KINETIC ENERGY 231 



= L_, etc., 

 2jnn 



so that V mx = 0. It follows that V m = 0, or Yraw = 0. 

 ^ ^ at ^ 



Similarly ^mv = and ^?mw = 0. The whole of the second 

 line of expression (80) is now seen to disappear, so that we find 

 for the kinetic energy the expression 



which proves the theorem. 



187. Next, suppose that the coordinates of the center of gravity 

 at any instant, referred to an imaginary set of fixed axes, are 

 x, y, z, the velocity of the center of gravity having, as before, com- 

 ponents Uy V, W. 



We have supposed that, relative to the center of gravity, the 

 particle m x has coordinates x lt y l} z lf and components of velocity 

 u v v v w r Thus, referred to the imaginary fixed axes, the coordi- 

 nates of the particle m^ will be 



while its components of velocity, as before, are 

 u -f- u lf v + v lt w + w r 



Let the force applied to the particle mjhave components X lf T v Z{. 

 As in 141, the work done on this particle by the external forces 

 is equal to minus the work performed by the particle against these 

 forces. Thus the work done on the particle while it moves over 

 any small element of its path is, by 118, equal to 



The total work done on all the particles in any small displace- 

 ment is therefore 



x,) + Y,d(y + y]) + Z,d(z + 

 and this can be separated into two parts as follows: 



