Rotatory Motion of Bodies. 17 
But the force on m parallel to Om’ arises entirely from the 
LI ‘ 
tension of mm’, and is = a Similarly, the force 
. tension of mm’ 
on m! parallel to Om is = ———7,—— . Hence, we have 
Xe + Vy + 32 =Xa + Vy + Ss. 
Similarly, X'g!' 4+- Y'y’ + 8/3" — X"z! + Y"y' + B"'z!, 
and Xx + Y"y + B"z = Xx" + Yy" = Sz" 
Or, putting for X, Y, 3, X’, &c. their values from (c), 
m (a'd'x + y/d'y + edz) =m (ad'a' + yd’y + 2d’) 
m! (ala! + yy! + 2'dx!) = ml! (dea! + Ly! meal @, 
m! (ada +yd?y"+ zd°2") =m (ade + yay + 2d’z) 
which are three equations of motion. 
Differentiating the first of equations (a), we have 
ada’ + ydy' + zdz' + adz +y'dy + dz =0. 
Hence, if we make* 
vdxe + y'dy + dz =rdt, we have xda' + ydy' + zdzv =—rdt 
Sim’. if 2’da'+y"dy'+2"dz'=pdt........ vd’ +y'dy"+2'dz"= =a (e) 
and if eda” +ydy" +2dz"=qdt ........ a"dx +y"dy +2"dz = —qdt 
Now, take the three equations, 
rdx + ydy + zdz =0 
a’dx + ydy + zdz = rdt } GS). 
ade +y"dy + 2"dz = — qdt 
Multiply by 2, 2’, x’,.and add; we then have, observing equa- 
tions (a’) and (0’), 
* This notation is the same as that used by Lagrange, Poisson, &c. with very little 
alteration. 
Vol. (1. Part I. (e 
