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# Dynamics ME 6441

GPA 3.62

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This 0 page Class Notes was uploaded by Chloe Reilly on Monday November 2, 2015. The Class Notes belongs to ME 6441 at Georgia Institute of Technology - Main Campus taught by Aldo Ferri in Fall. Since its upload, it has received 12 views. For similar materials see /class/234237/me-6441-georgia-institute-of-technology-main-campus in Mechanical Engineering at Georgia Institute of Technology - Main Campus.

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Date Created: 11/02/15

Example Using Rotational Transformation Matrices ME6441 Dr Ferri Z 20 Start by de ning several coordinate systems XYZ Inertial Frame Z vertical xoyozo Fixed to body AB rotates by angle 5 about the vertical axis 20 is always vertical and arm AB remains in the x020 plane at all times x0 coincides with X when i 0 xlylzl Also fixed to AB but with the x1 axis aligned with AB at all times yo and y1 coincident xzyzzz Fixed to body BC and rotated relative to xlylzl by angle 7 about x1 When 7 0 xzyzzz is alligned with xlylzl X3y323 Fixed to body CD such that X3 is along CD and rotated relative to xzyzzz by angle 77 about yz Vectorsum VDA rBA rCB rDC Or rAB rCB 7CD rm1 0 0 0 0 xlylzl 0 x2y222 0 x3y323 Know that x1 x0 X M Ry250yo Ry250RZ3Y Z1 20 H4 Z Rm Rylt 250gtRzlt gt x2 x1 X wRxltygty1RxltygtRylt250Rzlt gtY 22 21 Z a RM RxltygtR1 RxltygtRylt 250gtRzlt gt x3 x2 X 6 192y2 Rx 2RxltygtRylt 250Rz Y z3 22 Z j V Rai Rylt 2gtR2 Rx 2RxltygtRylt 250gtRzlt gt rAB rCB 7CD T T T VDAXYZR1i 0 R2i 0 R3i 0 0 xlylzl 0 x2y222 0 x3y323 Engineering Dynamics Jerry H Ginsberg Errata t0 the First Printing Prepared September 9 2008 This is the second version of an errata for Engineering Dynamics Corrections reported here will be incorporated into the second printing of the book If you find additional corrections please inform J H Ginsberg at jerryginsberg megatechedu Thank you Pg 15 eq 123 P 0Q P P 0Q P P OQOQ P 0Q Pg P 0Q P P OQOQ F F MLT2 16 eq 127 9 9807 ms2 16 eq 128 9 3217 fts2 or g 3860 ms2 31 third line after eq 211 path Hence it is one of the path variables 43 rst line 8 BF 11562 Substitution 43 rst set of equations a 067953 051415 00058 p 1500 m 43 second set of equations a 5388 2279314321 ms2 49 eq 4 Fix unknown vector 0 0 1 5390 701613 1 7103 0 y 0 x y 792 239 0 61 eq 2324 fourth line delete the unit vector 6 in the second bracket that is 63 second line from bottom F 3820 N 69 eq 2350 had a Pg 70 eq 2353 index for the third sum should be u that is 39 ahA a Z 2 AM Tyexhxmgt ka w Mama Pg 72 sixth equation line from bottom Delete R g 7 sin 0 7 cos 0 Pg 72 fourth equation line from bottom BER 1 BhR 7 1 BhR 7 7 ez 0 E be 00 EVE Pg 73 second equation line from top 8amp2 1 Bhg 7 80 hz 82 Pg 76 Example 211 rst line Change 1V1 to lower case of a mixed kinematical Pg 88 bottom of page gure for Exercise 245 Label the pivot as A and the upper end of the rotating arm as B Pg 89 Exercise 246 second line from bottom values of 7quot 7 1 gt1 Pg 104 rst line after eq 3129 treated and Rz as ie lower case subscripts Pg 111 rst equation 701059 701789 709782 Z X 0 408671 714980 700028 g Z Z Pg 111 second equation 708941 701789 TX 7 09782KZZ 708671 714980 ZZy 7 00028KZZ Pg 117 last equation in Example 35 MB 1 02010 702010 MB g R 70 3354 703354 In MB K 06708 06708 Pg 121 second line after eq 337 de ned by Eq 337 This Pg 124 third equation MB 1 71342 A773 Z 1116 mm APB 39 K 0 Pg 127 last line general a speci c auxiliary Pg 138 second equation line from bottom 7L2 2 c0s 7 L29 sin 7 L1 7 L2 cos 0 92 77 2L299 sin j Pg 144 rst equation line Close space 17F 170 17Pmyz 97 X fPG Pg 146 mid page last line of answer for 17p QRsinzbsin i 7 QLcoszbsin 7 QRcos 7 7 QLsinz sin lg Pg 148 paragraph following Figure 312 line 8 radius 7 5 6370 km Pg 148 paragraph