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# ENGMECHANICSDYNAMICS MEAM211

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This 169 page Class Notes was uploaded by Toby Hoeger on Monday September 28, 2015. The Class Notes belongs to MEAM211 at University of Pennsylvania taught by Staff in Fall. Since its upload, it has received 34 views. For similar materials see /class/215361/meam211-university-of-pennsylvania in Mechanical Engineering at University of Pennsylvania.

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Summary Newton s laws Beyond Force Balance a Force balance a Moment balance a Linear momentum N I mass times velocity of center ofmass FZF m as dtG ZMG u Angular momentum 1 recap 74 M2 Work a Energy k M ii i i mumquot Angular momentum about 0 and G Angular momentum about 0 r Angular momentum about G Angular momentum of particle ofmass m at G Kinetic Energy ofRigid Bodies undergoing Planar Motion D Kinetic energy ofin nitesimal element dm omegmtvm W n Kinetic energy ofthe body kinetic energy miss 7 mi menu 15 muting abmn tire centemrmass cemexafmass mum Example 1 Determlne the angular yeluelty quhe mu alterthe halls haye muyeu tu A andB39 letus rst an apmblEm m ealeulaung the angular mumentum ufa system Twu sullu spheres ramus t 3 m mass m W Z ll are man u 7 ea un a G Spmmng hunzuntal mu R wth angularveluclty u a rausee as shuwn Themass mument uflnema uf 0 rod A the ma ls 1K The halls are helutugether hy astnng whlehls suuuenly eut fl 0251h s2 Kiawm Amu Angularmomentum ofa system ofrigid bodies G Angularmomentum about 0 ma Angularmumentum ufmd ahuut a Angular mumentum ufpamcle uf massufthemuat a t Angular mumentum ufsphere abuutA Angular mumentum ufpamcle uf mass ufthe sphere atA u Furcebalance u Llnearmumentum 2 N 1721 7mL VG erIFdrvatZ vatl tel d n u Mumenthalanee dlI d u Angular mumentum 2 G MG A1517 IMGdtHGtZ HGtl 1 ur fur a xedpamt 0 ur fur a md pm 0 M d 2 A1014 IMO HOtZ HOtl 1 mmm mmm LinearAngular Impulse and Momentum Example 1 Determlne the angular veluclty ufthe ruu alterthe halls haye muyeutu A and 539 Key ohseryauun 7 Nune ufthe external fumes pruuuee a Twu suhu spheres radlus t 3 m mumem shunt may ms the angular mass m W 2 lb are muunteu un a mumemum ls Bunserved splnnmg hunzuntal mu R wth angular valuch u a radsec as shuwn Them s mu ent uflnema The angular mumentum eunslsts quhe ufthe mu ls 1K The halls are held angular mumenta quhe spheres and the tugetherhyastnngwhrehls larmumentum quhemd suuue y eut f 70251b 52 Wm rmymmms quot5 P kdky snhx minuupmay My an mum Aath my a m zv39vwrwu WWW memwmrmg Mrm xwxy Ma 39MuxauXF Emusz sysm I AWImmmnw mLewawam mmvmm mm mmm mmm m 7mm Work am bums ma mmms w amwnmummgnp any a mmm Work done for a rigid body xiiw mmzu Example 2 A sphere cylinder andhoop each havingthe same mass andradws are om re an no me Detamme the valency ofeach body a er n has rolled through a dutance conespond ng to a change ofelevauonh ng mmzu Example 3 rrassm releasedvmhzemvelumy am mate g mums MEAM 211 Synthesis of Simple Planar Linkages Professor Vijay Kumar Department of Mechanical Engineering and Applied Mechanics University of Pennsylvania January 15 2006 1 Introduction Planar linkages are used in a wide range of applications for transforming a given input motion into a desired output motion for transforming a speci ed input force or torque into a desired output force or torque or for guiding the motion of a rigid body or a speci c point on a rigid body through a desired path or trajectory For example the linkage of a casement window transforms rotary motion into the desired motion of a window sash so it can be moved from a fully open to a fully closed position The damper on a door offers a resistance to linear motion that is transformed into resistance to rotary motion The slider crank mechanism transforms the force from the high pressure gases inside a closed cylinder into a torque on a crankshaft in an internal combustion engine The rotary motion of an input crank is converted into the rocking motion that is suitable for a lawn sprinkler We will consider the special class of linkage design problems associated with rigid body guidance and solve this class of problems using four bar linkages The design problem is stated as follows Problem 1 Giuen n successiue positions and orientations of a rigid body synthesize a planar four bar linkage such that its coupler link assumes these n positions and orientations during the course of the crank motion 2 Two position synthesis The two position synthesis problem is the following design problem Problem 2 Giuen two positions and orientations of a coupler link synthesize a planar four bar linkage such that its coupler assumes these two positions and orientations during the course of the motion of the linkage The design procedure for this problem is best described graphically Consider the rigid body guidance problem in Figure 1 The design procedure is as follows 1 Select suitable candidates for points Q and P on the coupler Figure 2 Mark the locations of the corresponding points Q1 and P1 in Position 1 and Q2 and P2 in Position 2 Note that this step involves choosing two points and should be guided by practical considerations to Draw the line segments 62ng and P1P2 Draw the perpendicular bisector of each line segment Figure 3 Call these perpendicular bisectors L042 and L312 respectively 03 The center point or the xed pivot7 07 must lie on the line L042 why7 Similarly7 the center point R7 must lie on the line L312 See Figure 4 Chose O and R on the lines L042 and L312 respectively Once again this choice might need to be guided by practical considerations Once these points are chosen7 the four bar linkage is uniquely determined See Figure 5 Position 1 Position 2 Figure 1 Rigid body guidance through two speci ed positions Circle points on coupler Coupler link Figure 2 The choice of two circle points7 P and Q on a coupler for rigid body guidance 3 Three position synthesis The three position synthesis problem is the following design problem Problem 3 Giuen three positions and orientations of a coupler link synthesize a planar four bar linkage such that its coupler assumes these three positions and orientations during the course