the force the station wall ground pushes against his feet Thus the FN provides the sensation of weight So if I want it to feel like I m standing on earth when I m in the space station then I want FN mg where g 98 msz 2 This is the speed I need to spin the mg mv gt outer edge of the station in order to V I produce earthlike weight 57 Motion on a Vertical Circle Let s consider an object moving on the inside of a vertical circle Assume the circle sits on the earth s surface For example a motorcycle performing a looptheloop stunt 3 What are the forces on the motorcycle as it travels around the circle There are 4 places along the circle where the 4 F N3 WF 2 forces are easily determined J 4 2 a 39 Let s draw the FBD at each of the four W M I W positions a Since he s moving in a circle we should be able to identify the centripetal force from the 1 W FBD The centripetal force at each point depends on the vector sum of the radial forces 2 2 mvl mv3 1FCFN1 W 3FCFN3W 1quot 1quot mvz mvz 2 4 239FCFN2 4 FFN4 Note The centripetal force is not constant since his speed changes as he moves around the circle and the contribution to the centripetal force by his weight also depends on his position 3 Consider position 3 for a moment i There is a minimum speed I must have at N3 W position 3 in order to make it around the 4 4 F 4 F 2 2 loop 3 Another way to say this is that I need to M A stay in contact with the track at position W W 3 or in other words FN3 can t be zero 1 That is the condition that sets the W minimum speed right when FN3 0 0 2 2 2 3FCI3WMV3 gtW g v3 gtv31lgr r r r I must be going at least this fast at the top of the track in order to stay in contact with it and complete the loop Anything less than this and I will fall down toward the center of the loop D Example NASA s 09 formerly KC135 NASA ies a 4engine turbojet plane along a parabolic trajectory At the top of the trajectory the arc is approximately circular Passengers in the plane experience zerogquot conditions weightlessness for a period of 2030 seconds If the plane is ying with a speed of215 ms what is the maximum radius of curvature ofthe ight path that the pilot must follow so that the passengers experience zerog Draw the FBD that acts on a passenger in his seat 2 FEW r Weightless means Ngoes to zero I m no longer in contact with my eat Ch 4 Forces and Newton s Laws If have seen far it is because have stood on the shoulders of giants Isaac Newton In the previous chapters we used the concepts of displacement velocity and acceleration to study the motion of objects Now we want to know how the object moves and why it moves or just as importantly why an object doesn t move To do this we need to develop the concept of a Force Simply put a Force in physics is a push or a pull on an object We will discover that there are several types of forces that can act on an object but they all fall into two broad categories Contact forces and Non contact forces But there is one commonality among all forces They each have the ability to change the motion of an object When an object falls and hits the ground it remains at rest Aristotle 350 BC It wasn t until the 1600 s that these ideas were expanded upon Objects continue with uniform motion if no external forces act on them Galileo 1609 1687 Newton publishes his Principia or The Mathematical Principles of Natural Philosophy Newton was born on Christmas in 1642 the same year Galileo died He was born premature and not expected to survive He enrolled in Trinity College at Cambridge when he was 18 There he studied math and optics After graduation he lived in seclusion in the countryside to avoid the second Plague in Europe Over the next year and a half he did the following Developed the binomial theorem of mathematics Began his study of mechanics Developed the Universal Law of Gravitation Analyzed the decomposition of light into its spectrum Developed both forms of calculus integral and differential When he finished he was 25 In this course we will be interested in his studies of mechanics and gravitation Newton expanded on Galileo s ideas about the motion of an object in the absences of forces An object at rest tends to stay at rest and an object in motion tends to stay in motion at a constant speed moving in a straight line unless acted upon by some external force Newton s First Law The property of objects that make them obey Newton s First Law is something called Inertia Newton s First Law is often referred to as the