Physics for Engineers & Scientists I
Physics for Engineers & Scientists I PHY 2048
University of Central Florida
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This 36 page Class Notes was uploaded by Elmira Macejkovic on Thursday October 22, 2015. The Class Notes belongs to PHY 2048 at University of Central Florida taught by Saiful Khondaker in Fall. Since its upload, it has received 9 views. For similar materials see /class/227624/phy-2048-university-of-central-florida in Physics 2 at University of Central Florida.
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Date Created: 10/22/15
Recap 0 linear momentum 9 2 mV 0 Newton s Second Law 21 2 ma 2 m dv 2 d mv 51 13 dt dt dt 0 The momentum of an isolated system is conserved P1i quot39 32i P1r quot39 M o The impulse of the force F acting on a particle equals the change in the momentum of the particle f Apzpf pl 2L th21 0 Elastic collision momentum and kinetic energy are conserved o Inelastic collision kinetic energy is not conserved although momentum is still conserved o If the objects stick together after the collision it is a perfectly inelastic collision Recap 2D collision mmmmmmmmmmmmmm re b After the collision 2004 ThomsonBrooks Cola Xdirection ern errrCOS 6 m2V2rCOS ydirection 0 m1v1fsin 6 mzvzfsm The Center of Mass There is a special point in a system or r object called the center of mass that 3 moves as if all of the mass of the system is y concentrated at that point l The system will move as if an external force M were applied to a single particle of mass M V located at the center of mass m M is the total mass of the system y iing Iquot lt XCM L I quot32 m x m x b T l l 1 1 2 2 y L 1 9 7 XCM V CM m 4 9C2 gtI moon ImmsmBmoks Colo The center of mass is closer to the particle with the larger mass Center of Mass of a system of particles mlx1 1712x2 m3x3 me n For a system of n particles xCM m1m2 m3 m 2 mix 2 mix x i 139 CM 2 mi M where M is the total mass of the system 2 miyi Z miZi Similarly yCM 2T and ZCM 2T The center of mass can be located by its position vector rCIVI Emir r i CM M r is the position of the ith particle defined by r1 xilyiJZik Center of Mass of an extended object y An extended object can be considered a distribution of small mass elements Am Am The center of mass is located at position 2139 rCM CM r1quot 2 xiAml rCM xCM M Z If we let the number of elements approachmimntainity Z xiAml 1 W gigoTijdm 1 1 yCM Z j ydm ZCM ZEJ de 1 The center of mass of any symmetrical I CM E dm object lies on an axis of symmetry and on any plane of symmetry Center of Mass Rod Find the center of mass of a rod of mass M and length L lt L o The location is on the xaxis or yCM ZCM 0 O x 4 Let A be the linear mass density AML dx dmxtalx L 1 1 L 2 x2 2L2 xcm2 dem2 Ix dx M M 0 M 2 0 2M Since AML x L2 M cm 2M L 2 Now let us assume that the rod is non uniform and A varies linearly with x Adx L L L 3 xcm zijxdm zijxgdxzijmxdx Zgszdx aL M M 3M 39 A rod of length 300 cm has linear density massperlength given by 2 500 gm 200 X gmz where X is the distance from one end measured in meters a What is the mass of the rod b How farfrom the X 0 end is its center of mass Motion of a System of Particles We can describe the motion of the system in terms of the velocity and acceleration of the center of mass of the system We can also describe the momentum ofthe system and Newton s Second Law for the system Assume the total mass M of the system remains constant The velocity of the center of mass of a system of particles is 2mm drCM I VCM M o The momentum can be expressed as MVCM Zm 2P PM The total linear momentum ofthe system equals the total mass multiplied bythe velocity of the center of mass The acceleration of the center of mass aCM dVCM i 2m 3 dt M o The acceleration can be related to a force MaCM 2E Newton s Second Law for a System of Particles Since the only forces are external the net external force equals the total mass of the system multiplied by the acceleration ofthe center of mass EFeX M aCM The center of mass of a system of particles of combined mass M moves like an equivalent particle of mass Mwould move under the influence of the net external force on the system The total linear momentum of a system of particles is