KINETIC ENERGY OF A RIGID BODY 291 



235. THEOREM. Let k be the radius of gyration about any axis 

 through the center of gravity, and let k 1 he the radius of gyration 

 about a parallel axis distant a from the former, then 



Let PQ be any axis through the center of gravity G, and let 

 P'Q' be any parallel axis distant a from the former. Suppose that 

 the rigid body has a motion of rotation 

 about P'Q', the angular velocity being fl. 



Then the velocity of G is aft, and the 

 motion may be regarded as compounded 

 of a motion of translation of velocity aft 

 together with a rotation ft about the 

 axis PQ. By formula (120), the kinetic 

 energy is 



It is also ^ Jf&' 2 ft 2 , where k f is the radius of gyration about P'Q'. 

 Hence we have 



2 = $ M (a 2 ft 2 + 



and the result follows on dividing through by \ Jfft 2 . 



236. Alternative proof. This theorem may also be proved geometrically. 



Let L be any particle of the body, and let the plane of fig. 142 be 

 supposed to be the plane through L at right angles to the two axes of 

 , rotation, these axes cutting the plane in 



the points A, A' respectively. Let LA =p, 

 LA' = p', and let LN be drawn perpendicu- 

 lar to A A'. Then Mk 2 = ^mp 2 , and also 



Mk' 2 = %mp' 2 



__^_M = &"& + 

 142 



- 2p AA' cos 0) 



Now ^N is the projection of the line from L to the center of gravity, 

 upon the line A A'. Hence ^ra AN = 0, and we have 



giving the result to be proved, after division by M. 