Law of Inertia lnertia is a measure of how much mass an object has In other words how much matter it contains For example a lead brick has a lot more mass and thus inertia than does a penny The quantitative measure of inertia is Mass ln SI units mass comes in kilograms kg Key point If the net force on an object is zero then the object will remain at rest or continue on at constant velocity Thus if there is a net force acting on an object then the object s velocity WILL change Av Thus the object must be accelerating a T So forces are synonymous with accelerations In other words a force is a push or a pull on an object that can overcome the object s inertia and produce an acceleration Now for a given force what determines how much acceleration I m going to get It s the object s INERTIA The more inertia and thus mass an object has the more it resists a change in its motion 2 F mZz Newton s 2quot Law This is Newton s 2nd Law In words it says The sum of the forces acting on an object is equal to its mass times its acceleration Notice that forces are vectors They require both magnitude and direction to completely describe them Unit5 MassIAcceleration kg22 kg N Memoquot 5 S It takes about 1 N of force to push a stamp on an envelope Frames of reference Newton s First Law and 2nd Law may be invalid depending on an observer s frame of reference Your frame of reference or reference frame is your perspective from which you observe a physical system In physics we attach a set of coordinate axes to our frame of reference so that we can measure things like position velocity and acceleration For example let s say you are standing at rest watching a passenger train go by you at constant speed I m sitting in my seat on the train with my lap tray down as I enjoy my favorite cold beverage An ice cube from my drink is sitting on the tray in front of me As the train goes by you we both agree that the ice cube is obeying Newton s First Law You see the ice cube continue to move in a straight line at constant speed Relative to me see the ice cube remain at rest But now the train slams on its brakes What do we see The train comes to a halt but the ice cube continues to move at the same speed before the train s deceleration now sliding forward on my tray You standing next to the train would say that the ice cube continues to obey Newton s First Law it continues on in straight line motion at constant speed I however would say that some force has been applied to the ice cube causing it to accelerate and move forward on my tray When actually there is no net force on the ice cube So who is right How can the ice cube obey Newton s First Law in one reference frame but not the other It turns out that Newton s Laws are only valid in Inertial Reference Frames An Inertial Reference Frame is one in which the acceleration of the frame is zero You are standing on the Earth which is a good approximation of an inertial reference frame so when the trains stops you see the ice cube keep moving at constant speed thereby obeying Newton s First Law I am attached to the reference frame of the train which has a nonzero acceleration when the brakes are applied From my point of view it looks like a force has been applied to the ice cube when actually there is no force I am in a noninertial reference frame In practice we often refer to Newton s 2nd Law as just F ma We do this however with the following understanding 1 That this is a vector equation 2 That F means the net the total the vector sum of all the forces that act on the object We can rewrite Newton s 2nd Law this way a m Now it is easy to see that the more force I apply to an object the greater its acceleration But the more massive an object is the more it resists acceleration Example A 71 Okg crate is pulled along a horizontal frictionless surface The puller applies a force of 300 N at an angle of 40 wrt the horizontal What is the acceleration of the crate Draw a figure m710kg Since there is no friction the crate will slide in the positive x direction when the force is applied So to calculate the acceleration in the x direction ax we need the component of the force in the x direction Let s break F down into its x and ycomponents F F cos 40quot 300 Ncos 40quot 230 N Fx 230N m 39 710kg ax Example Two forces F1 and F2 act on the 700kg block shown in the drawing The magnitudes of the forces are F1 450 N and F2 250 N What is the horizontal 39 quot quot quot and quot39 quot of the block Again assume the