666 Prof. H. H. Jeffcott on the Whirling Speeds of a 



(fig. 4). Let A be the centre of the load. Consider an 

 element or mass dm at any point P, and let f, r] be the co- 

 ordinates o£ P with respect to the principal axes of the body. 

 Then AQ = f, PQ = *?. Let the angle between AQ and ST 

 be 6. The centrifugal force on dm acts along TP and has 

 magnitude co 2 . PT . dm. The moment of this force about an 

 axis through A perpendicular to the plane of bending is 

 or . PT . AB . dm. Hence - M = jV . PT . AB . dm. 



Now PT=AS + PB=«+fsin0 + ?7Cos0, 



and AB = £cos# — rj sin 6. 



.*. — M = a 2 ( (f cos 6 — t? sin #)(« + £ sin <9 + ?7Ccs 6) dm 



= co 2 u cos 0\fjdm — co 2 a sin 6\ rjdm -f co' 2 sin cos d\fpdm 

 -co 2 sin cos j r) 2 dm-t- co 2 (cos 2 6— sin 2 0) \ grjdm, 



where the integrals are taken throughout the load. 



Now the solid being one of revolution, the integrals xgdm, 

 iTjdm, \£r]dm vanish. 



Hence 



M = co 2 sin cos \_\rfdm - Jf*im] 



= o) 2 sin cos [2§ V 2 dm- j"(f 2 4V>Zw] . 



But 2(*97 2 <im = I 1 , or the moment of inertia of the load about 

 the shaft axis: and \ (f; a + nf)dm==l a , or the moment of 

 inertia of the load about a diameter through the centre 

 perpendicular to the plane of bending. 



Accordingly M = (L — X 2 )eo 2 sin cos 



= (I 1 — I 2 )te 2 0, since 6 is small, 



or =(1!— j 2 )«i) -^. 



When the load is a thin disk I 1 = 2I 2 and Ij — I 2 = I 2 = mk 2 . 



For a solid circular cylinder of axial length l t and 

 diameter d 1 , fixed at its centre to the shaft (supposed thin), 

 we have approximately I 1 = ^md ] 2 and I 2 = ? 1 g ^(3i 1 2 + 4/ 1 2 j : 



S0 that I 1 -I 2 = i 1 8 m(3^i 2 -4/i 2 ). 



When I 2 is greater than I x a vibration of oscillatory 

 character becomes possible. 



Thus, for a single load for which both the centrifugal 

 force X and the centrifugal moment M are appreciable, the 



