Cinética Plana de Corpos Rígidos - Equações do Movimento e Momento de Inércia

Parte 6 Cinética Plana de Corpos Rígidos Equações do movimento F ma Soma das forças que atuam sobre o corpo rígido Diagrama de Corpo Rído Força Resultante Diagrama de Forças Resultantes m massa do corpo rígido a aceleração do centro de massa G do corpo rígido MG Iα Soma dos momentos das forças com relação ao eixo que passa pelo centro de massa do corpo rígido Diagrama de Corpo Rígido Momento resultante em torno do eixo que passa pelo centro de massa do CR Diagrama de Forças Momentos Resultantes Nesse caso F ma como antes MP Iα ρ ma vetorialmente MP Iα ma d escalarmente I momento de inércia de massa do corpo rígido com relação ao centro de massa G Vide Tabelas a seguir α Aceleração angular do corpo rígido Escolha de um sistema de eixos para escrever as Equações do Movimento válido tanto para a cinemática como para cinética do problema Muitas vezes é interessante fazer a soma dos momentos de forças com relação a um eixo que passa por um ponto P diferente de G Ponto P um ponto onde algumas forças não produzem momento 4 Momento de Inércia de Massa Definição rα dm diferencial da força resultante tangencial Raio de Giração 5 Teorema dos eixos paralelos 𝐼 𝑚𝑜𝑚𝑒𝑛𝑡𝑜 𝑑𝑒 𝑖𝑛é𝑟𝑐𝑖𝑎 𝑐𝑜𝑚 𝑟𝑒𝑙𝑎çã𝑜 𝑎𝑜 𝑒𝑖𝑥𝑜 𝑞𝑢𝑒 𝑝𝑎𝑠𝑠𝑎 𝑝𝑒𝑙𝑜 𝑐𝑒𝑛𝑡𝑟𝑜 𝑑𝑒 𝑚𝑎𝑠𝑠𝑎 𝐼 𝑚𝑜𝑚𝑒𝑛𝑡𝑜 𝑑𝑒 𝑖𝑛é𝑟𝑐𝑖𝑎 𝑐𝑜𝑚 𝑟𝑒𝑙𝑎çã𝑜 𝑎𝑜 𝑒𝑖𝑥𝑜 𝑞𝑢𝑎𝑙𝑞𝑢𝑒𝑟 𝑎 𝑢𝑚𝑎 𝑑𝑖𝑠𝑡â𝑛𝑐𝑖𝑎 𝒅 𝑑𝑜 𝑒𝑖𝑥𝑜 𝑞𝑢𝑒 𝑝𝑎𝑠𝑠𝑎 𝑝𝑒𝑙𝑜 𝑐𝑒𝑛𝑡𝑟𝑜 𝑑𝑒 𝑚𝑎𝑠𝑠𝑎 TABLE D4 PROPERTIES OF HOMOGENEOUS SOLIDS m mass of body shown BODY MASS CENTER MASS MOMENTS OF INERTIA Circular Cylindrical Shell Ixx 12 mr2 112 ml2 Ixx1 12 mr2 13 ml2 Izz mr2 Half Cylindrical Shell x 2rπ Ixx Iyy 12 mr2 112 ml2 Ixx1 Iyy1 12 mr2 13 ml2 Izz mr2 Ixz 1 4π2 mr2 Circular Cylinder Ixx 14 mr2 112 ml2 Ixx1 14 mr2 13 ml2 Izz 12 mr2 Semicylinder x 4r3π Ixx Iyy 14 mr2 112 ml2 Ixx1 Iyy1 14 mr2 13 ml2 Izz 12 mr2 Ixz 12 169π2 mr2 Rectangular Parallelepiped Ixx 112 ma2 l2 Iyy 112 mb2 l2 Izz 112 ma2 b2 Iyy1 112 mb2 13 ml2 Iy2y2 13 mb2 l2 TABLE D4 PROPERTIES OF HOMOGENEOUS SOLIDS Continued m mass of body shown BODY MASS CENTER MASS MOMENTS OF INERTIA Spherical Shell Ixx 23 mr2 Hemispherical Shell x r2 Ixx Iyy Izz 23 mr2 Iyy Izz 512 mr2 Sphere Izz 25 mr2 Hemisphere x 3r8 Ixx Iyy Izz 25 mr2 Iyy Izz 83320 mr2 Uniform Slender Rod Iyy 112 ml2 Iyy1 13 ml2 TABLE D4 PROPERTIES OF HOMOGENEOUS SOLIDS Continued m mass of body shown BODY MASS CENTER MASS MOMENTS OF INERTIA Quarter Circular Rod x y 2rπ Ixx Iyy 12 mr2 Izz mr2 Elliptical Cylinder Ixx 14 ma2 112 ml2 Iyy 14 mb2 112 ml2 Izz 14 ma2 b2 Iyy1 14 mb2 13 ml2 Conical Shell z 2h3 Iyy 14 mr2 12 mh2 Iyy1 14 mr2 16 mh2 Iz 12 mr2 Iyy 14 mr2 118 mh2 Half Conical Shell x 4r3π z 2h3 Ixx Iyy 14 mr2 12 mh2 Ixx1 Iyy1 14 mr2 16 mh2 Izz 12 mr2 Ixz 12 169π2 mr2 Right Circular Cone z 3h4 Iyy 320 mr2 35 mh2 Iyy1 320 mr2 110 mh2 Izz 310 mr2 Iyy 320 mr2 380 mh2 TABLE D4 PROPERTIES OF HOMOGENEOUS SOLIDS Continued m mass of body shown BODY MASS CENTER MASS MOMENTS OF INERTIA Half Cone x rπ z 3h4 Ixx Iyy 320 mr2 35 mh2 Ixy1 Iyy1 320 mr2 110 mh2 Izz 310 mr2 Izz 310 1π2 mr2 Semielipsoid z 3c8 Ixx 