Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar
    Science Advisor
    Gold Member

    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


    User Avatar
    Science Advisor
    Gold Member

    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

    User Avatar

    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.
  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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook