Finding the Center of Mass of a Hollowed Out Sphere

AI Thread Summary
The discussion focuses on calculating the center of mass of a hollowed-out sphere made of styrofoam with a cavity filled with a denser material. The user initially calculates the center of mass for the full sphere and realizes that the mass density changes due to the cavity. They explore treating the problem as two spheres: the original sphere and the filled cavity, leading to a clearer understanding of mass ratios. After calculating the volumes and masses of both spheres, they successfully derive the center of mass using the appropriate formula. The final result confirms their calculations align with the answer provided in the reference material.
Blingo
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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.
 
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Blingo said:
Hey guys, this is my first thread here. I'm just looking for a hint or two (no answer please).

Maybe, you could treat this situation as two constant density spheres instead of one sphere and one sphere with a hollow somehow?
 
hmm, thanks. I'll try that now.
 
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
 
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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.
 
R/2 - [Edit] 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!
 
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Blingo said:
Y = (m1y1 + m2y2)/(m1 + m2)
That's fine. (I would measure the distances from the center of the large sphere.)

however, i don't know how to determine the distance between the two spheres.
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..."

i determined the masses by multiplying the density by volume, but perhaps that ain't the way to do it.
That's fine. All you need is the ratio of the masses.
 
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.
 
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