Analytical mechanics for students of physics and engineering . Iv=ifQ !J2»+f0 ^dm — 7 I y4dz -\-ttt I z2y2 dz [dm = mry-dz] 4 J—a J —a =f r ^ - ^2 dz+2 ^ i>2 - *> ?*? and Sttto515 = I TOO2 Iy> = Iy + ma-= I ma2. 124. Theorem III. — 77/r moment of inertia of a homogeneousright cylinder about a transverse axis equals the moment of in-ertia of two lamince which fulfill the following conditions, (a)Each lamina has a mass equal to that of the cylinder. (6) Onelamina occupies the entire area of the transverse section of thecylinder through the given axis, while the other lamina occupiesthe
Analytical mechanics for students of physics and engineering . Iv=ifQ !J2»+f0 ^dm — 7 I y4dz -\-ttt I z2y2 dz [dm = mry-dz] 4 J—a J —a =f r ^ - ^2 dz+2 ^ i>2 - *> ?*? and Sttto515 = I TOO2 Iy> = Iy + ma-= I ma2. 124. Theorem III. — 77/r moment of inertia of a homogeneousright cylinder about a transverse axis equals the moment of in-ertia of two lamince which fulfill the following conditions, (a)Each lamina has a mass equal to that of the cylinder. (6) Onelamina occupies the entire area of the transverse section of thecylinder through the given axis, while the other lamina occupiesthe < ntire area of the longitudinal section of the cylinder through(he j . CENTER OF MASS AND MOMENT OF [NERTIA 159 Let Y, Fig;. 00, be the axis with respect to which it isdesired to find the moment of inertia of the cylinder. Leidlvdenote the moment of inertia of an element bounded bytwo transverse sections relative to 1 he K-axis, and ///,.. denotethe moment of inertia of the same element relative to the. Fig. 90. F-axis, a parallel axis through the center of mass of theelement. Then, by theorem II, we havedly= dly»+ (x2+z2)clm, where (x2+z2) is the square of the distance between tin-two axes. Similarly the moment of inertia of the elementabout the F-axis which is parallel to the F-axis and inter-sects the same elements of the cylinder, is given by dlv>= dlv- + dlv- between the last two equations we obtain dly= dly- + x*dm= Ki-d/n + x-drn, 160 ANALYTICAL MECHANICS where Ai is the radius of gyration of the element of massabout the F-axis. Integrating the last equation we have Iv= I Kx*dm+ I x-dm. Jo Jo Each of the elements of mass has its own F-axis similarlyplaced. Therefore ki is the same for all the elements of massand remains constant during the integration. Hence Iv = Ki2m+ I x-dm where h = Kx-m and 72 = / #2 dm. It is not difficult to see that h is the sum of the moments of inertia of all the elementsof mass relative to their respective F-ax
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