Advanced Engineering Dynamics 2010 Part 13 docx

Advanced Engineering Dynamics 2010 Part 13 docx

Advanced Engineering Dynamics 2010 Part 13 docx

... The speed of the particle relative to the fixed frame will be u = J(uf + us. + uf) (9.79) so that 2 2 -1’2 y=(l - u/c) for a frame of reference moving with the particle. The proper ... If we now consider a group of particles then the four-momentum will be T (0 = Cm,y,(c 11, 4 UJI It is convenient to write this four-vector matrix in a partitioned form such as (0 =...

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Advanced Engineering Dynamics 2010 Part 6 docx

Advanced Engineering Dynamics 2010 Part 6 docx

... longer. The result of the integration is )] (5.55) 131 2 [ eJ(e2 - 1)sinO -ln( J(e + I) + {(e - 1)tan(6/2) t= JK($ - 1131 2 1 + e cos 8 J(e + 1) - {(e - l)tan(6/2) ... conditions when o = oo. Thus 1 12 Dynamics of vehicles (5.96) 1 2 where C, is the drag coeficient. The drag coefficient is the sum of two parts, the first being the sum o...

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Advanced Engineering Dynamics 2010 Part 8 docx

Advanced Engineering Dynamics 2010 Part 8 docx

... approximation can be made Constant force applied to a long bar 137 Now let the bar be of finite length L, as shown in Fig. 6 .13. At x = L the strain has to be zero. Therefore at ... (6.43) (n + 1)X nX +-I EA v=c ( EA CX EA - (2n + 1) PAE Fig. 6 .13 Impact of two bars 135 v z,/z, (ZJZ, - 1) c2 (1 + Z,/Z,) (Z,/Z, + 1) F', = -...

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Advanced Engineering Dynamics 2010 Part 2 doc

Advanced Engineering Dynamics 2010 Part 2 doc

... a group of particles 15 1 .13 Newton's laws for a group of particles Consider a group of n particles, three of which are shown in Fig. 1.1 1, where the ith parti- cle has ... on the particle is the vector sum of the forces due to each other particle in the group and the resultant of the external forces. If & is the force on particle i due to particle ....

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Advanced Engineering Dynamics 2010 Part 3 ppt

Advanced Engineering Dynamics 2010 Part 3 ppt

... then the time integral will be integrated by Parts because 6u = 0 at t, and t2. The second term in equation (3.18) is Integrating by parts gives (3.19) 50 Hamilton S principle ... integrating by parts, h 6x 1; - It:mi 6x dt - kx 6x dt = 0 4 (3.10) Rotating fiame of reference and velocity-dependent potentials 37 It is interesting to note that for a charged par...

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Advanced Engineering Dynamics 2010 Part 9 potx

Advanced Engineering Dynamics 2010 Part 9 potx

... exponen- tially, (6 .135 ) U" = U(- 1)" e" e-Kn Substituting equation (6 .135 ) into equation (6 .130 ) leads to o = 2J (dm) cosh (k'/2) (6 .136 ) for o > oca. ... motion is S(U,+~ - u,) - s(u, - n,-,) = mu, (6 .130 ) Let us assume (6 .13 1) Substituting equation (6 .13 1) into equation (6 .130 ) and dividing through by the common...

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Advanced Engineering Dynamics 2010 Part 11 pps

Advanced Engineering Dynamics 2010 Part 11 pps

... CB, -S0, 0 a,C0, C8, -SO3 0 &e3 *[A13 = [ i' c: ! a37 ] The overall transformation matrix is o[A13 = o[A]i ~[Alz z[A13 The elements of this matrix are A,, = ... shall only be con- cerned with the overall dynamics and not with the detail. This is a vast subject area of which dynamics is a substantial and vital part. 8.2 Typical arrangements 8.2.1...

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