conserved if no net external force is acting on the system MVCM pm constant when EFeX 0 41 A ZOOkg particle has a velocity 200 3003 m s and a 300kg particle has a velocity Loo 600jm 5 Find a the velocity ofthe center of ma s and b the total momentum of the system Recap Ar 0 Instantaneous 0 Average velocnty V E velocity V V AV gcAC leerfgtieon a o Instantaneous 39 if ti At acceleration o Kinematic equations 17f 217 a7 ffZt7it12ampt2 o Projectile motion An object in projectile motion will follow a L parabolic path 4 V5 32 Ar dr 22357 mm dt Vi v It 6 ytant9ix x2 2v cos 9 mac mmmmmmmmmmm le Lecture 9 Agenda for Today Chapter 4 Motion in Two Dimensions Projectile motion The particle in uniform circular motion Tangential and radial acceleration Relative velocity and relative acceleration Analyzing Projectile Motion 0 Consider the motion as the superposition of the motions in the X and y directions The Xdirection has constant velocity ax O The ydirection is free fall ay g The actual position at any time is given by rf r vt 12th o The final position is the vector sum of the initial position the position resulting from the initial velocity and the position resulting from the acceleration q ig quot vr 7 x y Range and Maximum Height of a Projectile o The range and height of a projectile can be y expressed in terms of the initial velocity vector 1 0 2 v2 sinz a v 111 26 h g V R g 2 g h 9 o This is valid only for symmetric trajectory 0 i x R o The y component of the velocity is zero at the maximum height of the trajectory o The accleration stays the same throughout the trajectory At point A maximum height y 17 VyA 0 vyf vyl aytA gt t Ovl31n6 l gtA A g yf y vyit 12ayt2 Substituting for t gives h 2m smegvl s1n191 1 38 v1 sing 2 gt Range tB 2tA xA R R xf x1 vxitB 12axtl23 2v sin 61 R0vlcos6i 0 v2 sin2191 g R R is maximum when sin 26 1 gt 26 900 gt 61 450 The maximum range occurs at 49 45C mmmmm masonBrows Data 11 50 ms XIll NonSymmetric Projectile Motion Ex 44 A stone is thrown from the top of a building upward at an angle of 30 degree to the horizontal with an initial speed of 20 ms as shown in figure The height of the building is 450 m A How long does it take the stone to reach the ground B What is the speed of the stone just before it strikes the ground 0 Follow the general rules for projectile motion 0 Break the y direction into parts up and down or symmetrical back to initial height and then the rest of the height 0 May be nonsymmetric in other ways Li 1 2001115 n 0 A I i 450 m I x r l y 450111 39 b 2004 ThomsnnJEmoks Cole X pl 4 21 A soccer player kicks a rock horizontally off a 400m high cliff into a pool of water If the player hears the sound of the splash 300 s later what was the initial speed given to the rock Assume the speed of sound in air to be 343 ms Uniform Circular Motion 0 Uniform circular motion occurs when an object moves in a circular path with a constant speed c The change in the velocity vector is due to the change in direction 0 The vector diagram shows Av vf v I Centripetal Acceleration oThe acceleration always points toward the center of the circle of motion b oThe magnitude of the centripetal acceleration VZ vector is given by 2 A V V ac 4 7 V The period T is the time required for one f complete revolution 2 7T r T E V Tangential Acceleration 0 The tangential acceleration causes the change in the speed of the particle o The radial acceleration comes from a change in the direction of the velocity vector M o The tangential acceleration at d t o The radial acceleration a V o The total acceleration 2 2 Magnitude 61 ar at Total Acceleration ln Terms of Unit Vectors 0 Define the follownng unit vectors f and r lies along the radius vector 6 is tangent to the circle 0 The total acceleration is 431 Figure P435 represents the total acceleration of a particle moving clockwise in a circle of radius 250 m at a certain of time At this instant find a the radial acceleration b the speed of the particle and c its tangential acceleration A tire 0500 m in radius rotates at a constant rate of 200 revmin Find the speed and acceleration of a small stone lodged in the tread of the tire on its outer edge Relative Velocity 0 Two observers moving relative to each other generally do not agree