15 mb2 c2 Iyy 15 ma2 c2 Izz 15 ma2 b2 Ixx 15 mb2 1964 c2 Iyy 15 ma2 1964 c2 Elliptic Paraboloid z 2c3 Ixx 16 mb2 12 mc2 Iyy 16 ma2 12 mc2 Izz 16 ma2 b2 Ixx 16 mb2 13 c2 Iyy 16 ma2 13 c2 Rectangular Tetrahedron x a4 y b4 z c4 Ixx 110 mb2 c2 Iyy 110 ma2 c2 Izz 110 ma2 b2 Ixx 380 mb2 c2 Iyy 380 ma2 c2 Izz 380 ma2 b2 Half Torus x a2 4R22πR Ixx Iyy 12 mR2 58 ma2 Izz mR2 34 ma2 SAMPLE PROBLEM 65 A metal hoop with radius r 6 in is released from rest on the 20 incline If the coefficients of static and kinetic friction are μs 015 and μk 012 determine the angular acceleration α of the hoop and the time t for the hoop to move a distance of 10 ft down the incline Solution The freebody diagram shows the unspecified weight mg the normal force N and the friction force F acting on the hoop at the contact point C with the incline The kinetic diagram shows the resultant force m a through G in the direction of its acceleration and the couple I a The counterclockwise angular acceleration requires a counterclockwise moment about G so F must be up the incline Assume that the hoop rolls without slipping so that a ra Application of the components of Eqs 61 with x and yaxes assigned gives ΣFx m ax mg sin 20 F m a ΣFy m ay 0 N mg cos 20 0 ΣMG I a Fr mrα Elimination of F between the first and third equations and substitution of the kinematic assumption a ra give a g2 sin 20 3222 0342 551 ftsec2 Alternatively with our assumption of a ra for pure rolling a moment sum about C by Eq 62 gives a directly Thus ΣMC I a m ad mgr sin 20 mr2 ar m r a a g2 sin 20 To check our assumption of no slipping we calculate F and N and compare F with its limiting value From the above equations F mg sin 20 m g2 sin 20 01710 mg N mg cos 20 0940 mg But the maximum possible friction force is Fmax μs N Fmax 0150940 mg 01410 mg Because our calculated value of 01710 mg exceeds the limiting value of 01410 mg we conclude that our assumption of pure rolling was wrong Therefore the hoop slips as it rolls and a ra The friction force then becomes the kinetic value F μk N F 0120940 mg 01128 mg The motion equations now give ΣFx m ax mg sin 20 01128 mg m a a 0229 322 738 ftsec2 ΣMG I a 01128 mgr m r2 α α 01128 322612 726 radsec2 The time required for the center G of the hoop to move 10 ft from rest with constant acceleration is x 12 a t2 t 2xa 210738 1646 sec Ans Helpful Hints 1 Because all of the mass of a hoop is a distance r from its center G its moment of inertia about G must be mr2 2 Note that a is independent of both m and r 3 Note that α is independent of m but dependent on r SAMPLE PROBLEM 66 The drum A is given a constant angular acceleration α0 of 3 rads2 and causes the 70kg spool B to roll on the horizontal surface by means of the connecting cable which wraps around the inner hub of the spool The radius of gyration kg of the spool about its mass center G is 250 mm and the coefficient of static friction between the spool and the horizontal surface is 025 Determine the tension T in the cable and the friction force F exerted by the horizontal surface on the spool