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Moment of inertia for a thin rod

  1. Apr 3, 2013 #1
    1. The problem statement, all variables and given/known data
    The thin rod shown has a length L = 100 cm, and a density that varies from 26 g/cm at the origin to 3 g/cm at the far end. Determine a) the moment of inertia about axis 1 (passing through the center of mass of the rod), and b) the moment of inertia about axis 2 (passing through the heavy end of the rod).


    picture attached


    2. Relevant equations

    I already found
    Mass of rod = 1.45*10^3 g
    Center of mass of rod = 36.8 cm from the more dense end

    The density as a function of position I found to be: ρ(x)=.23x + 3



    3. The attempt at a solution
    For the first part:

    I used I = ∫ r^2 * m/L dr
    where -36.8 < r < 63.2

    and got 1.46*10^6 g*cm^2

    For the second part I know it has to do with the parallel axis theorem:

    I = I_cm+mh^2

    where h = 36.78 cm

    and got 3.4*10^6 but that's not right.

    Any ideas?
     

    Attached Files:

    Last edited: Apr 3, 2013
  2. jcsd
  3. Apr 3, 2013 #2
    Since the density is not uniform, you can't just assume it is m/L. Use the density function you found above. Pay attention to the relation between r in the integral and x in the density function.
     
  4. Apr 3, 2013 #3

    SammyS

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    attachment.php?attachmentid=57444&d=1365042546.gif

    Your density function gives a linear density of 3 g/cm at the origin rather than at x = 100 cm ,

    and a linear density of 26 g/cm at x = 100 cm rather than at the origin.
     
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