of the motion of the linkage Figure 4 The center points or the xed pivots7 O and R7 must lie on the perpendicular bisectors of the line segments 62ng and P1P2 respectively Thus 0 and R7 can be chosen to lie anywhere on the lines I Position 1 Position 2 Figure 3 The locations of the circle points7 P and Q at the two speci ed positions Positionl f Posi onz 0 R L012 rL R 12 L042 and L312 respectively Once again we will solve this problem graphically Consider the rigid body guidance problem in Figure 6 The design procedure is as follows 1 Select suitable candidates for points Q and P on the coupler Figure 2 Mark the locations of the to 03 F Draw the line segments Q2ng and Q2623 Draw the line segments P1P2 and P2P3 corresponding points Q1 and P1 in Position 17 Q2 and P2 in Position 2 and Q3 and P3 in Position 3 Figure 7 Call these perpendicular bisectors L042 and L033 respectively The center point or the xed pivot7 07 must lie on the intersection of lines L042 and L023 Figure 7 Call these perpendicular bisectors LRJZ and L323 respectively The center point or the xed pivot7 R7 must lie on the intersection of lines L342 and L323 Once the xed pivots O and R are determined7 the four bar linkage is uniquely determined See Figure 8 Draw the perpendicular bisector of each line segment Draw the perpendicular bisector of each line segment Figure 5 Once the location of O and R on the lines L042 and L312 is chosen7 the four bar linkage is uniquely determined Position 2 Position 1 Position 3 Figure 6 Rigid body guidance through three speci ed positions Position 2 Position 3 Figure 7 The xed pivot 0 must lie on the perpendicular bisectors of the line segments 62ng and Q2623 Similarly the xed pivot R must lie on the perpendicular bisectors of the line segments P1P2 P2P3 Thus the intersection of L042 and L033 determines the location of O and the intersection of L312 and L323 determines the location of R Figure 8 The choice of P and Q leads to the unique determination of O and R and the four bar linkage OQPR amp i MEAM 211 Four Bar Linkages 1 Types of linkages 2 Synthesis Design MEAM 211 Four Bar L1nkage El Simplest dof linkage El Used very commonly Types of four bar linkages Grashof T e l o Crank rocker s I lt1 q R 777 o Doublerocker o Draglink s smallest l1nk length NonGrashof or Grashof TVpe H l longest link length P q 2 other two lengths o Triplerocker s lgtp q m amp 1533 MEAM 211 Facts El Grashof linkages 0 There exists one link that completes a full 360 degree rotation relative to another link 0 This is the shortest link 0 The pair of joints connecting this link to other links are such that they rotate through 360 degrees 0 Three types of linkages gt Crank rotates through 360 deg relative to frame gt Coupler rotates through 360 deg relative to frame gt Frame rotates through 360 deg relative to let us say crank El Non Grashof linkages o No link completes a full 360 degree rotation relative to another link Peml erxrk39u ng Ega n MEAM 211 Geometric significance of Grashof Inequality I5 u39 39 e u 39 I 39 3 r I h o 39 o o u 0 e I o h o I y I I a I n e n 39 39 G I 39 l I 3 39 39 MEAM 211 quotu MEAM 211 I Crank Rocker slltpq Crank is the shortest link Position 2 I Path of the moving i pivot on quot the crank Position 1 ampa n OPEN CLOSED OPEN C OSED MEAM 211 Synthesis PROBLEM DOOR IN OLD DESIGN SWINGS OUT NEED DOOR IN NEW DESIGN SWINGS OUT 40 LESS amp h MEAM 211 E H 33975quot Door swing area 3925 Open position 1 Closed I 3 0 a iosoquot 5 i Ii 225quot 3925 395 25 E P n Lrum m MEAM 211 TwoPosition Synthesis of Linkages Find a four bar linkage whose coupler when attached to a speci ed rigid body guides the rigid body between two given positions and orientations Position 1 Position 2 ampan MEAM 211 Circle points on coupler Coupler link ampa n MEAM 211 Position 1 Position 2 auxmg MEAM 211 I Position 1 Position 2 0 393 f R L012 quot LR 12 ampa 3 MEAM 211 41 39 g Penn rumTm MEAM 211 ThreePosition Synthesis of Linkages Position 2 Position 1 Position 3 Penn rumTans MEAM 211 Position 2 Position 1 L 0 23 L0 12 LR 2 Posmon 3 PPeml erxrk39un39ng MEAM 211 Egg3 nz MEAM 211 Mechanism Any device that transforms motion 0 Ampli cation 0 Change type of motion gt rotation lt gt translation gt uniform lt gt reciprocatingoscillating Sprocket and Chain connecting Cam piston follower crank slider n Slider crank linkage an amt MEAM 2 1 1 I Go 211 El Understand kinematics of mechanisms 0 Focus on motion El Apply particle kinematics 0 Focus on motions of particles attached to parts of mechanisms 0 Interested in constrained motion Note TecMcally machines are deVices that transmit or modify energyforces Mechanisms transmitmodify motion FEga3nHz MEAM 211 Position Vectors Position Vector s Triangle law of vector addition gx Note No need for a x y z coordinate system amp h MEAM 211 Velocity and Acceleration Vectors What do these vectors really mean 1 r 12 d d d Er EI39 Er 111 dt dt J dt dt dt dt 399 MEAM 211 Differentiation of Vectors Assume o All coordinate systems are parallel for now 0 O is xed to an inertial frame Justified because we are dealin with particles and not rigid bodies 4 x Emn MEAM 211 3 Velocity and Acceleration Vectors 0 is fixed to an inertial frame d d d Er Erar d d d d d d EErj 11 1 Why have dropped the sub script as 0 from V A 0 And erw39l39mz 4K 0 from VB 0 MEAM 211 Constraints Constraints on the position configuration of a particle or system of particles 0 Particle in 3D but constrained to lie on a plane AxByCzD0 o A particle suspended from a string in three dimensional space x Ut2 y b2 z c2 r2 0 o A particle on spinning platter carousel x a cos03t 1 y a sin03t 1 Egg3n I w 200 lbs American Heritage Emn MEAM 211 Pulleys cables blocks and tackles Block and Tackle assembly El An assembly of blocks pulleys and cables that allow a single cable to hoist a large weight or apply a large force httpwwwjimloycomcindyb10ckhtm MEAM 211 Example If she moves to the left at 2 msec how fast does the mass m move Ie What is the velocity of the Exercise 2510 p 85 mass Egg3 nz E MEAM 211 xample A is pulled down at 3 ms What is the motion of B Exercise 2511 p 85 Ega n MEAM 211 Four Bar Linkages coupler follower frame F ampa n MEAM 211 rBA zncos6i1ris1n6ij amn MEAM 211 