358 Prof. J. Parser on the Applicability of Lagrange's 

 should have, were continuous forces applied to the body, 



d dT dT „„ 



— • — .- — — = ^'-component 01 iorces, 



(It till/ (lib 



-z -. 7-7 = moment of forces round CKZ,, 



dt dyjr d^ n 



O'Z, being a parallel through 0' to OZ. 



It follows, integrating through the short interval of time 

 during which the instantaneous impulse would act, that 



dT 



-Ts- = component of impulse along OX, 



dT 



— — OY 

 dy ~ " " U1 ^ 



dk= » " 0Z ^ 



dT 



— r = component of moment of impulse round O'Z,. 



dyjr 



In the actual motion 



dt dx dx ' 



dT 

 ^-component of impulse = ■— = constant, 



.*., since T does not depend on x, y, z, 

 ^-component of impulse = -= 



and similarly for the ^/-component and the ^-component. 



dT 

 Let us now proceed to find the physical meaning of *rr-j 



and interpret the equation 



d dT _dT 



dt d^r d^fr ~~ 



It is clear from Green's theorem that T can be expressed as a 

 quadratic function of p, q, r, u, v, w with constant coefficients ; 



dyjr dp dyfr dq dyfr dr d-ty 



,<Wfa<W m faa^<to 



du dyjr dv dyjr dw (Ity 



When p, q, r are expressed in terms of <£, yjr, 0, their coeffi- 

 cients do not involve the precessional angle ty. Therefore the 

 first three terms vanish. 



