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Moment of Intertia, proving Parralel axis theorum. Help please.

  1. Jun 10, 2009 #1
    Imagine we have a Equilateral triangle in the xy plane with one vertex at the origin. The triangle will be rotated about the z axis. The depth of the triangle, d, will be constant.[tex]B_r=B(r)[/tex]

    From the picture I see that [tex] B_r=\frac{r}{h}B , V=\frac{B_rhd}{2}=\frac{Bdr}{2}[/tex]

    First, I need to find I.

    [tex]
    I=\sum M_i*r_i^2=\int r^2 dm. p=M/V, so, dp=p dv[/tex]

    [tex]
    I=p\int r^2 dv[/tex] Substituting using[tex] dp=p dv and dv=\frac{Bd}{2}dr[/tex], we get:

    [tex]
    I=\frac{pBd}{2}*\int r^2 dr=\frac{pBd}{2}*\frac{r^3}{3}.[/tex] Using p=M/V:
    [tex]
    I=\frac{Mr^3}{3h}=\frac{Mr^2}{3}.[/tex] R_total/h=1.

    Now, lets say I want to find I for the same shape, but now there is a distance d from the vertex to the z axis.

    Everything is the same as the above, except, [tex]B_r[/tex] is now a piecewise function, where B(r)=0 for r<d and B(r)=[tex] B_r=\frac{r-d}{h}B for r>d. [/tex]

    Reworking it with the new B_r, I can get the new I. After that I just solve for x in [tex] I_2=I_1+x[/tex] to get the parallel axis theorem.

    The thing is, I don't know how to generalize this one example to work for all shapes. Any help is appreciated.
     
  2. jcsd
  3. Jun 14, 2009 #2
    Lets pretend that whoever answers my question get free cookies.

    Also, now that I think about it, I am not sure this method is valid.
     
  4. Jun 15, 2009 #3

    Cyosis

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    Homework Helper

    Yes picking a shape will not help you prove the theorem. What is the definition of the moment of inertia? Note that it is a vector quantity. Now displace the axis a distance d away from the original axis, note that d is a vector too.
     
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