following Figure 312 line 9 rotation is wire 0034 ms2 Pg 149 second line after eq 363 path is 7 5 cos A and Pg 150 second line above eq 368 to approximate sin A 7 sinA in the Pg 151 paragraph preceding Example 315 second line required by Eqs 369 then the Pg 171 Exercise 356 The labels A and B were confused The problem statement should have been Airplane A travels eastward at constant speed 11A 560 km hr while airplane B executes a constant radius turn p 32 km in the horizontal plane at constant speed 113 1440 kmhr At t 0 the angle 0 locating airplane B was zero and SA at that instant was 52 km Radar equipment on aircraft B can measure Pg 179 last full line preceding eq 427 obtained by premultiplication according Pg 189 eq 4311 17A 1713 DAB X fAB Pg 200 eqs 12 First line is incorrect and lines should be realigned 01 b2 sin cos til I 7 gt1 sin b2 sin 02 7 gt1 cos b1 cos b2 7 sin b1 sin b2 cos 7 gt2 cos 6 cos b1 cos b2 7 sin b1 sin 2 cos 7 sin 6 7 sin b1 sin b2 sin E 7 gt2 cos 6 cos b1 cos b2 7 sin b1 sin b2 17A 1713 DAB X fAB Pg 200 answer for I 7 sin cos cos 7 7 a 1 11 6 6 1 2 cos 1J1 7 s1n 1K1 sin 02 cos 6 cos 1 P 206 rst line after eq 4417 The angular acceleration is given by the second P 209 rst gure Swap labels C and D so that the pin is C Pg 219 Exercise 413 fourth line and when 0 120 P may 230 eq 518 symbol in second sum should be fjO that is 7 N 7 N d2 d2 N 3F ij fjO E WWW j1 j1 j1 0Q P 238 rst line of last equation Delete overbar on 31 OF Mshaft dHC a i 2 1 W 7 01 72mR2w2 sm 0 cos 0 2mR2w2 s1n0 k Pg 241 Figure 53 caption relative to a moving L yz P 244 rst line above Example 52 regardless of its symmetry Pg 244 third line of Example 52 statement to determine HO Pg 244 third line of Example 52 solution of change of H0 The two P Pg 245 rst equation rst line HO 277132 iwl P 245 second rst line HO 72m32w1i 7 P 245 change last sentence of Example 52 This is the same as the expression that would have been obtained if Example 51 had evaluated the moment of momenta relative to point 0 rather than point C OD 244 third line from bottom of page To evaluate HO we recall from 0Q OQOQ Pg 253 rst paragraph third line properties of other shapes Pgs 254 255 text between eq 5210 and eqs 5214 The following provides a fully vectorized derivation of the parallel axis theorems that is quite useful for software imple7 mentation 1t also corrects an error in eqs 5214 where the subscripts in the second term of each equation should have had a prime 7730 135 31135 213 5210 To proceed we invoke an identity for the vector triple product ax13x5213a575a13 Which changes Eq 5136 to H3 HG m fBG 39 7730 I 7 fBG PBG 39 Let the inertia properties With respect to the parallel coordinate systems L yz and x y z be 0 and 3 respectively Then the matrix form of the preceding is 13 a Io w m VBGP TBG U 7 TBG TBG w 521110 Where U is a 3 X 3 identity matrix and rBG 3 343 2ng This relation must apply for any to so the factor of w on each side of the equality must match The result is the parallel amis transformation of inertia properties 13 w m VBor TBo U1 TBo TBor 34123 2123 7503343 7x323 5212 Isl m i3y3 g 2123 103213 75513213 734323 g 34123 The rst form is useful for computer applications While using the second form to match like elements yields the scalar transformations The diagonal terms govern the moments of inertia Iy y 1111 105 2123 5213 1 In in g 31123 While the off diagonal terms transform the products of inertia Im y Iym my mxByB Im z Izm m2 mBZB Iy z 2y Iyz myBZB Pg 264 second line of Example 57 parallel to the orthogonal edges and Pg 268 eq 5234 subscript for e3 1 0 0 fellT 0 12 0 MT Wm 62 63H 0 0 3 63T Pg 275 eq 531 Remove box highlighting the equation Pg 276 eq 534 the overdot above the rst occurrence of HA needs to be centered Pg 280 rst equation the overdot above HA needs to be centered Pg 297 rst paragraph fth and sixth lines