Solution The freebody diagram and the kinetic diagram of the spool are drawn as shown The correct direction of the friction force may be assigned in this problem by observing from both diagrams that with counterclockwise angular acceleration a moment sum about point G and also about point D must be counterclockwise A point on the connecting cable has an acceleration at ra 0253 075 ms2 which is also the horizontal component of the acceleration of point D on the spool It will be assumed initially that the spool rolls without slipping in which case it has a counterclockwise angular acceleration α aDxDC 075030 25 rads2 The acceleration of the mass center G is therefore a ra 04525 1125 ms2 With the kinematics determined we now apply the three equations of motion Eqs 61 ΣFx m ax F T 701125 a ΣFy m ay N 70981 0 N 687 N ΣMG I a F0450 T0150 70 02502 25 b Solving a and b simultaneously gives F 758 N and T 1546 N Ans To establish the validity of our assumption of no slipping we see that the surfaces are capable of supporting a maximum friction force Fmax μs N 025 687 1717 N Since only 758 N of friction force is required we conclude that our assumption of rolling without slipping is valid If the coefficient of static friction had been 01 for example then the friction force would have been limited to 01687 687 N which is less than 758 N and the spool would slip In this event the kinematic relation a ra would no longer hold With aDx known the angular acceleration would be α a aDxGD Using this relation along with F μk N 687 N we would then resolve the three equations of motion for the unknowns T a and α Alternatively with point C as a moment center in the case of pure rolling we may use Eq 62 and obtain T directly Thus ΣMC I a m ar 03T 700252 25 701125045 T 1546 N Ans Helpful Hints 1 The relation between a and α is the kinematic constraint which accompanies the assumption that the spool rolls without slipping 2 Be careful not to make the mistake of using 12 mr2 for I of the spool which is not a uniform circular disk 3 Our principles of relative acceleration are a necessity here Hence the relation aDx GDa should be recognized 644 If the system is released from rest while in the horizontal position shown determine the angular acceleration of the lightweight rightangle shaft The sphere of radius r has mass m Neglect friction at the bearing O Problem 644 667 The robotic device consists of the stationary pedestal OA arm AB pivoted at A and arm BC pivoted at B The rotation axes are normal to the plane of the figure Estimate a the moment MA applied to arm AB required to rotate it about joint A at 4 rads2 counterclockwise from the position shown with joint B locked and b the moment MB applied to arm BC required to rotate it about joint B at the same rate with joint A locked The mass of arm AB is 25 kg and that of BC is 4 kg with the stationary portion of joint A excluded entirely and the mass of joint B divided equally between the two arms Assume that the centers of mass G1 and G2 are in the geometric centers of the arms and model the arms