y F Ega n MEAM 211 mm vs MEAM 211 MEAM 21 1 httpwwwseasupennedumeam211 Introduction Professor Vij ay Kumar httpwwwseasupennedukumar E 2 MEAM 211 What is Dynamics The Big Picture Three parts El Geometry Shaes curves Types of Problems 0 Describe motion 0 Given motion what forces caused motion 0 Given forces predict motion mamg MEAM 211 What is Dynamics And why should we study it Motivating Examples 0 Cruise missile 0 Internal combustion engine 0 Mechanisms h pMautohowstuffworks comengineS htm R mmmi MEAM 211 Mechanisms httpwww hrnckeno um 39 indwrhfm 39 httgwww krevcom I ht 3 F ALLEZ CHJKMME WRU ELLIEH 539 mm u olleminn Emlcpnlylcchniquu htt vvvvvvornitho terdeen hshcrankhtm I MEAM211 Mechanisms iThei Papto graph Mechanism niml ul um Mapch Suspmsinn chiclu Mm 0 mm on 1 mu m m mane hudy m l l x m m mklc pm I mm x movcs nn 1 mm mm 1 hnlimnmi mum1 x 1E7 mm 1 along a hmzunr path me MEAM 211 Mechanisms Oscillating Water Sprinkler rocker Planetary Gear 500 1 reduction MEAM 211 Headcontrolled Feeding Device for Quadriplegics El Head motion controls a passive mechanical feeder MEAM 211 Dynamics Historical Perspective El Chapter I O Newtonian mechanics gt Kepler planetary motion gt Galileo importance of acceleration gt Newton Principia Mathematica 1687 100 years later mmmg MEAM 211 Historical Perspective Continued El Chapter II o Lagrangian mechanics gt Bernoulli principle of Virtual work statics gt Euler dynamics of rotating rigid bodies gt D Alembert extension of Virtual work to dynamics gt Lagrange Mechanique Analytique 1788 Analytical mechanics gt Analytical because it is based on a few fundamental principles gt Lagrange describes it as an approach which does not require drawing any diagrams e g free body diagrams ulnarm MEAM 211 Focus in MEAM 21 l El Newtonian mechanics El No relativistic effects no quantum mechanics 0 Reasonable velocities o Reasonable length scales El Practical problems El Analysis and Design El Projects linked to MEAM 247 Minxmg MEAM 211 Lectures and Recitation Lectures Instructor 0 New material Dr Vijay Kumar 0 Problem solving kumarmeupennedu Recitation 1 Forum for solving problems 2 Computer AidedEngineering labs Teaching Assistants Spring Berman Nathan Michael Jimmy Sastra meamZ 1 1seasupennedu Graders Olga Epshteyn Megan Arensen allrun E MEAM 211 xample from last lecture A is pulled down at 3 ms What is the motion of B Exercise 2511 p 85 Ega n MEAM 211 I KEY Identify 3 particles Let rl be the radius of pulley i b constraints There are 3 particles and 2 cables Whose lengths cannot change x1 a7cr1 x2 a7cr2 x2 2 L1 x1 x3 b xz7cr3 x3 b L2 Differentiating these constraints we obtain uu39K39unng MEAM 211 Discussion on linkages and mechanisms will resume on Wednesday MEAM 211 Dynamics of Particles T8 Chapter 3 Three Keys El Calculate acceleration andor force 0 Cartesian coordinates 0 Polar coordinates 0 Path coordinates El FreeBody Diagram FBD and Inertia Response Diagram 1RD El Force balance 0 Total resultant force mass Acceleration Fma in the appropriate coordinate system Egg3n MEAM 211 I Summary Transformations between unit vectors El Understanding the relationship between sets of unit vectors is very important 0 Visualize 0 Write down the dot products e1 e2 e3 ampaa MEAM 211 I Three Steps El Find the force andor acceleration of the particle in appropriate coordinate system 0 You may have to use variables for unknown quantities El Draw the FreeBody Diagram FBD and the Inertia Response Diagram 1RD o FBD must show all force vectors no components and the particle and nothing else 0 1RD must show mass times acceleration vector no components and the particle and nothing else El Commit to a suitable coordinate system 0 Redraw FBDIRD with components 0 Write force balance with components 39395 Penn ruirK39l39rinz MEAM 211 I Example 1 The carousel rotates about an axis perpendicular to the plane of this paper passing through 0 at a constant rate of 115 rotations per minute 12 radssec Your friend mass 50 kgs is walking radially outward as seen by someone fixed to the carousel with a speed of 2 mph 089 meterssecond at the point shown OP 55 m What are the forces acting on your friend s foot El Find the acceleration andor acceleration of the particle in appropriate coordinate system El Draw the FreeBody Diagram FBD and the Inertia Response Diagram 1RD El Commit to a suitable coordinate system 0 Redraw FBDIRD with components 0 Write force balance with components MEAM 211 Example 2 A r39 A C m quot79 FBD 52 57 Z 77lt57 T Z B 9 lt9 m 0 o gt L A M xv rva 1RD amp 15911 MEAM 211 Example 3 A car is climbing a hill and accelerating at 03 g ltgt10 deg What is the steadystate value of 9 MEAM 211 Example 36 Acceleration for a weird trajectory Calculate the forces on the payload if the arm follows a path given by rt at1 azt a3t2 9t b1t Ith Problem 3310 Speed 2 20 mph Weight 2 3600 lbs Radius of curvature 50 ft lquot Four springs 800 lbin spring constant W What is the compressionextension on each spring ampa n MEAM211 I Problem 3113 Two Masses Given mA 2 28kg 1713 28kg 39g 36A 713ms2 l Equations of Motion m 3 3T m g F X t I B IHBJ B 1111 21 B Constraint Equation 3x A 2xB constant m amp 1533 HE MEAM 2 1 1 I Kinetics of Rigid Planar Bodies m 48 gagging MEAM211 I Types of motion Rectilinear translation Curvrlinear translation a 1 General planar motion Rotation abou a fixed point C d mPenn L6 Elam MEAM 21 1 Kinetics of a System of Particles The center of mass for a system of particles accelerates in an inertial frame as if it were a single particle With mass m equal to the total mass of the system acted upon by a force equal to the net external force N d V F 2 F1 2 m r m G il dt The rate Of Change Of angular The rate of change of angular mgme tum fa16 WStem about momentum of the system about 3 leed 9011110 15 equal to the the center of mass G is equal resultant 1110111th 9f all to the resultant moment of all eXtemal forces aCtmg 0 the external forces acting on the system abOUt 0 system about G d H d H O M G M G 0 dl MEAM 2 1 1 I Rectilinear Translation a dHG dl Peg mym K MEAM 21 1 Rectilinear Translation The rigid body can be treated as a single particle of mass m at the CM a dHG dl No rotation gt No change in angular gt No net