the angular momentum requires application of the required moment The same is true for linear momentum The Pg 307 top of page x answers for and I components of MA EMA mg L sin0 way 7 Izz 7 I wme 1 2 2 77111 Lgtlt 0 gtQs1n gt 1 m ZR2 7 L2 7 Z 7 Qc0s gt Q s1n0 EMA I Msino My c0s Izzozz 41m 7 Iyywmwy l 2 2 7quot 39 m4R Lgtlt 0mcos0 7m GP 7 L2 7 7 9 We 9 Pg 312 second gure lnsert label G at center and show 5L axis 2 horizontal diameter and y axis 2 vertical diameter Free body diagram of the rolling disk Pg 314 paragraph following rst set of equations rst line unknown so the moment Pg 317 line following second equation to xyz so we need to express 7 in terms Pg 318 rst equation Change Z to 0 in rst line and change sign in third line H A 7 7100003 1 7 I sin Z cos Z sin02 I sinzZ sin cos 0 if 7 Um 7 Iyy cos gtsin0 cos 0 I cos cosEffsin f 2 7 19090 W 7 In c0s gtsin0 2 cos 0 1M 002 j LuzZ Um 7 Iyy sin Z sin c0s 7 I sinzZ cos Z sin 2 2 1 7 1 I sin gtsin 0 W I P 318 last equation ID 0 mymglRl sinw m icosw P 320 eq 623 second line 0Q EMA j 71W 7 0sz P P 325 rst paragraph second line of the box has no obvious 329 eqs 4 third line SWIG1 F sinQ NA c0s gt 7 337 eqs 10 rst line 1nsert comma before N2 to separate equations OQOQ P P Pg 339 third paragraph fth to fourth lines preceding eq 642 instantly from time t to time t2 While the Pg 345 eq 6416 both path integral symbols should have C rather than 0 OF 339 third paragraph second line of such forces are those generated OF 2 2 WM lmsde lm w 1 1 Pg 345 eq 6417 W142 W241 Pg 346 Eq 6420 Path integral should have C rather than 0 2 W112 f mgK dXi dYJ dZK 1 Pg 347 Eq 6423 rst line The path integral should have C rather than 0 that is 2 G M W112 jf 7 7 d7 r rd gt rd0sin gt g 1 Pg 348 eq 6434 The path integral should have C rather than 0 that is 2 1 2 2 e m 1102 57710101 2F er 1 Pg 350 eq 6439 delete the path integral symbol N 73 Z yc jEFmEBEMgcw j1 Pg 353 fth line after gure diagram exert any moment about Pg 353 rst line after eq 1 The angular velocity of these bodies is the sum of the Pg 358 fourth equation from top of page second line Insert 103 factor preceding rads 717920321 23868431 40000T103 rads Pg 359 fourth line from bottom of page Delete clause at end of rst sentence to change simultaneously This is a fundamental Pg 362 third paragraph replace sentence preceding eq 6444 in Eqs 6440 Also because the centers of mass and the contact point lie in the 8 plane for planar motion only the I components of the cross products of the moments of F about points A and B are required They are given by Pg 362 eq 6444 Delete condition for equation to apply that is I foi Ff ix 1 iFfj Pg 363 rst line after eq 6450 In view of Eq 6442 the velocity Pg 365 eq 6453 lnsert unit vector 7 after F that is e 170A 397 17013 397 f iJkFi 7 0 r 7 0 T lltUCA0391 11013 39 1 Pg 395 last line the is repeated Pg 404 second line preceding Figure 76 then 11391 0 aflae This Pg 406 Solution to Example 72 fth line we de ne ql X3 Q2 YB Pg 411 rst paragraph after Figure 79 fth line is tangent to the constraint surface at the instant Pg 420 eq 738 rst line All subscripts P should be capitalized Pg 420 rst paragraph following indented text in italics fourth line 7 9 sin 59 There is no pt term Pg 427 second line after eq 746 couples allows us to collect Pg 434 Figure 710 a Change 0 label to Pg 438 seventh line after second gure Delete second 7 we have f if with f Mstgn C7 As we Pg 443 fourth text line from bottom of page to SW of F proceeds in Pg 443 last text line R in Eqs 7 gives Pg 445 Example 712 fourth line of problem statement corresponding to 0 being the Pg 447 second line after eq 753 partial derivatives Bin g depend that is delete the dot above qj Pg Pg Pg Pg P 0Q P 0Q P P OQOQ P 0Q 452 midpage line preceding equation for SW 0 by 60 so the virtual work is d 8T 460 fth line of equation for a at bottom of page