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# GEN PHYSICS A STUDIO PHY 2048C

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This 181 page Class Notes was uploaded by Garett Kovacek on Thursday September 17, 2015. The Class Notes belongs to PHY 2048C at Florida State University taught by Simon Capstick in Fall. Since its upload, it has received 37 views. For similar materials see /class/205523/phy-2048c-florida-state-university in Physics 2 at Florida State University.

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Physics 2048 Chapter Two One dimensional 1 D motion Prof Chris Wiebe Prof Simon Capstick Fall 2006 Chapter Two 1D motion 0 Kinematics the study of motion we don t care about what causes the motion ie forces that is dynamics o One dimension in a straight line eg updown rightleft northsouth etc o Coordinate system an x axis with an origin x0 positive direction and negative direction Galileo History Greek world view 300 BC Aristotle Modern world view 1500 modern day V Newton Chapter Two 1D motion Position a measure of where you are relative to some pre defined point Position can be positive or negative in 1D and is usually denoted as Distance traveled a running total of the change in position between your starting point and your ending point This is always positive Displacement difference in position between your starting point and end point denoted AX Xf Xi An example if you start and end at the same point your displacement is zero but the distance traveled could be large Arrow indicates positive direction 39 Origin g I I 39 l 1 I l 39 ll 39 r Initial position Final position Xi Xf Example car on a race track after one lap the displacement is zero but the total distance traveled is not O 39 I 0 Chapter Two 1D motIon 0 Speed and velocity 0 Speed distance traveled divided by the total time it took always positive Units lengthtime eg ms or kmhour 0 Velocity change in position over a time interval Units lengthtime 0 Average speed total distanceelapsed time 0 Average velocity vav displacementelapsed time AxAt 0 Let s do an example to discuss these new concepts 249 0 Physics problem solving strategy 1 What is given 2 Draw a picture 3 What are the equations I need 4 Solve 5 Check your answer 6 Units Chapter Two 1D motion Instantaneous velocit velocity at a particular instant in time 0 eg Driving to this class this morning what was my instantaneous velocity at 742 am Ax dx this is actually the derivative Vt leAZ gtO At dt What does this mean In the limit as At gt 0 this is from calculus it is the slope of a tangent line at some time t o It means that we need the velocity at an instant of time or when At 0 We can do this by looking at slopes closer and closer to At O 0 An example say that a particle follows the following position vs time graph What is the instantaneous velocity at t 1 s 0000 00000 33 Chapter Two 1D motion v 100 ms instantaneous velocity at t 1 s 30 25 average veIOCIty g 20 vav121 ms from interval V K H 025 s to 175 s g 15 5 v 137 ms g aquot average veIOCIty L 10 from interval 0 s 5 y to 200 s l l l l l l o 05 1 15 2 25 3 Time 1 s o What is my instantaneous velocity at t 1s 0 We can find this by taking average velocities closer and closer to t 1 s ie At 0 s If we knew the position xt we could also take the derivative Chapter Two 1D motion Analogy Velocity is the rate of change of displacement with time Acceleration is the rate of change of velocity with time Average acceleration Av Vf V1 aaverage At if t1 Instantaneous acceleration Av dv at LimAHO At dt Note If the average acceleration is zero then the velocity is constant over that time interval vf vi 0 or vi vf Chapter Two 1D motion 0 Typical accelerations in ms2 Ultracentrifuge 3 x 106 Batted baseball 3 x 104 Bungee jump 30 Acceleration of gravity on Earth 981 Emergency stop in a car 8 Acceleration of gravity on the 162 Moon Chapter Two x vs t v vs t a vs t 00 Visualizing these graphs is the key to success in this chapter Be careful that you do not confuse velocity with acceleration Examgles 1 An air track with an object moving at constant velocity 2 An air track with an object moving at constant acceleration ChapteriZ Static Equilibrium and Elasticity in i f J i r i i i 1 i 3 i v iii Conditions for Equilibrium oihere are two necessary conditions for an object to be in static equilibrium 1 The net external force acting on the body must remain zero 2H 2 The net external torque about any point must remain zero 27H 122 t CONDiiiONS FOR EQUltiBRiUiii i i i It r i i 9 i 4 i i ii if i i i 3quot salt I J l ii 3 139 3 t 5 quoti kiwi ifquot i i 3 ii 1 at g n in i it it i 39 i i it RivJ t i Center otgravity oihe net torque due to gravity about any point can be calculated as it the entire weight W were applied at a single point the center of gravity CENTER OF GRAVITY DEFlNED The net torque about some axisO due to gravity can be calculated ii the entire weight is concentrated at the point cg the center of ravity Center otgravity l o t point 0 is directly above the cg nen rcg and W are both in the same irectton so the net torque rch W is zero r I h This mobile is being supported from it s cg so there is no net torque Demonstration oil we supportan object away from is cg a torque will be exerted on hat object by gravity until the cg lies reneath the pivot point 0 Static equilibrium problems oStrategy for solving these problems o Sum of the forces zero h 1 o2 Sum oit e torques zero about some axis oibese two conditions must be true for static euilibrium Force on an Elbow problem 9 You hold a 6 kg weight in your hand with your forearm making a 90 degree angle with your upper arm Your biceps muscle exertsan upwards force that acts 34 cm from the pivot pointO at the elbow Model the forearm and hand asa 30cm long unilrm rod witha mass oi 1 Find the manitude ot the Frce me him the n the irection the Force on an Elbow problem 9a Draw a free body diagrao oApply he 2 conditions for stazic equilibrium oZTO 0 ouao mhgbZ omo 39 ml Force on an Elbow problem ob ZF 0 do the xdirection F 0 0 0 uax 0 Therefore FuaX 0 do the ydirection Fuay Fm mhg mg 0 Fuay m mh Fm 7 kg91 k 53 N 4 N Therefre F 44 N o o oar Demonstration We can calculate what forces act on the lever arm to provide the condition of static equilibrium about pointO i The net external force acting on the body must remain zero Erin 2 The net external torque about any point must remain zero 230 no CONDtttONSFOiEQUttIBRtUtr j roblem 48 ran a free body diagram P Frnd the components of each iorce perpendicular to the pivot arm a Find the sum of the torques o 9D o For max static 8 2 had friction F Static equilibrium 2 0 About the pivot point 9moosdidgSmoosdnrg iromsuoirpo Problem 48 Solve 9mM5mm F W 10msin Solve for Fw 0089 Solve fortheangle 1 m using t ig 9003 Problem 48 oUse the sum of the forces SOlvefor 70kg22kg981n sz 39 normal force 3 29025N 90m Chapter 3 Two and three dimensional motion Prof Chris Wiebe Prof Simon Capstick The Northern Lights charged particles accelerating in our atmosphere giving off light Warmup problem I Relative motion problem adding 2D vectors I Problem 363 I Key to these problems drawing the right picture I Make sure that you have the right frame of reference a perspective from which a system is observed The monkey hunter PRS A hunter sees a monkey in a tree above her She takes aim on the monkey with her dartgun At the moment she fires her gun the monkey let s go and falls Where should she aim to hit the monkey a Above the monkey b Right at the monkey c Below the monkey Answer b Right at the monkey Both objects fall at the same rate In the projectile s frame of reference where the objects are accelerating at the same rate the target is always right on track l 1 Range and launch angle y m I For a given initial 3 3 velocity what is the maximum range R 15 I This is 45 degrees 10 I Launch demo I Can we show that this is true I Break up into components and then solve for when R is a maximum 0 10 20 30 40 50 60 70 A m v0 vO cos GO v0y vO sin 60 Calculating Range I The time of flight T at top of flight we are at time T2 and vy 0 Use the equation v v0 at l l vy v0y gT2 v0y gT2 O l l T 2 v0y g 2 v0 sinGo g I The maximum height h at T2 U y O v0y T2 12gT22 U h V0y VOy g VOy V0y2g