1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Science Advisor

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