as slender rods Problem 667 SAMPLE PROBLEM 63 The concrete block weighing 644 lb is elevated by the hoisting mechanism shown where the cables are securely wrapped around the respective drums The drums which are fastened together and turn as a single unit about their mass center at O have a combined weight of 322 lb and a radius of gyration about O of 18 in If a constant tension P of 400 lb is maintained by the power unit at A determine the vertical acceleration of the block and the resultant force on the bearing at O Solution I The freebody and kinetic diagrams of the drums and concrete block are drawn showing all forces which act including the components Ox and Oy of the bearing reaction The resultant of the force system on the drums for centroidal rotation is the couple I α I0 αx where I k02 m I I0 18122 322322 225 lbftsec2 Taking moments about the mass center O for the pulley in the sense of the angular acceleration a gives ΣMG I α 4002412 T 1212 225 α a The acceleration of the block is described by ΣFy may T644 644322 α b From at r α we have a 1212α With this substitution Eqs a and b are combined to give T 717 lb α 367 radsec2 a 367 ftsec2 Ans The bearing reaction is computed from its components Since α 0 we use the equilibrium equations ΣFx 0 Ox 400 cos 45 0 Ox 283 lb ΣFy 0 Oy 322 717 400 sin 45 0 Oy 1322 lb O 2832 13222 1352 lb Ans Solution II We may use a more condensed approach by drawing the freebody diagram of the entire system thus eliminating reference to T which becomes internal to the new system From the kinetic diagram for this system we see that the moment sum about O must equal the resultant couple I α for the drums plus the moment of the resultant ma for the block Thus from the principle of Eq 65 we have ΣMO I α mad 4002412 6441212 225 α 644322 1212 With a 1212α the solution gives as before α 367 ftsec2 We may equate the force sums on the entire system to the sums of the resultants Thus ΣFy Σmay Oy 322 644 400 sin 45 322322 0 644322 367 Oy 1322 lb ΣFx Σmax Ox 400 cos 45 0 Ox 283 lb Helpful Hints 1 Be alert to the fact that the tension T is not 644 lb If it were the block would not accelerate 2 Do not overlook the need to express k0 in feet when using g in ftsec2 SAMPLE PROBLEM 64 The pendulum has a mass of 75 kg with center of mass at G and has a radius of gyration about the pivot O of 295 mm If the pendulum is released from rest at θ 0 determine the total force supported by the bearing at the instant when θ 60 Friction in the bearing is negligible Solution The freebody diagram of the pendulum in a general position is shown along with the corresponding kinetic diagram where the components of the resultant force have been drawn through G The normal component On is found from a force equation in the ndirection which involves the normal acceleration r ω2 Since the angular velocity ω of the pendulum is found from the integral of the angular acceleration and since Ot depends on the tangential acceleration r α it follows that α should be obtained first To this end with I0 k02 m the moment equation