I The range R at time T we are back down l lx0vXT2va g 0 R 2 302 ginwo 00360 g R 2 v02 sineo 00360 g v02 sin2GOg 2sinAcosA sin2A Maximum Range I We need to take the derivative of this to find the maximum range dRdGO O dv02 sin260gd00 2 This occurs for when 260 90 2 V0 COS200g 0 degrees or 60 45 degrees When is cos260 O Also note that for some angles 970 the Range R is the same such as 70 and 20 degrees or any combo that gives 90 degrees as a total This is because sin270 sin 1 m 140 sin 40 sin 220 sin 2A sin 290A Test PRS question Two students throw snowballs off of a roof at the same velocity but one throws the snowball at 40 degrees below the horizontal the other throws at 40 degrees above the horizontal Which snow ball has a faster final speed just before hitting the ground a Snowball thrown upwards b Snowball thrown downwards c Snowballs land with the same final velocity Answer c Why The snowball thrown upwards has the same velocity at point in the picture as the vi of the snowball thrown downwards Thus it has the same final velocity New PRS question 2D motion I A cart moving along this airtrack in the x direction launches a projectile in the ydirection If the cart continues to move at the same constant speed where does the projectile land I a In front of the cart I b Behind the cart I c In the cart I Answer c In the cart If it is moving at the same constant velocity in the xdirection then it has to land in the cart II Last topic Uniform circular motion I Uniform circular motion motion at constant speed I Speed is constant but velocity is changing l1 Similar triangles Avv Ar r Av vr Ar so AvAt vr ArAt l3 Take limit of At gt 0 centri etal acceleration avrvv2r In this limit Av points towards center of circle Ht Wt 37t v t 39 quot27t At An example of circular motion I A star running out of fuel can expode Into a supernova leaVIng behind a dense core called a neutron star I Question A neutron star composed of neutrons has a radius of about 20 km If a neutron star rotates once every second the period or time per revolution is 1 s I 1 What is the speed of a particle at the equator I 2 What is the magnitude of the particle s centripetal acceleration I Bonus question If the star rotates faster do the answers to a and b increase decrease or stay the same A fast spinning neutron star at the center of the Crab Nebula Practice problems I Chapter 3 43 45 47 49 53 55 57 61 65 73 75 77 79 81 83 93 Chapter 19 The Second Law Wodynamics Ludwig Boltzmann 1844 1906 S kB log W Entropy and the Second Law V ofThermodynamics The second law of thermodynamics tells us about entropy The entropy of the universe system plus its surroundings can never decrease What is entropy Entropy symbol 8 is a measure of disorder in a system The total entropy tends to increase for any process Entropy is sometimes called Time s Arrow because it tells us which way time flows in physics equations the entropy increases A f t 39th I n examp e 0 en ropy Wl gas mo ecu es The Situation on the left corresponds to a more ordered state than the right High temperature If we let heat flow between these two systems the resulting state is more disordered a random distribution of speeds in both containers at the same T Intermediate temperature Low temperature a b c Examples of entropy 1 Salt shaker demo 2 Drop of food coloring into water 3 Free expansion of a gas One definition of entropy AS AQT For an isothermal expansion AEAQWO SoAQW 80 A0 nRT nVfVi and AS nR lnVfVi The expansion of this gas is an example of the increase of entropy it would be odd if the molecules wouldn t expand out of the cylinder 4 The melting of an ice cube on a table top AS QTtable mass L T fusion a The free expansion of a gas is driven by the tendency for entropy to increase for any given process Q The melting of an ice cube at room temp is an example of entropy increase for a process why The Second Law and Heat Engines Heat engines work through the flow of heat from a hot reservoir Oh to a cold reservoir Q with the extraction of some energy as work W in the process What does the second law of thermodynamics say about this process We know that the change in entropy must be greater than or equal to zero for any process In terms of heat flow this means that heat tends to flow from hot to cold It would be odd if we had heat flow from cold to hot eg ice reforming in a cup of water So the second law tells us that it is impossible to have a process where the net result is heat flow from a cold object to a hot one We need to do work to do this this is how a refrigerator works Cold reservoir at temperature Tc Efficiency of heat engines For one Cycle AE 0 The Otto cycle like your internal 39 t combustion engine I Since Qin AEint Wb Wby rewritten 1St Law s Compression a gt b 0 Work done by engine is Ignition adds heat b 39gt C Power stroke 0 gt d l W Qh Qc net heat Qin Exhaust and intake d gt 8 added to system Efficiency of engine is amount of work done divided by amount of heat absorbed at Th lose heat QC 8 Qh QC Qh 1 QC Qh Note This is always less than one like a The Carnot cycle The most efficient engine is called a Carnot cycle Every step is a reversible process What do we mean by reversible For any process AS 2 0 For a reversible process AS 0 no entropy change so the process is reversible with respect to the second law of thermodynamics Examples of reversible processes quasistatic process ice and water at the same temperature Carnot cycle a 1 TCTh TC and Th are the temperatures of the hot and cold reserviors isothermal Q in expansion 1t T h Adiabatic expansion Isothermal compression 20th at T C Ad iaba tic compression V The Carnot cycle for an ideal gas An example problem 48 Isothermal entro chan e py g ASnR1nE 1 2 mol83 14 Jmol K1n80 L 40L This is the entropy change for the gas What is the entropy change for the universe Use for a quasistatic process AStotal 0 So the total entropy change is Astotal ASgas ASuniverse So Asumverse 115 JK mol Refrigerators Fridges work like engines in reverse use work to drive heat from a cold reservoir to a hot reservoir Qc W Qh Coefficient of performance for a fridge COP QcW The higher the COP the better the fridge most fridges have COP of 56 Note perfect fridge has an infinite COP no work Hot reservoir at39tempera ture Th 1 Low pressure liquid High pressure liquid Condenser coils outside refrigerator Qil C to out51de from inside refrigerator Hi h ressure to coils g p vapor Work Compressor motor Cooling coils V inside refrigerator Low pressure Compressor Vapor motor PRS question I Suppose your AC breaks in your apartment in Tallahassee in the middle of the summer Your nonscience major friend suggests to leave the fridge door open to cool down your place What would be the result of leaving the fridge running in your apartment and leaving the doors and windows closed A The temperature of the apt would decrease B The temperature would stay the same C The temperature would increase Answer C The temperature would increase Remember a fridge takes out heat energy from inside the fridge and then dumps it to the back into the coils Leaving the fridge open would require the fridge to do more work to cool down the inside of the fridge and therefore the heat dumped out of the back would increase increasing the temperature of the room Entropy and probability a Let s go back to the free expansion example AS nR nVfNi lf Vf 2V then ASnR n2 NkB n2 Boltzmann S kB In W where W is the number of states Since the volume increases the number of available states increases This is an irreversible process entropy of the system increases Can we calculate the probability that the system will remain in the same state AS 0 Let s say that we have 10 molecules initially in the left hand side LHS What is the probability that one molecule will be in the left hand side About 50 or 12 for random motion of the molecules The chance for any two particular molecules to be in the LHS 12 x 12 11 same as the chance that a coin flipped twice will come up heads each time for example The chance that all 10 will be in the LHS is therefore 1210 11024 For a large number of molecules this is very small 12N 0 It could happen but the odds are so small that it doesn t statistical physics Entropy and the availability