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Calculating moment of inertia for nonuniform sphere

  1. Oct 21, 2014 #1
    1. The problem statement, all variables and given/known data
    A sphere with radius R = 0.200 m has density ρ that decreases with distance r from the center of the sphere according to [tex]\rho = 3.00 \times 10^3 \frac{kg}{m^3} - (9.00 \times 10^3 \frac{kg}{m^4})r[/tex]

    a) Calculate the total mass of the sphere.

    b) Calculate the moment of inertia of the sphere for an axis along the diameter.

    2. Relevant equations
    Moment of inertia for a solid sphere: [tex]\frac{2}{5}MR^2[/tex]

    Density: [tex]\rho = \frac{M}{V}[/tex]

    Volume of a sphere: [tex]V = \frac{4}{3} \pi R^3[/tex]

    3. The attempt at a solution

    I managed to figure out part a). I started with [itex]\rho = \frac{dM}{dV}[/itex] and [itex]V = \frac{4}{3} \pi r^3[/itex]. Solving for dM gets you [itex]dM = \rho dV[/itex] and taking the derivative of V yields [itex]dV = 4 \pi r^2 dr[/itex]. Substituting that back into the expression for dM yields [itex]dM = 4 \rho \pi r^2 dr[/itex]. I then substituted the expression for ρ given in the problem statement and integrated from 0 to R to find the mass of the sphere, 55.3 kg. According to the back of the book this is correct.

    For part b) I know there's going to be another integral involved but I can't seem to get it right. I know the moment of inertia for a solid sphere (stated above under the relevant equations) so I thought I could plug my expression for dm back into it and integrate from 0 to R but that didn't get me a correct answer. I'm not quite sure where to go next. Any help would be greatly appreciated!
     
  2. jcsd
  3. Oct 21, 2014 #2

    haruspex

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    Please post your working for your attempted solution.
     
  4. Oct 21, 2014 #3

    SteamKing

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    To calculate the MOI of a non-uniform sphere, you have to go back to the mathematical definition:

    http://hyperphysics.phy-astr.gsu.edu/hbase/isph.html#sph2
     
  5. Oct 21, 2014 #4
    Like I said in the original post, I took the equation for the moment of inertia for a solid sphere, plugged in my expression for dM and the expression for ρ from the problem and integrated from 0 to R. The answer I got was not even remotely correct according to the back of the book. I ended up with an integral that looked like: [tex]\frac{8 \pi}{5} \int_0^R (3.00 \times 10^3 \frac{kg}{m^3})r^4 - (9.00 \times 10^3 \frac{kg}{m^4})r^5 \, dr[/tex]
     
  6. Oct 21, 2014 #5
    I'm really not sure what to do with this. I think I get the part about using the expression [itex]\frac{1}{2}y^2 dm[/itex] to sum the moments of small disks about the z axis but I have no clue how to apply that to this problem. How do I find y? And when I plug the expression given for ρ back into this integral I'll have two variables, y and r. I have no idea how to deal with multiple variables.
     
  7. Oct 21, 2014 #6

    haruspex

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    Yes, you described your method in the OP, but that's no substitute for the actual working.
    You can check your answer by throwing away the variable density part. Does it give the right result for a uniform density sphere - no.
     
  8. Oct 22, 2014 #7

    SteamKing

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    Remember, the distance r from the center of the sphere is such that r2 = x2+y2+z2

    For each slice of the sphere located at z from the x-y plane, the sphere is going to have a certain radius. This radius can be called x or y depending on your preference.
     
  9. Oct 22, 2014 #8
    I'm really trying to understand this but I'm just not wrapping my head around what you're trying to tell me. Where does [itex]r^2 = x^2 + y^2 + z^2[/itex] come from? If I solve that for r and plug it back into the given expression for ρ I have three variables - x, y, and z. I'm not at all familiar with multivariable Calculus so I haven't the foggiest idea how to tackle that.

    Thanks anyway but after 2 days of trying to figure this out perhaps I'm just destined to not understand this.
     
  10. Oct 22, 2014 #9

    SteamKing

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    The distance between two points in three dimensions is related to the difference in the x, y, and z coordinates of these points by the Pythagorean relation, which is expressed as [itex]r^2 = x^2 + y^2 + z^2[/itex]. If you take the center of the sphere as the origin of the coordinate system, then the location of any point on or inside the sphere can be expressed by its (x,y,z) coordinates, and you can also calculate the distance of that point from the center of the sphere.

    By careful application of the definition of the moment of inertia for the sphere, you don't need to know any multivariable calculus.
     
  11. Oct 22, 2014 #10
  12. Oct 23, 2014 #11
    Thank you so so much for this! This is exactly the kind of thing I was looking for. The whole process is much more clear to me now.
     
  13. Oct 23, 2014 #12
    It is to me, too! No problem!
     
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