block is sitting on a horizontal frictionless surface We have the figure let s choose axes y directions F1 The acceleration in the horizontal direction F15 is ax To calculate this we need the net force in the xdirection 7900 77 Then we can use Newton s 2nd Law 1x F 700 k I 2 g g ax 2P m x Break F1 down into its A L V L F1 Ecos70quot 154N Fsin70quot 423N F2 already lIes along an aXIs F2 F 250N ON 2 F 154N250N 1 37m 2 The horizontal acceleration is x T 39 39 S 2 m g 137 ms In the negative x direction Newton s 3erI Law Law of Action and Reaction When an object exerts a force on a second object the second object exerts an equal but opposite force on the first Example Two ice skaters are standing facetoface on frictionless ice The skater on the left has a mass of 500 kg and pushes the skater on the right with a force of 400 N The skater on the right has a mass of 700 kg What is the acceleration of both skaters m5o0kg m700kg F40N F40N Buy Newton s 3rOI Law the skater on the right pushes back against the skater on the left with an equal but opposite force g aright m 70 0k right 39 g Eeft 400 N aleft 500 kg left 057 ms2 080 ms2 Skater on the left has a larger acceleration since her mass is smaller 46 Fundamental Forces Currently there are 3 fundamental forces 1 Gravitational Force 2 Strong Nuclear Force 3 Electroweak Force With the exception of the gravitational force all the forces we will study this semester are related to the third fundamental force the electroweak force When I was in middle school there were considered to be 4 fundamental forces gravity electrical strong nuclear and weak nuclear Many theoretical scientists have spent much of their careers trying to reduce the of fundamental forces down to one Now let s study the gravitational force 47 The Gravitational Force lt s May 16 1666 and Sir Isaac Newton is outside contemplating how the moon revolves around the Earth While thinking he notices an apple fall from a nearby tree and strike the ground At that moment he realized that the same pull that made the apple fall to the ground must also apply to the moon The moon moves in a nearly perfect circle around the earth Since it s moving in a circle its velocity is changing Since its velocity is changing there must be an acceleration Since there is an acceleration there must be a force The same force that pulls the apple toward the earth must also pull the moon toward the earth This is the If gravity didn t pull on the moon it would continue to move in a straight line The gravitational force from the earth pulls the moon away from its straightline motion Newton s Universal Law of Gravitation So How do you calculate the gravitational force between two objects Let s consider two objects with masses m1 and m2 and separate them by some distance r 1E FrF V x F G mlmz Then r2 Notice that the force is directly proportional to the product of the masses of the two objects and inversely proportional to the square of the distance between them r is the distance between the centers of each mass The gravitational force equation is called an inverse square law Nature loves inverse square laws and we will encounter them again G is the universal gravitational constant Properties of Newton s Law of Gravitation 1 It grows weaker with distance It gets stronger for increasing masses 2 3 It is always attractive 4 2 G 667259x1039 N m kg Note The forces on the two masses are equal but opposite as dictated by Newton s 3lrd Law It s directed along the line containing the two masses D What is the acceleration of a tennis ball falling near the surface of the earth It s 9398 mSz but Mass of the ball Mass of the earth m m B E We can calculate the force on the ball F G Z Radius of the earth But we can also write the force on the ball using Newton s 2nd Law F mBa m These two forces have to be equal a G B E 1quot E m E a G 2 This must be the acceleration of the ball Let s calculate it Iquot E 11 Nm2598x1024 kg a667x1039 6 2 kg 638x10 m 98 ms2 g COOL Thus the acceleration due to gravity depends on the mass of the earth and the distance I am from the earth s center If I m standing on the earth s surface then this distance is just the radius of the earth Now we can calculate the acceleration of gravity on any planet or body m 735x1022 kg Let s take the moon for example amquot G G 16 ms2 r300 174x106 m2 A amoon 6 aearth Notice also that the acceleration decreases with the square of the distance On the surface of the earth the acceleration