of energy I Another way of looking at entropy it tells us about wasted energy or energy that we cannot use for work I An example Let s say that you drop an object a height h and it hits the floor The potential energy of the object mgh is converted to thermal energy when it hits the floor and stops moving What is the change in entropy for this process I I Heat increase A0 AST mgh The temperature T is the temperature of the cold reservoir ie the floor I Therefore the entropy increase is AS mghT Note that we can write W T A8 lost For an engine with a cold reservoir T the work lost is VVI TC ASCycle TC QTC QTh Q1 TcTh I For a Carnot engine this tells us about the efficiency W SQ where a 1 TCTh ost Does the process of life violate the second law of thermodynamics I The development of living organisms requires that organic molecules order themselves into complicated structures that have to interact in a certain way I Does this violate the second law of thermodynamics I No because we use energy to create these structures which produces byproducts that increase the entropy of the universe A final fun example in statistics problem 83 There are 26 letters and four punctuation marks space comma period and exclamation point used in the English language so we have a grand total of 30 characters to choose from This fragment is 330 characters including spaces long there are then 30330 different possible arrangements of the character set to form a fragment this long We can use this number of possible arrangements to express the probability that one monkey will write out this passage and then an estimate of a monkey s typing speed to approximate the time required for one million monkeys to type the passage from Shakespeare I Assuming the monkeys type at random express the probability P that one monkey will write out this passage P 1 10495 30330 I Assuming the monkeys can type at a rate of 1 character per second it would take about 330 s to write a passage of length equal to the quotation from Shakespeare Find the time Trequired for a million monkeys to type this particular passage by accident T 330990495 106 I Express the ratio of Tto Russell s estimate 2 3 30X1049S 1y 484 k316x107s Y Russell End of chapter and end of the course I Skip sections on heat pumps I The Third Law of Thermodynamics S gt O as T gt O K As you approach zero Kelvin the entropy of the system approaches zero I Review problems 21 23 25 31 neatl 33 35 47 49 51 55 59 61 I Next class Review for final I Awards ceremony I Something fun more prizes An example Problem 97 Off center collision in 2D Break up into x and y components Momentum is conserved if there are no external forces How do we know if energy is conserved Work it out for homework y Xdirection Ydirection pxiszf pyizpyf X Of 01 O O mv mv1 cos 30 mv2 cos 60 0 mvl gm 30 mvz gm 60 or or v 2 v1 cos 30 v2 cos 60 0 2 v1 sin 30 v2 sin 60 2 equations 2 unknowns solve for one substitute into the other Problem 97 How do we know if the collision IS elastic or not KEi KEf Calculate this homework Note that the angle after the a Before collision After collision collision between the two at objects is 90 degrees This only works for elastic collisions You can show this by looking at ltbgt Or v1i v1f v2f Also from KE 12 mv1i2 12 mm2 12 mV2f2 80 v1 v1 v2 Draw a right triangle angle between v1f and v2f IS 90 By Pythagorean theorem this must be a right triangle Chapter 9 Rotational motion Prof Chris Wiebe Prof Simon Capstick The rotation of the Earth about it s axis can be demonstrated through the rotation of stars about Polaris North Star Rotational motion Positive direction Let s look at a particle moving on a rotating disk We first need to define the concept of an angle 6 and angular displacement A6 Rotational motion describes how this angle changes not distance as a function of time Convention 6 gt O counterclockwise G lt 0 clockwise Units Radians 1 revolution 360 degrees 2n radians ln SI units we usually just drop the radians units so for radius 39s of rotation exampe an ange of 130 If I rotate from point A to point B my degrees can be expressed as angular displacement would be positive Tr radians orjust1T GB eA AG efinal einitial Rotational motion The angular velocity is the rate of change of the angle d6 a dz Units rads degreess revolutions per minute rpm etc The time rate of change of angular velocity is the angular acceleration 6126 d a Units rads2 wgt0 wgt0 0lgt0 otlt0 1 1 angular speed angular speed increases decreases o lt O u lt O ot lt 0 0L gt 0 Be careful about sign conventions Constant angular acceleration For constant angular acceleration you can use the same equations that you use before for constant linear acceleration x gt 6 Analogous variables v gt w a a or Constant Linear Acceleration Constant angular acceleration 1 1 d v rar2 a A9wor a39r2 2 2 1 1 0a1 4 mw c f v2 1 2307 4 m2 mg 2a Constant angular acceleration Demonstration of a rotating object with a constant angular acceleration What should the angle angular velocity and angular acceleration curves look like as a function of time What about tangential motion Tangential velocity at Every point on a rotating object also Point A has a tangential velocity and acceleration How are these related to angular velocity and angular acceleration The tangential velocity is vt dxdt We can define x in terms of the arc length x r6 or dx r d6 constant r Therefore vt rdGdt rw There is also a tanential acceleration at Similarly ea 39 3939 centripetal acceleration towards the center Wh39Ch 395 a0 v2 Centripetal acceleration By substitution we thenhave that the at point A centripetal acceleration IS aC rw2r r002 at how fast vt is changing with time Two types of acceleration The total acceleration of the system is the sum of these two accelerations at and aC An example a discus a thrower that is rotating faster and faster before he releases the discus a total acceleration of the discus Rotational motion PRS CONCEPTUAL CHECKPOINT 10 1 Two children wide on a marrngovmund with child 1 at a greater distance from the axis of mtaticm than child 2 Is the angular speed of child 1 a greater than M less than or c the same as the angular speed of child 2 Rotational motion PRS CONCEPTUAL CHECKPOiNT 101 Two children ride on a merryvgmmund with child 1 at a greater distance from the axis of rotation than child 2 Is the angular speed of child 1 a greater then b less than or c the same as the angular speed of child 2 Reasoning and Discussion At any given tithe the angle I for child 1 is the same as the angle for child 2 as shown Therefore when the angle for child 1 has gone through 2rr for example so has the angle for child 2 As a result they have the same angular speed In fact oath and every point on the merry go rotmd has exactty the same angular speed The tangential speeds are different however Child 1 has the greater tangential speed since he travels around a larger circle in the same time that mild 2 travels around a smaller circle This is in agreement with v rm since child 1 has the larger radius Answer cl The angular speeds are the same In Rotational motion is complicated Calvin and Hobbes Note that for rotating objects the angular velocity is the same for all points on a disk or any rotating object at any given point in time The tangential velocity COMPARE A VOlNT ON THE WT NE FONT ON THE changes with r the farther Remsamenean you are away from the amalgam center the faster you S mt m fmm WE SNA E REVMTKMS PER have to go to complete a 3am revolution which is a distance of 2m An example Problem 29 Starts from rest to O A a wo am 0 Att58 a55 8radsz55 Tangential acceleration at 5 s O 12 m8 rad2 Centripetal acceleration aczmz 61053O12m400rads2 192ms2 wzaAt Rotational Kinetic Energy What is the kinetic energy of a rotating object If you had a system of particles rotating about a common axis with the same angular velocity then the kinetic energy of the ith particle is 1 2 K1 2 mlvl 2 We can simplify this to get K 12 mivj2 Z The moment of inertia l of a system for a particular axis of rotation is I Zmrz Thisdepends on the axis of I i rotation and the mass distribution Moment of inertia The moment of inertia depends on the mass and the radius quot i A change in moment of inertia