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Homework Help: Moment of Inertia question

  1. Oct 5, 2008 #1
    1. The problem statement, all variables and given/known data
    Show that the rotational inertia of the rod about an axis through its center and perpendicular to its length is ML^2/12, where M is Mass and L is length from axis of rotation.


    2. Relevant equations
    ∫r^2dm=Mr^2


    3. The attempt at a solution

    Well, you see the integral that I did in the relevant equations? I got most of the work done there, assuming that R=L. My problem is understanding where the /12 comes from in the question. There were no numbers given in the initial question, just M and L.
     
  2. jcsd
  3. Oct 5, 2008 #2

    Dick

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    Great. dm should have a dr in it. What's R? What are the limits on your integral? Take rho to be the density of the rod per unit length. What's M in terms of rho and L? What's the moment of inertia in terms of rho and L? Now eliminate rho.
     
  4. Oct 6, 2008 #3
    well, dm=pdL is what im assuming...

    so ∫dm=∫pdL
    M=pL

    but shouldnt the right side of the above equation be a definite integral while the left is an indefinite integral?

    i think the right side of the above equation should go from -L/2 to L/2, but that raises the question of the limits of the left side of the equation...

    either way, assuming both sides are indefinite integrals, substituting pL for M in ML^2 gives me

    pL^3=I

    but thats as much as i think i understood of your post.

    thanks for the quick reply.
     
  5. Oct 6, 2008 #4

    Dick

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    Now you want to integrate r^2*rho*dr for r from -L/2 to +L/2 to get the moment of inertia. Then use M=rho*L to get rid of the rho.
     
  6. Oct 6, 2008 #5
    Okay, so....

    ∫(-L/2,L/2) pR^2dr= (1/3)R^3p|(-L/2,L/2)
    =(1/3)(L/2)^3p-(1/3)(-L/2)^3p
    =(1/3)(L/2)^3p+(1/3)(L/2)^3p
    =(2/3)(L^3/8)p
    =L^3/12p

    p=M/L

    =(L^3/12)(M/L)
    =ML^2/12


    Thank you so much for your help.
     
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