290 Theorems and Problems 
absolute velocities of G andr, consequently 
the vis viva of the point R may be resolved 
into the vires vive of the points G and r; be- 
cause the quantities of matter, supposed to 
move with these three points, are equal. | 
Propitem V._ Required the centre of 
oscillation of the system 6, 7 and k? ti 
Soxrution. Let OS (plate 4, fig. 5,) be 
the length of a simple pendulum, which 
vibrates through similar arcs in equal times 
with the system b, & and J, vibrating upon 
the point 0; and let the matter in the point S 
be equal to ‘all the matter in b, k and 1; make 
os=OS8; and s is the centre of oscillation 
required. Now to find the length of OS or 
os, we are to consider that the matter in the 
system acts by its weight at G perpendicular 
to the horizon to give the point R a certain 
angular velocity ; and the matter in the pen- 
dulum acts at S in the same direction to give 
GS the same angular velocity ; therefore put 
O S=s; and we have as g:s:: R?:s* by 
cors. 1 and 2, prob, 3; henceasg: BR: is 8. 
9.E.T. : 3 
Cor. 1, R*=gs. 
Cor. 2. If the system b, &, and J, revolve 
about the point 0; put t= the time of revolu- 
tion, m= the matter in b, k andl: and the 
m Zs 
m R* 
vis viva of the system | is as™—, or as ~. 
¢2 t 
ae 
2 
———seerercerr 
i 
