338 GENEBALIZED COOBDINATES 



horizontally, and % b cos vertically. The whole velocity v of the center of 

 gravity of the rod is therefore given by 



U 2 = (a sin 6 + 1 6 sin 0) 2 + (| 6 cos 0) 2 

 = a 2 sin 2 2 + ab sin sin 00 + 6 2 2 . 



The angular velocity of the rod is 0, and its radius of gyration k is given by 



H6H- 



Thus, if m is the mass of the rod, the kinetic energy of the rod is 



= | m (a 2 sin 2 2 + a6 sin 6 sin 00' + | & 2 2 ). 



Lastly, the horizontal distance of the end of the piston rod from the center 

 of the flywheel is a cos + 6 cos 0, so that the velocity of the piston and piston 



rod is 



a sin 6-6 1) sin 0. 



If J\f is the mass of the piston and piston rod, the kinetic energy of this part 

 of the engine is W a sin + 6 sin 0) 2 . 



We now have, for the whole kinetic energy T, 



2 T = Id* + m (a 2 sin 2 6 2 + ab sin 6 sin 00' + i 6 2 2 ) 



+ M (a sin + 6 sin 0) 2 . (a) 



The potential energy TF, measured from a standard configuration in which 



(6) 



Here 3fC is the total mass of flywheel and crank, and A, e are the polar coor- 

 dinates of its center of gravity when 0=0. 



The changes in and are not independent. A glance at the figure shows 

 that we must always have 



a sin = 6 sin 0, (c) 



and on differentiating this, we can see that if and are taken as generalized 

 coordinates, we must suppose them connected by 



a cos 50 6 cos 50 = 0. 

 Thus Lagrange's equations will be 



0, (d) 



d /dL\ dL 



( )-- -- X6cos0 = 0. 