due to gravity is 98 msz but if I move farther away from the surface the acceleration decreases So the acceleration due to gravity on top of Mount Everest is less than it is on the surface at sea level We often refer to gravity as a force but actually it is an acceleration The force due to gravity is called weight You weigh less on the moon since the acceleration due to the moon s gravity is less than that of the earth You weigh less on the top of Mount Everest too Near the surface of the earth where g 98 msz an apple falls 49 m in 1 second However where the moon is the acceleration due to earth s gravity is only about 00027 msz so the moon only falls about 1 mm toward the earth each second Mass and Weight Remember your weight W mg is the force on you due to gravity It can change depending on the value of g ie whether you re on earth or the moon etc Your mass m is a measure of how much matter your body contains how many atoms you are made of Your mass never changes It s the same on the moon on the earth on mars etc Example The drawing below shows one possible alignment forthe sun earth and moon The gravitational force FSM that the sun exerts on the moon is perpendicularto the gravitational force FEM that the earth exerts on the moon Determine the magnitude and direction ofthe net gravitational force on the moon l rSM 150 x1011m Moon FSM mM 735 x 1022 kg Sun 39 7 FEle mS199gtlt103 kg rEM385gtlt108m m 598 gtlt1024 kg Earth rSM 150 x1011 m Moon FSM mM 735 x 1022 kg Sun 8 FNET FEM mS199gtlt103 kg rEM385gtlt108m mg 598 x 1024 kg Earth First let s draw in the resultant vector FNet To calculate FNet we need to find FSM and FEM and then use Pythagorean Thm FSM 0 434x1020 N r SM Fm FSZM FEzM 477x1020 N mEmM 20 FEM G rEZM 13998 d0 N FNETis477gtlt1020N directed at 2450 below F a the horizontal line DIrectIon tan0 0 tan 1 between the sun and 5 SM moon Before we study more forces let s discuss Free Body Diagrams FBD A FBD is a drawing or sketch that shows the fundamental forces that act on an object The object or body is drawn free from the rest of its surroundings Often times the body is represented as a single point located at its center of mass In dynamical problems with multiple forces acting on a body the FBD is extremely important So we should really amend our list of problem solving steps as follows 1 Relax breathe It s just a word problem 2 Draw a figure for the problem 3 Draw a FBD for the problem and label the forces 0 O 48 The Normal Force So we ve been saying that it s important to understand how and why an object moves under the influence of forces Just as important is why an object doesn t move Let s take the example of a concrete block sitting at rest on the top of a table Since the block is not moving and thus not accelerating there must be no forces acting on if the block right Wrong 39 There are forces acting on the block Newton s 2quotquot 39 Law tells us that if the object is not accelerating then the sum of the forces 2F must be zero Therefore whatever forces there are acting on the block must sum to zero ie they cancel each other out So what forces do we have We must have the weight of the block There must be another force one to cancel the weight What is it It s the table pushing back up on the block This is the Normal Force FA The Normal Force cancels the weight so the net force on the block is zero Normal Force The component of a force that a surface exerts on an object which is perpendicular to the surface Whenever there is an object in contact with a surface that surface will apply a normal force to the object Let s look at the block on the table some more What would the FBD look like FN F N quot 3 Define y upward We only have forces in the ydirection Newton s 2nd Law says 2F may 0 since there is no acceleration 2 Fy FN W 0 3 FN W Does the normal force always equal the weight No The normal force equals the weight if the object is sitting on a horizontal surface and no other vertical forces act except gravity What if the surface is not horizontal Example A60kg box is placed on a frictionless ramp that is inclined at an angle of 30 wrt the horizontal a What is the normal force acting on the box b What force must be applied to the box along the inclined plane to hold the box in place 4 39 What forces do we have acting on the box Weight which is always downward And the normal force which is always perpendicular to the surface of contact We can orient our axes however we wish but a real convenient