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Homework Help: Center of Mass Problem

  1. Nov 1, 2004 #1
    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. jcsd
  3. Nov 1, 2004 #2


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    Maybe, you could treat this situation as two constant density spheres instead of one sphere and one sphere with a hollow somehow?
  4. Nov 1, 2004 #3
    hmm, thanks. I'll try that now.
  5. Nov 1, 2004 #4


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    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'.

    Last edited: Nov 1, 2004
  6. Nov 1, 2004 #5
    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.
  7. Nov 1, 2004 #6


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    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!
    Last edited: Nov 1, 2004
  8. Nov 1, 2004 #7

    Doc Al

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    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.
  9. Nov 1, 2004 #8

    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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