LAGKANGE'S EQUATIONS 



333 



ILLUSTRATIVE EXAMPLE 



Common pendulum. As a simple example of the use of Lagrange's equations, 

 let us consider the problem of the motion of the common pendulum. A rigid 

 body is constrained to move so that one point remains fixed, while the line 

 06? joining to the center of gravity moves in a vertical plane. Let be the 

 inclination of OG to the vertical; then the position of the system is entirely 

 fixed as soon as the value of is known. The kinetic and potential energies are, 

 in the notation of 245, 



so that 



T = 

 L = 



W = Mgh (I - cos 0) , 

 - Mgh (1 - cos 0). 



dL 



Thus = MWd. and Lagrange's equation, 

 50 



^ /5IA _ dL 



~ ' 



becomes 



Mk 2 = - Mgh sine, 

 dt 2 



FIG. 153 



the same equation as was obtained in 246, and from this the 

 motion can be deduced. 



We notice, however, that Lagrange's method shows that the motion is inde- 

 pendent of the method of suspension of the pendulum, provided only that it is 

 constrained to move in the way described. For instance, the result is true if 

 there is no pivot at all at 0, the constraints being imposed by a suspension of 

 strings. 



272. Let us now consider the second alternative to that exam- 

 ined in 271. It may be that if we assign arbitrary values to 

 86 j, 8# 2 , , $0 n , the new configuration obtained is not in every 

 case a possible one. It may be that there are certain relations 

 which must be satisfied, in order that the constraints imposed by 

 the mechanism may not be violated. 



For instance? in the illustration already employed, let there be two 

 ropes hanging from a ceiling of a room, such that on pulling one down one 

 inch the mechanism compels the second to rise two inches. Let lt 2 

 denote the lengths of ropes below the ceiling. Then a displacement in 

 which 50J = -j^ inch, 50 2 = -$ inch, is not a possible displacement ; such 

 a displacement is not permitted by the mechanism above. We must always 

 have 50 l5 50 2 connected by the relation 



501 + \ 50 2 = 0. 