moment momentum about CM m Engaging NIEAM211 I Curvilinear Translation b dHG dz Pwm K MEAM 21 1 CurVilinear Translation The rigid body can be treated as a single particle of mass m at the CM No rotation gt No change in angular gt No net moment momentum about CM 5 Peg mm Pwm t I NIEAM211 I Rotation about a xed point fixed point 0 N MEAM 2 1 1 I General Planar Motion dHG dz d Pwm K MEAM 211 Angular Momentum of a Rigid Body about its Center of Mass Recall oExpression for angular momentum about 0 J rdmo de b2 bl dm body V 39 quotdinG A similar derivation 0 9H 1 a O 0 IO oExpression for angular momentum about G 9H6 IGco G rdmGYdm body 1G usually available in tables it is a property of the rigid body unlike IO which also depends on the choice of 0 Penn Eugenwing MEAM 2 1 1 I General Planar Motion M dz G d HG IGOJZIGeb3 Mass An 1 net moment of gu at an al moment acceleration extern of inertia forces about G about G Peg mym i MEAM 211 General Planar Motion for a Rigid Body El Force balance G 2 rd de N V 39 body F E m rdmG G i1 El Moment balance d H G dt ZMG M2 1G Xb3 ma b2 b1 dm M 1 2 b1 dm G V W rdInG r G G F 1 JFz m MEAM 2 1 1 Example 1 V0 C A uniform sphere of mass m and radius r is projected along a rough horizontal surface with a linear velocity v0 The coefficient of kinetic friction between the sphere and the surface is yk Determine a the time t1 at which the sphere will start rolling without sliding and b the linear and angular velocities of the sphere at time t1 Peim SOLUTION l 2 Draw the FBD with external including reaction forces on the sphere and the 1RD and write the three scalar equations Solve the three corresponding scalar equilibrium equations for the normal reaction from the surface and the linear and angular accelerations of the sphere Apply the kinematic relations for uniformly accelerated motion to determine the time at which the tangential velocity of the sphere at the surface is zero ie when the sphere stops sliding Engnm39nng Example 2 400 mm E 00289kgm2 mm 3 kg g The portion AOB of the mechanism is actuated by gear D and at the instant shown has a clockwise angular velocity of 8 rads and a counterclockwise angular acceleration of 40 rads2 Determine a tangential force exerted by gear D and b components of the reaction at shaft 0 MEAM 211 SOLUTION 1 Draw the FBD with external including reaction forces on A03 and the 1RD and write the three scalar equations 2 Evaluate the external forces due to the weights of gear E and arm OB and the effective forces associated with the angular velocity and acceleration 3 Solve the three scalar equations derived from the freebodyequation for the tangential force atA and the horizontal and vertical components of reaction at shaft 0 j m Penn 4 immian Mass moment of inertia about any point 0 Mass moment of inertia about center of mass G MEAM 2 1 1 I How to calculate mass moments of inertia Fundamental Result Parallel Axis Theorem 0 G mrdmG2 Mass moment of inertia about 0 of a single particle of mass m concentrated at the center of mass G m Penn 49gt Engnnnng MEAM 2 1 1 Free vectors and bound vectors Bound vectors Free vectors 0 Characterized by an axis 0 Not characterized by an 0 Wrong to associate with a 3X15 single point 0 But referenced to a point El Examples El Examples 0 Angular velocity of a rigid Linear velocity of a point P body characterized by an axis Moment about a point P 0 Force applied to a rigid body tied to a line of action of the force In general VP VQ MP MQ m Penn A Engnnnng MEAM 21 1 Relating moments about two different points MO 2MP er gtltF rPO XF MQ 2M0rOQgtltF 0r MQ 2MP rPQszrPQXF un MEAM 211 y MEAM 211 P MEAM 211 Definitions 0 Degrees of freedom of a system The number of independent variables or coordinates required to completely specify the configuration of the system gt 130th 0 a plane 2 o 2 planar links connected by a pin joint 4 gt Point in 3D space 3 Human shoulder gt Line on a plane 4 9 Car 0 Kinematic chain A system of rigid bodies connected together by joints A chain is called closed if it forms a closed loop A chain that is not closed is called an open chain Serial chain If each link of an open chain except the first and the last link is connected to two other links it is called a serial chain MEAM 211 Definitions continued Joints Joints are connections between links gt Revolute rotary or pin joint R gt Prismatic sliding or telescoping joint P gt Helical or screw joint H gt Spherical or ball joint S Planar kinematic chain All the links are constrained to move in or parallel to the same plane A planar chain can only allow prismatic and revolute joints In fact the axes of the revolute joints must be perpendicular to the plane of the chain while the axes of the prismatic joints must be parallel to or lie in the plane of the chain Connectivity of a joint The number of degrees of freedom of a rigid body connected to a fixed rigid body through the joint El ampnni MEAM 211 Degrees of Freedom and Constraints Number of Degrees of Freedom for planar mechanisms 3 number of moving rigid bodies Constraints due to jointsconnections E1 im MEAM 211 The Planar 3R manipulator o Planar kinematic chain 0 All joints are revolute with connectivity 1 o What is the number of degrees of freedom ENDEFFECTOR Link 3 Joint 3 Link 1 ACTUATORS Joint 1 P 3ampnni MEAM 211 Connectivity Mobility and Degrees of Freedom El Connectivity of a joint The number of degrees of freedom of a rigid body connected to a xed rigid body through the joint gt prismatic joint 1 o spherical joint 3 gt revolute joint 1 o helical joint 1 El Mobility of a chain Number of degrees of freedom of the chain 0 Serial chain gt Examples do 3R chain El M P 3ampnnt MEAM 2 1 1 I Examples Mobility Degrees of Freedom The Adept 1850 Palletizer The Bckman Coulter ORCA M 4 Five degrees of freedom J MEAM 211 Connectivity Mobility and Degrees of Freedom 9 Mobility of a chain Number of degrees of freedom of the chain 0 General expression 1 M 3n J 1Zf il n number of links j number of joints between the links f connectivity of joint 139 32u MEAM 211 Most Commonly Encountered Linkages follower coupler connecting piston crank slider crank 4 frame Fourbar linkage Slider crank linkage Pb3amp nz I MEAM 211 Closed Chain