Ssl

3Xj dy/ 3z,i . Mi[xi ' - y • -z, . <V/; dqi dqi dqi,

/ 3Xj dy/ 3it , ■ Mj ( Xi —- + }'; fr- + Z; —- ) Sq2 dq2 dq2 dq2

Xj ---h Yi — + Zi — 8qi dqi dqi dqi, v dx> , v dW , 7 dz' 1 £ Xj ---h Yi---hZ,— 8q2

dq2 dq2 dq2

Now let us shift our attack on the problem and consider the kinetic energy of the system. This is z 1=1

Now calculate P- and P- to obtain

dK dqi dK dqi i=l

dqi dqi dqi

We next calculate the time derivative of for which the chain rule gives d2Xj d2X:

0 0

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