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# ENGR MECH DYNAMICS ME 16

UCSB

GPA 3.94

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This 17 page Class Notes was uploaded by Daren Beatty Jr. on Thursday October 22, 2015. The Class Notes belongs to ME 16 at University of California Santa Barbara taught by Staff in Fall. Since its upload, it has received 42 views. For similar materials see /class/227086/me-16-university-of-california-santa-barbara in Mechanical Engineering at University of California Santa Barbara.

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Date Created: 10/22/15

Exampl Problem lb collar is constrained to travel along a rectilinear and frictionless bar of length L l m E 5 v a S n E n 0 to the right With a speed of 11fts deterr L L reaches D Wlth zero speed Road Map e know dwe speed of the collar at two different positions and we know the forces acting on the collar as it moves between those positions 0 Applying t e BWE should then give us the one equation we need to determine the stiffness of the springs I l Example Problem 438 continued Modeling The FBD of the collar at an arbitrary position 5 as PM W WM a Conswa N k mg 90 Ao WM mm Wrong NL MA A cm L PL 7a4 f 20 Example Problem 438 continued Modeling The FBD of the collar at an arbitrary position is as shown mg l D The collar does not move in the y direction so mg does l no work 0 P1 and P2 do work and they are both conservative forces Governing Equations NewtonEulerBalance Equations Since all the forces doing work are conservative we will use the following form of the BWE TA VA TD VD Material Models The potential energy at each position is the sum of the individual potentials energies of the springs Therefore v v1 v2 where v1 lkaf lkL17Lo2 and v2 Ems lkurLo 2 2 2 2 and where 7 L1 the current length of spring 1 L2 the current length of spring 2 Lo the unstretched length of springs 1 and 2 I l Example Problem 438 continued Evaluating the potential energies at A and B we have 1 1 1 1 VOA 5M V1D EMMEM V2A 5 45 V2D k523 Kinematic Equations For the kinematic equations we need to find the spring lengths the LS the spring 39 the 6 s and the kinetic energies The lengths of the two springs at positions A and D are recall that L 5ft L2 L2 LilV fi TL L2 xE L2 9L2 13 L 7 2 7 7 7 7 7 L1D7 L 4 7 2 L7 L2A7 4 16 7 4 L7 L2D7 2 Using these eqns for the lengths the deformations become 5 5 13 L 60A L L07 61D gL Lm 62 4L 7 Lo 62D E 7 Lo Exampl Problem 438 c The unstretched length of the springs IS Finally me kmeIIC energies at A and D are 7 1 2 7 1 6b 2 7 TA 7 5va 7 E 11fts 71127ft7ib TD Oftrlb Computation Substituting the maternal and kinemauc equations into the BWE we get l 2 2 2 or plugging In numbers 1127ftrlb k0 74593 k2 54m gt k 5 yamft I l A motfinal A 44 z A P x Vt MAW inD a 96m M 41 D in WM LIN9 fang gdmcg 960 M aw D Example Problem 438 continued Discussion 4 Verification 0 All dimensions work out properly 0 This problem would have been much harder to solve using Newton s 2nd law 0 The five step procedure really shines on a problem like this with all the details that are involved in the solution A Stack o wo Books in Thrown on a Table You throw a stack of two books on a table 9 Given the friction coefficients between A and B and between A and the table what are the final XA and X3 if A and B start at X 0 With an initial horizontal velocity of v0 075 ms7 We can answer this question by applying the BWE to each book 7 Bjil B separately V3 7V 239 f5 For B we have L Na 7 A a a 2 731 v51 kwm g Mum where U14p is the work done on B by all M x forces for which we do not have a potential NA energy function In addition quotPositionsquot in the Balance of Work and Energy for Systems For T31 V31 U172an T32 V32 mm 0 D is the position at which the books V rst strike the table and LEE x i3 0 is the position at which the books NB i come to rest i Now for two important observations NB mAg 0 D Q etc are not the positions ofjust one element of the system D is the ifigtblt iA initial position of each book is the NifAii A final position ofeach book etc 9 Nothing can be said about when A and B achieve ie their final positions A and B may get to their respective at different times I i Goi Back to Balance of Work and Ene Going back To the balance of work and energy for may 3 T81 V81 U172p T32 V32 wnh z E PM B 1 A T81 Emavgy T82 07 V81 07 V82 07 NE 7 a B X32 HI 114 O 48 A9 a Lt EA dig Lama W74 NA A M Mt 715 Mia Goi Back to Balance of Work and Ene Going back To The balance of work and energy for may 3 7131 V81 U172p T32 V32 wnh 1 A T81 Emavgy T82 07 V81 07 V82 07 NE 7 a g B m 11172 0 f8 dXB fBXBQ Mum S ngai A Similarly for A we have NA TA1 VA1 Mam TA2 VA27 9 wnh 1 TA1 57724ng TA2 07 VA1 07 VA2 07 XAZ 1114 fairom fairom o y Accounting for Friction and Doing a Bit of Computation Our material models for this system are simply the friction laws f3 MABkNB MABkmBg7 fA MAMVA MOMMA meg7 where NA and NB are found by applying Newton39s 2nd law in the y direction Next using fA and f3 in the BWE for B and A we get 1 EmBVOZ MABkmBgXBZ 0 1 EmAVoz MAmasg PAkmA msgXA2 0 These are 2 equations in the 2 unknowns XAZ and X32 whose solution is m 0952 13 m and X32 0095 57 m Now let39s look at what we would have obtained had we tried to apply the BWE to the system as a whole I i Total Balance of Work and Energy for the System We now speculate that for the system as a whole we can say T1 V1 U172np T2 V2 For the FBD shown 1 T1 7 EmAV02 EmBVoZ7 T2 07 V1 XA2 U172np 0 7739 dXA fXAZ MAM mA m13gXA27 where we have used f ukN IMAM mA l mgg So overall we get 1 2MAV02 mBV02IAltMA msgXA2 0 5 We got XAZ 0052 13 m before so this is clearly wrong Let39s see Why a Comparing the Two Work Energy Balance Statements rag Going back to the BWE for A and B and summing the two IndIVIdual BWE statements we have quotMy 1 2 1 2 ll L EmAvo Emgvo MABlltmggX82 XA2 4 MAMmA m8gXA2 0 N Now recall that the wrong expression we obtained earlier was 1 1 EVENo2 EmBVg MAM quotM m8gltA2 0 Notice that the wrong expression is missing the term in blue MAB mBMXm XA2 This is the work done by friction between A and B due to the motion an relative to B How is the BWE correctly applied to a system of particle57 I l Putting Ideas on a Firm Foundati a We have the usual system of n particles 3 is the total external force acting on partic e i a In addition particle i With mass m Interacts With the other n 71 particles through internal forces labeled 770 17n7i7 j Now applying Newton39s second law to the i particle gives u Eng ma 1 j1 Computing the Total Kinetic Ene As we did for a single particle a we now doT The equaTIon E 21 m WITh 17 and o InTegraTe along The paTh o m o Summing all n of These equaTIons gives and using a dEdt i u if 2quot d7 F fr 417 mi dF 1 37172 I j1 U I 1 37172 I dt I Now looking aTjust The righTehand side In d7 n quot 1 2 mi dF mde 77774 g rzldt I 2 I 221 n n 1 2 7 25 37 Th 1 1 where T 2f1mv2 Is The ToTal lltneTIc energy of The sysTem I l

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