way is to set x up the ramp and then y will be perpendicular to the ramp and ycomponents be broken down x Now break the vectors down in the FBD into their x FN is done since it lies along the yaxis but Wcan xcomponent ycomponent FN O FN Wsin3 0 Wcos3 0 Now we can apply Newton s 2nd Law for each direction xcomponent ycomponent FN 0 FN W Wsin3 0 Wcos3 0 Since there is no acceleration in the ydirection the sum of the forces in that direction will be zero EFy FN Wcos30quot 0 gtFN Wcos30quot mg cos30quot 60 kg98cos 30 m b In order to keep the box in place I must apply a force F that is equal but opposite to the component of the weight directed down the ramp in this case W F Wx Wsin300 mg sin 30quot 60 kg98 sin 30quot 294 N Physical insight from previous example In the previous example we found out that when you have an object on a frictionless inclined plane the normal force on the object is equal to the y component of the weight and the force required to keep the box from sliding down the ramp is equal to the xcomponent of the weight F W COS 0 Let s see if our answers make sense in the limiting N cases where 6 0 and 0 90 F W sin 6 0 90 0 0 FNWcos0Wcos0W FNWcos6Wcos9O O FWsin6Wsin00 FWsin6Wsin90 W The normal force goes to zero It s no The normal force equals the we39QhF longer in contact with the ramp The object 5 herlzehfa The Obleet WOh t Shde is in freefall so the force required to down a herlzehtal ramp maintain its position is its entire weight Apparent Weight We have all had the experience of feeling heavier or lighter when traveling in an elevator If the elevator is at rest and then accelerates upward we feel heavier If the elevator is at rest and then accelerates downward we feel lighter The motion of the elevator can give rise to an apparent weight which is different from our actual weight Why The sensation of weight that we feel comes from the force that the floor of the elevator applies to our feet That is it s the normal force from the elevator floor a i i 1 fti are i 700 V quotE 14 321 40013 m x r 0 w 7 y it a h f 0 i i o v i i O z i 39 i i i f i vii milH Hu iI c Ji Ki MW 700 N 1 n my 1 WW 700 N i h J MW 2 700 N 700 N 1 i 039 La U f I Qquot 3 39 i J i D a i 1 V i a No acceleration b Upward acceleration c Downward acceleration d Freefall How do we calculate apparent weight We use Newton s 2nd Law Let s look at the FBD for the guy in the elevator Iy We have his weight W downward and the normal force FN from the floor upward Newton s 2nd Law tells us that 2F FN W may Here ay is the acceleration of the man and elevator FN is the apparent weight that we feel It s the normal force against our feet FN W apparent W mg Wapparent W may mg may gt Wapparent ay I Notice if ay 0 Le the elevator is not accelerating then WWWquot mg your true weight Remember g is always a positive number but ay could be positive or negative If I cut the rope holding the elevator then it accelerates downward at ay g 39 l and your apparent weIght goes to zero Youre in freefaquot Einstein s Theory of General Relativity The Principle of Equivalence I measure the same acceleration on the apple Opposite the direction of the elevator s acceleration If I drop an apple on the surface of the earth I measure its acceleration to be equal to g 98 msz w 9 r Q i E E t l l 39 Repeat the same experiment in an elevator that is moving upward with an acceleration ga8mbK You can t tell the difference between the acceleration of gravity and the acceleration in some other reference frame like an elevator rame The Principle of Equivalence A uniformly accelerated reference f gtn is exactly equivalent to a gravitational field in the opposite directi 49 Frictional Forces We ve been discussing forces on objects who have surfaces in contact The Normal Force is perpendicular to the surface There is another force due to surfaces in contact in which the force is parallel to the surface This is call the Frictional Force We will often just refer to this force as Friction Static Friction Frlctlon comes In two types Kinetic Friction The Frictional Force tends to oppose the direction of motion ie it tries to slow the object down But not always Consider the situation of a block sitting on a table If I want to slide the block across the table then I have to overcome a frictional force just to get it to move static friction And then I have to overcome another frictional force to keep it moving That is the kinetic friction or friction of motion Let s say I want to try