about O gives ΣMO I0 α 75981025 cos θ 0295275α α 282 cos θ rads2 and for θ 60 ω dω α dθ from 0 to ω ω dω from 0 to π3 282 cos θ dθ ω2 488 rads2 The remaining two equations of motion applied to the 60 position yield ΣFn mr ω2 On 75981 sin 60 75025488 On 1552 N ΣFt mr α Ot 75981 cos 60 75025282 cos 60 Ot 1037 N O 15522 10372 1556 N Ans The proper sense for Ot may be observed at the outset by applying the moment equation ΣMG I α where the moment about G due to Ot must be clockwise to agree with α The force Ot may also be obtained initially by a moment equation about the center of percussion Q shown in the lower figure which avoids the necessity of computing α First we must obtain the distance q which is q k02r q 02952 0250 0348 m ΣMq 0 Ot0348 75981cos 600348 0250 0 Ot 1037 N Ans Helpful Hints 1 The acceleration components of G are of course an r ω2 and at r α 2 Review the theory again and satisfy yourself that ΣMO I0 α I α mr2 α mfr α 3 Note especially here that the force summations are taken in the positive direction of the acceleration components of the mass center G 694 The robotic device of Prob 667 is repeated here Member AB is rotating about joint A with a counterclockwise angular velocity of 2 rads and this rate is increasing at 4 rads2 Determine the moment MB exerted by arm AB on arm BC if joint B is held in a locked condition The mass of arm BC is 4 kg and the arm may be treated as a uniform slender rod Problem 694 6105 The connecting rod AB of a certain internalcombustion engine weighs 12 lb with mass center at G and has a radius of gyration about G of 112 in The piston and piston pin A together weigh 180 lb The engine is running at a constant speed of 3000 revmin so that the angular velocity of the crank is 30002π60 100π radsec Neglect the weights of the components and the force exerted by the gas in the cylinder compared with the dynamic forces generated and calculate the magnitude of the force on the piston pin A for the crank angle θ 90 Suggestion Use the alternative moment relation Eq 63 with B as the moment center Problem 6105 16 Cinética do Ponto Material Um exercício interessante de Merian J L DINÂMICA LTC 1976 A separação dos termos em i e j conduz às duas equações escalares x ω2 0 e N m bω2 2ω x Observase que a solução geral da primeira equação é por substituição direta x A sinh ωt B cosh ωt As constantes de integração A e B são determinadas pelas condições x x0 e x 0 quando t 0 que dá A 0 e B x0 Assim a coordenada x é x x0 cosh ωt Resp A expressão para a força normal N contém a velocidade relativa x x0 ω senh ωt que pode ser expressa em função de x pela substituição da identidade cosh2 ωt senh2 ωt 1 Então ωt sqrtcosh2 ωt 1 sqrtxx02 1 e x x0 ω sqrtxx02 1 A expressão para a força normal N tornase N m ω2 b 2 sqrtx2 x02 Resp a qual é limitada aos valores de x x0 Probl 3185 Chapter 6 61 a g3 62 a 3g 63 B 15 lb 64 P μe M mg cos θ 65 a 0268g 66 d μt k2 67 a and b P mg cb 68 P 3M mg 69 a 1306 ms2 right 610 a 0706 ms2 right 611 FA 1110 N Ox 45 N right Oy 667 N down 612 a 414 ms2 613 T 273 N Ax 1834 N