on the outcome of an experiment 0 For example observers A and B below see different paths for the ball I Jsillwrli 7 39 muml S S 7 V0 Reference frame Sis stationary Reference frame 8 is moving at V0 This also means that 8 moves at V0 relative to 8 Define time t O as that time when the origins coincide The positions as seen from the two reference frames are related through the velocity r r vO t The derivative of the position equation will give the velocity equation v v vO These are called the Galilean transformation equations Acceleration in Different Frames of Reference 0 The derivative of the velocity equation will give the acceleration equation 0 The acceleration of the particle measured by an observer in one frame of reference is the same as that measured by any other observer moving at a constant velocity relative to the first frame 435 A river has a steady speed of 0500 mls A student swims upstream a distance of 10 km and swims back to starting point If the student can swim at a speed of 12 mls in still water how long does the trip take Com pare this with the time interval required for the trip if the water were still Recap 0 linear momentum p 2 mV 0 Newton s Second Law 21 ma mdv M dp dt dt dt 0 The momentum of an isolated system is conserved p1i p2i p1f p2f o The impulse ofthe force F acting on a particle equals the change in the momentum ofthe particle tf Ap pf pl Jtx th I 0 Elastic collision momentum and kinetic energy are conserved o Inelastic collision kinetic energy is not conserved although momentum is still conserved o lfthe objects stick together after the collision it is a perfectly inelastic collision 18 A railroad car of mass 250 X 104 kg is moving with a speed of 400 ms t collides and couples with three other coupled railroad cars each of the same mass as the single car and moving in the same direction with an initial speed of 200 ms a What is the speed of the four cars after the collision b How much mechanical energy is lost in the collision 25 A 120g wad of sticky clay is hurled horizontally at a 1009 wooden block initially at rest on a horizontal surface The clay sticks to the block After impact the block slides 750 m before coming to rest If the coefficient of friction between the block and the surface is 0650 what was the speed of the clay immediately before impact TwoDimensional Collisions The components of the momentum is conserved in each directions x y independently If the collision is elastic use conservation of kinetic energy as a second equation Particle 1 is moving at velocity v1 and Cl particle 2 is at rest Xdirection the initial momentum is m1v1 y direction the initial momentum is O After the collision the momentum in the Xdirection is m1v1fcos 6 mzvzf cos p the momentum in the y direction is m1v1fsin 6 mzvzfsin Xdirection ern errrCOS 6 m2V2rCOS b After the collision 2004 ThomsonBrows Cole y direction 0 m1v1fsin 6 mzvzfsin TwoDimensional Collision Example Before the collision the car has the total momentum in the X direction and the van has the total momentum in the y direction After the collision both have X and y components e mun momma we 29 Two shuffleboard disks of equal mass one orange and the other yellow are involved in an elastic glancing collision The yellow disk is initially at rest and is struck by the orange disk moving with a speed of 500 ms After the collision the orange disk moves along a direction that makes an angle of 370 with its initial direction of motion The velocities of the two disks are perpendicular after the collision Determine the final speed of each disk 33 A billiard ball moving at 500 ms strikes a stationary ball of the same mass After the collision the first ball moves at 433 ms at an angle of 300 with respect to the original line of motion Assuming an elastic collision and ignoring friction and rotational motion find the struck ball39s velocity after the collision Lecture 16 Chapter 6 Circular Motion and Other Applications of Newton s Laws Today s agenda 63 Motion in accelerated frames 64 Motion in the presence of resistive forces EUCF Dr Saiful l Khondaker Motion in accelerated frames Fictitious Forces Newton s second law is applied in a non inertial frame of reference a reference frame that is accelerating