and pull the block to the right As start to pull with some force F the block doesn t move because of the static frictional force fs Notice that fs is parallel to the surface of contact and it opposes the direction that I m trying to pull the block So I keep making my pulling force F bigger and bigger fs keeps getting bigger and bigger too canceling out my pulling force untilfs reaches its maximum value Once I go past this point so that F is now greater than fs the block breaks free and starts to slide Now that the block is sliding no longer have to overcome the static frictional force fs but I have to overcome the kinetic frictional force fk to keep the block moving Static Frictional Force Force you must overcome to get an object at rest to move maximum static frictional force Kinetic Frictional Force Force you must overcome to keep an object moving fk MkF N us and Mk are the coefficients of static and kinetic friction respectively us and Mk are material dependent ie it depends on what the two surfaces are that are in contact l3 For exam 396 Rubber on concrete Mk 080 Rubber on wet concrete Mk 025 It requires more force to get an object at rest to move then it does to keep it moving Now let s revisit the block on the inclined plane again but this time let s include friction Example A60kg block is at rest on an inclined plane that makes an angle of 30 wrt the horizontal What is the coefficient of static friction between the block and the plane Let s draw the FBD The block is not accelerating so I know the following must be true by Newton s 2nd Law W 2mmaxo 2Fx W0 y 2Fymay0 2FyFN Wy0 gt SF WsinH FN Combine the two equations us cos 9 Wsin6 gt us tan6 gt us 058 410 The Tension Force Forces are often applied to objects by ropes or cables Pulling a box for example with a rope This gives rise to a forceT called tension We will assume that the ropes are massless unless otherwise stated That way the force we apply to the rope is equal to the force transmitted to the object In other words no force is needed to accelerate the rope Massless ropes transfer the pulling force to the object Example A block with a weight of 422 N sits on a horizontal table connected to a hanging mass M The blocks are connected by a massless rope over a frictionless and massless pulley lfthe coefficient of static friction between the block and the table is 075 what is the maximum mass for M2 such that the 422N block remains stationary Set up a coordinate system M1 39422 N 5 Now we need to draw the FBD 39 5 Whenever you have a mechanical system involving l multiple masses objects you should draw a FBD y for each ob ect Let s call the mass on the table M1 x Notice the tension on each mass is the same since they are connected by the same rope and our pulleys are Mass on the table Hanqlnq mass ideal T Now we can apply Newton s 2nd law to each mass and in each direction 1 E W2 Mass on the table X Hanging mass T W2 The sum of the forces in the ydirection for mass 1 is zero since the mass does not accelerate in that direction 2FxT fs0gtTfs IT 5FN1 SWl The sum of the forces in the xdirection for mass 1 is also zero since we don t want the mass to move so T fs sets that condition The sum of the forces in the ydirection for mass 2 is zero since the mass does not accelerate in that direction From the two boxed equations on the right they both equal the tension so I can set them equal to each other m 075422 WM gtM 323k Ms 1 28 2 98 g DOES THE SCALE READ 1OON ZOON ZERO 2 we 96mm gt2 gt256 uOnnm O 60 z gtnnmrmn2mm 41m 692 90 11 095 CW ma QWQZT V I 411 Equilibrium Conditions An object is in eguilibrium when its acceleration is zero Thismeansthat 2Fmc70 In 2 dimensions then 2 Fx 0 E y The forces acting on an object must balance in each direction Remember zero acceleration does not necessarily mean no motion As long as the velocity does not change in time which means that a system moving at constant speed could be in equilibrium 412 Nonequilibrium Conditions If a system is in nonequilibrium then it must be accelerating By Newton s 2nd Law then 2 F ma ln 2 dimensions then 2 Fx ma All the problems we will do can be classified as either an equilibrium or a nonequilibrium problem If it s an equilibrium problem then the sum of the forces in both the x and y direction will be zero If it s a nonequilibrium problem then the sum of the forces in the xdirection will be max and the sum of the forces in the ydirection will be may Example The steel lbeam in the drawing