290 



MISCELLANEOUS METHODS AND APPLICATIONS. [CHAP. XII. 



Let OA, OB be the two wires, a the angle which each of them makes with 

 the horizontal, AB the horizontal position of equilibrium of the rod. A'B' a 



Fig. 76. 



displaced position, 6 the angle between AB and A'B'. Then 6 is the angular 

 velocity, and 6 the angular acceleration of the rod. The instantaneous centre 

 in any position is the point of intersection of perpendiculars to OA, OB 

 drawn from the ends of the rod. We denote by /, /' the positions of the 

 instantaneous centre corresponding to AB and A'B', and by G, O' the corre- 

 sponding positions of the centre of inertia. 



The moment of the kinetic reaction about /' is m (k 2 +I'G" 2 ) 0, where m is 

 the mass of the rod and k its radius of gyration about its centre of inertia. 

 With sufficient approximation we may put IO for I'G'. 



The forces acting on the rod are pressures at its ends, whose lines of 

 action pass through 7', and its weight. Now 01' is a diameter of a circle of 

 which A'B 1 is a chord subtending an angle 77 2a at the circumference, and 

 thus 01' is of constant length and 11' is therefore ultimately at right angles 

 to 01 and horizontal. Also GG' being ultimately at right angles to IG is 

 horizontal, and thus the moment of the weight about /' is - mg (II' GG'}. 

 Hence we have the equation of moments 



m (F + 7# 2 ) 6=-mg (IP - GO"). 



Now let 2a be the length of the rod. We find 



//' = BB' sec a = IBB sec a aQ cosec a sec a, 



GG' = lG6=ad cot a, 

 and the equation becomes 



ma? ( + cot 2 a) 6= - mgaB (sec a cosec a - cot a), 