Inertial obscrvcr According to the inertial observer a the tension is the centripetal force mv2 The noninertial observer b sees r mv2 T F ctiti0us T r O Dr Saiful l Khondaker Fictitious Forces A small sphere of mass m is hung by a cord from the ceiling of a boxcar that is accelerating to thearight i u i i 7 it i t The inertial observer a sees Tf U nhst ru l ZszTsin zma a ZFszcos mgzO m V The noninertial observer b sees 22le Z T 511149 FfjctitiouS E ZF39yTcost9 mg0 u u 533391 minarwtmrg ii39i hi a 2006 Thanm am cor a9 Up Dr Saiful Khondaker 621 A 05 kg object is suspended from the ceiling of an accelerating boxcar as shown in Figure prev slide Taking a3 msZ find a the angle that the string makes with the vertical and b the tension in the string Motion with Resistive Forces Motion can be through a medium Either a liquid or a gas The medium exerts a resistive force R on an object moving through the medium and is opposite to the direction of motion The magnitude of R can depend on the speed in complex ways W nnnnnnnnnnnnnn a a We will discuss only two R is proportional to v Good approximation for slow motions or small objects R is proportional to v2 Good approximation for large objects Dr Saiful l Khondaker R Proportional To v The resistive force can be expressed as R b v 62 b depends on the property of the medium and on the shape and dimensions of the object The negative sign indicates R is in the opposite direction to v dquot 63 mg bvzmasz r 2 64 A V J Initially v O and dvdt Q As tincreases R increases and a decreases The acceleration approaches 0 when R mg At this point v approaches the terminal speed of the object U 117 60 21 2004 ThomsonBrooks o a Dr Saiful l Khondaker 27 A small piece of Styrofoam packing material is dropped from a height of 200 m above the ground Until it reaches terminal speed the magnitude of its acceleration is given by a g bv After falling 0500 m the Styrofoam effectively reaches terminal speed and then takes 500 s more to reach the ground a What is the value of the constant b b What is the acceleration at t 0 c What is the acceleration when the speed is 0150 ms Summary NonUniform Circular Motion 52F zFr th Motion in accelerated frames Fictitious Forces Resistive force mg Rzma Small object slow motion b v La rg e 0 hi ect R 12 DrA v2 mg a 0 v vT terminal speed Chapter 5 The Laws of Motion Force 0 Forces are what cause any change in the velocity of an object A force is that which causes an acceleration o The net force is the vector sum of all the forces acting on an object Also called total force resultant force or unbalanced force Zero Net Force 0 When the net force is equal to zero The acceleration is equal to zero The velocity is constant c Equilibrium occurs when the net force is equal to zero The object if at rest will remain at rest lfthe object is moving it will continue to move at a constant velocity Classes of Forces Contact forces involve physical contact between two objects l Field forces act through empty space 6 No physical contact is required Fundamental Forces l Gravitational force A 6 Between two objects l Electromagnetic forces 6 Between two charges Nuclear force 6 Between subatomic particles Weak forces 6Arise in certain radioactive decay processes Newton s First Law 0 If an object does not interact with other objects it is possible to identify a reference frame in which the object has zero acceleration 6This is also called the law ofinertia 6 It defines a special set of reference frames called inertial frames We call this an inertial frame of reference 0 A reference frame that moves with constant velocity relative to the distant stars is the best approximation of an inertial frame 6 We can consider the Earth to be such an inertial frame although it has a small centripetal acceleration associated with its motion Newton s First Law Alternative Statement In the absence of external forces when viewed from an inertial reference frame an object at rest remains at rest and an object in motion continues in motion with a constant velocity Newton s First Law describes what happens in the absence of a force Also tells us that when no force acts on an object the acceleration of the object is zero Inertia and Mass The tendency of an object