has a weight of 80 kN and is being lifted upward at a constant velocity What is the tension in each cable attached to its ends Is this an equilibrium or nonequilibrium problem It is an equilibrium problem since the beam is not y accelerating Therefore 2F 0 EFy 0 L Choose a coordinate system x Draw the FBD 700 700 Now break the vectors down into their components W is i along y so its done but T needs to be broken down 2FT T0 lbeam 2FyTyTy W0 gt2TyW 70 From the figure Ty Tsin70quot W 8000N 2Tsin70quot W gtT4 4260 25in 70 25in 70 Example Ignore all frictional effects and assume the rope and pulley are massless What is the acceleration of the two blocks and the tension in the rope WI 7422 N W y W1 When the 185N block is hung on the rope this system will begin to accelerate since there is no friction E Therefore this is a noneguilibrium problem W2 Thus the nonequilibrium conditions apply 185 N Choose coordinate axes 1 2 W2 7 ma 2 my Draw the FBD for each mass W1 accelerates in the xdirection and W2 accelerates in the ydirection Thus the following must be true Ki 2 Fy may KL EL EFX m1ax2Fx 2Fy02FyFm mo Must include a minus sign here since W E 2 is accelerating in ydirection 2n 0 Since the two masses are connected by the rope we know that their accelerations will be the same Thus ax y Cquot T mg Since 711 Wg we can write them Now our 3 equations look like this FN1 W1 ke thIS T W11a T W mza g FNl WI I can set the two equations for tension T equal to each Wza other and solver for a the acceleration T W2 g g g g g WlWz 185N98 2 a 298ms 422N185N Do the limits on the expression for the acceleration a make sense If W2 0 then a 0 And if W1 0 then a g Now that we have a we can go back and calculate the tension T T 422 N8298 128N g 9 Notice that the tension does not equal the weight of the hanging mass 185 N since the system is accelerating Example Two blocks are connected by a rope as shown The coefficient of static friction between the block and the incline is 045 The incline makes an angle of 42 wrt the horizontal What is the maximum value for m such that the system remains in equilibrium Chapter 4 Forces and Newton s Laws of Motion Newton s 2nd law 2 F ma Newton s gravitational force mlmg F G 712 G 6673 x 10 11Nm2kg weight on earth W mg g 98ms2 apparent weight Wappmm m9 W for free fall ay g frictional forces fsU SFN7 fkU kFN translational equilibrium 6 0 2120 ZFy0 Chapter 5 Dynamics of uniform circular motion velocity period T i 27W 7 T centripetal acceleration v2 ac r centripetal force F5 mac satellites in circular orbits me 7 Chapter 6 Work and energy Work W Fdd Fcos0d Work Energy theorem K E mv2 W AKE KEf KEG Work done by gravity near earth s surface ngv m9h0 hf Gravitational potential energy near earth PEG mgh Non conservative work WNC AK E APE WNC E f E0 Power 13 E orP t t 7 Chapter 7 Impulse and momen tum Impulse de ned f Fm Linear momentum of one particle 17 m1 Impulse momentum theorem A17 quot41 Newton s 2nd again A A F At total momentum 13 Stat when E Fen 0 momentum con served so AP0 or P0Pf l dz m center of mass Z m z Z mil239 950M 7 CM mtot mtot P tot vOM mtot Chapter 8 Rotational Kinemat ics angular displacement A00f 00 average angular velocity E At Q average angular acceleration amp At a equations when angular acceleration is constant wf we at 1 2 A0 wot Eat a Lug ZQAO tangential variables 5r0 UTrw aTroz centripetal acceleration Chapter 9 Rigid objects in equi librium torque r distance from axis to F TFrsin6 Fl rotational equilbrium the 3rd equa tion 2 739 0 center of gravity aka center of mass moment of inertia I mr2 point I 1mg rigid see table I tab mr2 rigid point etc Newton s 2nd law for rotation 27104 Rotational work WRZTQ Rotational kinetic energy 1 KER gm total mechanical energy 1 1 Etot imi Iw2 mgh if WNC O and 27m 0 then tot energy conserved Etot 0 Etot f angular momentum Lzlw if 27 O angular momentum is conserved Dado Ifwf TABLE a1 Moments of lnertia tor Various Rigid Objects of Mlass IVl lliinwallecl hollow cylinder or lioop Rl 1 MR2 Solid cylinder or dislr Rl 1 git1R2 Thin rod axis perpendicular to rod and passing tilrongll center L t l l I 11 12ML2 Thin rod axis perpendicular to red and passing through one end l 1 ML2 MR2 a T quot mlm Solid sphere axis tangent to surface firr R 7 2 1 EMR Thinwalled spherical shell axis through center 2 2 I O R1 Thin rectangular sheet axis parallel to one edge and passing through center of other edge 2

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