508 
ME. A. COHEN ON THE HIEEEEENTIAL COEEEICIENTS 
But as the axes of coordinate axes are principal axes, %(mxy)=0, X{mxz)=0. 
Therefore 

A being the moment of inertia about Ox. 
We have still to investigate the expression %m{f'^y—fyZ) vAiexe and 
We obtain by substitution 
tm{f^y —fy z)=7!r^Xm{Vyy-\-v^) — 7!Tyt{mv^y)—T!TX{mv^). 
But since 
v^=z^y—y7!T,, 
'^z—y^x—x^x, 
it is evident that for principal axes we have 
%{mvjj)= —v!^{my% X(mv^)=7ffy^(mz^). 
Hence 
'^m(f:y-fyZ)=r!TyZ!TXS(my^)-'X(mz^)) 
= TSyTssJ^ '2m{y‘^ + x^) — 'Zm{z^-{-0[^)) 
= VTyZjj^C—B), 
where C and B are the moments of inertia about the axes of z and y. 
Substituting then this last expression and the expression in (4.) in equation (3.), we see 
that the sum of the moments of the moving forces about equals 
A “Ir+CC-BK^.. 
Hence by D’Alembert’s principle, if L, M, N be the sum of the moments of the 
impressed forces about the axes of a?, y, z, we have 
L=A^'+(C-BK^.. 
Similarly, 
M=B J>'+(A-CK=.., 
N=C^*+(B-AK^,. 
I will now show the full import of each of the steps in the above analytical proof. 
We have seen, in section 39 of Chapter III., that the acceleration of the particle is 
the result of the acceleration det (0^(0), E), and the acceleration det (O, V). If then 
the particle’s mass be m, its moving force is represented by 
m det (0^(0), W)-\-m det (O, V). 
