339 
1908-9.] On Lagrange’s Equations of Motion. 
19. As a first and very simple example, we take the motion of a particle 
of mass m in a plane curve. If at time t the radius- vector drawn from a 
fixed point be of length r, and make an angle d with an axis of x drawn 
from the same origin, the co-ordinates of the particle are 
x = r cos 0 , y = r sin 6. 
Hence for the kinetic energy T we have 
2T = m{(r cos 0 - rd sin d) 2 + (r sin d 4- rd cos d) 2 ) . . (41), 
or 
2T = m(r 2 + r 2 d 2 ) ..... (45). 
In applying (13) to the problem of finding the r, 6, equations of motion 
of the particle, we have to take the first expression for the kinetic energy. 
We obtain 
d 3T 
By (13) we have to subtract from this 
m(r cos d — rd sin 0 )— (cos 6) + m(f sin 6 -\-r0 cos 0)-^(sin 0 ) ; 
dt dt 
d 
that is, mrd 2 . The same result would, of course, be obtained by calculating 
0T /dr. Hence the r-equation of motion is 
m{r — r6 2 ) = K ; 
where It is the applied force in the outward direction along r. 
For the d-equation we have 
d 0T 
dt dd 
+ 2 rfd). 
By (13) we have to subtract from this 
d • d 
- m(r cos d - rd sin 6) — {r sin 6) + m(r sin 6 + rd cos d)~(r cos 6) 
v 'dt dV 
or zero, thus the d-equation of motion is 
m(r 2 d -1- 2 rfd) — © ; 
where 0 is the applied force perpendicular to the radius- vector. 
This method, if it had been applied to the value of T in (45), would 
have failed. T is here a sum of squares referred to a set of axes so 
specialised that in the formation of T the quantities cos d, sin d have taken 
the special values 1 and 0 ; and unless we go back to the fundamental 
expressions, for the velocities along the unspecialised axes Ox, O y, it is not 
apparent how the process is to be carried out. 
It will be observed that from (44) we have 
3 / n\ 3 / 
—(cos 6) = - — (r sin 
00 v ’ dr 
-L(sin d) = ^-(r cos d) 
dd ’ dr ' 
