# Center of mass of a solid

• Apr 3rd 2011, 07:58 PM
bazingasmile
Center of mass of a solid
Hi,

I'm confused on how to calculate the center of mass of a solid that's located out of one sphere and into another. The problem goes on like this:

Determine the center of mass of a solid located out of a sphere, of radius 1, centered at the origin, and inside a sphere, of radius 1, centered at point $\left(0,0,1\right)$.

All I get from my textbook is the mass formula:
$m=\int\int\int_{E} \rho(x,y,z)\mathrm{d}V$

And then if the center of mass is actually the same thing as a center of inertia, I would have to calculate the moments related to each plan of coordinates.

How can I do that when I've got two spheres one above the other. I believe their equations are respectively:
$x^2+y^2+z^2=1$
and
$x^2+y^2+(z-1)^2=1$

As plotted below.

Can you please give me a hint? I'm blocked there.

Thanks a lot,

Bazinga
• Apr 4th 2011, 11:08 AM
running-gag
Hi,

The center of mass G of a solid is given by the formula

$\int\int\int_{E} \rho(x,y,z) \vec{OM}\mathrm{d}V = \left(\int\int\int_{E} \rho(x,y,z) \mathrm{d}V\right) \vec{OG}$

If the solid is homogeneous, $\rho(x,y,z)$ is constant (independent with respect to x, y and z) and therefore can be pulled out of the integral

$\int\int\int_{E}\vec{OM}\mathrm{d}V = \left(\int\int\int_{E} \mathrm{d}V\right) \vec{OG}$

In your example, as far as I have understood, the solid is axysimmetric with respect to the z axis therefore G is on the z axis

The projection of the formula on the z axis gives :

$\int\int\int_{E} z r \mathrm{d}r \mathrm{d}\theta \mathrm{d}z = \left(\int\int\int_{E} r \mathrm{d}r \mathrm{d}\theta \mathrm{d}z) z_G$