l 39 lt19 ll of any body causes a change 1P in the rotational motion We ll get back to this later g g when we talk about angular KC 3 momentu m l 2 M The kinetic energy of a 7 J rotating body is K 12 l b2 a by This is similar to translational motion K 12 mV2 A figure skater changes her I acts like the mass of the moment of inertia by pulling her arms inwards reduces the radius of the COM System of her arms J Prof Chris Wiebe Prof Simon Capstick Graphical representation of SHM What do these functions look like xt A cosoo t 5 Shown here for 6 0 VXt 03A Sin03 t 5 note how this is a max when xt is a min and a min when xt is a max This is also true for for at vt is a max when at is a min a aXt 032A cosoo t 5 notice how it is like xt with negative sign DZA link I 4 NIH a SHM demo We can plot xt vt and at for an object on a spring Notice how the phase 5 just depends on what we call t O s We can determine this from our initial conditions ie at t O s x A cos 5 You can then solve for 5 if you can measure x and the amplitude A Also notice how the period is the same for all three motions The period is independent of the amplitude We will be using these equations over and over again An example A block with a mass of 200 g is connected to a light spring for which the force constant is 500 Nm The block is displaced A500 cm from equilibrium on a frictionless horizontal surface a Find the period of motion 0 lkm 5 Nm200 kg 5 rads So the period is T 2Tr03 126 s b What is the max speed of the block vmax max of 03A Sin03 t 8 03A 0250 ms c What is the max acceleration amax MA 125 ms2 An example d What is xt vt at xt A cosco t 5 0050 m cos500t The phase is zero because xt0s 0050 m vXt 0A sinco t 5 0250 mssin500t aXt 02A cosco t 5 125 ms2cos500t What happens when the phase is not 0 Equilibrium lt x gt A x 3 A A A MA 1 I J 1 J 1 v M A more complicated example where phase is NOT zero Problem 96 in Chapter 14 a Find the frequency f and period T Note to 21TT so 3 rads 21TT and T 209 s Frequency f 1T 0477 s1 1 Hertz 1 Hz 1 4 This is how many eventsunit time b Where is the particle at t O s xt A cosoo t 6 04m cos3t Tr4 At t0s x 04 m cos1T4 0283 m Be sure that either your calculator is in radians or you can convert between radians and degrees 0 Where is the particle at t 05 s x 04 m cos305Tr4 04 m cos 228 rad 0264 m The vertical spring A vertical spring acts just like a horizontal spring but the equilibrium position is different The new equilibrium position is defined by mg kyo or y0mgk Even though gravity is acting on the system it just acts to change the equilibrium position so the new displacement is y y yo m 3 n 7 E ln gt gt 1 I l V F5 539 li gt li39i Posmon w1th l w i spring gt I unstretchcd gt gt i 3 quot g l K i in Equilibrium position with mass m attached Spring stretches an amount iO mg k Object oscillates around the equilibrium position with a dis placement i39 i 7 30 Vertical spring demo Connection between SHM and circular motion Why is 00 called the angular frequency There is a subtle connec on between Circular motion and simple harmonic motion Read page 433 for more details not test material but it is pretty cool httpwwwphysicsuoguelphcatutorialsshmphaseOhtml Energy of a simple harmonic oscillator The total mechanical energy of a simple harmonic oscillator is K U 12 mv2 12 kx2 for a spring for example E 12mcoAsinco t 5212kACOSo t 52 E 12 mmA2sincot5212 kA2COSot52 But 02 km or mco2 k so we have E 12 kA2Sinot5212 kA2COSot52 Or E 12kA2 a constant The total amount of energy is a constant and proportional to the square of the amplitude Energy of a SHO The energy oscillates between potential and kinetic energy but the total energy is a constant Xt for one oscillation A U Etotal LI 1E av 2 total 0 K Etotal Kay Etotal 0 c The potential and kinetic energy as a function of time PRS Question The Pendulum The motion of a pendulum is described by SHM How does the period of the swing depend upon the amplitude for small amplitudes A The Period increases with increasing A longer swings longer period B The Period decreases with increasing A longer swings shorter period C The Period is independent of the amplitude Answer C For SHM the period does not depend on the amplitude of the swing Why is this true The Simple Pendulum Newton s law a d2sdt2 mg sin m This is the component which causes acceleration in the direction of the oscillation Arc length sL d2sdt2 L d2ldt2 So Newton s law reads L d2ldt2 g sin d2ldt2 gL sin mg cos q The arc length is s The Simple Pendulum d2lgtdt2 gL sin For small angles 1 we have sin 39 1 try this on your calculator with angles set to radians For small oscillations around equilibrium d2lgtdt2 gL 1 Compare to d2Xdt2 km X 032 X Just simple harmonic motion with o gL 2 T 27 03 27 Lg 2 s mg cos q Note the period does NOT depend on the amplitude I this is why you can use them to measure time Chapter 7 Conservation of Energy Prof Chris Wiebe Prof Simon Capstick How fast does a rocket need to go to escape Earth s gravity The kinetic energy must overcome the gravitational potential energy of the Earth quotIt s time we face reality my friends We39re not exactly rocket scientists 9 Conservation of Energy III Putting it all together with an example problem 82 What is frictional force 910 Block travels L4m If 4 f HKN uKmgcos6 1quot Xe before hitting spring quot 2 x3 L r 0 Spring is compressed a distance X part a Bi 2 Ef Part b Friction fAs Ef Ei UiKiUfKf fASUfKfZUiKi mgAh 12 sz fAs mgAh 12 kX2 What is Ah Ah 2 L XSine What is Ah Ah L Xsin9 Therefore we have mgLXsin6 12 kX2 HKmgcoseLjXmgLjXSin9 kxz I kx2 mgsin6x mgL Sing O I Use quadrat1c equatlon to get X 0793 m Use quadratic equation to solve for X 0989 m Part c homework L 154 m Mass and energy El There is somewhere else that energy can go energy can be converted into mass or mass into energy El Einstein in 1905 and his theory of relativity c speed of light 3 x 108 ms A small mass can create Large energy a huge amount of energy The massive amount of energy stored in a mass as small as a nucleus is predicted by Einstein s equation Quantum Mechanics El Energy cannot be transferred in a continuousway it is transferred 1n d1screte un1ts called quanta El These are so small that we usually can t see the effects of quantum mechanics in the everyday world El In the atomicnuclear world you have to use d1fferent laws of phys1cs to descr1be how energy 1s transferred between objects El This leads to strange things happening such as the possibility of The laws of phys1cs change objects existing in two states at the at the atomic level same time particles acting like waves and even breaking the law of conservation of energy if you can do it fast enough for no one to notice End of chapter I Practice problems Chapter 7 3 7 9 17 19 21 25 27 33 35 43 45 47 49 51 Chapter 10 Angular Momentum Prof Chris Wiebe Prof Simon Capstick More on crossproducts Crossproducts are used extensrvely In vector calculus 2 Just for fun let s try a calculation What is the crossproduct of 1i x 2j i andj are unit vectors 2j Cross product 1 12sin90 2 k 39 The direction of this vector negative sign you get x from the right hand rule Fingers in direction of x X392139392k curl hand towards the y direction thumb is in z direction From last lecture Angular Momentum L l w r x p a Torque T r x F 3 Cross product lfC U A x B then C is a perpendicular to A and B and C f ABsincp Right hand rule for 7 direction of angular velocity and torque Right hand rules for angular velocity torque PRS question right hand rules Using the right hand rule which way is the angular momentum vector pointing for a barrel rolling straight towards you A To your right B To your left C Towards you D Away from you Answer A To your right Conservation of Angular Momentum If there are no net external torques then dLdt O 39before Lafter Last lecture we used this to explain why a figure skater s speed increases as she pulls her arms in Iwbefore Iwafter Last class we had a student pull weights towards herself and her angular velocity increased Our student had an initial ri 077m and let s say that her final rf 025 m By how much does her speed increase Assuming that the weights act like point masses mri2wimrf2wf gt oofuiirir2 77252 rotating 9 times as fast Conservation