276 The Rev. A. Thacker on the Motion of a Free Pendulum: i 

 Hence 





dt^ 



I 



d^u R y 



-y^ --((J^co^a) cos X, sin <at-\ y =0 



dt^ \./ / ,^ / 



d^z 

 df' 



—g sinX 



R ^ ^ 



m I 



(1-) 



Let the position of the ball be now referred to other coordi- 

 nates X, Y, Z, the axis of X being taken due east, that of Y due 

 north, and that of Z towards the centre of the eai-th. We then 

 ,have 



X = a; sin ft)^ — y cos ft)/ 



i Y = .rsin\cos&)/ + ?/sinXsinft)/— ^cosX 



Z — x cos X cos (ot + y cos X sin cot + z sin X 



a?=X sin «/ + Y sin X cos cot + Z cos Xjcos cot 



?/= —X cos ft)/4- Y sin X sin Q)t-\-Z cos X sin cot 



z= — YcosX + ZsinX. 



EUminating x, j/, and z from equations (1.), we shall find 



d^X ^ . ^dY ^ ^dZ 2Y , I^ X ^ "^ 



•3ft) smX-TT — 2ft) cos X -7- — ft>^AH r =0 



at ml 



ft)^ Y sin^X -f &)^ sin X cos \.[a^z) 



co^Y sin X cos X + ft>^ cos^X(a — ; 



(2.) 



1, Finally, wx will suppose the horizontal motion of the ball re- 

 ferred to axes which revolve about the vertical with an uniform 

 angular velocity co sin X. If x^ y be the coordinates of the ball, 

 we have 



0?= X cos (cot sin X) — Y sin {cot sin X) 



7/ = X sin {cot sin X) + Y cos {cot sin X) 



z=z 



X = a? cos {cot sin X) + y sin {cot sin X) 



Yi:^ i^'a? sin {cot sin X) + y cos {cot sin X) . 



