Dynamics of Mechanical Systems 2009 Part 1 ppsx

... AAn An An=++ 11 22 33 BBn Bn Bn=++ 11 22 33 AB A A A B B B AB AB AB AB AB AB AB AB ⋅= + + () ⋅++ () =⋅+⋅+⋅ +⋅+⋅+⋅ +⋅+⋅ 11 22 33 11 22 33 11 1 1 12 1 2 13 1 3 212 1 32 3 2 23 2 3 313 1 32 3 2 nnnnnn nn ... Forces 375 11 .9 Generalized Forces: Inertia (Passive) Forces 377 11 .10 Examples 379 11 .11 Potential Energy 389 11 .12 Use of Kinetic Energy to Obtain Genera...

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... Eqs. (2 .11 .6) and (2 .11 .7). By substituting from Eqs. (2 .11 .6) and (2 .11 .7) into Eqs. (2 .11 .1) and (2 .11 .2), we obtain: (2 .11 .8) and (2 .11 .9) Hence, we have: (2 .11 .10 ) Observe that Eq. (2 .11 .10 ) ... =− AAA BBB CCC BBB CCC AAA CCC AAA BBB mp BBB AAA CCC AAA CCC BBB CCC BB 12 3 12 3 12 3 12 3 12 3 12 3 12 3 12 3 12 3 12 3 12 3 12 3 12 3 12 3 12 3 12 3 12...

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... ⋅       +⋅       n n nn n nn n nn n n nn n nn 1 2 31 3 12 1 23 1 3 13 1 22 nn nn 11 13 10 ⋅= ⋅= and d = 0 and 1 n n n nn n 11 31 3 dt d dt d dt ⋅⋅=−⋅ RB d dt d dt d dt ωω× = ⋅       +⋅       +⋅       n n nn n nn n nn 1 1 11 1 22 1 33 RB ddtωω× ... obtain: (4 .13 .10 ) RB ωω= + +ωωω 11 22 33 nnn R GRB RGRB rrVn V n=× =×− () ωωωω 21...

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... aNN G XZ 1 111 2 11 11 2 1 2= () − () +− − () [] l ˙˙ cos ˙ sin ˙˙ sin ˙ cosθθθθ θθθθ vn ann OO 22 1 11 1 11 1 2 13 ==−lll ˙˙˙˙ θθθ and vNN O XZ 2 11 1 =− () l ˙ cos sinθθ θ aNN O XZ 2 11 1 2 11 11 2 1 =− () +− ... + () ==−− 12 334 567 13 311 2 32 2 13 223 0 3 10 4 15 3 8 4 8 0 12 12 16 30 60 24 32 11 31 144 19 2−+nn Mnnn O =− 16 0 30 16 8 12 3 ftlb 059...

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... ab ab 11 22 33 11 22 33 11 11 1 212 13 13 212 1 2222 2323 313 1 3222 3333 0593_C07_fm Page 203 Monday, May 6, 2002 2:42 PM 230 Dynamics of Mechanical Systems where we have made use of Eqs. (7 .12 .4) ... z III III III =++ =++ =++ Innnn Innnn Innnn 1 11 1 12 2 13 3 1 2 21 1 22 2 23 3 2 3 31 1 32 2 33 3 3 PO jj PO jj PO jj III I III I III I =++= =++= =++= In i PO ij i...

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... 8 .10 .4 Free-body diagram of rod B 2 . nnnnnn nnnnnn 11 1 1 1 2 21 2 1 2 2 12 1 1 1 2 22 2 1 2 2 =+ =+ =− + =− + cs cs sc sc , , nc s nc s ns c ns c 1 1 11 1 12 1 2 21 2 22 2 1 11 1 12 2 2 21 2 22 =− =− =− ... nnnnnn nnnn nn 11 2 1 21 2 1 22 21 2 1 11 2 1 12 12 2 1 21 2 1 22 22 2 1 11 2 1 22 =− =+ =+ =−+ −− −− −− −− cs cs sc sc , , Fa...

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... m y ω vvv v 02 10 2 0 2 20 21 1 2 212 2 11 2 12 2 11 2 −− () +− () = () − () [] + () − () −− () [] + () ll l l llll ωω ω ω ωωωω 12 2 11 2 02 1 1 () −− () [] = () v ll lωω ω v 01 2 512 11 12= () + () llωω v 012 23= () +llωω ωω ... ii i N i i N i Gi =+⋅×+× () =       +⋅×       +× () =++ === === ∑∑∑ ∑∑∑ 1 2 1 2 1 2 1 2 1 2 0 1 2 2 11 2 1 1 2 11 2 2 vv...

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... be: (11 .7 .1) and (11 .7.2) (11 .7.3) (11 .7.4) where n 1 , n 2 , and n 3 are the unit vectors of Figures 11 .7.2 and 11 .7.3. From Eqs. (11 .4.4) and (11 .4 .15 ) we obtain the partial velocities and partial ... , x P x P x PP x P x P x P P x P x P x PP x x 1 1 2 1 3 11 1 2 2 2 3 2 3 1 3 2 3 3 3 11 2 12 2 1 00 002 00 ====+ () == ==+ () === θ θ θ l l 33 3 32 =+ () l...

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... Generalized Dynamics: Kane’s Equations and Lagrange’s Equations 419 or (12 .2 .16 ) and or (12 .2 .17 ) In Sections 11 .10 and 11 .12 , Eqs. (11 .10 . 21) , (11 .10 .22), (11 .12 .14 ), and (11 .12 .15 ), we found ... Figure P 11. 11. 9. Repeat Problems P 11. 11. 6 and P 11. 11. 7 finding the potential energies for the relative orientation angles. P 11. 11. 10: Use the results of P...

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... F(A, ψ) is: (13 .10 .14 ) Then, Eqs. (13 .10 .4) and (13 .10 .5) become: (13 .10 .15 ) and (13 .10 .16 ) Upon integration of Eqs. (13 .10 .15 ) and (13 .10 .16 ) we obtain: (13 .10 .17 ) and (13 .10 .18 ) where φ 0 is ... (13 .8 .11 ), (13 .8 .12 ), and (13 .8 .13 ) in the form of Eqs. (13 .8 .14 ), (13 .8 .15 ), and (13 .8 .16 ). The derivatives of ξ 1 , ξ 2 , and ξ 3 may be e...

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