to resist any attempt to change its velocity is called inertia Mass is that property of an object that specifies how much resistance an object exhibits to changes in its velocity Mass is an inherent property of an object Mass is independent ofthe object s surroundings Mass is independent ofthe method used to measure it Mass is a scalar quantity The SI unit of mass is kg Mass vs Weight Mass and weight are two different quantities Weight is equal to the magnitude of the gravitational force exerted on the object Weight will vary with location Newton s Second Law 0 When viewed from an inertial frame the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass 0 Algebraically 2F m a 0 BF is the net force This is the vector sum of all the forces acting on the object o Newton s Second Law can be expressed in terms of com ponents eZFX m ax Table 51 Units of Mass Acceleration and lT39orcea 2F m a y y System of Units Mass Acceleration Force SI ko 11182 N k0 11182 6 2F2 m 82 US cuchnnary 3115 L2 lb slang FL52 I N 022511 2004 ThornsonlEruols Cole 2 A force F applied to an object of mass m1 produces an acceleration of 300 msZ The same force applied to a second object of mass m2 produces an acceleration of 100 msZ a What is the value of the ratio m1m2 b If m1 and m2 are combined find their acceleration under the action of the force F 14 Three forces acting on an object are given by 1162 ooi 200N F2 sooi r 300N and F3 45 oiN TL 39 r 39 accelerati nn 39 5 Insz a What is the direction of the acceleration b What is the mass of the object c If the object is initially at rest what is its speed after 100 s d What are the velocity components of the object after 100 s Recap Forces are what cause any change in the velocity of an object A force is that which causes an acceleration When the net force is equal to zero The acceleration is equal to zero The velocity is constant In the absence of external forces when viewed from an inertial reference frame an object at rest remains at rest and an object in motion continues in motion with a constant velocity The tendency of an object to resist any attempt to change its velocity is called inertia Mass is that property of an object that specifies how much resistance an object exhibits to changes in its velocity o Newton s Second Law 2F rn a o in component form 0 Weight Fg mg Recap EFXrnaXY EFyrnay EFzrnaZ 14 Three forces acting on an object are given by 1162 ooi 200N F2 sooi r 300N and F3 45 oiN TL 39 r 39 accelerati nn 39 5 Insz a What is the direction of the acceleration b What is the mass of the object c If the object is initially at rest what is its speed after 100 s d What are the velocity components of the object after 100 s Gravitational Force The gravitational force F9 is the force that the earth exerts on an 0 j ct This force is directed toward the center of the earth lts magnitude is called the weight of the object Weight Fg mg Because it is dependent on g the weight varies with location 9 and therefore the weight is less at higher altitudes Weight is not an inherent property of the object Newton s Third Law 0 If two objects interact the force F12 exerted by object 1 on object 2 is equal in magnitude and opposite in direction to the force F21 exerted by object 2 on object 1 F12 39 F21 6Note on notation FAB is the force exerted by A on B n F1111 l 536 l l I I l l l l I 7 777 A Fg Flim mm 4 r n o m w m a m o m m l a e m m r ham mmmmmmmmm ale Free Body Diagram Newton s Third Law Alternative Statements Forces always occur in pairs A single isolated force cannot exist The action force is equal in magnitude to the reaction force and opposite in direction 6 One of the forces is the action force the other is the reaction force 6 It doesn t matter which is considered the action and which the reaction 6 The action and reaction forces must act on different objects and be of the same type Free Body Diagram 0 In a free body diagram you want the forces acting on a particular object 23 I z W a 7 l gtT Fg a b aquot Physical Freebody b picture diagram 1 Example 1 Equilibrium T o A lamp is suspended from a chain of negligible mass o The forces acting on the lamp are I the force of gravity Fg the tension in the chain T o Equilibrium gives g ZFy 0 gtT Fg 0 TFg Objects