Mechanisms j M3n j 1ZJ i1 5 4 MEAM211 The Pantograph Mechanism MEAM 211 Examples Mobility Degrees of Freedom DPlanar serial chain DPlanar parallel manipulator 0 number of links n 4 0 number of links n 8 0 number of joints j 3 0 number of joints j 9 o connectivity f l a connectivity f l 4 ENDEFFECTOR ENDEFFECTOR lt ACT UATORS M 3nJ391i il P 3ampnnt MEAM 211 Four Bar Linkage El n4 El j4 El M4 Types of four bar linkages Grashof Type I o Crank rocker s I lt1 q o Doublerocker o Draglink NonGrashof or Grashof TVpe II o Tn39plerocker Eltg2 ng MEAM 211 quotu E l ampnnz MEAM 211 Useful Websites El httpWwwpowertransmissioncom O Gearing El httpWwwhowstuffworkscom 0 General particularly good for automotive mechanisms El httpwww vingpigcoukmechanisms 0 Tutorial on mechanisms El httpwwwbtinternetcomhognosesamgcsepageZhtml 0 Tutorial on mechanisms 3 Elta1ampm MEAM 211 Mobility of 3dimensional mechanisms El Ingersoll Rand machine tool Stewart Platform 0 number of links n 20 0 number of joints j 24 o connectivity f l or 3 a 4 M6n j 1if i1 M619 24366 E Pb3amp nz hrm WWW hi tn I MEAM 211 Double Acting Steam Engine Immm Iva29 m mammal Sum rnr he fer Izs as High pmasm 3153mm Emma 513mm http travel howstuffworks com steam 1 htm J a p m MEAM 211 l Six Degreeof Freedom Manipulator Micromanipulator PI Hexapod Six degrees of freedom Pai im MEAM 21 1 Dynamics of Particles Beyond FBDIRD What else can we learn from Newton s 2nd and 3rd laws 0 Linear momentum and Linear impulse o Angular momentum and Angular impulse 0 Application gt Impacts between particles T8 Chapter 3 m Penn 49gt Engnm39nng MEAM 21 1 Linear Momentum El Newton s 2 Law F 2 ma total forcx F GD 1 4 linear momentum F dL L m V dr El Linear Impulse F 2 ma tJZFdr tJZmadt Ft L112 ImadtzLZ L1 quotw LI 1 t1 11 MW IS a vector t 2 t time I 1 Pem1 tngnnnng MEAM 2 1 1 Angular Momentum El Newton s 2nd Law t0tal forcx GD P D r 1 ma 1 x F 1 x ma Moment of the force 9 DEFINE about 0 angular momentum M Mozer H0rxmv 0 N0te really I PO MEAM 2 1 1 Angular Impulse quotquotquot J g P time t2 L112 Imadt 2L2 L1 t1 t2 1412 JModt H0 2H0t1 t1 m Penn 49gt Emmimg MEAM 21 1 Principles of Conservation of Momentum El Conservation of Linear Momentum t2 L112 Imadt L2 L1 11 If LI 2 0 L2 2 L1 El Conservation of Angular Momentum t2 141012 2 iModt 2 H0 t2 H001 t1 If AI 0 0 H02 2 H01 m 9 5313311 MEAM 21 1 Example 1 Forceimpulse associated with a swing Golf club swing Velocity 150 ftsecond 46 meterssecond 0 Duration 1 millisecond 0 Mass 162 oz 0046 kg 0 Total linear impulse 21 kg meterssecond 0 average force 2100 Newtons 210 kg 1 ng MEAM 21 1 Example 2 ForceImpulse mass ofthe ball 0145kg mass ofthe bat 09 kg velocity of swing 30 msec contact time lt l msec fastball velocity 100 mph 444 ms UUUUU El calculate average force impulse Pegggmw i MEAM 21 1 Impact 0 Linear impulse Change in Linear momentum o Angular impulse Change in Angular momentum 0 Impact deals with finite impulses in very very small time T8 Chapter 37 m Penn 49gt Engnnnng MEAM 21 1 Linear Impulse with Small Time Intervals El Large force small time interval t1At thl t1At let Lt1At Lt1At t1 El Impulse area under the curve 0 Same impulse for all three cases F F F t1At t1At n 51ellil t1At t1 t1 MEAM 21 1 Linear Impulse with Infinitesimal time El Large force in nitesimal time 3 key attribUteS interval 0 Time 1nterval 1s almost ZGTO 0 lim Al tends to zero El Impulse area under the curve Force is almost in nitely large 0 But impulse is nite F t1 m a MEAM 21 1 When is the zero time impulse model appropriate 05 15128535 PLR FUDZ oooo cameo n 00 i moo END Z 191 l l l l SjUE1Pv re 139813 00009120390 Not always m Penn 49gt Engnnnng MEAM 21 1 Are these impacts 02 135933 PLRV00170 FUDSD 0001OOseo g o 51ml Yes for some calculations No for some calculations What if you have to estimate the What if you have to maximum hei ht of a bouncin predictanalyze the shape of the g g balloon basketball over successwe bounces m Penn 49gt Engnnnng M39EAM211 Real 1ty Collisions have three stages or states 0 Deformation compression gt Deformation increases 0 State of maximum deformation gt Rate of change of deformation is zero o Restitution expansion gt Deformation decreases M39EAM211 Collisions in l dimension El Change in momentum V 0 Let LI denote the linear impulse acting on 2 7 LI mlv1 mlv1 V2 7 LI mzv2 m2v2 i V1 N te all quantities positive gt Adding gpem M39EAM211 Real 1ty Collisions have three stages or states 0 Deformation compression gt Deformation increases 0 State of maximum deformation gt Rate of change of deformation is zero o Restitution expansion gt Deformation decreases MEAM 21 1 How can we model impacts El Newton s laws 0 Accelerations are in nite if forces are in nite El Conservation of energy 0 Only conserved for elastic collisions o Inelastic El Conservation of momentum o For each particle o For the system of particles El Need an impact model 0 Poisson s model of restitution and compression deformation impulses o Newton s model of approach and separation velocities m Penn A Engnnnng M211 The amusement park ride shown in Figure 2 consiets of two rotating platforms 1 and 2 with the riders on seats attached to 2 The rider in question is sitting at point P The carousel body 1 in the gure rotates at the uniform angular Velocity wl lr adsec k with the axis of rotation through point 0 The smaller platform body 2 in the gure is pinned to carousel 1 at point P and it rotates at the uniform angular velocity wz 27390d5ec k If you want to think about relative angular velocities the angular velocity of the platform 2 is 1 rad second relative to an observer 011 platform 1 The length OP is 2 meters While PQ is 1 meter Find the acceleration of the rider Q Peim Engnanng MEAM 21 1 A trehuchet with a telescopic link one of the many designs exhibited on April 17 has the projectile secured in a cup as shown in Figure 1 At the instant shown the length OP 05 meters but extending outwards at a uniform rate of 1 metersec The angle 6 30 17 ad s and 01739ad52 Vhat the acceleration vector ap for the projectile Clearly draw the unit vectors used in your solution procedure m e MEAM211 Kinetics of planar rigid bodies Impulse and Momentum 0 Linear momentum o Angular momentum dVG N i1 Inertia Response