right Ay 1557 N up 614 a NA 1280 lb 40 NB 1920 lb 60 b NA 1908 lb 596 NB 1292 lb 404 615 a NA 1920 lb 60 NB 1280 lb 40 b NA 2550 lb 796 NB 652 lb 204 616 A 1192 N 617 Ay 1389 N down 618 a 161 ftsec2 619 TA 1299 lb TB 390 lb α 805 radsec2 620 P 1187 lb θ 492 B 410 lb Bst 417 lb 621 t 341 s 622 M 1960 Nm CCW 623 F 783 N M 287 Nm 624 μ 0598 625 D 234 N α 587 rads2 CW 626 N 257 kN 627 a θ 513 b θ 248 a 54 g 628 a θ tan1 v2gr b Slips first if μ b2h and μ tan β v2 gr μ tan θ1 μ tan θ Tips first if μ b2h and tan β b2h v2 gr b2h tan θ1 b2h tan θ 629 a D 1714 N b D 2178 N 630 θ 0964 nose up 631 B 1883 N 632 A 202 kN 633 R 490 N 634 FA FB 245 N 635 α 1193 rads2 CCW FA 769 N 636 Io 1453 lbftsec2 637 α 912 radsec2 CW 638 a α g2r CW O mg2 b α 2g3r CW O mg3 639 R 357 lb 640 A 563 N 641 α 8g3πb CW Ox 32mg9π2 left Oy 1 329π2 mg up 642 β π2 α 8g3bπ CW β π α 8g3bπ CW 643 AA α 32g5b BB α 32g7b 644 α 1547 gπ CW 645 α 0389 gb CW 646 M ωρdτ 12 π r4 4lt 13 r2 rl r2 647 On 2mg sin θ Ot mg2 cos θ 648 a α 785 rads2 CCW b α 628 rads2 CCW 649 R 181b 650 a α 846 rads2 CCW b α 1116 rads2 CCW 651 Ox 866 lb at all times 652 Mf 01045 lbft Mmot 0836 lbft 653 M mr2α R 22 mnπ α2 ω4 654 x 123 α g3l CW 655 x b6 α 32 gb CW 656 α 67 l g 127 km 5 3 CW 657 t 786 s 658 O 14 mgcos2 θ 100 sin2 θ 659 α g2r CCW A 0593 mg 660 b 407 mm R 1678 N 661 A 221 N B 1103 N 662 FA 1083 N FB 1416 N 663 αB 255 rads² CCW 664 T 987 N A 1007 kN 665 α 384 radsec² CW t 349 sec 666 a F mlα6 b A 106 mlα c ω 3 667 a MA 1098 Nm CCW b MB 551 Nm CCW 668 R 1013 N 669 α μs 01880 b θ 531 670 μs 0229 b θ 546 671 aB P2m 3i j 672 aA P10m 37i 9j 673 α 488 rads² CW aG 5j ms² 674 α 861 rads² CCW aG 5j ms² 675 α 0310 rads² aG 687 ms² ay 274 ms² 676 aO 702 ms² α 908 rads² CCW 677 ω 0775 678 a 1380 ftsec² F 1714 lb 679 a 1331 ftsec² F 0693 lb 680 aO 241 ms² up incline α 938 rads² CW 681 ω 1kO 2πir 682 A αA gr sin θ μs 0 B αB g2r sin θ μs 12 tan θ 683 F 23 mg N 1316 mg 684 T 208 N 685 T 2313 mg 686 N mg r²ω²R r 687 α 228 rads² CW μs 0275 688 α 212 rads² CW α 0425 ms² right F 1938 N 689 α 0295 rads² CCW a 1027 ms² right F 1762 N 690 α 12bg7b² 3h² CW TA 3mgb² h²7b² 3h² up 691 BC 403 N tension 692 B 364 N 693 aO 373 ms² down incline 694 MB 355 Nm CCW 695 αA 1143g down incline 696 α 5α7r CW ω 107r g1 cos θ α sin θ 697 aA 593 ms² left 698 α 84α65L CCW 699 ω 297 radsec CCW 6100 αA 14109 g down incline 6101 s 1866 ft 6102 θ FkO² α cos θ g sin θ Ot mg sin θ α cos θ1 F²kO² On mg cos θ α sin θ1 2F²kO² 2gF²kO² θmax 531 6103 v 1173 ms mg 6104 N 6g sin θ1 3 sin² θ α CW L1 3 sin² θ 6105 A 347 lb 6106 a m 350M b α 000581 Fr CCW μsmin 0589 3g2l 6107 α 721 CW 6108 α 1818 radsec² CCW RA 1128 lb right RB 0359 lb 255 aA 650 ftsec² down aB 569 ftsec² down incline 6109 ω 24g7L CW uG 314 gL 6110 ω 48g7L 6111 v 297 ftsec 6112 ω 0839 gb CW 6113 v 301 ms 6114 θ 332 6115 O 9127 mg up 6116 A vA 2gx sin θ B vB gx sin θ 6117 h 545 mm 6118 ω 1319 rads 6119 N 3240 revmin 6120 k 926 Nm ω 242 rads CW 6121 l₀ 900 mm 6122 ω 331 rads 6123 M 2841 lbin CW 6124 v 6gb sin θ2 6125 ω 459 rads 6126 vA 245 ms 6127 ΔE 0435 ftlb 6128 N 346 lb