center of gravity
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The sphere can be represented in two dimensions by the circle x2+y2=a2 and the plane by the vertical line x=c. 

Because of spherical symmetry the CoG lies along the x-axis at some point G(g,0).

Consider a vertical disc thickness t with its centre at (x,0). Its radius is y=√(a2-x2) so its circular area is π(a2-x2) and its volume π(a2-x2)t. If the disc has a density ρ, then its mass=ρπ(a2-x2)t. Its distance from G is |g-x|. Each disc has a CoG on the x-axis and all the mass of the disc can be regarded as focussed on to a point x on the x-axis, i.e., at (x,0).

The sum of the moments around CoG is zero, and the moment is mass × distance from CoG=ρπ(a2-x2)t(g-x). If t is infinitesimally small we can write t=dx, therefore:

∫ρπ(a2-x2)(g-x)dx=ρπ∫(a2g-a2x-gx2+x3)dx=0. The limits for the integration are [c,a].

ρπ[a2gx-a2x2/2-gx3/3+x4/4]ca=ρπ[a2g(a-c)-½a2(a2-c2)-⅓g(a3-c3)+¼(a4-c4)]=0 (note that a-c is a factor and a≠c),

a2g-½a2(a+c)-⅓g(a2+ac+c2)+¼(a3+a2c+ac2+c3)=0,

⅓g(2a2-ac-c2)=⅓g(2a+c)(a-c)=½a2(a+c)-¼(a3+a2c+ac2+c3)=¼(a3+a2c-ac2-c3)=¼(a-c)(a+c)2 (note that a-c is again a factor on both sides),

g=¾(a+c)2/(2a+c). In 3D this is the point (¾(a+c)2/(2a+c),0,0).

(Note that when c=0, the CoG=¾a2/2a=⅜a from the centre, which is the CoG of a hemisphere.)

by Top Rated User (1.2m points)

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