Diagram MEAM 2 1 1 I LinearAngular Impulse and Momentum El Force balance N Fzz zdeG i1 611 El Moment balance dHG dl G or for a xed point 0 dHO m MO El Linear momentum t2 L112 I l ZVG 2 I l ZVG 1 t1 El Angular momentum t2 AIG12 IMGdI HGI2 HG1 t1 or for a xedpoint 0 t2 14101 2 IMOdI H0l2 H001 t1 m Penn 49gt Engnm39nng MEAM 21 1 Angular momentum about 0 and G Angular momentum about 0 Angular momentum about G Angular momentum of particle of mass m at G m Penn 49gt Engnm39nng MEAM 21 1 Example 1 Two solid spheres radius r 3 in mass In W 2 lb are mounted on a spinning horizontal rod R with angular velocity 0 6 radsec as shown The mass moment of inertia of the rod is IR The balls are held together by a string which is suddenly cut iR 0251bfts2 Determine the angular velocity of the rod after the balls have moved to A and B m Penn 49gt Engnnnng MEAM 2 1 1 I spinning horizontal rod R with angular velocity 0 6 radsec as together by a string which is suddenly cut 1R 2 0251bfts2 Two solid spheres radius r 3 in mass In W 2 lb are mounted on a Example 1 Determine the angular velocity of the rod after the balls have moved to A and B Key Observation None of the external forces produce a moment about the y axis the angular momentum is conserved shown The mass moment of inertia The angular momentum consists of the of the rod is IR The balls are held angular momenta of the spheres and the angular momentum of the rod m Penn 5 Engirkx nng MEAM 2 1 1 Example 2 A Zkg sphere with an initial velocity of 5 ms strikes the lower end of an 8 kg rodAB The rod is hinged atA and initially at rest The coefficient of restitution between the rod and sphere is 08 Determine the angular velocity of the rod and the velocity of the sphere immediately after impact mg MEAM 2 1 1 FBDIRD for the system m Penn 19 hawdung MEAM 2 1 1 FBDIRD for the system m Penn 19gt Engnm39nng MEAM 211 Work Energy and Power 0 Newton s Second Law 0 Momentum 9 Linear momentum o Angular momentum 0 Work 0 Energy 9 Kinetic Energy 9 Potential Energy V University of Pennsylvania l MEAM 211 I Work Work done by the force F on the particle P over the path from Q to R is given by R W 2 IF dr Q Note 0 is a point fixed in an inertial frame szF39dr miquotdr 25m drr The work done by F is equal to the change in the kinetic energy of the particle R Only Ft WQR Fdr mv v does work V University of Pennsylvania Mechanical Power F acts on the particle 0 Work done by F dW F dr 0 Power developed by F dW dr 2 F ch ch F VP 0 Units Metric Q Watts Newton metersecond British Q Lb ftsecond Q Horsepower o lHP 5501b stec 1 HP 746 w V University of Pennsylvania I NIEAM 211 MEAM 211 l Boeing 777200 2 PampW turbofan engines providing 74000 lbs of thrust Maximum takeoff weight MTOW 23 0000 kg 1 mile runway 1600 m What do you estimate the speed to be at the end of a 1 mile runway Estimated speed at the end of runway without drag 96 ms But takeoff speed 300 kmph 8333 ms What is the estimated drag force KE at the speed of 8333 ms 798610000 Joules Estimated average force through the length of the runway 496 kN Thrust 2 X 329 kN Drag force 161 kN m University of Pennsylvania 4 I NIEAM 21 1 Example A particle of mass m slides along a horizontal frictionless track which is shaped like a logarithmic spiral r r0 expa9 39 r0 If the initial speed is vowhen 90 nd 0 the speed of the particle and the magnitude of the track force acting on a mum pmqlg as a function of 9 a ma e ma r T E C x n o n u I o o o e u o o 1RD University of Pennsylvania 5 I NIEAM 211 Example Horizontal frictionless track which is shaped like a logarithmic spiral r r0 expa9 The initial speed is v0 when 90 Find the speed of the particle and the magnitude of the track force acting on the particle r r b1 VP re a b1 b2 e1 2 M unit normal V1 a2 Track force N e1 ab1 b2 62 2 Same as e Force in the e2 direction V1 a University of Pennsylvania 6 I lVIEAM 211 Example Since N is normal to the track dr is tangential to the track r WINwh 70 Therefore the speed of the particle is E e ll gt E 1 2 1 6 Newton s Laws VPNe1 Vla2 W d rb gtltmv rb gtltNe dt 1 P l l University of Pennsylvania 7 I NIEAM 211 Conservative Force Field F is conservative o F is a function only of the position of the particle and the work done by the force F on the particle P to get it from Q to R is independent of the path F is a function only of the position of the particle and the work done by the force F on the particle P is zero along any closed path For multiple coordinates say x and y Fx 2 13 64 x a 39 Four equivalent de nitions 0 There exists a scalar function I 112 such that dWFdr d oThere exists a scalar function 1 PE and a coordinate s such that d ds F Fa y 3 University of Pennsylvania 8 I NIEAM 211 Conservation of Mechanical Energy F is conservative 0 There exists a scalar function I such that dW F dr 2 d a Work done by F R 1 WQR gFdr Emv v Q R oTotal energy is constant mv2 IR m University of Pennsylvania 9 I NIEAM 21 1 Example Find the magnitude of the force T which acts on the midpoint of the crank given the force F acts on the piston Neglect inertia friction and gravity 3 Velocity Equations from before 93 sin 93 926 7 T x T Q Or ame 1 2 COS63 2 39 F 7 C08 9 3 I 3 O 92 V4 94 f University of Pennsylvania mmzn mmzn Lecture 2 Kinematics of Panicles Rectilinear motion 11 StraightLinemotion 21 ldirnensionalrnotion 1 Analytical solutions forpositionvelocity 21 o Position x0 1 Solving equations ofrnotion o Velocity vt o Analytical solutions 1 7D review 21 o Acceleration at o Nurnerical solutions 21 o Jerk yo 1 Numerical integration Appendix A 21 a snap st 1 Position vectors 22 1 Cartesian coordinate systern 22 Two types ofproblems 11 VelocityAcceleration vectors 22 1 Given forces frndrnotion 1 State vector state space Extensions to 2 and 3 dimensions 1 Given motion nd forces igaun39 ism MEAIvnu MEAIvnu External motion is known nd force Extemal forces are known nd motion Consider parnele wrtn rnonon given by x 6t2 43 v 393 1217 3B 39139 Free body diagrarn Inertia response diagram 2 a Q s 12 s 6 dt dt2 Special cases 1 Acceleration is a given function oftirne ot f F ma altrgt rgtconstam 2 Acceleration is a given function ofposition ax x Me can own mama mars now 3 Acceleration is a given function