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Mass Moment of Inertia Help

  1. Nov 8, 2004 #1
    Can someone please help me with this problem...I need to find the mass moment of inertia about the y-axis (see picture)

    I know this can be done as two composite areas/volumes and I know that doing this through integration is not the "best" or "easiest" way...but this is how I want to do it...with that said, here is what I have done/tried...but I get the wrong answer...


    [tex]I_{yy}[/tex] = mass moment of inertia about the y axis

    [tex]I_{yy}=\int{\int{\int{(x^2+z^2)}}}dm[/tex]

    [tex]I_{yy}=\int{\int{\int{(x^2+z^2)}}}\rho dV[/tex]

    [tex]I_{yy}=\rho*\int{\int{\int{(x^2+z^2)}dy}dx}dz[/tex]

    [tex]I_{yy}=\rho*\int{\int{\int{(x^2+z^2)}rdr}d\theta}dz[/tex]

    [tex]I_{yy}=\rho*\int{\int{\int{((r\cos{\theta})^2+z^2)}rdr}d\theta}dz[/tex]

    [tex]I_{yy}=7830*\int_{0}^{.09}{\int_{-\pi}^{0}{\int_{.04}^{.08}{(r^3\cos^2{\theta}+z^2r)}dr}d\theta}dz[/tex]

    [tex]I_{yy}=.0355779...[/tex]


    BTW, as you can see, rho is 7830 for this problem


    Answer = .0249726... (I found this answer through calculating the mass moment of inertia by composite areas, and I know this is correct)
     

    Attached Files:

  2. jcsd
  3. Nov 8, 2004 #2
    btw, in case you can't tell from the limits...I took theta to be the angle measurd from the positive x-axis to the positive y-axis
     
  4. Nov 8, 2004 #3
    Try evaluating that last integral again and see if you get the right answer.

    --J
     
  5. Nov 8, 2004 #4
    I used a calculator to do it...and I plugged it in like 100 times or more lol....I haved checked it all many times over....I also tried doing it all seperately (like doing each part on the calculator seperately)..just to make sure the calc was not messing up...but it still did not work...I got the same answer everyway I did it....
    :(
     
  6. Nov 8, 2004 #5
    Mathematica's spitting out the correct answer from your integral.

    --J
     
  7. Nov 8, 2004 #6
    hm............thanks for evalutating that for me!!! So I was right then! hahaha....hm....now I wonder why my calc gave me the wrong answer (ti-89)....oh well...I will go through it the long way...which I did plan on doing anyways....I just didn't wanna spend all the time on figuring it out without knowing I was doing it right...

    Oh well...thanks for using Mathematica..my ti-89 failed me for the first time :/
     
  8. Nov 8, 2004 #7
    You got your 89 in degrees mode. Set it back into radians and you're good.

    -J
     
  9. Nov 8, 2004 #8
    DoH!!!!!!!! ALL that time and that was my problem....sigh....thanks for pointing out my stupidity :/

    I just did it again any my calc did not fail me..haha..got the right answer

    thanks again!
     
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