experiencing a net force 0 Forces acting on the crate 6 A tension the magnitude of force T 6 The gravitational force F9 6 The normal force n exerted by the floor ZFX Tmax ZFyzn FgO gtnFg If T is constant then a is constant and the kinematic equations can be used to more fully describe the motion of the crate b o 2001 Thomsnnl mcla Cola 17 The distance between two telephone poles is 500 m When a 100kg bird lands on the telephone wire midway between the poles the wire sags 0200 m Draw a freebody diagram of the bird How much tension does the bird produce in the wire Ignore the weight of the wire 500 m Lecture 8 Agenda for Today Chapter 4 Motion in Two Dimensions 0 The position velocity and acceleration vectors 0 Two dimensional motion with constant acceleration o Projectile motion Position and Displacement o The position of an object is described by its position vector r o The displacement of the object is if r defined as the change in its position Kka E Ar rf ri ri Path of particle Average VeIOCIty 17 DC 0 The average velocity is the ratio of 0 the displacement to the time interval for the displacement A 139 At The direction of the average velocity is the direction of the displacement vector Ar V Instantaneous Velocity 1 o The Instantaneous veIOCIty IS the limit x 1 Directionofvat of the average velocity as At quot approaches zero Ar dr Vhm 7 Al gtO o The direction of the instantaneous velocity vector at any point in a particle s path is along a line tangent to QMWMW the path at that point and in the direction of motion 0 The magnitude of the instantaneous velocity vector is the speed The speed is a scalar quantity Average Acceleration 0 The average acceleration of a particle as it moves is defined as the change in the instantaneous velocity vector divided by the time interval during which that change occurs Vf V AV 3 If t At 0 As a particle moves Av can be found in different ways 0 The average acceleration is a vector quantity directed along Av a 2004 tmmsawamole Dela Instantaneous Acceleration 0 The instantaneous acceleration is the AV dV limit of the average acceleration as a 11111 A 7 AvAz approaches zero At gt0 t t Kinematic Equations for TwoDimensional Motion 0 For constant acceleration 2D equation of motion is similar to those of onedimensional kinematics 39 39 gt 7 d dx A A A A POSItlon r Xl y Velocity V 2 2 1 J Vx1vy dt dt dt a constant ax and ay are also constant We can use 1D kinematic equations vxf vxi axt and vyf vyl ayt at W at 13 17 vxiaxti vyiaytj inivyijaxiayjt I 7 i Ux a e 2904 nunsunBmols Cale Similarly 2 2 xf 2x1 vxit12axt andyf yl vyit12ayt 77f xfiyfj x1 vxit 12 elf yl vylt 12 aytz A A A A A A 2 xl1ylvm1vylt12ax1ayt Ffi17it12a t2 T 1 2 Eat F f is the vector sum of W i the original position i 39 Vquot ii displacement 71 arising from the initial velocity iii and a displacement 1255t2 resulting from the constant lt xi gt acceleration 1 2 lt x gtlt uwl Eaxl mmmmmmmmmmmmmmmm ls Kinematic Equations Component Equations 0 vf v at becomes vxf vx at and vyf vy ayt o r r V 1 2 at2 becomes 1 2 Xf X vx 1 2 at and yf y vy 1 2 ay 6 The vector position of a particle varies in time according to the expression r 300i 7 60012jm a Find expressions for the velocity and acceleration as functions of time b Determine the particle39s position and velocity at t 100 s 712m ms dv 39 d r A aim 7120f1 712 0 msZ d dr Hi n 7 1 a Ve el jioool 50m b 1300 7a00 m v7120i ms Projectile Motion 0 projectile motion object may move in both the X and ydirections simultaneously 0 The freefall acceleration g is constant over the range of motion and is directed downward o The effect of airfriction is negligible y 0 Reference frame chosen y is vertical with upward positive 0 Acceleration components ay g and aX O t0 xiyi0 velocity v and angle 91 vxi 2 vi cos 9 and vyl 2 vi sm 9 wwwwwwwww t0 x 2y 0 vxi vl 0056 andvyl 2 vi sinBl xf x1 vxit12axt2 0vl cosHltO x x 2 v1 cosHit gt t v1 cosHi yf yl vylt12ayt2 0vl sinBlt 12gt2 Substituting for t gives X X vsint9 12 2 y I Vicosgi gvicosgi L 2 y taHQM ZVZZCOSWJX Olt6i lt 7H2 This is in the form of y ax bx2 which is the standard form of a parabola An object in projectile motion will follow a parabolic path
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