of Angular Momentum We can demonstrate the law of conservation of angular momentum We ll drop two identical disks on to one another one is rotating the other is initially stationary What is the final angular velocity of the system Frictionless shaft l39 116 i I r 11 12 60f I V 2 At rest Conservation of Angular Momentum Li 391 Di points up Frictionless shaft I Lf I1 of I2 of both up Ilr wi I r I I1 IZ 0pomt3 up 12 wf I 391 Di I1 I2 Df At rest 03f103i1I2 Ifthe disks are identical then I1 IZ and of 12 coi Let s do the demo Another example of angular momentum conservation At the end of last class I had a complicated example of sitting on a stool with a rotating wheel Can we use conservation of angular momentum to predict the motion of the person on the stool as they tilt the rotating wheel The component of angular momentum in the z direction is conserved before and after the tilt Problem 55 a Conservation of angular momentum Ii 1101 Ifwf wf wi f 6kgnr12 wf ml5reVS b Change in kinetic energy M18kgm25revx27rradjz S I39CV AKK K11 2112 1reVZ7r1rad2 f i E fwf Eiwi 6kgm215 x j S I39CV 0 Where did this increase come from The person does work What happens if there is a net torque If there is a net torque in a system then angular momentum is not conserved dLdt Texternal JIlexternaICHFIdLAL 39final39Linitial Demo what makes a wheel stand when it rotates Since there is a torque on the i systemTrXmg it produces a df change in the direction of the angular momentum vector L M L We get precession Collision problems For linear momentum we looked at lots of situations where there were collisions Linear momentum is conserved if there are no net forces acting Angular momentum works in the same way for collisions it is conserved if there is no net torque acting on the system The collision can still be elasticinelastic Problem 69 an inelastic collision Treat these problems just like linear momentum problems this time the rod has mass so it has angular momentum At the collision itself Li Lf 08di w any 0892me rxmv Rod point mass Solve for angular velocity right after collision Md2 064de Problem 69 Conservation of energy after collision EiEf39gt UiKiUfKf 2 Ia2 Mgimng1 coslt9 2 CM of Rod ball The moment of inertia of the system is 2 2 1 m08d Md 064de Md2 Now we can substitute the moment of inertia 3 into the conservation of energy equation 2 and then i substitute the angular velocity Mg 2 mgd 1 0 56 equation 1 into 2 to get 0 32dmv2 Md2 064de Chapter 9 Rotational motion Prof Chris Wiebe Prof Simon Capstick The rotation of the Earth about it s axis can be demonstrated through the rotation of stars about Polaris North Star Rotational motion vs linear motion TABLE 92 Analogs in Rotational and Linear Motion Rotational Motion Linear Motion Angular displacement A6 Displacement Ax d9 dx An uhr velocit w Veloci g y dr ty dt do 120 Lil d 1x An ular acceleration a a Acceleration 1 g m dt dt 2 Constant angular a z aquot a Constant acceleration quotu on at acceleration equations Ag a At equations Ax i A av E139 1 l wav 3a w 6 60 wot at2 02 aquot ZaAB vdV 51quot l 2 7 A trot 2at 12 223 2n Ax Torque T Force F Moment of inertia 1 Mass m Work dW 716 Work dW F5115 Kinetic energy K 2quotle Kinetic energy K imv2 Power P no Power P Fv Angular momentum L w Momentum p mv Newton s second law Tm a LL Newton s second law F J mn tip dt quotd dr Jr Angular momentum is introduced in Chapter 10 Moment of inertia The moment of inertia depends on the mass and the radius quot i A change in moment of inertia l 39 lt19 ll of any body causes a change 1P in the rotational motion We ll get back to this later g g when we talk about angular KC 3 momentu m l 2 M The kinetic energy of a 7 J rotating body is K 12 l b2 a by This is similar to translational motion K 12 mV2 A figure skater changes her I acts like the mass of the moment of inertia by pulling her arms inwards reduces the radius of the COM System of her arms PRS Demonstration rolling objects Which object will make it to the bottom of the ramp first A The solid circle B The hoop C The circle with the small wheels The rotating object will take longer than the object on small wheels Why There is the same gravitational potential energy for both objects that is converted to kinetic energy Two kinds of kinetic energy translational and rotational Larger radius more rotational energy higher moment of inertia means less translational energy at the bottom of the ramp Calculating the moment of inertia How can we calculate the moment of inertia For a system of particles 1 Zmrz For a continuous object I Irzdm r is the distance from the axis to mass element dm Note units kg m2 Let s do an example of a moment of inertia calc Moment of inertia calculation Example 93 Uniform rod of mass M length L rotating about an axis perpendicular to the rod and through one end Since the rod is uniform we can break it up into pieces of length dx with mass dm What is the moment of inertia for this axis K Q L l dm dx I I I I l til V I I I4 1 gt lt dx I dm total mass X fraction in dx M dxL 1 rzdmzi x2MdxL MLj dex ML 78313 1 ML23 We can calculate moments of inertia through any object and about any axis Table 91 Moments of Inertia of Uniform Bodies of Various Shapes Thin cylindrical shell about Thin cylindrical shell about Thin rod about perpen Thin spherical shell about axis diameter through center dicular line through diameter center I MR1 MR IzgzviRul JIL1 II2ML2 Solid sphere about diameter Solid cylinder about axrs Solld cylinder about diameter through center Thm rod about Perpen dicular line through one end R I 31er2 I EM2 717 39 I 3 v 39 t quotg I 39 MR F ML Solid rectangular parallel I 1 ML a Hollow cylinder about axis Hollow cylinder about eP Ped abou 3X15 through diameter through center center perpendicular to face 39v R I r t J 7 391 39 I MRR3 I 1MR R IIZMLl 1 ZMnquotl 1 A disk is a cylinder whose length l is negligible By setting I 0 the above ionnulagt for cylinders hold for disks Revisit the rolling race Moment of inertia for a wheel lwheel 12 mR2 ll Moment of inertia for a hoop lhOOID mR2 The moment of inertia of the hoop is larger than the wheel The smaller the wheel s radius the smaller the moment of inertia This is why the small wheels object makes it to the end first it has the smallest moment of inertia and therefore the largest amount of translational kinetic energy compared to rotational kinetic energy The parallel axis theorem What happens when we rotate an object about an h a axis that does not 39 i intersect the center of mass we can use the parallel aXIs theorem If an object of mass M is U rotating about an axis parallel to the CM axis but a distance h away we can find the new moment of inertia using I lcm Mh2 39Cm An example of the parallel axis theorem rut Thin rod about perpen dicular inc th rough center M 1 3ML3 Thin rod about perpen dicular line through one 0 k Thin spherical shell about iameter Solid rectangular parallel epiped about axis through center perpendicular to face 391 b IIR n a 1113A4nli l1 we formulas for cylinde hold for disks What is the moment of inertia of a rod for an axis through one end I lCIVI mh2 112 ML2 mL22 l112 ML2 1AML2 112ML2 312ML2 l 412 ML2 13 ML2 This is what we calculated before Newton s Second Law for Rotation In order to rotate an object from rest we need to apply a force If this force is applied v radially from the CM there is no rotation The object will start moving a if it is applied in a tangential direction Another way of saying this is that we need to apply a torque A torque causes an angular acceleration on an object I I Torque rF1 rma rmro mr2o I Newton s Second Law for Rotation Another way of saying this is that if the sum of forces on an object cause it to rotate then a net torque is applied to an object The sum of the torques can be written as Z torques Zr T mr20i lo net Analogy Torque rotational force Moment of inertia mass x Angular acceleration acceleration Calculating torques a Torque rF sin90 161 P rF To calculate a torque you need to know the applied force and the angle the force is being applied with respect to the point of l contact r l Remember the torque is a maximum when applied tangentially If the angle between the radius and the force is less than 90 Idegrees then the