Consider a halved solid torus (half a donut). The radius of the torus are \$R_1\$ and \$R_2\$. I need to find its center of mass. The hint they give is that the center of mass of a homogeneous solid object \$Omega subset Bbb R^3\$ is calculated as\$\$overlinex=int_Omegaoverlinerdoverliner.\$\$ I really don"t understand this formula, I don"t know what \$overliner\$ means and what is \$Omega\$ in this case. I"d appreciate that someone explains what this formula means and how to apply it in this problem. Thanks in advance.

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asked Oct 19 "13 at 0:01 TwinkTwink
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I found it easiest to use cylindrical coordinates to set up the integrals needed for the center of mass. Before we do so, however, I set my coordinate system up as follows. I have positive \$x\$ coming out of the screen, positive \$y\$ going to the right, and positive \$z\$ up. In cylindrical coordinates \$(r,phi,z)\$:

\$\$x = r cosphi\$\$\$\$y=r sinphi\$\$

where we have the limits defining the region \$Omega\$:

\$\$phi in left < -fracpi2,fracpi2 ight>\$\$\$\$z in <-R_2,R_2>\$\$\$\$r in left < R_1 - sqrtR_2^2-z^2, R_1 + sqrtR_2^2-z^2 ight>\$\$

Also, for an object of constant mass density, the expression for the \$x\$ component of the center of mass is

\$\$arx = fracdisplaystyle int_Omega d^3 vecx , xdisplaystyle int_Omega d^3 vecx\$\$

(Note that, by symmetry, we have \$ary=0\$ and \$arz=0\$.) Let"s first evaluate the denominator:

\$\$eginalign int_Omega d^3 vecx &= int_-pi/2^pi/2 dphi , int_-R_2^R_2 dz , int_R_1 - sqrtR_2^2-z^2^R_1 + sqrtR_2^2-z^2 dr , r \ &= fracpi2int_-R_2^R_2 dz left \ &= 4 pi R_1 int_0^R_2 dz , sqrtR_2^2-z^2\ &= pi^2 R_1 R_2^2 endalign\$\$

Now we evaluate the center of mass:

\$\$eginalignarx &= frac1pi^2 R_1 R_2^2 int_-pi/2^pi/2 dphi , int_-R_2^R_2 dz , int_R_1 - sqrtR_2^2-z^2^R_1 + sqrtR_2^2-z^2 dr , r^2 cosphi \ &= frac43 pi^2 R_1 R_2^2 int_0^R_2 dz , left \ &= frac83 pi^2 R_1 R_2^2 int_0^R_2 dz ,left <3 R_1^2 sqrtR_2^2-z^2 + left (R_2^2-z^2 ight )^3/2 ight > \ &= frac83 pi^2 R_1 R_2 left ( frac3 pi4 R_1^2 R_2 + frac3 pi16 R_2^3 ight )endalign\$\$

Simplifying, I get

\$\$arx = frac4 R_1^2+R_2^22 pi R_1\$\$