262-264] 



ROTATING FRAME. 



299 



rectangular axes of x and y. Also let u, v be component velocities of the 

 particle parallel to the axes of x and y . 



Fig. 80. 



We have x x cos - / sin 0, y y cos + x sin 0, 

 whence # = (# y 0) cos (y + 3/0) sin 0,) 



i/ = (j/ + .2/0) cos + (#/ y 0) sin 0.J 



Also # = M cos0 v sin0, y -= -y cos -f w sin 0. 



Hence we find u = x - y0, v =y -f ^ 0. 



Now, if we write &amp;lt;o for 0, o&amp;gt; is the angular velocity of the moving axes, 

 and the resolved parts parallel to the moving axes of the velocity of the 

 particle whose coordinates are x\ y are 



x -a&amp;gt;y and y +a&amp;gt;x . 



We may prove in precisely the same way that, if a, /3 are the resolved 

 parts of the acceleration of P parallel to the axes of # , y, then 

 a =u G&amp;gt;V and 8=v- 



In the problem of Article 262, taking axes through M along and perpen 

 dicular to AJ3, the angular velocity of the moving axes is 6, and the co 

 ordinates of P are a + 1 cos x and I sin x- From these the component velocities 

 of P relative to M otherwise obtained in that Article might be deduced. 



*264. Examples. 



1. Two equal circular rings each of radius a and radius of gyration k 

 about its centre are freely pivoted together at a point of their circumferences 

 so that their planes are parallel, and the rings are so thin they may be 

 regarded as in the same plane. The system being at rest on a smooth table 

 with the pivot in the line of centres, the pivot is struck by a blow perpen 

 dicular to the line of centres so that it moves off with velocity V. Prove 

 that the angle 6 which either radius through the pivot makes with its initial 

 direction at any subsequent time is given by the equation 



