Rotatory Inertia on the Vibrations of Bars. 47 



VI. 1. Pivoted-free Bar. 

 /= d cos 7,f -f d' sin 7^ + d n cosh 7 2 £ + d'" sinh 7^. 



Origin at pivoted end. 

 Hence, by (11), 



#£_ ^when£=0; 



d + d" = 0, 



and y l 2 d = y. 2 2 d" , 



d=d"=0. 

 /= ^ sin 7l f+^ // sinh 7 2 f. . . . (33) 

 Also, by (12), 



X d% \ a? U j 



-7 1 Vsiny 1 / / =-7 2 2 ^"smh7 2 /', . . (34) 



d ^(\ 4 7i-7i 3 ) cosy/ +^' // (\ 4 72 + 7 2 3 ) cosh 7 2 Z / = 0, 



or, since V72 + 72 3 ^ 4 + 7 2 2 72 _/ 7if\ Y2__7i 

 X 4 7l — 7^ — V— 7! 2 '7! "" ^ "~ 7 2 2 / y~ 7 2 ' 



^'72 cos 7iZ'— </ / "7! cosh y 2 Z'=0. . . (35) 



Eliminating <i and ^' between (34) and (35) 



7i 3 sin 7/ cosh y 2 l' = y. 2 * cos 7/ sinh y 2 l\ . (36) 



or 7 X 3 tan7 1 Z / = 7 2 3 tanh7 2 r (Gr) 



By means of (35) we obtain for (33) 



/=D r 7l ^ +72 !H*Ml. . . . (H) 



J L cos yd' 'cosh 79/' J 



an 



7it cosh y. 2 i 



2. Correction for Potatory Inertia in a Pivoted-free Bar. 



Before proceeding to find the correction from (Gr) we 

 require the roots of 



tanm=tanhwi (37) 



These may be found 



57T 97T 



(i.) by noticing that these roots lie very close to -j- ,—r ,... 

 and putting m= -r- +e, where e is small, for the 1st root. 