torque will be ess Definition Torque F r sincp V where p is the angle between the radius vector and the applied force Torque rF sin 0 O 3 Calculating torques a more complicated example What if the force is applied at a different angle and at a different contact point r Measure r from the axis of rotation 0 Find the angle that the applied force makes Torque FF 339 P with respect to r Use torque rF sin p An example of using torque An Atwood machine problem 74 We can now look at problems where the pulley has mass it can then have a moment of inertia and rotational kinetic energy a What is the acceleration of the two objects b What is the tension in the string supporting m1 T in m2 c What would your answer be if you neglected the mass of the pulley already know the answer to this T1 T2 The Atwood machine Positive direction a Draw free body diagrams 39 Define a positive direction Use Newton s Second Law for linear and rotational motion mrg T1 T1 ZszTl m1g2mla 1 270 T2 T1r 2100 2 ZFX m2g T2 m2a 3 The Atwood machine For the pulley IO 12 mr2 and o ar So we then have for equation 2 Tz Tm mrzl 2702T2T1r210a r Now we have 3 equations 3 unknowns T1 T2 and a We can solve for the acceleration to 99quot 510g 500g9810ms2 a m2 m1g a i g510g250g m1m2m 2 94780ms The Atwood machine b What is T1 and T2 From equation 1 we can solve for T1 Zm 05001ltg981ms2 009478ms2 We can also solve for T2 T2m2ga 05101ltg981ms2 009478ms2 So the difference between the two is ATg 49ymN 49MN The Atwood machine c What would the answers be if the pulley had no mass Do this for homework You should find that T1 T2 49536 N Pulleys give a slight difference in tension if they have mass Rolling without slipping You will see the statement in problems that an object ros without slipping What does this mean This means that there is continuous contact with the object and the surface without sliding in other words the center of mass moves with the same instantaneous velocity as a point on the edge of the Asia n taneous rotation axis rotating object The velocity of the CM is wR This is just like an instantaneous distance traveled in 1 rotation rotation about the contact point 21TR Therefore VCM 21TRT where In this case vedge vCM 00R T is the period for one rotation Nee the center of mass 21TT is the same as to so vCMwR accelerates at the same rate as a point on the edge aCM dR Power of rotational motion Remember Power Work donetime What is the power for rotational motion how much power is needed to rotate an object at a certain rate Linear motion for an object moving at velocity v under the influence of a force F in the direction of v we have P Fv Rotational motion P Too for an object under a torque T and moving at angular velocity 00 Chapter 10 Angular Momentum Prof Chris Wiebe Prof Simon Capstick Why does her motion Change when she pulls her arms inwards Angular Momentum In linear motion we defined linear momentum as p mv For rotational motion we will define angular momentum as L It This is a vector with units kg m2s What direction does this vector point in Same as no What direction does on point in We need to define a convention The right hand rule for angular velocity We can use the right hand rule to define the direction of on Right hand rule take yournghthandcud your fingers so that they point in the rotation direction Your thumb is the direction of w and also of L The right hand rule for rotation The cross product and torque Torque has a direction as well It is a vector This can be defined by using 7 another right hand rule 7 Take your right hand lay it flat along r with your palm in the direction of Fyour fingers curl in this direction The torque is in the direction The right hand rule for torque Fingers of right hand along r Of Your thumb F is along direction that fingers curl ThlS Can be expressed In Torque vector is in thumb direction another way Torque r x F The cross product of vectors r and F What is the cross product Vector A x B is perpendicular to A and B The magnitude is AB sin p where p is the angle between A and B The right hand rule for any vector A B Fingers in direction ofA curl them towards B palm is in direction of B Your thumb defines A x B C Note that for torque Torque rF sin p Definition of Angular Momentum Another definition of angular momentum in terms of cross products is L r x p Lrxmvmrvsinltp An example For a particle moving in the xy plane L is in the zdirection Lmrvkmr2wk mr2 w w Conservation of Angular Momentum In linear momentum if ZFext 0 no external forces dpdt0 and linear momentum is conserved For angular momentum if ZTeXt 0 no external torques then dLdt O and angular momentum is conserved before afterw or Iwbefore Iwafter An example this is why the figure skater spins faster when her arms are pulled in her moment of inertia is smaller so her angular velocity increases Chapter 11 Gravity Prof Chris Wiebe Prof Simon Capstick I 39P E JE Gravity There are 4 fundamental forces in the Universe The electromagnetic force The strong nuclear force The weak nuclear force The gravitational force Gravity is the weakest force out of the four and not well understood Gravity J The force of gravity can be seen only with massive objecm such as the orbit of the moon around the earth or the earth around the sun l It is only attractive in nature an ing with mass is attracted to ano er object with mass J Two major developments led to the theory of gravity 4 Tycho Brahe 16th century h measured the motions of e planets J Johanes K 39 ler later put these together 1 laws which describe Timquot Bram these motions Kepler s Laws 1 All planets move in vellitical orbits With the Sun at one focus 2 A line joining any planet39to the Sun sweeps out equal areas in equal times planets move faster near the SUn 3 The square of the period of any planet is proportional to the cube of the semimajor axis of its orbit T2 Ca3 ararp2 Kepler s Laws The average distance of a planet from the Sun focus is a rarp2 ra is the aphelion furthest distance from the Sun rp is the perihelion closest distance T2 C a3 C is a constant I For the Earth T 1 yr a 1 AU astronomical unit 15 million km C 1 yr2AU3 Data for Kepler s Laws I T2 C83 01 gtx quotC Q 3 g 0quot to quotC O 395 0 Q4 0 0 5000 10000 15000 20000 25000 30000 Mean orbital radius cubed AU3 19 This brings us to Newton r Kepler39s laws were em irical rules based on the careful observations 0 Brahe Newton Showed that if the force holding the planet in its orbit igained with the inverse square of its distance from the un Leads to Kepler39s laws Assumed that it acts between any two massive ODJECL S Einstein showed that even light energy but no mass is attracted to the Sun Newton s Law of Gravity 1686 Force exerted on pointlike body 1 due to pointlike body 2 is Cm 11712 A F12 392 1 12 12 1 12 l 12I3912 G667gtlt10 11 N mZkg2 measured by Cavendish 1798 Cavendish experiment demo Measures the constant G by measuring the attraction between 2 bodies i Equilibrium Torsion fiber Equilibrium i m2 Position 1 I position Position 2 What happened to F mg I For an object of mass m near the surface of the Earth Assume that for calculating gravity all of the mass of the Earlh can be Concentrated at its center Newton developed calculus and used it to prove this for spherically symmetric bodies F GMEmRE2 mg g GMeREZ We have assumed that the m in Newton39s gravity law is the same as the inertial mass m in Newton s dynamical law F ma Principle of equivalence of gravitational and inertial mass tested to 1 part in 1012 PRS question What is the work done by gravity for the moon completing 12 of an orbit around the earth assuming a nearly circular orbit A Work mmg distance of 12 orbit B Work GmeammmWJrZXdistance of 12 orbit C Work O D Work mmoonngistance between earth and moon Answer C Work 0 The force and distance are always perpendicular to eachother so F39dx 0 and Work 0 An example The 155 J The 153 is h385 km up from Earth39s surface Use Kepler s third law to nd its period if RE 6370 km J We rst have to use Kepler s law T2Ca3 J What is C This changes for different bodies J F ma mvzlr GMEmrZ J V2 n For a circle i V2 27E rT2 J GMEIr 21 rT2 T2 4n2r3GME Cr3 T2 4 12 r3GME rREh6755km6755x105m T 2 Reh3 Re 9 5529 5 921 min Gravitational Potential Energy