of velocity av v Fma12m76mt Special case 3 aV v ientt Example Viscous Damping ray nu mum mun xltrgt enanism usedto reouee gun A st orificesi i i using piston and nderto oeeeierate at rate proportional to tneiryeioeity Determine ya x0 and vx u mum Solution W Integrate a dvdt rkv to find ya Integrate ya nixalt to find x0 x 49 t s 7 v m vue X0 2 Idx on aquotquotdt u u gn nit Pznwu l l Solution continue d Integrate a v dvdx rkv to find vx l ekv dv rkdx Idv ekjdx dz n vn ye on 4m x VEI 1 Special case 2 force Fx L106 x Many passive systems a Simple pendulum i Springmass system Control systems for positioning a Guidance systems formissiles D Car 0 Imagine a car being accelerated or decelerated tomrd an intersection t nnn mm m mm m Fx kx Increasing k What at is an arbitrary function of r Mm p12 2 5 TS in 3013mm pumion n on snnein mm In mm an Example Suppose velocity is known function oftime Gven X u 20 ininai condition nan Find X Numerical Integration u x0 X 0111 Needtosolve old pm See Appendix A dz Wm mum Basic idea Approximate E R 1m 7x911 t truly pld 39 ne mmzu mam AlgoIiLhm Numerical Imegmiion of ODEs x I Inmal value problem leen tlne lnlual state at v m to eompute tlne whole tnateetonvm In steps ofat seconds Kan Ina WWW X w 25 on as w Kora y 23 39 x quot W y 12yy0ll to z I z quot W quot W Explicit Euler or BMW Emquot AppendixA ynl y fy un mm m mm In Euler s method Truncation eIrors u Explicit evaluate denvatlve usmg values at tlne beglnnlng ofthe tune step o Not very accurate requlres small tune steps for stabllltv Loca Humane em G Oba Humane error 2 ylyhfxny0h Y o Global accuracy 0h u Implicit Evaluate denvauve uslng values at tlne end oftlne tune step y Y 2 l ynl y hfxslyu0h o May requlre lteratlon slnee tlne answer depends upon vvlnatls ealeulated xx x x o x H KM x at tlne end x1 o Stlll notvery accurate global accuracy 0m o Uneondtlonallv stable for all tune step slzes nsa mm In mm 11 Stability 2Dimensional motion Need more powerful representation a Positions require two coordinates Cartesian coordinates A nurnerieal rnethodis smkifenors oeeurnng at one stage of the proeess o rt do not tend to e magni ed atlater stages ya eri lrn thodis unstable if errors oeeurring at one stage of the u Need position vectors proeesstendto be magnifiedat later stages I I I n In general the stability ofa nurnerieal seherne depends on the step size r X 39 y 1 Usually large step sizes leadto unstable soluaons e i Need areference fram u Irnplieit rnethods are in general more stable than explieit rnethods 39 d reference frames gt Frame xedto i c assrnum Eanh CantEr ufthe Eanh CantEr ufthe Sun CantEr ufthe Universe a Moving reference frarnes gt F rarnes flxedto moving bodles El Need to be able to differentiate vectors WM j V0 aI 1390 sl u K MEAIvnu Differentiation of VectoIs Transfonnations between unit vectors a Understanding the relationship between sets ofunit vectors is very irnportant 39 39 u x My M k o Visualize Differentiate with respect to time 0 Wn39te down the dot products el u 4 el 141 g 143 g i Le Lg Lg vectors Cosme ofangle Dilferentiate with respect to time j M 11 11 between vectors x k k c kg k g quota quotquotquot quotv MEAM 21 1 Kinematics of Planar Rigid Bodies Chapter 6 23 University of MEAM 21 1 Planar motion Does the couplercar rotate Translate University of MEAM 211 Planar Rigid Body Motion For any two points say A B fixed to a rigid body Position at 1 Ar 0 VBA citquot 7dr W 2V VA Position at I 0 Position OfA and BI Rigid body constraint Velocity ofA and B in 5quot University ofP 39 39 3 MEAM 211 Expression for V3 V A For any two points say A B fixed to a rigid body dr dr r B 113 VBA ngz Bdt AVB VA 1391 2 rBAcoseisin9j 1 dt d g uuuuuu n d I d I grBA grBA acoselgsmej a constant VBA rBA sin COS a University of T 39 39 4 V L MEAM 211 De nition of Angular Velocity VBA rBA sin6icosej9 B 52 Can rewrite as vB o r3 1 35 VBA karBAcosel31n93 4 e 1 A1 Define angular velomty W 0 6k So the relative velocity for points A B VBA a X rBA 13 University of P 39 39 5 MEAM 211 Kinematics of Planar Rigid Bodies Key Fact Relative velocity between any two points fixed on any rigid body VQP body B Q VQP ChiP JDX rQP angular velocity of the rigid body 23 University of 7 39 6 MEAM 211 l Slider Crank Linkage Velocity Analysis Before by solving velocity equations 7 7quot 93 1 2 00563 2 Q 0 frame 7 2cos92 93 W62 7 piston 7 3 00539 3 r2 3 a slider 0 92 V4 94 Alternative method solve by writing vector equations representing rigid body constraints m n 1 y Universn y ofi 7 y 93 QcoL r3 r2 0 92 VPQ 033 X rPQ Magnitude of 033 unknown but direction is known m 5 MEAM 21 1 Example Given crank angular velocity 02 solve for piston velocity VP frame VQO 032 X rQO piston slider r4 94 R X VP in this direction VP 2 VQ vPQ Magnitude of 033 VP unknown University of 39 MEAM 21 1 Example y Given crank angular velocity 032 1 rads solve for piston velocity VP 93 Solve closure equations to get r3 93 30 deg V2 40 695 r1 802 r1 0 92 60 deg a x r4 All dimensions in cm 23 University of 7 39 9 MEAM 211 Examples Transmissions 39 Gears S 1 gears Need gearstransmlss10ns to o Hellcal gears o H oid ears yp g Decrease Increase speeds Gear reductlons Geartrains Increase decrease torques O Worm O Planetary 12 032 O Harmonic output Chaln amp Chaln Dr1ves Transmlssmn T1 031 input can 5 University of 7 39 1 Spur gear 0 Loud Each time a gear tooth engages a tooth on the other gear the teeth collide and this impact makes a noise 0 Wear and tear Helical gears 0 Contact starts With point contact to line contact Crossed helical gears o Shaft angles need not be parallel MEAM 211 Spur and Helical Gears University ofPermsylvum39u MEAM 211 Rolling Contact Contact points P1 and P2 coincident instantaneously contact quot I normal BodyA rolls on boob E VPl sz 9E University of 7 39 1 39 MEAM 211 Modeling of Gears 39J AAv 1 L A A 4 c 7 b 3 I Mun The kinematics of rotation of a pair of meshing gears can be modeled as a rotation of the corresponding pitch circles m University of 7 39 3 MEAM 211 Rack and Pinion 0 Similar to a Wheel on a ground With friction 0 But positive engagement 0 Rack is a gear With in nite pitch circle radius 0 Converts rotary motion to linear motion 0 Linear speed 9 Proportional