By Cantrell J.C.

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Let us calculate the change in potential energy. -,--- ..... , F' ___ F W' - W = Ph, where P is the weight of the solid and h the increase in the height of its center of gravity. Clearly, h = AA' = BB'. Fig. 12 • , . 32 The Center of Gravity, Potential Encray, and Work Furthermore, supposing the solid to be homogeneous, we can write p = Vi'. <--~ where V is tile volume of the solid and i' its specific weight. vyh. 23) ~d, 'W' =- W""BDF' +. tA,,,,', and, therefote. 24) t of , , that is, W' - W is equal to the differetce the potential energy of the bodies BB' D' D and AA.

S) is the moment of area of the rectangle with respect to the axis of rotation. ~ Let us now substitute an arbitrary figure Q for this rectangle (fig. 24). ". 2 4 z ,. h C He P- H, R2 Fig. 23 approximate each of these strips by the rectangle inscribed in each strip. If n denotes the number of strips, and we allow this number without. bound, the approximations become successively better. We then have " v = lim (,Vl + Vll + ... + Vn) , t' " .... 61), • 51 J • ,The Center of. ~vity. Potential Energy.

S = ,21T' DC· sin fJ· 2Ra . 21), however, tells US that DC s= = R(sin a/a). Consequently, 21TR sin ex sin,8. 2Ra = 21TR· 2R sin a sin,8 . a . Since , " 2R sin a = I, we now have Again referring to the sketch, we note that the 'second factor of this product is. equal tp the altitude of the spherical strip (that is, the projectron of the chdtd"A B onto the diameter PQ). Denoting this altitude by H, we finally' obtain' the formula ' , " ... , --. 3. 30 rotates about the axis 00'. The surface area ,of the resulting solid'is equal to , " ' ( + a)2 ·4 ay'(2) ~ '= (my'(2~ .