This is just mgh if we stay near the surface of the Earth h ltlt RE Further away the gravitational acceleration decreases as GMEmnZ decreases Move ds in gravitational eld F of Earth 11 F IS 2 Fi IS F dr M11711 A CIAEm 1 r2 r2 I I Integrate Gravitational Potential Energy Choose U0 Oat r RE U An example Escape velocity I In order for something moving away from the Earth to escape its gravitational pull it must eventually reach very large r where Uf O u If its velocity there is small it has the minimum required energy Kf 0 mm0 Escape velocity Ef Kf Uf 0 Ei Ui VZmViZ GMEmRE Er 0 vi ZGMEREV2 29 2 x 981 W52 X537X106 mv2 112 kms Chapter Four Newton s Laws Today39s lecture Kinds of forces Prof Chris Wiebe Prof Simon Capstick Fundamental There are onl fundamental Universe four types of orces In the Gravity The Electromagnetic Force electricity an magnetism The Stron Nuclear Force holds nuc ei together The Weak Nuclear Force nuclear decay This is amazing Everything can be broken down to only 4 types of interactions Forces neutron a gmvilaliun squot u alaclrornngnnliu Iona 9 c swung nuelear Imemuon a weak nuclear inleraclinn Electron and proton in hydrogen atom Quarks inside protons neutrons Electron Proton Neutrino Review from last lecture Newton s Laws First Law An object at rest remains at rest as long as no net force acts upon it Or An object moving with a constant v continues to move at a constant v as long as no net force acts upon it Second Law An object of mass m has an acceleration a equal to the net force ZF divided b the mass m The direction of a is the same direction as t e net force F ma Third Law For every action there is an equal and opposite reaction Free fall in an elevator The Normal force tells you about your apparent weight loss or gain What happens if the cord is cut and the elevator falls The Normal force zero This is a situation when there is free fall and the only force acting on you is mg downwards Whenever ou see a problem as ing for a free fall situation it means that the Normal force Ils zero ll V v 7 Tower of Terror ride at Disney Orlando when you drop your apparent weight also drops although you aren t quite in free fall g Practice PRS Force of c s wheels on the road Where is the equal and opposite force for a car moving at a constant velocity A There is no equal and opposite force in this case B Friction C The force of gravity on the car alone Answer B Friction If there was no friction then the wheels would just spin on the car and it would not move We will talk more about friction later Real PRS Two students of different masses are pulling towards each other on frictionless carts using a rope Which one of the following statements is true A The acceleration of the larger student is the same as the smaller B The acceleration of the smaller student is greater than the larger student C The acceleration of the larger student is greater than the smaller student Answer B This is from Newton s Third Law Kinds of Forces you will see in this course 1 Contact Forces Contact forces exist as action reaction forces between objects when hey touc Not one of the four fundamental forces gguestion You push on two objects M1 gt M which are touching each other ere is no friction in this problem Is the contact force i the same ii larger when ou us on the large box or iii larger when Kou push on the smaller box if you pus with the same force F Answer iiig larger when you push on the smaller ox This is an important application of Newton s second an third laws Let s see how they are used together g Example of contact forces F200N 12 11 500 kg Physical picture Box 1 F Box 2 i5 i5 Free b 0 dy diagrams 2 Let s say that M1 is 10 kg M is 5 kg and 0 N F1 is the contact force that 2 exerts on 1 F2 is the contact force that 1 exerts on 2 By Newton s 3rd law they are equal and opposite g Example of contact forces Now let s push with the same force on the smaller mass and calculate the contact forces 2 100 kg F 200 N 500 kg I E1 f2 Note F1 and F2 are larger in this scenario Equilibrium position 2 Spring Forces 1 Force constant k units Nm spring stiffness JY M WVVVV I For small oscillations x springs follow Hooke s xXo Law PX k Ax is negative because Ax is positive Fr F kAX Displacement kJf V vWMquot JJ x This law WOrkS for 1 1 IltAxispositivebecauseAxis negative sorts of things in the Fx real world including J interactions between quot LAX atoms in a solids 39 s which act like springs x0 3 Strings and cables Ta Raising a bucket I Strings always pull at constant v The magnitude of the force along a string is called the tension T For a pulley we assume that the tension T is the same on both sides of the pulley another way of saying this is that the pulley is massless Physical picture 3 Strings and cables Strategy for solving problems y y T Isolate each body 2 T11 2 Draw a free body diagram 7 is 3 Write down the T1 1 1T1 sum of the forces 1W in different direCtionS 2 F o T2 2T1 2 F o Tl W 4 Solve Forces acting Forces acting on the pulley on the bucket Putting it altogether with an example end of chapter 4 Problem 4 96 Practice Problems Chapter 4 29 49 57 63 69 77 Did this guy need a huge force to move these heavy things No he just needed an acceleration the force could be small Problem 4 96 System is accelerating What is the tension T Mass 1 ZFXT mlgsinem1a Mass 2 ZF ng T m2a Solve for a I a mz mlsinegm1m2 e x m1gCOSe al 171 ms2 ie down the inc ine x qlgsine ngl T m2g a 0863 N What is the time it takes to move 1 m Ax vot 12at2 Solve for t with v0O t V2Axa 108 s Chapter Eight Systems of Particles and Conservation of Linear Momentum Prof Chris Wiebe Prof Simon Capstick Objects in the real world move along trajectories determined by their center of mass PRS question Linear mo mentu m o A particle s linear momentum is defined as p mv units kg ms It is a vector 0 The change in momentum is important for many processes 0 Throw a ball and an eraser downwards with velocity vi They both have approximately the same mass 0 Which has the larger change in momentum after it hits the floor 0 A The ball 0 B The eraser o C The change in momentum is the same for the same objects 0 Answer A The ball since it bounces upwards and has a final velocity in the opposite direction Conservation of Linear Momentum 0 Consider a system of particles Mch m1V1 m2V2 2i mivi 2i pi Psys 0 Take the time derivative dPSysdt M dvcmdt M acm The center of mass acceleration is given only by the external forces on the system M acm 2i Fiext 0 So the momentum of a system is changed only by the sum of the external forces dP dt 2 Fiext sys Conservation of Linear Momentum 0 When the sum of the external forces on a system is zero the total momentum of the system does not change dPSYSdt Z Fiext 0 PSys 2 mivi Mvcm a constant vector This is a very important result the momentum of a system is conserved if no external forces act on a system This is the law of conservation of linear momentum Conservation of linear momentum 0 Cart demonstration 0 Problem 850 0 Use conservation of momentum Before collision not moving After collision both objects moving i vcanoe vgirl 25 ms 2mivi mgirlvgirl meanoevcanoe 0 55 kg25 msi 75 kg17canoe 0 183ms Collisions We need the concept of momentum to start looking at collisions Why Consider the following example A 10000 truck moving at 1 ms has the same kinetic energy as a 100 kg football player moving at 10 ms If these two objects collided however their resulting motion would be very different Sometimes we talk about a concept called impulse to describe collisions The impulse is equal to the change in momentum after a collision Ap p pi This is also equal to FaIverage At since FeIverage ApAt So the impulse tells us about the amount of force during a collision The concept of impulse o In any collision the force usually varies as a function of time so we use the average force in calculating the impulse 0 You can think about the impulse as the area under a force vs time graph for a collision the integral ofl ldp F dt A golf ball being hit by a club experiences a large force over a short period of time which gives the ball a final momentum Impulse Why are airbags used in cars Why do they have sandfilled pylons on roads to use for safety In a collision your