to pinion speed V rp 0 P x r 139 University of 3 MEAM 21 1 Analysis of Spur Gears 0 PinionP 0 Gear G 0 Number of teeth n o Radius r o Angular velocity 03 0 G rP nP n r n r DP rG nG P P G G oThe maximum reduction in a single stage is limited 0T0 get higher reduction oMultiple stages oBut olead to bulky package and weight oSpur gears have high wear and tear and are noisy my 1 y University of ll MEAM 211 Analysis of Planetary Gears Simple Example 7 7 r 7 Ring gear R Sun gear S Canier aim C Planet gear P Frame F I iunrr LIE A simple pinnxlun near Him Waldron and Kinzel 1999 If carrier is stationa But suppose the ring gear S W stationary and the carrier is 0S 2 rP 60 5 ri not stationary 0 r5 0 R 75 3 University of 7 39 6 MEAM 211 Analysis of Planetary Gears 39 Simple Example Ring gear R stationary Sun gear S Canier aim C Planet gear P Frame F IfrP2rS2rR6 andcoR0z Fiunrr LIE A slVlelL pimixlm slcur Irrm Waldron and Kinzel 1999 0 S 4 Assume pos1t1ve counter 03 C clockwise directions 3 University of 7 39 17 MEAM 21 1 Planetary Gears Planet gears share the 9095 ef ciency load usmg swmg links Planet gear meshing with internal gear 3 05 0 N111 rnegoalror STANDARD mgmcu39s 31 EC GEARHEAD PLANETARV GEARHEAD Thumsun Mlcnm 3 lbs 110 Ronknnkoma NYwaslu develo a me ml 20 are mm mme warheads that New geameads are offered In laur lame sizes 40 60 90 and 115 mm with three from ll mug l 39lquot AHsAnlIalm 39 f h Fa Augc i l n e 9 h i an easy dmp n n Excess replace em for parallel ex nacom load backlash sh a swtgeameads and 6b University of 7 39 8 Aerodynamics of Sports Balls Bruce D Ko rhmann January 2007 Unless OTher39wise Specified All Dam From MehTa RD Aerodynamics of Spor Ts Balls Annua Rev6w of FudMeczam cs 198517 15 Spherical Sports Balls 39 um rquot i I 5 1 mm 0 5 J 7 A A e 11 Complex Aerodynamics Ro ia rion amp NonUniform Surface Begin wi rh NonRo ra ring Uniform Sphere Spherical Symme rry Implies Aerodynamic Force MUST be Opposi re To Direc rion of Fligh r Drag Then Consider Seams amp Spin Asymme rry LH Drag on NonRotating Uniform Sphere Dimensional Analysis Drag FD Depends on S eed V Damefer D FDfV9D9p9 lak Density p ViscosiTyu F Roughness LengTh k D 6 Parame rers Minus 3 D 050V2A Dimensions Mass Leng rh Time 3 Dimensionless ParameTers Re D DVD 5 D i Significam ly Reduce ExperimenTal D TesTing Required 1 MaTch Re and a To Find CD and Then CompuTe Drag CD f Re8 Experimental Data for Drag on NonRotating Uniform Spheres 6 km x 105 O GOLF BALL 900 1250 500 150 V ROUGH SPHERES 395 ACHENBACH 1974 7515 O U 3 i SMOOTH SPHERE 2 ACHENBACH 1972 1 PostCritical Drag of Rough Spheres I Much Higher Due to l Increased Friction I 6 105 2 4 6 106 2 See Below Dramatic Reduction in Drag Coefficient Occurs at Critical i Reynolds Number Re Figure 21 Variation of golf ball and sphere drag where k is the sandgrain roughness height and d is the ball diameter Bearman amp Harvey 1976 Two Sources of Aerodynamic Drag Friction amp Pressure FRICTION ON SURFACE DUE TO VISCOSITY OF AIR HIGH PRESSURE ON FRONT OF BALL DUE TO ONCOMING FLOW LOW PRESSURE WAKE BEHIND BALL BECAUSE FLOW SEPARATES FROM SURFACE Pressure Drag Much Larger Than Friction Drag for Spheres PreCritical Reynolds Numbers High Re or Surface Roughness Turbulent Boundary Layer Separation Delayed Much Lower Pressure Drag Separation Surface Roughness Trips Boundary Layer to urbulent Turbulence amp Gives Lower Critical Reynolds Number aminar boundary layer e w mun llmaexnspaczneh nqq estxnnuxndynamcsq h 5mm axllexmlhlfen 6 TWO Sources of Flow Asymmetry SEAMS SPIN gt a gt 2 309 0320 0000 u 00 SEAMS OR SPIN q PROMOTE TRANSITION amp DELAY Q q SEPARATION ON UPPER SURFACE 00 003 amp U DELAY TRANSITION amp PROMOTE SEPARATION ON THE LOWER SURFACE Effect of Seams Depends on Both Re and Orientation of Seams Difficult to Represent With Simple Models Flow Visualization Of Wake Asymmetry 0n Spinning Golf Ball http WWW scielobrscielophppidSO1 0247442004000400003ampscriptsci7a1 ttextampthigen Asymmetry amp Generation of Lift FLOW REMAINS ATTACHED amp MOVING RAPIDLY LOW PRESSURE OVER UPPER SURFACE FLOW SEPARATES amp MOVES MORE SLOWLY SMALLER REGION OF LOW PRESSURE OVER LOWER SURFACE Pressure Asymmetry Causes Lift Force Normal to Velocity Vector Lift amp Drag Coef cient Data for Spinning Dimpled Golf Ball LifT STrongly DependenT 394 L on Spin C 0 Insufficien l39 Data To Clearly D 9 Es l39ablish Effect of Reynolds D 393 Number 1 Similar Trends Observed For 0 Soccer Ball 2 Reasonable Approxima l39ion Lif l39 Coefficien l39 Depends Only on Spin Drag Weakly DependenT on Spin General Trend is for Modes l39 0 I Increase In Drag wrl39h Spin 0D Reasonable Approxima l39ion Drag SPIN NUMBER S D Independen r of Spin Important Detail Consideration of Relative Orientation of Spin Axis 39 Most Test Data Gathered with Angular Velocity Normal to Linear Velocity Simulates Backspin on Golf Ball or Topspin on Tennis Ball For Uniform Ball Symmetry Guarantees Lift Force Normal to Both Vectors Actual Trajectories Involve Arbitrary Orientation of n and V Ignore Aerodynamic Moments Assume Direction of on Fixed in Inertial Space Assume Only onquot Affects Lift Compute Effective Spin Number Based MD on oquot s 2V SPIN NUNBER n Final Recommended Simple Model Reasonable for Soccer Golf Tennis Baseball Curveball CD f Re8 5 BCCER RE CD 30000 0 40 100 000 0 50 quotI 00 000 0 51 200 000 0 5 39I 250 000 0 4 3 300 000 0 443 330 000 0 30 350000 0 20 325 000 0 00 400 000 0 0 500 000 0 0 B00 000 0 10 2 000000 0 15 4 000000 010 GDLF RE CD 30000 040 45000 035 50000 030 00000 024 240000 020 4000000 030 B lSEB lLL RE CD 40000 040I 00000 050 110000 032 1 1100000 035 TENNIS C0 00 CL 5 IL 000 000 004 010 010 010 020 023 040 033 Soccer39 CD Es rima red fr om Smoo rh Sphere Tennis Ball quotFuzzquot Yields High Drag Nearly IndependenT of Re Baseball CD Es rima red fr39om ModeraTe Roughness Sphere Some Typical Numbers TYPE D CITI SUCCER 222 VOLLEYBALL 210 TENNIS 55 BABEBALL 75 GDLF 43 BASKETBALL 243 CRICKET 22 PING PDNG 33 httpWWWscielobrscielophppidSO10247442004000400003ampscript V n05 30 30 45 40 00 10 20 5 sciiarttextamptlngen Re E143 444 425 155 195 172 152 055 013 I remds 10 72 30 00 S 023 033 01 014

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