292 MOTION OF KIGID BODIES 



237. From the theorem just proved, it follows that the radius 

 of gyration about any axis can be found as soon as we know that 

 about a parallel axis through the center of gravity, and vice versa. 

 We now give some examples of the calculation of radii of gyration. 



CALCULATION OF EADII OF GYRATION 



. 238. Uniform thin rod. Let the rod AB be of length 2 a, and 

 let k be its radius of gyration about an axis through A perpendic- 

 ular to its length. Let r be its mass per unit 

 1 1 ' length, and let x be a coordinate which meas- 

 ures distances from A. The element which 

 extends from x to x -f dx is of mass r dx t and its perpendicular 

 distance from the axis of rotation is x. Hence 



7,2 _ ^ _ __ JO '"" __ 1 ^2 



?a , " 2ra 3 

 rax 



so that the radius of gyration is = 



V o 



About the center of gravity, which is distant a from A, the 

 radius of gyration is given by 



*=$-= | , 



so that the radius of gyration about the center of gravity is -= - 



V3 



239. Rectangular lamina. Let us suppose the lamina to be of 

 edges 2 a, 2 6, and let us find its radius of gyration about an axis 

 through its center at right angles to its plane. Let us take axes as 

 in fig. 144, and let cr denote the mass per unit area. Then 



