# Center of Mass Problem

1. Nov 1, 2004

### Blingo

Hey guys, this is my first thread here. I'm just looking for a hint or two (no answer please).

A sphere of styrofoam has radius R. A cavity of radius R/2 centered a distance R/2 directly above the center of the sphere is hollowed out and filled with a solid material of density five times the density of styrofoam. Where is the center of mass of the new sphere?

I figured that to do three dimensional center of mass calculations, I just break it down for each dimension and solve, so I was able to get the center of mass for the pre-cavity styrofoam sphere, which is just R.

I think that, if the new sphere's center of mass were to be calculated in the same manner, the mass density would no longer be constant and would therefore be a function. However, that function is eluding me.

Last edited: Nov 1, 2004
2. Nov 1, 2004

### NateTG

Maybe, you could treat this situation as two constant density spheres instead of one sphere and one sphere with a hollow somehow?

3. Nov 1, 2004

### Blingo

hmm, thanks. I'll try that now.

4. Nov 1, 2004

### Garth

Treat it first as three spheres:
the stryofoam sphere complete - density 'rho', the filled cavity density 5'rho' and a 'subtracted' cavity of density -rho.

In other words treat it as two spheres: the complete stryofoam sphere and a filled cavity of 4'rho'.

Garth

Last edited: Nov 1, 2004
5. Nov 1, 2004

### Blingo

thanks guys for the help. one final question -- when i treat the problem as two separate spheres, i thought to use the equation

Y = (m1y1 + m2y2)/(m1 + m2)

however, i don't know how to determine the distance between the two spheres. i determined the masses by multiplying the density by volume, but perhaps that ain't the way to do it.

6. Nov 1, 2004

### Garth

R/2 -  draw a diagram it is easy. Remember the mass of the smaller sphere is reduced by the amount of stryofoam you have taken out.

mass is density times volume - hint: density is mass/volume!

Last edited: Nov 1, 2004
7. Nov 1, 2004

### Staff: Mentor

That's fine. (I would measure the distances from the center of the large sphere.)

It's given in the statement of the problem: "A cavity of radius R/2 centered a distance R/2 directly above the center of the sphere..."

That's fine. All you need is the ratio of the masses.

8. Nov 1, 2004

### Blingo

Thanks!

I think I've got it now.

The volume of the first sphere is:

V1 = (4/3)(pi)R^3

The volume of the second sphere is:

V2 = (4/3)(pi)(R/2)^3 = (1/6)(pi)R^3

Comparing the two spheres gives:

V1 = 8 * V2

The mass of the first sphere is:

M1 = Rho * V1

The mass of the second sphere is:

M2 = (4 * Rho) * V2 = (4 * Rho)(V1/8)

Once again, comparing the two spheres gives:

M2 = (1/2) M1

Then using the equation

Y = [(M1 * Y1) + (M2 * Y2)]/(M1 + M2) = (M2 * Y2)/(M1 + M2)

Y = [(1/2)M1 * (R/2)]/[M1 + (1/2)M1] = [(1/2)M1 * (R/2)]/[(3/2)M1]

Y = (1/6)R

Which is what the answer book says.