284 MISCELLANEOUS METHODS AND APPLICATIONS. [CHAP. XII. 



*251. Illustrative problem. Two uniform rods AS, BC of masses m 15 

 w 2 and lengths a, b are freely hinged at B, and AB can turn about A in a 

 vertical plane. The system starts from rest in a position in which AB is hori- 

 zontal and BC vertical. To discuss the initial curvature of the path of any 

 point of BC. 



Let AB make an angle 6 with the horizontal, and BC an angle with the 



Fig. 75. 



vertical at time t. Since B describes a circle of radius a about A, and since 

 the centre of inertia of BC describes a circle of radius ^b relative to B, the 

 system of kinetic reactions will be as shown in the diagram. 



By taking moments about B for BC, and about A for the system, we 

 obtain the two equations 



m 2 (%b 2 + 1 3 2& 2 ) - m^a'e %b sin (0 + 0) - m 2 a0 2 %b cos (6 + 0) = - %m z gb sin 0, 

 m l (ia 2 + T V* 2 ) +m 2 ad{a+ $b sin (6+(f>)}+m 2 a0 2 \b cos 



\m^ga cos 0+m 2 g (a cos 6 4- \b sin 0). 



Adding the equations, and dividing out common factors, we have 

 sin (6 + 0) - |m 2 b<j> 2 cos (6 + 0) 



...... (1), 



and the first equation is 



|60-ia^'sin(^-|-0)-ia^ 2 cos(^ + 0)=-|^sm0 ......... (2). 



In the initial position = 0, = 0, = 0, = 0, and we have 



ri, _o A _ 3 i+ 6 2 ff 

 0o-0, 0o- - 



In any position we have, by Maclaurin's theorem, 



also 



6= 6 Q t +i<V 2 + ..., 0= 



Now, taking equation (2), we see that if were finite, would be of order 

 , and 6 of order t 2 , so that the terms would be respectively of orders 1, 2, 2, 