lnsert overdot above 0 mA mr sin0 i0 460 second equation line from bottom of page ST W 7 mA erre2 cos0sin0 mTB sin0l 02 462 equation for V at midpage V mgL cos0 462 second equation from bottom of page 8V W 7mgL s1n0 462 last equation 2 mLZ 7 2 mL2 7 1 sin0 cos0 7 mgL hint sin0 0 465 rst equation rst line Delete subscript from T 465 eq 7 second line 1 1 5 1 15 1 sin m 02 513 7 466 rst equation d 8T d 1 E E 1 1 Ii s1n gt2 0 7 5111b s1n0 sm 1g 15 1 sin m 9 7 1 sin0 sin 2w 1 1 sin 2 gt 94gt 7 ii c0s0 sin 2m w l1lsin9COS 2M 1 Pg 468 eq 766 lnsert 12 factor in second line after second equal sign 8T2 l a l 39n 39 n 39n 6539 n 65 85 2M 845W 2M lt q q l l n j n n P 0Q Pg P P OQOQ P 0Q P 0Q P 0Q P P P P P P Pg OQOQOQOQOQOQ P 0Q P P 0Q OF 475 gure at top of page Fix legend solid line is St dashed line is CUT dot dash line is 481 Exercise 724 last line Ifm 2 20 rads 497 sixth line on the motion is beg 498 second equation line from bottom lnsert H after 9 1 7 57711 m2 505 eq 3 ECXTYJZK 505 set of eqations above eqs 4 5 cos coswl sinwf isinQK jsil1 ICOSZ J 7 me cosilJIJrsinw Jc0s K ks 509 eq 823 d E z G 240 516 rst line after eq 8240 In view of Eq 8239 the result 522 rst full line above eq 8261 533 eq 6 Change G to lower case 533 last line q 0 0 0 71127001 of q 534 fourth line above gures the initial value 7T2 7 0010 We also nonholonomic equations Thus the 535 second line after gure puter models that do 539 rst equation after eqs 4 77130 E 563 last equation First term on right side needs a dcc factor tt1 dx tto 1 L 6Tdt M vw 6111 to 0 569 equation for V at bottom of page Deleted exponent 2 inside integrand on rst line 570 fourth equation line from bottom of page N MLquot Fem 712 Qn QR MU E Banj39quot 2 L 2 jil P 570 second equation line from bottom of page N ML F 7 p112 H271quot2 n 7 an n 2 9 PW 1951 19919 lt L 2 9 OD Pg 575 rst full line above Example 93 done numerically by solving q at Pg 584 eq 9229 rst line BE 1 for n K 1N yields 341 Pg 585 eq 9234 d 8 3 7 1K ldt 34139 391 Q17 7 7 7 Pg 586 rst paragraph third line so the slope Z E dZdR so that Pg 586 second paragraph third line axis and y about the y axis Therefore Pg 586 third equation from bottom of page B By 039 cos 6 Pg 587 second equation Change 0392 to JR in last term on second line 2 7 R 7 2 2T Em m 302 C086 02 RJrUsin 2 1 if 7 31 0Rsin Pg 592 second line after eq 939 must be suitably selected if the virtual displacement is to be kinematicallly admissible Pg 593 rst line are ql X3 qz Y3 and Pg 595 rst line form in Eq 939 We Pg 604 eq 4 10 130 X PGC IYSPGC Pg 607 second paragraph third line other forces of concern are the normal force exerted on the collar and the spring force We Pg 608 eq 8 rst line Insert a prime after 5 i 72 c0s quotyf39yg sin cos 0 sin 7 Wheel 7 R cos R cos R cos f 7 Pg 610 rst paragraph line following second equation on the page required to convert derivatives of V to generalized forces are Pg 610 second paragraph line following equation for 173 changing each quotydt to 57 Pg 611 line after eq 10 The second Gibbs Appell equation aSa g P2 is Pg 637 third line from top a speci ed motion 1n other situations the response Pg 639 paragraph following eq 1018 fth line is such that 0 gt 6 and has the same sign as M Cases Pg 641 eq 2 I 6 tan 1 tan gt 1068O Pg 642 rst equation a w sin 65 i cos w 0185265 7 098271 Pg 643 eq 10110 rst line K sinecoszZ Jr sin sin gtj C080 Pg 643 eq 10111 second line c0s I Pg 643 eq 10112 rst and third lines 1 components should have cos 0 rather than cos 1 that is HG HOE HG 7 sin cos sin sin cos 01 1 sin coszZ sin gtgt1 2 sin sin gt 9cos gtgt 13 c0s 1 Pg 643 equation at bottom of the page 1 2 Ho 1 mi 13 mi 13mm Pg 644 eq 10114 third line 3c0s0z1gtgt chos Pg 645 rst paragraph sixth line equations are identically satis ed Pg 645 rst paragraph seventh line and HoIg Thus

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