momentum changes The force that you feel is determined by Faverage I If you increase the time for the collision with airbags or sand filled containers you will reduce the net force you feel Airbags increase the collision time and therefore decrease the average force felt 0 Impulse question 0 In the movie Superman Returns Superman deflects bullets off of his chest Suppose that he is sprayed with 3 g bullets at the rate of 1000 bulletsmin and the speed of each bullet is 500 ms Also assume that the bullets bounce back with no change in speed What is the average force on Superman s chest I Faverage Ap mvf vi mvivi 2mvi In 1 minute F ApAt 10002mvi60 5 Therefore F 50 N small average force This is because of the small mass of the bullet and it is an average force ONT O Collisions There are two kinds of collisions Elastic Kinetic energy is conserved Inelasic KE is not conserved KEi KEf How can you tell if a collision is elastic or Inelastic Most collisions are inelastic energy lost to deformation of objects heat sound etc When 2 objects stick together completey inelastic collision The max amount of KE is lost allowed by the law of conservation of linear momentum Collisions o f drop a ball and watch it bounce on the floor what kind of collision is this 0 How do I know 0 Airtrack demonstration of nearly elastic and inelastic collisions 0 Can we calculate the final velocities of the twomass system OOOONT O 0 Elastic collisions If a collision is elastic we can use two pieces of information Conservation of linear momentum Zpbefore Zpa er Conservation of energy Ei E Forthe two masses in 1D we have 2P before ZP a er m1V1i m2V2i m1V1f m2V2f Ei Er 12 m1V1i2 12 m2V2i2 12 m1V1f2 12 m2V2f2 You now have 2 equations 4 unknowns assuming you know the masses In order to proceed any further you need to know another piece of information such as the initial velocities If one of the masses is stationary say m2 then v i 0 If you have the initial velocity V1 then you have 2 equations 2 unknowns V1f v2f You can solve for v1f and v2f for homework V1f m139m2m1m2v1i V2f 2m1m1m2v1i l Elastic 1D collision v2i 0 Let s show this is true for airtrack Elastic collisions 0 Two more tidbits about elastic collisions o If you take the two equations from last slide conservation of momentum and conservation of energy you can show that in 1D V2f39V1f39V2i39V1i o This means that the speed of approach of two objects is the same as the speed of recession 0 Test this take two masses and collide them on the airtrack Set v2i O 0 After the collision the two masses move apart from each other with the same speed as V1 Coefficient of restitution 0 Most collisions are somewhere between elastic and perfectly inelastic o A measure of how elastic a collision is can be made with the coefficient of reStltUtlon e VrecessionVapproach Elastic collision Inelastic collision V2f39V1f39V2i39V1i e39V2r39V1fV2i V1ilt1 e v2fv1fv2i v1i 1 energy is lost Perfectly inelastic collision o For the perfectly inelastic collision can we calculate the final velocity of the twomass system 0 Cannot use conservation of energy unless they tell you how much energy is lost 0 You can always use conservation of linear momentum unless there are external forces 0 Zp before Zp after m1V1i m2V2im1 m2 Vf o vf m1v1i m2v2im1 m2 Let s show this for the perfectly inelastic collision on the airtrack 0 The ballistic pendulum o A more complicated Problem 877 example the ballistic pendulum MM 0 Imagine a projectile hits a 9 pendulum that was initially stationary and L causes it to move 05quot o If we have the mass of the bullet m the mass of the bob at the end of 1 Q the pendulum M and the final angle that the system rises to can can g u 0 we calculate the initial v 3 velocity of the bullet vb The ballistic pendulum momentum for collision 1 mvb m MV You can t use conservation of energy at the collision itself why But you can use it after the collision Ei Ef initial right after collision final at final height ummm 12 mMV2 Mmgh V Z US0 MmgL Loose M V 2gL1 cos6 Sub 2 Into 1 to get vb 1 lszl1 cowl 450 ms Use conservation of linear Lcos 2D or 3D collisions In 2D the collisions become more complicated The key is to break things down into components in the x y or 2 directions You can still use conservation of momentum conservation of energy if the collision is elastic or if they tell you how much energy is lost Note how the center of mass moves if there are no external forces the Psyste before the collision is equal1 to the PSystem after the collision miVi Vlvcm p sys constant vector COM motion Time lapse photography for a 2D collision on an airtrack An example Problem 97 Off center collision in 2D 0 Break up into X and y components 0 Momentum is conserved if there are no external forces 0 How do we know if energy is conserved Work it out for homework Xdirection Ydirection pxiszf pyizpyf Of 01 O O mv mv1 cos 30 mv2 cos 60 0 mvl gm 30 mvz gm 60 or or v 2 v1 cos 30 v2 cos 60 0 2 v1 sin 30 v2 sin 60 2 equations 2 unknowns solve for one substitute into the other Chapter 14 Oscillations I What keeps time 39fquot in a watch Prof Chris Wiebe Prof Simon Capstick The Simple Pendulum d2lgtdt2 gL sin For small angles 1 we have sin 39 1 try this on your calculator with angles set to radians For small oscillations around equilibrium d2lgtdt2 gL 1 Compare to d2Xdt2 km X 032 X Just simple harmonic motion with o gL 2 T 27 03 27 Lg 2 s mg cos q Note the period does NOT depend on the amplitude I this is why you can use them to measure time Damped oscillations not on nal Friction and other non conservative forces can take energy away from a system in SHM This decreases the amplitude of the vibration we call this damped motion Shocks on cars naturally dampen out vibrations of the vehicle An example of damped motion Shocks on cars dampen oscillations Driven oscillations and resonance not on nal but still pretty cool Why do wine glasses shatter with certain sound frequencies The vibrations from sound waves match the natural frequency of the glass If you cause an object to vibrate with an external vibration driven oscillations you can increase the amplitude if your frequency matches the natural frequency 55 quotQ quot of the object Other examples of this pushing a child on a swing resonances in musical instruments r172 Hz Q ml min Hz Q 50 Resonances in a guitar Sound vibrations can shatter glass if they are right frequency End of Chapter Skip sections 144 and 145 Review problems 23 25 27 29 31 33 35 37 39 41 43 45 47 49 51 57 61 63 65 97 119 neat PRS Question Which of the following curves represent the total j energy of a simple harmonic oscillator as a C function of position Answer C E a constant 12 kA2 VVavcs Prof Chris Wiebe Prof Simon Capstick 7 r a 391 7 112 Why can t someone hear you scream In space 9 sf a 3 I Traveling Waves i Waves can transmit energy and momentum without moving matter Example Ocean waves A transverse wave disturbance is perpendicular to motion of wave i l l sound waves electromagnetic WWW waves lig ht o 77l um Two kinds of waves 13 39 llllll Can be transverse l Water waves i Light electromagnetic radiation Can be longitudinal l Like sound waves compression waves in air A longitudinal wave compression is along the same direction as the motion of the wave Traveling waves Can we represent a pulse on a spring mathematically Suppose shape of pulse is y fx at t 0 f is called a wave function Pulse moves at speed v At time t pulse still has shape y fX Now X X v t Pulse moving toward X y fx vt Pulse moving toward x y fX vt i l x l l 439 o My m l39 y lll39 quotWinnill Speed of waves on a string How does the speed of a wave depend upon the tension linear density of the string massunit length symbol p Solve by dimensional analysis Speed v has units of ms Tension has units of N kg ms2 Linear density kgm Therefore v Taps Looking at the units we find that we need s1 on the t LHS so or 12 For the kg to cancel we Dave Matthews then need 3 12 So we then have demonstrates that the Tension on a string Tensionp speed of a wave is See pages 469 439 a or ew on S proportional to root of the tension and inversely proportional to the root of the linear density more of this later Laws approach

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