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Complicated moment of inertia question

  1. Sep 1, 2009 #1
    1. The problem statement, all variables and given/known data
    a sphere of radius R, with centre at r=0, has a density of rho(r)=rho(0)(1-r/R), using sphericals.

    a. what is the mass of the sphere?
    b. write down the intergral for the moment of inertia, about the z axis.
    c. Solve it.

    2. Relevant equations
    I=[tex]\int[/tex][tex]\int[/tex][tex]\int[/tex]rho(r)*r^2 dV


    3. The attempt at a solution

    a. M=rho*V=rho(0)(1-r/R)*4*pi*r^3/3

    i need help on, i am getting somehting like..
    triple intergral form 0-2pi,0-pi,0-R of (rho(0)(1-r/R)*r^3*sin(theta) dr dtheta dphi

    ca anyone help me? by makign sure my intergal is right, or fixing it, ive only ever done uniform density before, so im a bit confused about this.
     
    Last edited: Sep 1, 2009
  2. jcsd
  3. Sep 1, 2009 #2
    srry. made a mistake.
     
  4. Sep 1, 2009 #3
    for that integral i get (pi/5)rho(0)(R^4)... this doesnt make sense, since to get the answer in terms of M and R, i need to sub in the value of mass.....

    wel maybe if I just take out all the things i know, i get (3/20)MR^2.... rofl I DOUBT THAT IS RIGHT..
     
    Last edited: Sep 1, 2009
  5. Sep 1, 2009 #4

    HallsofIvy

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    No, M is NOT "M=rho*V=rho(0)(1-r/R)*4*pi*r^3/3" because "r" is a variable. M is the integral of the density function over the volume of the sphere:
    [tex]M= \int_{\phi= 0}^{\pi}\int_{\theta= 0}^{2\pi}\int_r^{R} rho(0)(1-\frac{r}{R})dV[/tex].
     
  6. Sep 1, 2009 #5
    ok yes thankyou, that makes much more sense.
     
  7. Sep 1, 2009 #6
    ok i tried that integral, and i got (1/3)(pi*rho/R)(R^4-4R*r^3+3r^4).... i have no idea, what this means... im confused, because isnt mass constant... over the entire volume? r u sure i am meant to intergrate from r to R dr.. and not 0 to R?.. and if this this is right what shoudl my integral be for the moment of inertia?
     
  8. Sep 1, 2009 #7
    i think the moment of inertia should be what it is for a solid sphere.. 2/5MR^2.. but doign it the way you said, i get I=(1/5)(pi*rho/R)(R^5-5*R*r^4+4r^4) ... if i divide that for what I got form my M value... i just get a heap of mess...
     
  9. Sep 1, 2009 #8
    ok i got the right answer, being the right answer is 2/5MR^2... forgot then r=0 lol
     
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